LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS
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LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS LOS SOLUCIONARIOS CONTIENEN TODOS LOS EJERCICIOS DEL LIBRO RESUELTOS Y EXPLICADOS DE FORMA CLARA VISITANOS PARA DESARGALOS GRATIS.
Circuit Variables
1
Assessment Problems AP 1.1
To solve this problem we use a product of ratios to change units from dollars/year to dollars/millisecond. We begin by expressing $10 billion in scientific notation: $100 billion = $100 × 109 Now we determine the number of milliseconds in one year, again using a product of ratios: 1 year 1 hour 1 min 1 sec 1 year 1 day · · · = · 31.5576 × 109 ms 365.25 days 24 hours 60 mins 60 secs 1000 ms Now we can convert from dollars/year to dollars/millisecond, again with a product of ratios: 1 year 100 $100 × 109 · = = $3.17/ms 1 year 31.5576 × 109 ms 31.5576
AP 1.2
First, we recognize that 1 ns = 10−9 s. The question then asks how far a signal will travel in 10−9 s if it is traveling at 80% of the speed of light. Remember that the speed of light c = 3 × 108 m/s. Therefore, 80% of c is (0.8)(3 × 108 ) = 2.4 × 108 m/s. Now, we use a product of ratios to convert from meters/second to inches/nanosecond: 100 cm 1 in (2.4 × 108 )(100) 9.45 in 2.4 × 108 m 1s · · = = · 9 1s 10 ns 1m 2.54 cm (109 )(2.54) 1 ns Thus, a signal traveling at 80% of the speed of light will travel 9.45 in a nanosecond.
1–1
1–2 AP 1.3
CHAPTER 1. Circuit Variables Remember from Eq. (1.2), current is the time rate of change of charge, or i = dq In dt this problem, we are given the current and asked to find the total charge. To do this, we must integrate Eq. (1.2) to find an expression for charge in terms of current: q(t) =
t 0
i(x) dx
We are given the expression for current, i, which can be substituted into the above expression. To find the total charge, we let t → ∞ in the integral. Thus we have ∞
qtotal =
0
20e
20 −5000x ∞ 20 e (e∞ − e0 ) dx = = −5000 −5000 0
20 20 (0 − 1) = = 0.004 C = 4000 µC −5000 5000
= AP 1.4
−5000x
Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dq . In dt this problem we are given an expression for the charge, and asked to find the maximum current. First we will find an expression for the current using Eq. (1.2): i=
dq d 1 1 t = + 2 e−αt − 2 dt dt α α α
d 1 d t −αt d 1 −αt e = − − e dt α2 dt α dt α2
1 −αt t 1 = 0− − α e−αt − −α 2 e−αt e α α α
1 1 −αt = − +t+ e α α = te−αt Now that we have an expression for the current, we can find the maximum value of the current by setting the first derivative of the current to zero and solving for t: d di = (te−αt ) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0 dt dt Since e−αt never equals 0 for a finite value of t, the expression equals 0 only when (1 − αt) = 0. Thus, t = 1/α will cause the current to be maximum. For this value of t, the current is i=
1 −α/α 1 = e−1 e α α
Remember in the problem statement, α = 0.03679. Using this value for α, i=
1 e−1 ∼ = 10 A 0.03679
Problems AP 1.5
1–3
Start by drawing a picture of the circuit described in the problem statement:
Also sketch the four figures from Fig. 1.6:
[a] Now we have to match the voltage and current shown in the first figure with the polarities shown in Fig. 1.6. Remember that 4A of current entering Terminal 2 is the same as 4A of current leaving Terminal 1. We get (a) v = −20 V, (c) v = 20 V,
i = −4 A; i = −4 A;
(b) v = −20 V, (d) v = 20 V,
i = 4A i = 4A
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention, p = vi = (−20)(−4) = 80 W. Since the power is greater than 0, the box is absorbing power. [c] From the calculation in part (b), the box is absorbing 80 W. AP 1.6
Applying the passive sign convention to the power equation using the voltage and current polarities shown in Fig. 1.5, p = vi. From Eq. (1.3), we know that power is the time rate of change of energy, or p = dw . If we know the power, we can find the dt energy by integrating Eq. (1.3). To begin, find the expression for power: p = vi = (10,000e−5000t )(20e−5000t ) = 200,000e−10,000t = 2 × 105 e−10,000t W Now find the expression for energy by integrating Eq. (1.3): w(t) =
t 0
p(x) dx
1–4
CHAPTER 1. Circuit Variables Substitute the expression for power, p, above. Note that to find the total energy, we let t → ∞ in the integral. Thus we have w= =
AP 1.7
∞ 0
∞
5 −10,000x
2 × 10 e
2 × 105 −10,000x e dx = −10,000 0
2 × 105 2 × 105 −∞ 2 × 105 (e (0 − 1) = = 20 J − e0 ) = −10,000 −10,000 10,000
At the Oregon end of the line the current is leaving the upper terminal, and thus entering the lower terminal where the polarity marking of the voltage is negative. Thus, using the passive sign convention, p = −vi. Substituting the values of voltage and current given in the figure, p = −(800 × 103 )(1.8 × 103 ) = −1440 × 106 = −1440 MW Thus, because the power associated with the Oregon end of the line is negative, power is being generated at the Oregon end of the line and transmitted by the line to be delivered to the California end of the line.
Chapter Problems P 1.1
To begin, we calculate the number of pixels that make up the display: npixels = (1280)(1024) = 1,310,720 pixels Each pixel requires 24 bits of information. Since 8 bits comprise a byte, each pixel requires 3 bytes of information. We can calculate the number of bytes of information required for the display by multiplying the number of pixels in the display by 3 bytes per pixel: nbytes =
1,310,720 pixels 3 bytes · = 3,932,160 bytes/display 1 display 1 pixel
Finally, we use the fact that there are 106 bytes per MB: 3,932,160 bytes 1 MB = 3.93 MB/display · 6 1 display 10 bytes
Problems P 1.2
c = 3 × 108 m/s
so
1 c = 1.5 × 108 m/s 2
1.5 × 108 m 5 × 106 m = 1s xs P 1.3
1–5
so
x=
5 × 106 = 33.3 ms 1.5 × 108
We can set up a ratio to determine how long it takes the bamboo to grow 10 µm First, recall that 1 mm = 103 µm. Let’s also express the rate of growth of bamboo using the units mm/s instead of mm/day. Use a product of ratios to perform this conversion: 1 day 1 hour 1 min 250 10 250 mm · · · = = mm/s 1 day 24 hours 60 min 60 sec (24)(60)(60) 3456 Use a ratio to determine the time it takes for the bamboo to grow 10 µm: 10 × 10−6 m 10/3456 × 10−3 m = 1s xs
P 1.4
so
x=
10 × 10−6 = 3.456 s 10/3456 × 10−3
Volume = area × thickness 106 = (10 × 106 )(thickness) ⇒ thickness =
P 1.5
106 = 0.10 mm 10 × 106
300 × 109 dollars 100 pennies 1 year 1 day 1 hr 1.5 mm 1m · · · · · · 1 year 1 dollar 365.25 days 24 hr 3600 s 1 penny 1000 mm = 1426 m/s
P 1.6
Our approach is as follows: To determine the area of a bit on a track, we need to know the height and width of the space needed to store the bit. The height of the space used to store the bit can be determined from the width of each track on the disk. The width of the space used to store the bit can be determined by calculating the number of bits per track, calculating the circumference of the inner track, and dividing the number of bits per track by the circumference of the track. The calculations are shown below. Width of track =
1 in 25,400µm = 329.87µm/track 77 tracks in
Bits on a track =
1.4 MB 8 bits 1 side = 72,727.273 bits/track 2 sides byte 77 tracks
Circumference of inner track = 2π(1/2 )(25,400µm/in) = 79,796.453µm Width of bit on inner track =
79,796.453µm = 1.0972µm/bit 72,727.273 bits
Area of bit on inner track = (1.0972)(329.87) = 361.934µm2
1–6 P 1.7
CHAPTER 1. Circuit Variables C/m3 = 1.6022 × 10−19 × 1029 = 1.6022 × 1010 C/m3 C/m = (1.6022 × 1010 )(5.4 × 10−4 ) = 8.652 × 106 C/m Therefore, (8.652 × 106 )
Thus, average velocity =
P 1.8
m C =i × ave vel m s 1400 × 10−6 = 161.81 µm/s 8.652
20 × 10−6 C/s = 1.25 × 1014 elec/s 1.6022 × 10−19 C/elec 5280 ft 12 in 2.54 cm 104 µm · · · = 5.32 × 1012 µm [b] m = 3303 mi · 1 mi 1 ft 1 in 1 cm n = 23.5 Therefore, m [a] n =
The number of electrons/second is approximately 23.5 times the number of micrometers between Sydney and San Francisco. P 1.9
First we use Eq. (1.2) to relate current and charge: i=
dq = 20 cos 5000t dt
Therefore, dq = 20 cos 5000t dt To find the charge, we can integrate both sides of the last equation. Note that we substitute x for q on the left side of the integral, and y for t on the right side of the integral: q(t) q(0)
dx = 20
t 0
cos 5000y dy
We solve the integral and make the substitutions for the limits of the integral, remembering that sin 0 = 0: 20 20 sin 5000y t 20 sin 5000t − sin 5000(0) = sin 5000t q(t) − q(0) = 20 = 5000 5000 5000 5000 0
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value at t = 0, so q(t) = 4 × 10−3 sin 5000t C = 4 sin 5000t mC P 1.10
w = qV = (1.6022 × 10−19 )(9) = 14.42 × 10−19 = 1.442 aJ
Problems P 1.11
p = (6)(100 × 10−3 ) = 0.6 W; w(t) =
P 1.12
t 0
p dt
w(10,800) =
10,800 0
3600 s = 10,800 s 1 hr
0.6 dt = 0.6(10,800) = 6480 J
Assume we are standing at box A looking toward box B. Then, using the passive sign convention p = vi, since the current i is flowing into the + terminal of the voltage v. Now we just substitute the values for v and i into the equation for power. Remember that if the power is positive, B is absorbing power, so the power must be flowing from A to B. If the power is negative, B is generating power so the power must be flowing from B to A. [a] p = (20)(15) = 300 W [b] p = (100)(−5) = −500 W [c] p = (−50)(4) = −200 W [d] p = (−25)(−16) = 400 W
P 1.13
3 hr ·
1–7
300 W from A to B 500 W from B to A 200 W from B to A 400 W from A to B
[a]
p = vi = (−20)(5) = −100 W Power is being delivered by the box. [b] Leaving [c] Gaining P 1.14
[a] p = vi = (−20)(−5) = 100 W, so power is being absorbed by the box. [b] Entering [c] Losing
P 1.15
[a] In Car A, the current i is in the direction of the voltage drop across the 12 V battery(the current i flows into the + terminal of the battery of Car A). Therefore using the passive sign convention, p = vi = (−40)(12) = −480 W. Since the power is negative, the battery in Car A is generating power, so Car B must have the ”dead” battery.
1–8
CHAPTER 1. Circuit Variables [b] w(t) =
t 0
w(90) =
1.5 min = 1.5 ·
p dx;
90 0
60 s = 90 s 1 min
480 dx
w = 480(90 − 0) = 480(90) = 43,200 J = 43.2 kJ P 1.16
p = vi;
w=
t 0
p dx
Since the energy is the area under the power vs. time plot, let us plot p vs. t.
Note that in constructing the plot above, we used the fact that 60 hr = 216,000 s = 216 ks p(0) = (6)(15 × 10−3 ) = 90 × 10−3 W p(216 ks) = (4)(15 × 10−3 ) = 60 × 10−3 W 1 w = (60 × 10−3 )(216 × 103 ) + (90 × 10−3 − 60 × 10−3 )(216 × 103 ) = 16.2 kJ 2 P 1.17
[a] To find the power at 625 µs, we substitute this value of time into both the equations for v(t) and i(t) and multiply the resulting numbers to get p(625 µs): v(625 µs) = 50e−1600(625×10
−6 )
− 50e−400(625×10
i(625 µs) = 5 × 10−3 e−1600(625×10
−6 )
−6 )
= 18.394 − 38.94 = −20.546 V
− 5 × 10−3 e−400(625×10
−6 )
= 0.0018394 − 0.003894 = −0.0020546 A p(625 µs) = (−20.546)(−0.0020546) = 42.2 mW [b] To find the energy at 625 µs, we need to integrate the equation for p(t) from 0 to 625 µs. To start, we need an expression for p(t): p(t) = v(t)i(t) = (50)(5 × 10−3 )(e−1600t − e−400t )(e−1600t − e−400t )
Problems
1–9
1 = (e−3200t − 2e−2000t + e−800t ) 4 Now we integrate this expression for p(t) to get an expression for w(t). Note we substitute x for t on the right side of the integral. w(t) =
1 t −3200x (e − 2e−2000x + e−800x )dx 4 0
1 e−3200x e−2000x e−800x + − = 4 −3200 1000 800
t
0
1 1 1 1 e−3200t e−2000t e−800t + − − + − = 4 −3200 1000 800 −3200 1000 800
1 e−3200t e−2000t e−800t + − + 5.625 × 10−4 = 4 −3200 1000 800
Finally, substitute t = 625 µs into the equation for w(t): 1 w(625 µs) = [−4.2292 × 10−5 + 2.865 × 10−4 − 7.5816 × 10−4 + 5.625 × 10−4 ] 4 = 12.137 µJ [c] To find the total energy, we let t → ∞ in the above equation for w(t). Note that this will cause all expressions of the form e−nt to go to zero, leaving only the constant term 5.625 × 10−4 . Thus, 1 wtotal = [5.625 × 10−4 ] = 140.625 µJ 4 P 1.18
[a] v(20 ms) = 100e−1 sin 3 = 5.19 V i(20 ms) = 20e−1 sin 3 = 1.04 A p(20 ms) = vi = 5.39 W [b]
p
vi = 2000e−100t sin.2 150t 1 −100t 1 − cos 300t = 2000e 2 2 = 1000e−100t − 1000e−100t cos 300t
w
=
=
0
=
w
∞
1000e−100t dt − ∞
e−100t 1000 −100 0
∞ 0
1000e−100t cos 300t dt
e−100t −1000 [−100 cos 300t + 300 sin 300t] 2 + (300)2 (100) 100 = 10 − 1000 = 10 − 1 1 × 104 + 9 × 104 = 9 J
∞ 0
1–10 P 1.19
CHAPTER 1. Circuit Variables [a] 0 s ≤ t < 1 s: v = 5 V;
i = 20t A;
p = 100t W
i = 20 A;
p=0W
i = 20 A;
p=0W
i = 80 − 20t A;
p = −400 + 100t W
i = 80 − 20t A;
p = −400 + 100t W
i = −120 + 20t A;
p = −600 + 100t W
i = −120 + 20t A;
p = −600 + 100t W
i = 20 A;
p=0W
1 s < t ≤ 2 s: v = 0 V; 2 s ≤ t < 3 s: v = 0 V; 3 s < t ≤ 4 s: v = −5 V; 4 s ≤ t < 5 s: v = −5 V; 5 s < t ≤ 6 s: v = 5 V; 6 s ≤ t < 7 s: v = 5 V; t > 7 s: v = 0 V;
[b] Calculate the area under the curve from zero up to the desired time:
P 1.20
w(1) =
1 (1)(100) 2
= 50 J
w(6) =
1 (1)(100) 2
− 12 (1)(100) + 12 (1)(100) − 12 (1)(100) = 0 J
w(10) =
1 (1)(100) 2
− 12 (1)(100) + 12 (1)(100) − 12 (1)(100) + 12 (1)(100) = 50 J
[a] p = vi = (100e−500t )(0.02 − 0.02e−500t ) = (2e−500t − 2e−1000t ) W dp = −1000e−500t + 2000e−1000t = 0 dt 2 = e500t
so
ln 2 = 500t
so thus
2e−1000t = e−500t p is maximum at t = 1.4 ms
Problems
1–11
pmax = p(1.4 ms) = 0.5 W ∞
[b] w =
0
= P 1.21
−500t
[2e
−1000t
− 2e
∞ 2 −500t 2 e e−1000t ] dt = − −500 −1000 0
2 4 − = 2 mJ 1000 1000
[a] p = vi = 200 cos(500πt)4.5 sin(500πt) = 450 sin(1000πt) W Therefore, pmax = 450 W [b] pmax (extracting) = 450 W [c]
pavg
=
4×10−3 1 − cos 1000πt 450 450 sin(1000πx) dx = −3 −3 4 × 10 4 × 10 1000π 0
−450 −450 [cos 4π − cos 0] = [1 − 1] = 0 W 4π 4π −450 900 −450 [cos 15π − cos 0] = [−1 − 1] = = 71.62 W = 4π 4π 4π =
[d] pavg P 1.22
[a] q
=
area under i vs. t plot 6(5000) 6(10,000) 8(5000) = + 6(5000) + + 8(15,000) + 2 2 2 = 15,000 + 30,000 + 30,000 + 120,000 + 20,000 = 215,000 C
[b] w
=
pdt =
vi dt
0 ≤ t ≤ 20 ks v = 0.2 × 10−3 t + 8 0 ≤ t ≤ 5000s i = 20 − 1.2 × 10−3 t p
= =
w1
=
(8 + 0.2 × 10−3 t)(20 − 1.2 × 10−3 t) 160 − 5.6 × 10−3 t − 2.4 × 10−7 t2
5000 0
(160 − 5.6 × 10−3 t − 2.4 × 10−7 t2 ) dt
5.6 × 10−3 2 2.4 × 10−7 3 t − t 160t − = 2 3 5000 ≤ t ≤ 15,000s i
=
17 − 0.6 × 10−3 t
p
=
(8 + 0.2 × 10−3 t)(17 − 0.6 × 10−3 t)
= w2
= =
5000
= 720 kJ
0
136 − 1.4 × 10−3 t − 1.2 × 10−7 t2
15,000 5000
(136 − 1.4 × 10−3 t − 1.2 × 10−7 t2 ) dt
1.4 × 10−3 2 1.2 × 10−7 3 t − t 136t − 2 3
15,000 5000
= 1090 kJ
4×10−3 0
CHAPTER 1. Circuit Variables
1–12
15,000 ≤ t ≤ 20,000s i
=
32 − 1.6 × 10−3 t
p
=
(8 + 0.2 × 10−3 t)(32 − 1.6 × 10−3 t) 256 − 6.4 × 10−3 t − 3.2 × 10−7 t2
=
20,000
=
w3
15,000
6.4 × 10−3 2 3.2 × 10−7 3 t 256t − t − 2 3
=
P 1.23
[a]
p
dp dt
20,000
= 226,666.67 J
15,000
w1 + w2 + w3 = 720,000 + 1,090,000 + 226,666.67 = 2036.67 kJ
=
wT
(256 − 6.4 × 10−3 t − 3.2 × 10−7 t2 ) dt
=
vi = [104 t + 5)e−400t ][(40t + 0.05)e−400t ]
=
400 × 103 t2 e−800t + 700te−800t + 0.25e−800t
=
e−800t [400,000t2 + 700t + 0.25]
=
{e−800t [800 × 103 t + 700] − 800e−800t [400,000t2 + 700t + 0.25]}
=
[−3,200,000t2 + 2400t + 5]100e−800t
dp Therefore, = 0 when 3,200,000t2 − 2400t − 5 = 0 dt so pmax occurs at t = 1.68 ms. [b] pmax
[c] w
=
w
=
=
[400,000(.00168)2 + 700(.00168) + 0.25]e−800(.00168)
=
666.34 mW
t 0t 0
pdx 400,000x2 e−800x dx +
t 0
700xe−800x dx + t
t 0
0.25e−800x dx
400,000e−800x 4 2 + [64 × 10 x + 1600x + 2] = 6 −512 × 10 t t 0 −800x −800x 700e e (−800x − 1) + 0.25 4 −800 0 64 × 10 0 When t → ∞ all the upper limits evaluate to zero, hence (400,000)(2) 700 0.25 = 2.97 mJ. w= + + 6 4 512 × 10 64 × 10 800
P 1.24
[a] We can find the time at which the power is a maximum by writing an expression for p(t) = v(t)i(t), taking the first derivative of p(t) and setting it to zero, then solving for t. The calculations are shown below: p
=
0 t < 0,
p = 0 t > 40 s
p dp dt
=
vi = (t − 0.025t2 )(4 − 0.2t) = 4t − 0.3t2 + 0.005t3 W
=
4 − 0.6t + 0.015t2 = 0
0 < t < 40 s
Problems
1–13
Use a calculator to find the two solutions to this quadratic equation: t1 = 8.453 s;
t2 = 31.547 s
Now we must find which of these two times gives the minimum power by substituting each of these values for t into the equation for p(t): p(t1 ) =
(8.453 − 0.025(8.453)2 )(4 − 0.2 · 8.453) = 15.396 W
p(t2 ) =
(31.547 − 0.025(31.547)2 )(4 − 0.2 · 31.547) = −15.396 W
Therefore, maximum power is being delivered at t = 8.453 s. [b] The maximum power was calculated in part (a) to determine the time at which the power is maximum: pmax = 15.396 W (delivered) [c] As we saw in part (a), the other “maximum” power is actually a minimum, or the maximum negative power. As we calculated in part (a), maximum power is being extracted at t = 31.547 s. [d] This maximum extracted power was calculated in part (a) to determine the time at which power is maximum: pmaxext = 15.396 W (extracted) [e] w =
t 0
pdx =
w(0) = w(10) =
t 0
(4x − 0.3x2 + 0.005x3 )dx = 2t2 − 0.1t3 + 0.00125t4
0J
w(30) =
112.50 J
112.50 J
w(40) =
0J
w(20) = 200 J To give you a feel for the quantities of voltage, current, power, and energy and their relationships among one another, they are plotted below:
1–14
CHAPTER 1. Circuit Variables
Problems P 1.25
[a]
p dp dt
=
vi = (8 × 104 te−500t )(15te−500t ) = 12 × 105 t2 e−1000t W
=
12 × 105 [t2 (−1000)e−1000t + e−1000t (2t)]
=
12 × 105 e−1000t [t(2 − 1000t)]
1–15
dp = 0 at t = 0, t = 2 ms dt We know p is a minimum at t = 0 since v and i are zero at t = 0. [b] pmax = 12 × 105 (2 × 10−3 )2 e−2 = 649.61 mW [c] w
= =
P 1.26
12 × 105 12 × 105
∞ 0
t2 e−1000t dt
∞
e−1000t 6 2 [10 t + 2,000t + 2] 3 (−1000) 0
= 2400 µJ
We use the passive sign convention to determine whether the power equation is p = vi or p = −vi and substitute into the power equation the values for v and i, as shown below: pa
=
−va ia = −(−18)(−51) = −918 W
pb
=
vb ib = (−18)(45) = −810 W
pc
=
vc ic = (2)(−6) = −12 W
pd
=
−vd id = −(20)(−20) = 400 W
pe
=
−ve ie = −(16)(−14) = 224 W
pf = vf if = (36)(31) = 1116 W Remember that if the power is positive, the circuit element is absorbing power, whereas is the power is negative, the circuit element is developing power. We can add the positive powers together and the negative powers together — if the power balances, these power sums should be equal:
Pdev = 918 + 810 + 12 = 1740 W;
Pabs = 400 + 224 + 1116 = 1740 W Thus, the power balances and the total power developed in the circuit is 1740 W.
CHAPTER 1. Circuit Variables
1–16 P 1.27
[a] From the diagram and the table we have pa
=
−va ia = −(900)(−22.5) = 20,250 W
pb
=
−vb ib = −(105)(−52.5) = 5512.5 W
pc
=
−vc ic = −(−600)(−30) = −18,000 W
pd
=
vd id = (585)(−52.5) = −30,712.5 W
pe
=
−ve ie = −(−120)(30) = 3600 W
pf
=
vf if = (300)(60) = 18,000 W
pg
=
−vg ig = −(585)(82.5) = −48,262.5 W
ph
=
−vh ih = −(−165)(82.5) = 13,612.5 W
Pdel
=
18,000 + 30,712.5 + 48,262.5 = 96,975 W
Pabs
=
20,250 + 5512.5 + 3600 + 18,000 + 13,612.5 = 60,975 W
Therefore,
Pdel =
Pabs and the subordinate engineer is correct.
[b] The difference between the power delivered to the circuit and the power absorbed by the circuit is 96,975 − 60,975 = 36,000 One-half of this difference is 18,000W, so it is likely that pc or pf is in error. Either the voltage or the current probably has the wrong sign. (In Chapter 2, we will discover that using KCL at the top node, the current ic should be 30 A, not −30 A!) If the sign of pc is changed from negative to positive, we can recalculate the power delivered and the power absorbed as follows:
Pdel
=
30,712.5 + 48,262.5 = 78,975 W
Pabs = 20,250 + 5512.5 + 18,000 + 3600 + 18,000 + 13,612.5 = 78,975 W Now the power delivered equals the power absorbed and the power balances for the circuit. P 1.28
pa
=
va ia = (9)(1.8) = 16.2 W
pb
=
−vb ib = −(−15)(1.5) = 22.5 W
pc
=
−vc ic = −(45)(−0.3) = 13.5 W
pd
=
−vd id = −(54)(−2.7) = 145.8 W
pe
=
ve ie = (−30)(−1) = 30 W
pf
=
−vf if = −(−240)(4) = 960 W
pg
=
−vg ig = −(294)(4.5) = −1323 W
ph
=
vh ih = (−270)(−0.5) = 135 W
Problems
1–17
Therefore,
Pabs = 16.2 + 22.5 + 13.5 + 145.8 + 3 − +960 + 135 = 1323 W Pdel = 1323 W Pabs =
Pdel
Thus, the interconnection satisfies the power check P 1.29
pa
=
va ia = (−160)(−10) = 1600 W
pb
=
vb ib = (−100)(−20) = 2000 W
pc
=
−vc ic = −(−60)(6) = 360 W
pd
=
vd id = (800)(−50) = −40,000 W
pe
=
−ve ie = −(800)(−20) = 16,000 W
pf
=
−vf if = −(−700)(14) = 9800 W
pg
=
−vg ig = −(640)(−16) = 10,240 W
Pdel = 40,000 W Pabs = 1600 + 2000 + 360 + 16,000 + 9800 + 10,000 = 40,000 W
Therefore, Pdel = Pabs = 40,000 W
P 1.30
[a] From an examination of reference polarities, the following elements employ the passive convention: a, c, e, and f . [b]
pa
=
−56 W
pb
=
−14 W
pc
=
150 W
pd
=
−50 W
pe
=
−18 W
pf
=
−12 W
Pabs = 150 W;
Pdel = 56 + 14 + 50 + 18 + 12 = 150 W.
Circuit Elements
2
Assessment Problems AP 2.1
[a] To find vg write a KVL equation clockwise around the left loop, starting below the dependent source: ib ib so vg = + vg = 0 4 4 To find ib write a KCL equation at the upper right node. Sum the currents leaving the node: −
ib + 8 A = 0
so
ib = −8 A
Thus, −8 = −2 V 4 [b] To find the power associated with the 8 A source, we need to find the voltage drop across the source, vi . To do this, write a KVL equation clockwise around the left loop, starting below the voltage source: vg =
−vg + vi = 0
so
vi = vg = −2 V
Using the passive sign convention, ps = (8 A)(vi ) = (8 A)(−2 V) = −16 W Thus the current source generated 16 W of power. 2–1
2–2
CHAPTER 2. Circuit Elements
AP 2.2
[a] Note from the circuit that vx = −25 V. To find α write a KCL equation at the top left node, summing the currents leaving: 15 A + αvx = 0 Substituting for vx , 15 A + α(−25 V) = 0 Thus
α=
so
α(25 V) = 15 A
15 A = 0.6 A/V 25 V
[b] To find the power associated with the voltage source we need to know the current, iv . To find this current, write a KCL equation at the top left node, summing the currents leaving the node: −αvx + iv = 0
so
iv = αvx = (0.6)(−25) = −15 A
Using the passive sign convention, ps = −(iv )(25 V) = −(−15 A)(25 V) = 375 W. Thus the voltage source dissipates 375 W. AP 2.3
[a] A KVL equation gives −vg + vR = 0
so
vR = vg = 1 kV
Note from the circuit that the current through the resistor is ig = 5 mA. Use Ohm’s law to calculate the value of the resistor: vR 1 kV R= = 200 kΩ = ig 5 mA Using the passive sign convention to calculate the power in the resistor, pR = (vR )(ig ) = (1 kV)(5 mA) = 5 W The resistor is dissipating 5 W of power.
Problems
2–3
[b] Note from part (a) the vR = vg and iR = ig . The power delivered by the source is thus psource (−3 W) so vg = − =− psource = −vg ig = 40 V ig 75 mA Since we now have the value of both the voltage and the current for the resistor, we can use Ohm’s law to calculate the resistor value: 40 V vg = 533.33 Ω = R= ig 75 mA The power absorbed by the resistor must equal the power generated by the source. Thus, pR = −psource = −(−3 W) = 3 W [c] Again, note the iR = ig . The power dissipated by the resistor can be determined from the resistor’s current: pR = R(iR )2 = R(ig )2 Solving for ig , 480 mW pr = = 0.0016 R 300 Ω Then, since vR = vg i2g =
so
ig =
vR = RiR = Rig = (300 Ω)(40 mA) = 12 V
√
0.0016 = 0.04 A = 40 mA
so
vg = 12 V
AP 2.4
[a] Note from the circuit that the current throught the conductance G is ig , flowing from top to bottom (from KCL), and the voltage drop across the current source is vg , positive at the top (from KVL). From a version of Ohm’s law, 0.5 A ig = = 10 V G 50 mS Now that we know the voltage drop across the current source, we can find the power delivered by this source: vg =
psource = −vg ig = −(10)(0.5) = −5 W Thus the current source delivers 5 W to the circuit.
2–4
CHAPTER 2. Circuit Elements [b] We can find the value of the conductance using the power, and the value of the current using Ohm’s law and the conductance value: pg = Gvg2
so
G=
pg 9 = 2 = 0.04 S = 40 mS 2 vg 15
ig = Gvg = (40 mS)(15 V) = 0.6 A [c] We can find the voltage from the power and the conductance, and then use the voltage value in Ohm’s law to find the current: pg = Gvg2
so
vg2 =
8W pg = = 40,000 G 200 µS
Thus
vg =
40,000 = 200 V
ig = Gvg = (200 µS)(200 V) = 0.04 A = 40 mA AP 2.5
[a] Redraw the circuit with all of the voltages and currents labeled for every circuit element.
Write a KVL equation clockwise around the circuit, starting below the voltage source: −24 V + v2 + v5 − v1 = 0 Next, use Ohm’s law to calculate the three unknown voltages from the three currents: v2 = 3i2 ;
v5 = 7i5 ;
v1 = 2i1
A KCL equation at the upper right node gives i2 = i5 ; a KCL equation at the bottom right node gives i5 = −i1 ; a KCL equation at the upper left node gives is = −i2 . Now replace the currents i1 and i2 in the Ohm’s law equations with i5 : v2 = 3i2 = 3i5 ;
v5 = 7i5 ;
v1 = 2i1 = −2i5
Now substitute these expressions for the three voltages into the first equation: 24 = v2 + v5 − v1 = 3i5 + 7i5 − (−2i5 ) = 12i5 Therefore i5 = 24/12 = 2 A
Problems
2–5
[b] v1 = −2i5 = −2(2) = −4 V [c] v2 = 3i5 = 3(2) = 6 V [d] v5 = 7i5 = 7(2) = 14 V [e] A KCL equation at the lower left node gives is = i1 . Since i1 = −i5 , is = −2 A. We can now compute the power associated with the voltage source: p24 = (24)is = (24)(−2) = −48 W Therefore 24 V source is delivering 48 W. AP 2.6
Redraw the circuit labeling all voltages and currents:
We can find the value of the unknown resistor if we can find the value of its voltage and its current. To start, write a KVL equation clockwise around the right loop, starting below the 24 Ω resistor: −120 V + v3 = 0 Use Ohm’s law to calculate the voltage across the 8 Ω resistor in terms of its current: v3 = 8i3 Substitute the expression for v3 into the first equation: −120 V + 8i3 = 0
so
i3 =
120 = 15 A 8
Also use Ohm’s law to calculate the value of the current through the 24 Ω resistor: i2 =
120 V = 5A 24 Ω
Now write a KCL equation at the top middle node, summing the currents leaving: −i1 + i2 + i3 = 0
so
i1 = i2 + i3 = 5 + 15 = 20 A
Write a KVL equation clockwise around the left loop, starting below the voltage source: −200 V + v1 + 120 V = 0
so
v1 = 200 − 120 = 80 V
2–6
CHAPTER 2. Circuit Elements Now that we know the values of both the voltage and the current for the unknown resistor, we can use Ohm’s law to calculate the resistance: R =
AP 2.7
v1 80 = 4Ω = i1 20
[a] Plotting a graph of vt versus it gives
Note that when it = 0, vt = 25 V; therefore the voltage source must be 25 V. Since the plot is a straight line, its slope can be used to calculate the value of resistance: ∆v 25 − 0 25 R= = = = 100 Ω ∆i 0.25 − 0 0.25 A circuit model having the same v − i characteristic is a 25 V source in series with a 100Ω resistor, as shown below:
[b] Draw the circuit model from part (a) and attach a 25 Ω resistor:
To find the power delivered to the 25 Ω resistor we must calculate the current through the 25 Ω resistor. Do this by first using KCL to recognize that the current in each of the components is it , flowing in a clockwise direction. Write a KVL equation in the clockwise direction, starting below the voltage source, and using Ohm’s law to express the voltage drop across the resistors in the direction of the current it flowing through the resistors: 25 = 0.2 A −25 V + 100it + 25it = 0 so 125it = 25 so it = 125 Thus, the power delivered to the 25 Ω resistor is p25 = (25)i2t = (25)(0.2)2 = 1 W.
Problems AP 2.8
2–7
[a] From the graph in Assessment Problem 2.7(a), we see that when vt = 0, it = 0.25 A. Therefore the current source must be 0.25 A. Since the plot is a straight line, its slope can be used to calculate the value of resistance: ∆v 25 − 0 25 = = = 100 Ω ∆i 0.25 − 0 0.25 A circuit model having the same v − i characteristic is a 0.25 A current source in parallel with a 100Ω resistor, as shown below: R=
[b] Draw the circuit model from part (a) and attach a 25 Ω resistor:
Note that by writing a KVL equation around the right loop we see that the voltage drop across both resistors is vt . Write a KCL equation at the top center node, summing the currents leaving the node. Use Ohm’s law to specify the currents through the resistors in terms of the voltage drop across the resistors and the value of the resistors. vt vt −0.25 + + = 0, so 5vt = 25, thus vt = 5 V 100 25 p25 = AP 2.9
vt2 = 1 W. 25
First note that we know the current through all elements in the circuit except the 6 kΩ resistor (the current in the three elements to the left of the 6 kΩ resistor is i1 ; the current in the three elements to the right of the 6 kΩ resistor is 30i1 ). To find the current in the 6 kΩ resistor, write a KCL equation at the top node: i1 + 30i1 = i6k = 31i1 We can then use Ohm’s law to find the voltages across each resistor in terms of i1 .
2–8
CHAPTER 2. Circuit Elements The results are shown in the figure below:
[a] To find i1 , write a KVL equation around the left-hand loop, summing voltages in a clockwise direction starting below the 5V source: −5 V + 54,000i1 − 1 V + 186,000i1 = 0 Solving for i1 54,000i1 + 189,000i1 = 6 V
so
240,000i1 = 6 V
Thus, i1 =
6 = 25 µA 240,000
[b] Now that we have the value of i1 , we can calculate the voltage for each component except the dependent source. Then we can write a KVL equation for the right-hand loop to find the voltage v of the dependent source. Sum the voltages in the clockwise direction, starting to the left of the dependent source: +v − 54,000i1 + 8 V − 186,000i1 = 0 Thus, v = 240,000i1 − 8 V = 240,000(25 × 10−6 ) − 8 V = 6 V − 8 V = −2 V We now know the values of voltage and current for every circuit element. Let’s construct a power table:
Problems Element
Current Voltage (µA)
(V)
Power
Power
Equation
(µW)
5V
25
5
p = −vi
−125
54 kΩ
25
1.35
p = Ri2
33.75
1V
25
1
p = −vi
−25
6 kΩ
775
4.65
p = Ri2
3603.75
Dep. source
750
−2
p = −vi
1500
1.8 kΩ
750
1.35
p = Ri2
1012.5
8V
750
8
p = −vi
−6000
2–9
[c] The total power generated in the circuit is the sum of the negative power values in the power table: −125 µW + −25 µW + −6000 µW = −6150 µW Thus, the total power generated in the circuit is 6150 µW. [d] The total power absorbed in the circuit is the sum of the positive power values in the power table: 33.75 µW + 3603.75 µW + 1500 µW + 1012.5 µW = 6150 µW Thus, the total power absorbed in the circuit is 6150 µW. AP 2.10 Given that iφ = 2 A, we know the current in the dependent source is 2iφ = 4 A. We can write a KCL equation at the left node to find the current in the 10 Ω resistor. Summing the currents leaving the node, −5 A + 2 A + 4 A + i10Ω = 0
so
i10Ω = 5 A − 2 A − 4 A = −1 A
Thus, the current in the 10 Ω resistor is 1 A, flowing right to left, as seen in the circuit below.
2–10
CHAPTER 2. Circuit Elements [a] To find vs , write a KVL equation, summing the voltages counter-clockwise around the lower right loop. Start below the voltage source. −vs + (1 A)(10 Ω) + (2 A)(30 Ω) = 0
so
vs = 10 V + 60 V = 70 V
[b] The current in the voltage source can be found by writing a KCL equation at the right-hand node. Sum the currents leaving the node −4 A + 1 A + iv = 0
iv = 4 A − 1 A = 3 A
so
The current in the voltage source is 3 A, flowing top to bottom. The power associated with this source is p = vi = (70 V)(3 A) = 210 W Thus, 210 W are absorbed by the voltage source. [c] The voltage drop across the independent current source can be found by writing a KVL equation around the left loop in a clockwise direction: −v5A + (2 A)(30 Ω) = 0
so
v5A = 60 V
The power associated with this source is p = −v5A i = −(60 V)(5 A) = −300 W This source thus delivers 300 W of power to the circuit. [d] The voltage across the controlled current source can be found by writing a KVL equation around the upper right loop in a clockwise direction: +v4A + (10 Ω)(1 A) = 0
so
v4A = −10 V
The power associated with this source is p = v4A i = (−10 V)(4 A) = −40 W This source thus delivers 40 W of power to the circuit. [e] The total power dissipated by the resistors is given by (i30Ω )2 (30 Ω) + (i10Ω )2 (10 Ω) = (2)2 (30 Ω) + (1)2 (10 Ω) = 120 + 10 = 130 W
Problems
2–11
Problems P 2.1
P 2.2
Vbb
=
no-load voltage of battery
Rbb
=
internal resistance of battery
Rx
=
resistance of wire between battery and switch
Ry
=
resistance of wire between switch and lamp A
Ra
=
resistance of lamp A
Rb
=
resistance of lamp B
Rw
=
resistance of wire between lamp A and lamp B
Rg1
=
resistance of frame between battery and lamp A
Rg2
=
resistance of frame between lamp A and lamp B
S
=
switch
Since we know the device is a resistor, we can use Ohm’s law to calculate the resistance. From Fig. P2.2(a), v = Ri
so
R=
v i
Using the values in the table of Fig. P2.2(b), R= P 2.3
−80 80 160 240 −160 = = = = = 8kΩ −0.02 −0.01 0.01 0.02 0.03
The resistor value is the ratio of the power to the square of the current: 500 2000 4500 8000 12,500 18,000 = 2 = 2 = 2 = = = 500 Ω 2 2 1 2 3 4 5 62
P 2.4
Since we know the device is a resistor, we can use the power equation. From Fig. P2.4(a), p = vi =
v2 R
so
R=
v2 p
Using the values in the table of Fig. P2.4(b) R=
(−4)2 (4)2 (8)2 (12)2 (16)2 (−8)2 = = = = = = 20 Ω 3.2 0.8 0.8 3.2 7.2 12.8
CHAPTER 2. Circuit Elements
2–12 P 2.5
[a] Yes, independent voltage sources can carry whatever current is required by the connection; independent current source can support any voltage required by the connection. [b] 18 V source:
absorbing
5 mA source:
delivering
7 V source:
absorbing
[c]
P18V
=
(5 × 10−3 )(18) = 90 mW
P5mA
=
−(5 × 10−3 )(25) = −125 mW
P7V
=
Pabs =
(5 × 10−3 )(7) = 35 mW
(abs) (del)
(abs)
Pdel = 125 mW
[d] Yes; 18 V source is delivering, the 5 mA source is absorbing, and the 7 V source is absorbing P18V
=
−(5 × 10−3 )(18) = −90 mW
P5mA
=
(5 × 10−3 )(11) = 55 mW
P7V
=
(5 × 10−3 )(7) = 35 mW
Pabs =
(del)
(abs) (abs)
Pdel = 90 mW
P 2.6
Write the two KCL equations, summing the currents leaving the node: KCL, top node:
25A − 20A − 5A = 0A
KCL, bottom node:
− 25A + 20A + 5A = 0A
Write the three KVL equations, summing the voltages in a clockwise direction: KVL, left loop: KVL, right loop:
− v25 + v20 = 0 60V − 100V − v5 − v20 = 0
Problems
2–13
60V − 100V − v5 − v25 = 0
KVL, outer loop:
Note that since v5 , v20 , and v25 are not specified, we can choose values that satisfy the equations. For example, let v5 = −80V, v20 = 40V, and v25 = 40V. There are many other voltage values that will satisfy the equations, too. Thus, the interconnection is valid because it does not violate Kirchhoff’s laws. We can now calculate the power developed by the two voltage sources: pv−sources = p60 + p100 = −(60)(5) + (100)(5) = 200 W. Since the power is positive, the sources are absorbing 200 W of power, or developing −200 W of power. P 2.7
Write the two KCL equations, summing the currents leaving the node: KCL, top node:
− 30A − i8 + 10A = 0A
KCL, bottom node:
30A + i8 − 10A = 0A
Note that the value i8 = −20A satisfies these two equations. Write the three KVL equations, summing the voltages in a clockwise direction: KVL, left loop:
− v30 − 16V + 8V = 0
KVL, right loop:
− 10V + v10 − 8V = 0
KVL, outer loop:
− 16V − 10V + v10 − v30 = 0
Note that v30 = −8V and v10 = 18V satisfy the three KVL equations. The interconnection is valid, since neither of Kirchhoff’s laws is violated. We use the values of i8 , v30 and v10 stated above to calculate the power associated with each source: p30A = −(30)(−8) = 240 W
p16V = −(30)(16) = −480 W
CHAPTER 2. Circuit Elements
2–14
p8V = −(−20)(8) = 160 W
p10V = −(10)(10) = −100 W
p10A = (10)(18) = 180 W
Pabs =
Pdel = 580 W
Power developed by the current sources: pi−sources = p30A + p10A = 240 + 180 = 420 W Since power is positive, the sources are absorbing 420 W of power, or developing −420 W of power. P 2.8
The interconnect is valid since it does not violate Kirchhoff’s laws.
−10 + 40 + v5A − 50 = 0 15 + 5 + i50V = 0
so
v5A = 20 V
i50V = −20 A
so
p15A = −(15)(50) = −750 W p5A = −(5)(20) = −100 W
(KVL) (KCL)
p50V = (20)(50) = 1000 W p10V = (5)(10) = 50 W
p40V = −(5)(40) = −200 W
P 2.9
Pdev =
Pabs = 1050 W
First there is no violation of Kirchhoff’s laws, hence the interconnection is valid. Kirchhoff’s voltage law requires −20 + 60 + v1 − v2 = 0
so
v1 − v2 = −40 V
The conservation of energy law requires −(5 × 10−3 )v2 − (15 × 10−3 )v2 − (20 × 10−3 )(20) + (20 × 10−3 )(60) + (20 × 10−3 )v1 = 0 or v1 − v2 = −40 V Hence any combination of v1 and v2 such that v1 − v2 = −40 V is a valid solution.
Problems
2–15
P 2.10
The interconnection is invalid because KCL is violated at the right-hand node. Summing the currents leaving, −(−5A) − 3A + 8A = 10A = 0 Note that KCL is also violated at the left-hand node. P 2.11
Write the two KCL equations, summing the currents leaving the node: KCL, top node:
75A − 5v∆ − 25A = 0A
KCL, bottom node:
− 75A + 5v∆ + 25A = 0A
To satisfy KCL, note that v∆ = 10 V. Write the three KVL equations, summing the voltages in a clockwise direction: KVL, left loop: KVL, right loop: KVL, outer loop:
− v75 − 50V + vdep − 20V = 0 20V − vdep + v∆ = 0 − v75 − 50V + v∆ = 0
CHAPTER 2. Circuit Elements
2–16
Substitute the value v∆ = 10 V into the second KVL equation and find vdep = 30 V. Substitute the value v∆ = 10 V into the third equation and find v75 = −40 V. These values satisfy the first equation. Thus, the interconnection is valid because it does not violate Kirchhoff’s laws. Use the values for v∆ , v75 , and vdep above to calculate the total power developed in the circuit: p50V = (75)(50) = 3750 W
p75A = (75)(−40) = −3000 W
p20V = [5(10)](20) = 1000 W
pds = −(50)(30) = −1500 W
p25A = −(25)(10) = −250 W
P 2.12
Pdev = 3750 + 1000 = 4750 W =
Pabs
[a] Yes, Kirchhoff’s laws are not violated. (Note that i∆ = −8 A.) [b] No, because the voltages across the independent and dependent current sources are indeterminate. For example, define v1 , v2 , and v3 as shown:
Kirchhoff’s voltage law requires v1 + 20 = v3 v2 + 100 = v3 Conservation of energy requires −8(20) − 8v1 − 16v2 − 16(100) + 24v3 = 0 or v1 + 2v2 − 3v3 = −220 Now arbitrarily select a value of v3 and show the conservation of energy will be satisfied. Examples: If v3 = 200 V then v1 = 180 V and v2 = 100 V. Then 180 + 200 − 600 = −220 (CHECKS) If v3 = −100 V, then v1 = −120 V and v2 = −200 V. Then −120 − 400 + 300 = −220 (CHECKS)
Problems P 2.13
2–17
First, 10va = 5 V, so va = 0.5 V KVL for the outer loop: 5 − 20 + v9A = 0 so v9A = 15 V KVL for the right loop: 5 − 0.5 + vg = 0 so vg = −4.5 V KCL at the top node: 9 + 6 + ids = 9 so ids = −15 A Thus, p9A = −(9)(15) = −135 W
p20V = (9)(20) = 180 W
pvg = −(6)(−4.5) = 27 W
p6A = (6)(0.5) = 3 W
pds = −(15)(5) = −75 W
Pdev =
Pabs = 210 W
P 2.14
[a] Write a KVL equation clockwise aroud the right loop, starting below the 300 Ω resistor: −va + vb = −0
so
va = v b
Using Ohm’s law, va = 300ia
and
vb = 75ib
Substituting, so
300ia = 75ib
ib = 4ia
Write a KCL equation at the top middle node, summing the currents leaving: −ig + ia + ib = 0
so
ig = ia + ib = ia + 4ia = 5ia
Write a KVL equation clockwise around the left loop, starting below the voltage source: −200 V + v40 + va = 0 From Ohm’s law, v40 = 40ig
and
va = 300ia
CHAPTER 2. Circuit Elements
2–18
Substituting, −200 V + 40ig + 300ia = 0 Subsituting for ig : −200 V + 40(5ia ) + 300ia = −200 V + 200ia + 300ia = −200 V + 500ia = 0 Thus, 200 V = 0.4 A 500 [b] From part (a), ib = 4ia = 4(0.4 A) = 1.6 A. 500ia = 200 V
so
ia =
[c] From the circuit, vo = 75 Ω(ib ) = 75 Ω(1.6 A) = 120 V. [d] Use the formula pR = Ri2R to calculate the power absorbed by each resistor: p40Ω = i2g (40 Ω) = (5ia )2 (40 Ω) = [5(0.4)]2 (40 Ω) = (2)2 (40 Ω) = 160 W p300Ω = i2a (300 Ω) = (0.4)2 (300 Ω) = 48 W p75Ω = i2b (75 Ω) = (4ia )2 (75 Ω) = [4(0.4)]2 (75 Ω) = (1.6)2 (75 Ω) = 192 W [e] Using the passive sign convention, psource = −(200 V)ig = −(200 V)(5ia ) = −(200 V)[5(0.4 A)] = −(200 V)(2 A) = −400 W Thus the voltage source delivers 400 W of power to the circuit. Check:
P 2.15
[a]
Pdis = 160 + 48 + 192 = 400 W Pdel = 400 W
vo
=
800 = io
=
8ia + 14ia + 18ia = 40(20) = 800 V 10io 800/10 = 80 A
[b] ig = ia + io = 20 + 80 = 100 A [c] pg (delivered) = (100)(800) = 80,000 W = 80 kW
Problems
2–19
P 2.16
[a] Write a KVL equation clockwise around the right loop: −v60 + v30 + v90 = 0 From Ohm’s law, v60 = (60 Ω)(4 A) = 240 V,
v30 = 30io ,
v90 = 90io
Substituting, −240 V + 30io + 90io = 0 Thus
io =
so
120io = 240 V
240 V = 2A 120
Now write a KCL equatiohn at the top middle node, summing the currents leaving: −ig + 4 A + io = 0
so
ig = 4 A + 2 A = 6 A
[b] Write a KVL equation clockwise around the left loop: −vo + v60 = 0
so
vo = v60 = 240 V
[c] Calculate power using p = vi for the source and p = Ri2 for the resistors: psource = −vo ig = −(240 V)(6 A) = −1440 W p60Ω = 42 (60) = 960 W p30Ω = 30i2o = (30)22 = 120 W p90Ω = 90i2o = (90)22 = 360 W
Pdev = 1440 W
Pabs = 960 + 120 + 360 = 1440 W
CHAPTER 2. Circuit Elements
2–20 P 2.17
[a]
v2 = 2(20) = 40 V v8Ω = 80 − 40 = 40 V i2 = 40 V/8 Ω = 5 A i3 = io − i2 = 2 − 5 = −3 A v4Ω = (−3)(4) = −12 V v1 = 4i3 + v2 = −12 + 40 = 28 V i1 = 28 V/4 Ω = 7 A [b] i4 = i1 + i3 = 7 − 3 = 4 A
[c]
p13Ω
=
42 (13) = 208 W
p8Ω
=
(5)2 (8) = 200 W
p4Ω
=
72 (4) = 196 W
p4Ω
=
(−3)2 (4) = 36 W
p20Ω
=
22 (20) = 80 W
Pdis = 208 + 200 + 196 + 36 + 80 = 720 W
ig = i4 + i2 = 4 + 5 = 9 A Pdev = (9)(80) = 720 W
Problems P 2.18
2–21
[a]
vo
=
20(8) + 16(15) = 400 V
io
=
400/80 = 5 A
ia
=
25 A
P230 (supplied) = (230)(25) = 5750 W ib = 5 + 15 = 20 A P260 (supplied) = (260)(20) = 5200 W [b]
Pdis
Psup
=
(25)2 (2) + (20)2 (8) + (5)2 (4) + (15)2 16 + (20)2 2 + (5)2 (80)
=
1250 + 3200 + 100 + 3600 + 800 + 2000 = 10,950 W
=
5750 + 5200 = 10,950 W
Therefore, P 2.19
Pdis =
Psup = 10,950 W
[a]
v2 = 100 + 4(15) = 160 V; i1 =
v1 100 = 5 A; = 20 20
v1 = 160 − 30(2) = 100 V i3 = i 1 − 2 = 5 − 2 = 3 A
2–22
CHAPTER 2. Circuit Elements vg = v1 + 30i3 = 100 + 30(3) = 190 V vg − 5i4 = v2 Thus
i4 =
so
5i4 = vg − v2 = 190 − 160 = 30 V
30 = 6A 5
ig = i3 + i4 = 3 + 6 = 9 A [b] Calculate power using the formula p = Ri2 : p9 Ω = (9)(2)2 = 36 W; p10 Ω = (10)(2)2 = 40 W; p5 Ω = (5)(6)2 = 180 W;
p11 Ω = (11)(2)2 = 44 W p30 Ω = (30)(3)2 = 270 W p4 Ω = (4)(5)2 = 100 W
p16 Ω = (16)(5)2 = 400 W;
p15 Ω = (15)(4)2 = 240 W
[c] vg = 190 V [d] Sum the power dissipated by the resistors:
pdiss = 36 + 44 + 40 + 270 + 180 + 100 + 400 + 240 = 1310 W
The power associated with the sources is pvolt−source = (100 V)(4 A) = 400 W pcurr−source = −vg ig = −(190 V)(9 A) = −1710 W Thus the total power dissipated is 1310 + 400 = 1710 W and the total power developed is 1710 W, so the power balances. P 2.20
[a] Plot the v − i characteristic
From the plot: R=
∆v (125 − 50) = = 5Ω ∆i (15 − 0)
When it = 0, vt = 50 V; therefore the ideal current source has a current of 10 A
Problems
2–23
[b]
10 + it = i1
and
5i1 = −20it
Therefore, 10 + it = −4it so it = −2 A P 2.21
[a] Plot the v—i characteristic:
From the plot: 130 − (−30) ∆v = = 20 Ω ∆i 8−0 When it = 0, vt = −30 V; therefore the ideal voltage source has a voltage of −30 V. Thus the device can be modeled as a −30 V source in series with a 20 Ω resistor, as shown below: R=
CHAPTER 2. Circuit Elements
2–24
[b] We attach a 40 Ω resistor to the device model developed in part (a):
Write a KVL equation clockwise around the circuit, using Ohm’s law to express the voltage drop across the resistors in terms of the current it through the resistors: −(−30 V) − 20it − 40it = 0 Thus
it =
so
− 60it = −30 V
−30 V = +0.5 A −60
Now calculate the power dissipated by the resistor: p40 Ω = 40i2t = (40)(0.5)2 = 10 W P 2.22
[a]
[b] ∆v = 50 V;
∆i = 5 mA;
R=
50 V ∆v = = 10 kΩ ∆i 5 mA
Problems
2–25
[c]
10,000i1 = 2500is
so
i1 = 0.25is
0.02 = i1 + is = 0.25is + is = 1.25is Thus,
is =
0.02 = 0.016 A = 16 mA 1.25
[d] Predicted open circuit voltage: voc = vs = (0.02)(10,000) = 200 V [e] From the table, the actual open circuit voltage is 140 V. [f] This is a practical current source and is not a linear device. P 2.23
[a] Begin by constructing a plot of voltage versus current:
[b] Since the plot is linear for 0 ≤ is ≤ 24 A amd since R = ∆v/∆i, we can calculate R from the plotted values as follows: 24 − 18 6 ∆v = = = 0.25 Ω ∆i 24 − 0 24 We can determine the value of the ideal voltage source by considering the value of vs when is = 0. When there is no current, there is no voltage drop across the resistor, so all of the voltage drop at the output is due to the voltage source. Thus the value of the voltage source must be 24 V. The model, valid for 0 ≤ is ≤ 24 A, is shown below:
R=
2–26
CHAPTER 2. Circuit Elements [c] The circuit is shown below:
Write a KVL equation in the clockwise direction, starting below the voltage source. Use Ohm’s law to express the voltage drop across the resistors in terms of the current i: −24 V + 0.25i + 1i = 0 Thus,
i=
so
1.25i = 24 V
24 V = 19.2 A 1.25 Ω
[d] The circuit is shown below:
Write a KVL equation in the clockwise direction, starting below the voltage source. Use Ohm’s law to express the voltage drop across the resistors in terms of the current i: −24 V + 0.25i = 0 Thus,
i=
so
0.25i = 24 V
24 V = 96 A 0.25 Ω
[e] The short circuit current can be found in the table of values (or from the plot) as the value of the current is when the voltage vs = 0. Thus, isc = 48 A
(from table)
[f] The plot of voltage versus current constructed in part (a) is not linear (it is piecewise linear, but not linear for all values of is ). Since the proposed circuit model is a linear model, it cannot be used to predict the nonlinear behavior exhibited by the plotted data.
Problems
2–27
P 2.24
vab = 240 − 180 = 60 V; therefore, ie = 60/15 = 4 A ic = ie − 1 = 4 − 1 = 3 A; therefore, vbc = 10ic = 30 V vcd = 180 − vbc = 180 − 30 = 150 V; therefore, id = vcd /(12 + 18) = 150/30 = 5 A ib = id − ic = 5 − 3 = 2 A vac = vab + vbc = 60 + 30 = 90 V R = vac /ib = 90/2 = 45 Ω CHECK:
P 2.25
ig = ib + ie = 2 + 4 = 6 A pdev = (240)(6) = 1440 W
Pdis =
1(180) + 4(45) + 9(10) + 25(12) + 25(18) + 16(15) = 1440 W (CHECKS)
[a]
ib = 60 V/30 Ω = 2 A va = (30 + 60)(2) = 180 V −500 + va + vb = 0 so vb = 500 − va = 500 − 180 = 320 V ie = vb /(60 + 36) = 320/96 = (10/3) A id = ie − ib = (10/3) − 2 = (4/3) A vc = 30id + vb = 40 + 320 = 360 V ic = vc /180 = 360/180 = 2 A vd = 500 − vc = 500 − 360 = 140 V ia = id + ic = 4/3 + 2 = (10/3) A R = vd /ia = 140/(10/3) = 42 Ω [b] ig = ia + ib = (10/3) + 2 = (16/3) A pg (supplied) = (500)(16/3) = 2666.67 W
2–28 P 2.26
CHAPTER 2. Circuit Elements [a] Start with the 22.5 Ω resistor. Since the voltage drop across this resistor is 90 V, we can use Ohm’s law to calculate the current: 90 V = 4A i22.5 Ω = 22.5 Ω Next we can calculate the voltage drop across the 15 Ω resistor by writing a KVL equation around the outer loop of the circuit: −240 V + 90 V + v15 Ω = 0
so
v15 Ω = 240 − 90 = 150 V
Now that we know the voltage drop across the 15 Ω resistor, we can use Ohm’s law to find the current in this resistor: 150 V = 10 A i15 Ω = 15 Ω Write a KCL equation at the middle right node to find the current through the 5 Ω resistor. Sum the currents entering: 4 A − 10 A + i5 Ω = 0
i5 Ω = 10 A − 4 A = 6 A
so
Write a KVL equation clockwise around the upper right loop, starting below the 4 Ω resistor. Use Ohm’s law to express the voltage drop across the resistors in terms of the current through the resistors: −v4 Ω + 90 V + (5 Ω)(−6 A) = 0
so
v4 Ω = 90 V − 30 V = 60 V
Using Ohm’s law we can find the current through the 4 Ω resistor: 60 V = 15 A i4 Ω = 4Ω Write a KCL equation at the middle node. Sum the currents entering: 15 A − 6 A − i20 Ω = 0
so
i20 Ω = 15 A − 6 A = 9 A
Use Ohm’s law to calculate the voltage drop across the 20 Ω resistor: v20 Ω = (20 Ω)(9 A) = 180 V All of the voltages and currents calculated above are shown in the figure below:
Calculate the power dissipated by the resistors using the equation pR = Ri2R : p4Ω = (4)(15)2 = 900 W
p20Ω = (20)(9)2 = 1620 W
p5Ω = (5)(6)2 = 180 W
p22.5Ω = (22.5)(4)2 = 360 W
p15Ω = (15)(10)2 = 1500 W
Problems
2–29
[b] We can calculate the current in the voltage source, ig by writing a KCL equation at the top middle node: ig = 15 A + 4 A = 19 A Now that we have both the voltage and the current for the source, we can calculate the power supplied by the source: pg = −240(19) = −4560 W
[c]
pg (supplied) = 4560 W
Pdis = 900 + 1620 + 180 + 360 + 1500 = 4560 W Therefore,
P 2.27
thus
Psupp =
Pdis
iE − iB − iC = 0 iC = βiB
therefore iE = (1 + β)iB
i2 = −iB + i1 Vo + iE RE − (i1 − iB )R2 = 0 −i1 R1 + VCC − (i1 − iB )R2 = 0 Vo + iE RE + iB R2 −
or
i1 =
VCC + iB R2 R1 + R2
VCC + iB R2 R2 = 0 R1 + R2
Now replace iE by (1 + β)iB and solve for iB . Thus iB = P 2.28
[VCC R2 /(R1 + R2 )] − Vo (1 + β)RE + R1 R2 /(R1 + R2 )
[a] io = 0 because no current can exist in a single conductor connecting two parts of a circuit. [b]
−200 + 8000ig + 12,000ig = 0 so ig = 200/20,000 = 10 mA 3 −3 v∆ = (12 × 10 )(10 × 10 ) = 120 V 5 × 10−3 v∆ = 0.6 A 9000i1 = 3000i2 so i2 = 3i1 0.6 + i1 + i2 = 0 so 0.6 + i1 + 3i1 = 0 thus i1 = −0.15 A
2–30
CHAPTER 2. Circuit Elements [c] i2 = 3i1 = −0.45 A
P 2.29
First note that we know the current through all elements in the circuit except the 200 Ω resistor (the current in the three elements to the left of the 200 Ω resistor is iβ ; the current in the three elements to the right of the 200 Ω resistor is 49iβ ). To find the current in the 200 Ω resistor, write a KCL equation at the top node: iβ + 49iβ = i200 Ω = 50iβ We can then use Ohm’s law to find the voltages across each resistor in terms of iβ . The results are shown in the figure below:
[a] To find iβ , write a KVL equation around the left-hand loop, summing voltages in a clockwise direction starting below the 7.2V source: −7.2 V + 55,000i1 + 0.7 V + 10,000iβ = 0 Solving for iβ 55,000iβ + 10,000iβ = 6.5 V
so
65,000iβ = 6.5 V
Thus, iβ =
6.5 = 100 µA 65,000
Now that we have the value of iβ , we can calculate the voltage for each component except the dependent source. Then we can write a KVL equation for the right-hand loop to find the voltage vy of the dependent source. Sum the voltages in the clockwise direction, starting to the left of the dependent source: −vy − 24,500iβ + 9 V − 10,000iβ = 0 Thus, vy = 9 V − 34,500iβ = 9 V − 34,500(100 × 10−6 ) = 9 V − 3.45 V = 5.55 V
Problems
2–31
[b] We now know the values of voltage and current for every circuit element. Let’s construct a power table: Element
Current Voltage (µA)
(V)
Power
Power
Equation
(µW)
7.2 V
100
7.2
p = −vi
−720
55 kΩ
100
5.5
p = Ri2
550
0.7 V
100
0.7
p = vi
70
200 Ω
5000
1
p = Ri2
5000
Dep. source
4900
5.55
p = vi
27,195
500 Ω
4900
2.45
p = Ri2
12,005
9V
4900
9
p = −vi
−44,100
The total power generated in the circuit is the sum of the negative power values in the power table: −720 µW + −44,100 µW = −44,820 µW Thus, the total power generated in the circuit is 44,820 µW. The total power absorbed in the circuit is the sum of the positive power values in the power table: 550 µW + 70 µW + 5000 µW + 27,195 µW + 12,005 µW = 44,820 µW Thus, the total power absorbed in the circuit is 44,820 µW and the power in the circuit balances. P 2.30
[a] 12 − 2iσ = 5i∆ 5i∆ = 8iσ + 2iσ = 10iσ Therefore, 12 − 2iσ = 10iσ , so iσ = 1 A 5i∆ = 10iσ = 10; so i∆ = 2 A vo = 2iσ = 2 V [b] ig = current out of the positive terminal of the 12 V source vd = voltage drop across the 8i∆ source ig = i∆ + iσ + 8i∆ = 9i∆ + iσ = 19 A vd = 2 + 8 = 10 V
2–32
CHAPTER 2. Circuit Elements
Pgen
=
12ig + 8i∆ (8) = 12(19) + 8(2)(8) = 356 W
Pdiss
=
2iσ ig + 5i2∆ + 8iσ (iσ + 8i∆ ) + 2i2σ + 8i∆ vd
=
2(1)(19) + 5(2)2 + 8(1)(17) + 2(1)2 + 8(2)(10)
=
356 W; Therefore,
P 2.31
40i2 +
Pgen
=
Pdiss = 356 W
5 5 =0 + 40 10
so
i2 = −15.625 mA
v1 = 80i2 = −1.25 V 25i1 +
−1.25 + (−15.625 × 10−3 ) = 0 20
so
i1 = 3.125 mA
vg = 60i1 + 260i1 = 320i1 Therefore, P 2.32
vg = 1 V
VCC R2 (10)(60 × 103 ) = = 6V R1 + R2 100 × 103 R1 R2 (40 × 103 )(60 × 103 ) = = 24 kΩ R1 + R2 100 × 103 5.4 6 − 0.6 = = 0.18 mA 3 24 × 10 + 50(120) (24 + 6) × 103
iB =
iC = βiB = (49)(0.18) = 8.82 mA iE = iC + iB = 8.82 + 0.18 = 9 mA v3d = (0.009)(120) = 1.08 V vbd = Vo + v3d = 1.68 V i2 =
vbd 1.68 × 10−3 = 28 µA = R2 60
i1 = i2 + iB = 28 µA + 180 µA = 208 µA vab = (40 × 103 )(208 × 10−6 ) = 8.32V iCC = iC + i1 = 8.82 mA + 208 µA = 9.028 mA v13 + (8.82 × 10−3 )(750) + 1.08 = 10 Thus,
v13 = 2.305V
Problems P 2.33
2–33
[a]
[b]
P 2.34
From the simplified circuit model, using Ohm’s law and KVL: 400i + 50i + 200i − 250 = 0
so
i = 250/650 = 385 mA
This current is nearly enough to stop the heart, according to Table 2.1, so a warning sign should be posted at the 250 V source. P 2.35
P 2.36
[a] p = i2 R
parm =
250 650
pleg
250 = 650
ptrunk =
2
(400) = 59.17 W
2
250 650
(200) = 29.59 W 2
(50) = 7.40 W
CHAPTER 2. Circuit Elements
2–34
[b]
dT dt
= arm
tarm =
dT dt
dT dt
5 × 104 = 1414.23 s or 23.57 min 35.36
leg
tleg =
2.39 × 10−4 parm = 35.36 × 10−4 ◦ C/s 4
2.39 × 10−4 Pleg = 7.07 × 10−4◦ C/s = 10
5 × 104 = 7,071.13 s or 117.85 min 7.07
= trunk
ttrunk =
2.39 × 10−4 (7.4) = 0.71 × 10−4 ◦ C/s 25
5 × 104 = 70,677.37 s or 1,177.96 min 0.71
[c] They are all much greater than a few minutes. P 2.37
[a] Rarms = 400 + 400 = 800Ω iletgo = 50 mA (minimum) vmin = (800)(50) × 10−3 = 40 V [b] No, 12/800 = 15 mA. Note this current is sufficient to give a perceptible shock.
P 2.38
Rspace = 1 MΩ ispace = 3 mA v = ispace Rspace = 3000 V.
Simple Resistive Circuits
3
Assessment Problems AP 3.1
Start from the right hand side of the circuit and make series and parallel combinations of the resistors until one equivalent resistor remains. Begin by combining the 6 Ω resistor and the 10 Ω resistor in series: 6 Ω + 10 Ω = 16 Ω Now combine this 16 Ω resistor in parallel with the 64 Ω resistor: 16 Ω64 Ω =
1024 (16)(64) = = 12.8 Ω 16 + 64 80
This equivalent 12.8 Ω resistor is in series with the 7.2 Ω resistor: 12.8 Ω + 7.2 Ω = 20 Ω Finally, this equivalent 20 Ω resistor is in parallel with the 30 Ω resistor: 20 Ω30 Ω =
600 (20)(30) = = 12 Ω 20 + 30 50
Thus, the simplified circuit is as shown:
3–1
3–2
CHAPTER 3. Simple Resistive Circuits [a] With the simplified circuit we can use Ohm’s law to find the voltage across both the current source and the 12 Ω equivalent resistor: v = (12 Ω)(5 A) = 60 V [b] Now that we know the value of the voltage drop across the current source, we can use the formula p = −vi to find the power associated with the source: p = −(60 V)(5 A) = −300 W Thus, the source delivers 300 W of power to the circuit. [c] We now can return to the original circuit, shown in the first figure. In this circuit, v = 60 V, as calculated in part (a). This is also the voltage drop across the 30 Ω resistor, so we can use Ohm’s law to calculate the current through this resistor: 60 V =2A 30 Ω Now write a KCL equation at the upper left node to find the current iB : iA =
−5 A + iA + iB = 0
so
iB = 5 A − iA = 5 A − 2 A = 3 A
Next, write a KVL equation around the outer loop of the circuit, using Ohm’s law to express the voltage drop across the resistors in terms of the current through the resistors: −v + 7.2iB + 6iC + 10iC = 0 So Thus
16iC = v − 7.2iB = 60 V − (7.2)(3) = 38.4 V iC =
38.4 = 2.4 A 16
Now that we have the current through the 10 Ω resistor we can use the formula p = Ri2 to find the power: p10 Ω = (10)(2.4)2 = 57.6 W AP 3.2
[a] We can use voltage division to calculate the voltage vo across the 75 kΩ resistor: vo (no load) =
75,000 (200 V) = 150 V 75,000 + 25,000
Problems
3–3
[b] When we have a load resistance of 150 kΩ then the voltage vo is across the parallel combination of the 75 kΩ resistor and the 150 kΩ resistor. First, calculate the equivalent resistance of the parallel combination: 75 kΩ150 kΩ =
(75,000)(150,000) = 50,000 Ω = 50 kΩ 75,000 + 150,000
Now use voltage division to find vo across this equivalent resistance: vo =
50,000 (200 V) = 133.3 V 50,000 + 25,000
[c] If the load terminals are short-circuited, the 75 kΩ resistor is effectively removed from the circuit, leaving only the voltage source and the 25 kΩ resistor. We can calculate the current in the resistor using Ohm’s law: i=
200 V = 8 mA 25 kΩ
Now we can use the formula p = Ri2 to find the power dissipated in the 25 kΩ resistor: p25k = (25,000)(0.008)2 = 1.6 W [d] The power dissipated in the 75 kΩ resistor will be maximum at no load since vo is maximum. In part (a) we determined that the no-load voltage is 150 V, so be can use the formula p = v 2 /R to calculate the power: p75k (max) =
(150)2 = 0.3 W 75,000
AP 3.3
[a] We will write a current division equation for the current throught the 80Ω resistor and use this equation to solve for R: i80Ω = Thus
R (20 A) = 4 A R + 40 Ω + 80 Ω 16R = 480
and
R=
so
20R = 4(R + 120)
480 = 30 Ω 16
3–4
CHAPTER 3. Simple Resistive Circuits [b] With R = 30 Ω we can calculate the current through R using current division, and then use this current to find the power dissipated by R, using the formula p = Ri2 : 40 + 80 (20 A) = 16 A so pR = (30)(16)2 = 7680 W 40 + 80 + 30 [c] Write a KVL equation around the outer loop to solve for the voltage v, and then use the formula p = −vi to calculate the power delivered by the current source: iR =
−v + (60 Ω)(20 A) + (30 Ω)(16 A) = 0 Thus,
so
v = 1200 + 480 = 1680 V
psource = −(1680 V)(20 A) = −33,600 W
Thus, the current source generates 33,600 W of power. AP 3.4
[a] First we need to determine the equivalent resistance to the right of the 40 Ω and 70 Ω resistors: 1 1 1 1 1 Req = 20 Ω30 Ω(50 Ω + 10 Ω) + + = so = Req 20 Ω 30 Ω 60 Ω 10 Ω Thus,
Req = 10 Ω
Now we can use voltage division to find the voltage vo : 40 (60 V) = 20 V 40 + 10 + 70 [b] The current through the 40 Ω resistor can be found using Ohm’s law: vo =
20 V vo = = 0.5 A 40 40 Ω This current flows from left to right through the 40 Ω resistor. To use current division, we need to find the equivalent resistance of the two parallel branches containing the 20 Ω resistor and the 50 Ω and 10 Ω resistors: i40Ω =
(20)(60) = 15 Ω 20 + 60 Now we use current division to find the current in the 30 Ω branch: 15 (0.5 A) = 0.16667 A = 166.67 mA i30Ω = 15 + 30 20 Ω(50 Ω + 10 Ω) =
Problems
3–5
[c] We can find the power dissipated by the 50 Ω resistor if we can find the current in this resistor. We can use current division to find this current from the current in the 40 Ω resistor, but first we need to calculate the equivalent resistance of the 20 Ω branch and the 30 Ω branch: (20)(30) 20 Ω30 Ω = = 12 Ω 20 + 30 Current division gives: i50Ω = Thus, AP 3.5
12 (0.5 A) = 0.08333 A 12 + 50 + 10 p50Ω = (50)(0.08333)2 = 0.34722 W = 347.22 mW
[a]
We can find the current i using Ohm’s law: i=
1V = 0.01 A = 10 mA 100 Ω
[b]
Rm = 50 Ω5.555 Ω = 5 Ω We can use the meter resistance to find the current using Ohm’s law: imeas = AP 3.6
[a]
1V = 0.009524 = 9.524 mA 100 Ω + 5 Ω
3–6
CHAPTER 3. Simple Resistive Circuits Use voltage division to find the voltage v: 75,000 (60 V) = 50 V v= 75,000 + 15,000 [b]
The meter resistance is a series combination of resistances: Rm = 149,950 + 50 = 150,000 Ω We can use voltage division to find v, but first we must calculate the equivalent resistance of the parallel combination of the 75 kΩ resistor and the voltmeter: 75,000 Ω150,000 Ω = Thus, AP 3.7
vmeas =
(75,000)(150,000) = 50 kΩ 75,000 + 150,000
50,000 (60 V) = 46.15 V 50,000 + 15,000
[a] Using the condition for a balanced bridge, the products of the opposite resistors must be equal. Therefore, (1000)(150) = 1500 Ω = 1.5 kΩ 100 [b] When the bridge is balanced, there is no current flowing through the meter, so the meter acts like an open circuit. This places the following branches in parallel: The branch with the voltage source, the branch with the series combination R1 and R3 and the branch with the series combination of R2 and Rx . We can find the current in the latter two branches using Ohm’s law: 100Rx = (1000)(150)
so
Rx =
5V 5V = 20 mA; iR2 ,Rx = = 2 mA 100 Ω + 150 Ω 1000 + 1500 We can calculate the power dissipated by each resistor using the formula p = Ri2 : iR1 ,R3 =
p100Ω = (100 Ω)(0.02 A)2 = 40 mW p150Ω = (150 Ω)(0.02 A)2 = 60 mW p1000Ω = (1000 Ω)(0.002 A)2 = 4 mW p1500Ω = (1500 Ω)(0.002 A)2 = 6 mW Since none of the power dissipation values exceeds 250 mW, the bridge can be left in the balanced state without exceeding the power-dissipating capacity of the resistors.
Problems AP 3.8
3–7
Convert the three Y-connected resistors, 20 Ω, 10 Ω, and 5 Ω to three ∆-connected resistors Ra , Rb , and Rc . To assist you the figure below has both the Y-connected resistors and the ∆-connected resistors
(5)(10) + (5)(20) + (10)(20) = 17.5 Ω 20 (5)(10) + (5)(20) + (10)(20) Rb = = 35 Ω 10 (5)(10) + (5)(20) + (10)(20) = 70 Ω Rc = 5 Ra =
The circuit with these new ∆-connected resistors is shown below:
From this circuit we see that the 70 Ω resistor is parallel to the 28 Ω resistor: 70 Ω28 Ω =
(70)(28) = 20 Ω 70 + 28
Also, the 17.5 Ω resistor is parallel to the 105 Ω resistor: 17.5 Ω105 Ω =
(17.5)(105) = 15 Ω 17.5 + 105
Once the parallel combinations are made, we can see that the equivalent 20 Ω resistor is in series with the equivalent 15 Ω resistor, giving an equivalent resistance
3–8
CHAPTER 3. Simple Resistive Circuits of 20 Ω + 15 Ω = 35 Ω. Finally, this equivalent 35 Ω resistor is in parallel with the other 35 Ω resistor: 35 Ω35 Ω =
(35)(35) = 17.5 Ω 35 + 35
Thus, the resistance seen by the 2 A source is 17.5 Ω, and the voltage can be calculated using Ohm’s law: v = (17.5 Ω)(2 A) = 35 V
Problems
3–9
Problems P 3.1
[a] The 6 Ω and 12 Ω resistors are in series, as are the 9 Ω and 7 Ω resistors. The simplified circuit is shown below:
[b] The 3 kΩ, 5 kΩ, and 7 kΩ resistors are in series. The simplified circuit is shown below:
[c] The 300 Ω, 400 Ω, and 500 Ω resistors are in series. The simplified circuit is shown below:
P 3.2
[a] The 10 Ω and 40 Ω resistors are in parallel, as are the 100 Ω and 25 Ω resistors.
3–10
CHAPTER 3. Simple Resistive Circuits The simplified circuit is shown below:
[b] The 9 kΩ, 18 kΩ, and 6 kΩ resistors are in parallel. The simplified circuit is shown below:
[c] The 600 Ω, 200 Ω, and 300 Ω resistors are in series. The simplified circuit is shown below:
P 3.3
[a] p4Ω
=
i2s 4 = (12)2 4 = 576 W
p18Ω = (4)2 18 = 288 W
p3Ω
=
(8)2 3 = 192 W
p6Ω = (8)2 6 = 384 W
[b] p120V (delivered) = 120is = 120(12) = 1440 W [c] pdiss = 576 + 288 + 192 + 384 = 1440 W P 3.4
i2 = 8 A, is = 12 A [a] From Ex. 3-1: i1 = 4 A, at node x: −12 + 4 + 8 = 0, at node y: 12 − 4 − 8 = 0
Problems
[b] v1
=
4is = 48 V
3–11
v3 = 3i2 = 24 V
v4 = 6i2 = 48 V v2 = 18i1 = 72 V loop abda: −120 + 48 + 72 = 0, loop bcdb: −72 + 24 + 48 = 0, loop abcda: −120 + 48 + 24 + 48 = 0 P 3.5
Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Req = 6 + 12 + [4(9 + 7)] = 18 + (416) = 18 + 3.2 = 21.2 Ω [b] Req = 4 k + [10 k(3 k + 5 k + 7 k)] = 4 k + (10 k15 k) = 4 k + 6 k = 10 kΩ [c] Req = (300 + 400 + 500) + (6001200) = 1200 + 400 = 1600 Ω
P 3.6
Always work from the side of the circuit furthest from the source. Remember that the current in all series-connected circuits is the same, and that the voltage drop across all parallel-connected resistors is the same. [a] Req = 18 + (10025(22 + (1040))) = 18 + (20(22 + 8) = 18 + 12 = 30 Ω [b] Req = 10 k(5 k + 2 k + (9 k18 k6 k)) = 10 k(7 k + 3 k) = 10 k10 k = 5 kΩ [c] Req = 600200300(250 + 150) = 600200300400 = 80 Ω
P 3.7
[a] Req = 12 + (24(30 + 18)) + 10 = 12 + (2448) + 10 = 12 + 16 + 10 = 38 Ω [b] Req = 4 k30 k60 k(1.2 k + (7.2 k2.4 k) + 2 k) = 4 k30 k60 k(3.2 k + 1.8 k) = 4 k30 k60 k5 k = 2 kΩ
P 3.8
[a] 520 = 100/25 = 4 Ω 918 = 162/27 = 6 Ω Rab = 5 + 12 + 3 = 20 Ω
520 + 918 + 10 = 20 Ω 2030 = 600/50 = 12 Ω
3–12
CHAPTER 3. Simple Resistive Circuits [b] 5 + 15 = 20 Ω 2060 = 1200/80 = 15 Ω
36 = 18/9 = 2 Ω
15 + 10 = 25 Ω
36 + 3020 = 2 + 12 = 14 Ω
2575 = 1875/100 = 18.75 Ω
2614 = 364/40 = 9.1 Ω
18.75 + 11.25 = 30 Ω
Rab = 2.5 + 9.1 + 3.4 = 15 Ω
[c] 3 + 5 = 8 Ω
P 3.9
3020 = 600/50 = 12 Ω
6040 = 2400/100 = 24 Ω
812 = 96/20 = 4.8 Ω
24 + 6 = 30 Ω
4.8 + 5.2 = 10 Ω
3010 = 300/40 = 7.5 Ω
45 + 15 = 60 Ω
Rab = 1.5 + 7.5 + 1.0 = 10 Ω
[a] For circuit (a) Rab = 360(90 + 120(160 + 200)) = 360(90 + (120360)) = 360(90 + 90) = 360180 = 120 Ω For circuit (b) 1 1 1 1 1 1 30 1 = + + + + = = Re 20 15 20 4 12 60 2 Re = 2 Ω Re + 16 = 18 Ω 1818 = 9 Ω Rab = 10 + 8 + 9 = 27 Ω For circuit (c) 1530 = 10 Ω 10 + 20 = 30 Ω 6030 = 20 Ω 20 + 10 = 30 Ω 3080(40 + 20) = 308060 = 16 Ω Rab = 16 + 24 + 10 = 50 Ω
Problems
3–13
[b] Pa = (0.032 )(120) = 108 mW
P 3.10
Pb =
1442 = 768 W 27
Pc =
0.082 = 128 µ W 50
The equivalent resistance to the right of the 10 Ω resistor is (6 + 5(8 + 12)) = 6 + 520 = 6 + 4 = 10 Ω. We can use current division to see that the current then splits equally between the two 10 Ω branches. Thus the current through the 6 Ω branch in the original circuit is 5 A. This 5 A current splits between the branch with the 5 Ω resistor and the branch with the 8 + 12 = 20 Ω resistor and we use current division to determine the current in the 5 Ω resistor: i5Ω =
20 (5) = 4 A 20 + 5
Thus the power in the 5 Ω resistor is p5Ω = i25Ω (5) = 42 (5) = 80 W P 3.11
[a]
Req = 2 + 2 + (1/4 + 1/5 + 1/20)−1 = 6 Ω ig = 120/6 = 20 A v4Ω = 120 − (2 + 2)20 = 40 V io = 40/4 = 10 A i(15+5)Ω = 40/(15 + 5) = 2 A vo = (5)(2) = 10 V [b] i15Ω = 2 A;
P15Ω = (2)2 (15) = 60 W
[c] P120V = (120)(20) = 2.4 kW
3–14
P 3.12
CHAPTER 3. Simple Resistive Circuits [a] Req = RR = [b] Req
R R2 = 2R 2
RRR · · · R (n R’s) R = R n−1 R2 /(n − 1) R2 R = = = R + R/(n − 1) nR n [c] One solution: =
2000 2000 so n= =5 n 400 You can place 5 identical 2 kΩ resistors in parallel to get an equivalent resistance of 400 Ω. 400 =
[d] One solution: 12,500 =
100,000 n
so
n=
100,000 =8 12,500
You can place 8 identical 100 kΩ resistors in parallel to get an equivalent resistance of 12.5 kΩ. P 3.13
[a] We can calculate the no-load voltage using voltage division to determine the voltage drop across the 500 Ω resistor: vo =
500 (75 V) = 15 V (2000 + 500)
[b] We can calculate the power if we know the current in each of the resistors. Under no-load conditions, the resistors are in series, so we can use Ohm’s law to calculate the current they share: i=
75 V = 0.03 A = 30 mA 2000 Ω + 500 Ω
Now use the formula p = Ri2 to calculate the power dissipated by each resistor: PR1 = (2000)(0.03)2 = 1.8 W = 1800 mW PR2 = (500)(0.03)2 = 0.45 W = 450 mW [c] Since R1 and R2 carry the same current and R1 > R2 to satisfy the no-load voltage requirement, first pick R1 to meet the 1 W specification iR 1
75 − 15 = , R1
Thus, R1 ≥
602 1
Therefore, or
60 R1
2
R1 ≥ 3600 Ω
R1 ≤ 1
Problems
3–15
Now use the voltage specification: R2 (75) = 15 R2 + 3600 Thus, R2 = 900 Ω R1 = 1600 Ω and R2 = 400 Ω are the smallest values of resistors that satisfy the 1 W specification. P 3.14
Use voltage division to determine R2 from the no-load voltage specification: 6V=
R2 (18 V); (R2 + 40)
Thus,
12R2 = 240
18R2 = 6(R2 + 40)
so
so
R2 =
240 = 20 Ω 12
Now use voltage division again, this time to determine the value of Re , the parallel combination of R2 and RL . We use the loaded voltage specification: 4V=
Re (18 V) (40 + Re ) 14Re = 160
Thus,
18Re = 4(40 + Re )
so
so
Re =
160 = 11.43 Ω 14
Now use the definition Re to calculate the value of RL given that R2 = 20 Ω: Re =
20RL = 11.43 20 + RL
Therefore, P 3.15
so
8.57RL = 228.6
20RL = 11.43(RL + 20) and
RL =
226.8 = 26.67 Ω 8.57
[a] From the constraint on the no-load voltage, R2 (40) = 8 so R1 = 4R2 R1 + R2 From the constraint on the loaded voltage divider: 3600R2 3600 + R2 7.5 = (40) 3600R2 R1 + 3600 + R2 3600R2 3600 + R2 = (40) 3600R2 4R2 + 3600 + R2
CHAPTER 3. Simple Resistive Circuits
3–16
= So,
144,000R2 3600R2 (40) = 4R2 (3600 + R2 ) + 3600R2 4R22 + 18,000R2 144,000 = 7.5 4R2 + 18,000
.·. R2 = 300 Ω and
R1 = 4R2 = 1200 Ω
[b] Power dissipated in R1 will be maximum when the voltage across R1 is maximum. This will occur under load conditions. (32.5)2 = 880.2 mW 1200 So specify a 1 W power rating for the resistor R1 . The power dissipated in R2 will be maximum when the voltage drop across R2 is maximum. This occurs under no-load conditions with vo = 8 V.
vR1 = 40 − 7.5 = 32.5 V;
PR1 =
(8)2 = 213.3 m W PR2 = 300 So specify a 1/4 W power rating for the resistor R2 . P 3.16
Refer to the solution of Problem 3.15. The divider will reach its dissipation limit when the power dissipated in R1 equals 1 W So (vR2 1 /1200) = 1; Therefore,
Re (40) = 5.359, 1200 + Re
1200RL = 185.641 1200 + RL P 3.17
vR1 = 34.641 V
vo = 40 − 34.641 = 5.359 V and
Re = 185.641 Ω
.·. RL = 219.62 Ω
[a]
120 kΩ + 30 kΩ = 150 kΩ 75 kΩ150 kΩ = 50 kΩ vo1 = vo =
240 (50,000) = 160 V (25,000 + 50,000) 120,000 (vo1 ) = 128 V, (150,000)
vo = 128 V
Problems
3–17
[b]
i=
240 = 2.4 mA 100,000
75,000i = 180 V 120,000 vo = (180) = 144 V; 150,000
vo = 144 V
[c] It removes loading effect of second voltage divider on the first voltage divider. Observe that the open circuit voltage of the first divider is 75,000 vo1 = (240) = 180 V (100,000) Now note this is the input voltage to the second voltage divider when the current controlled voltage source is used. P 3.18
(24)2 = 36, R1 + R2 + R3
Therefore, R1 + R2 + R3 = 16 Ω
(R1 + R2 )24 = 12 (R1 + R2 + R3 ) Therefore, 2(R1 + R2 ) = R1 + R2 + R3 Thus, R1 + R2 = R3 ;
2R3 = 16;
R3 = 8 Ω
R2 (24) =6 R1 + R2 + R3 4R2 = R1 + R2 + R3 R2 = 4 Ω; P 3.19
so R2 = R3 /2 = 4 Ω
R1 = 16 − 8 − 4 = 4 Ω
Note – in the problem description, the first equation defines R1 not RL . [a] At no load:
R2 vs . R1 + R2 Re vo = αvs = vs , R1 + Re
vo = kvs =
At full load: Therefore k
=
α
=
R2 R1 + R2 Re R1 + Re
and and
where Re = (1 − k) R2 k (1 − α) Re R1 = α
R1 =
Ro R2 Ro + R2
CHAPTER 3. Simple Resistive Circuits
3–18
Thus
1−α α
Solving for R2 yields Also, [b] R1 R2
R2 Ro (1 − k) R2 = Ro + R2 k R2 =
(k − α) Ro α(1 − k)
(1 − k) (k − α) R2 .·. Ro R1 = k αk 0.05 = Ro = 2.5 kΩ 0.68 0.05 = Ro = 14.167 kΩ 0.12 R1 =
[c]
Maximum dissipation in R2 occurs at no load, therefore, [(60)(0.85)]2 = 183.6 mW 14,167 Maximum dissipation in R1 occurs at full load.
PR2 (max) =
PR1 (max) =
[60 − 0.80(60)]2 = 57.60 mW 2500
[d ]
P 3.20
PR1
=
PR2
=
(60)2 = 1.44 W = 1440 mW 2500 (0)2 =0W 14,167
[a] Let vo be the voltage across the parallel branches, positive at the upper terminal, then ig = vo G1 + vo G2 + · · · + vo GN = vo (G1 + G2 + · · · + GN ) It follows that
vo =
ig (G1 + G2 + · · · + GN )
Problems The current in the k th branch is ig Gk ik = [G1 + G2 + · + GN ] [b] io = P 3.21
ik = v o G k ;
3–19
Thus,
120(0.00125) = 30 mA [0.0025 + 0.0004167 + 0.00125 + 0.000625 + 0.0002083]
Begin by using the relationships among the branch currents to express all branch currents in terms of i4 : i1 = 2i2 = 2(10i3 ) = 20i4 i2 = 10i3 = 10i4 i3 = i 4 Now use KCL at the top node to relate the branch currents to the current supplied by the source. i1 + i2 + i3 + i4 = 8 mA Express the branch currents in terms of i4 and solve for i4 : 8 mA = 20i4 + 10i4 + i4 + i4 = 32i4
so
i4 =
0.008 = 0.00025 = 0.25 mA 32
Since the resistors are in parallel, the same voltage, 4 V appears across each of them. We know the current and the voltage for R4 so we can use Ohm’s law to calculate R4 : R4 =
vg 4V = 16 kΩ = i4 0.25 mA
Calculate i3 from i4 and use Ohm’s law as above to find R3 : i3 = i4 = 0.25 mA
.·. R3 =
vg 4V = = 16 kΩ i3 0.25 mA
Calculate i2 from i4 and use Ohm’s law as above to find R2 : i2 = 10i4 = 10(0.25 mA) = 2.5 mA
vg 4V = 1.6 kΩ .·. R2 = = i2 2.5 mA
Calculate i1 from i4 and use Ohm’s law as above to find R1 : i1 = 20i4 = 20(0.25 mA) = 5 mA The resulting circuit is shown below:
.·. R1 =
vg 4V = 800 Ω = i1 5 mA
CHAPTER 3. Simple Resistive Circuits
3–20 P 3.22
[a]
Using voltage division, 18 (40) = 15 V positive at the top v18Ω = 18 + 30 [b]
Using current division, 24 (60 × 10−3 ) = 20 mA flowing from right to left i30Ω = 24 + 30 + 18 [c]
The 9 mA current in the 1.2 kΩ resistor is also the current in the 2 kΩ resistor. It then divides among the 4 kΩ, 30 kΩ, and 60 kΩ resistors. 4 kΩ60 kΩ = 3.75 kΩ Using current division, 3.75 k (9 × 10−3 ) = 1 m A, i30 kΩ = 30 k + 3.75 k [d]
flowing bottom to top
Problems
3–21
The voltage drop across the 4 kΩ resistor is the same as the voltage drop across the series combination of the 1.2 kΩ, the (7.2 k2.4 k)Ω combined resistor, and the 2 kΩ resistor. Note that (7200)(2400) 7.2 k2.4 k = = 1.8 kΩ 9600 Using voltage division, 1800 (50) = 18 V positive at the top vo = 1200 + 1800 + 2000 P 3.23
[a]
First, note the following: 189 = 6 Ω; 205 = 4 Ω; and the voltage drop across the 18 Ω resistor is the same as the voltage drop across the parallel combination of the 18 Ω and 9 Ω resistors. Using voltage division, 6 (0.1 V) = 30 mV positive at the left vo = 6 + 4 + 10 [b]
The equivalent resistance of the 5 Ω, 15 Ω, and 60 Ω resistors is Re = (5 + 15)60 = 15 Ω Using voltage division to find the voltage across the equivalent resistance, 15 v Re = (10) = 6 V 15 + 10 Using voltage division again, 15 vo = (6) = 4.5 V positive at the top 5 + 15
CHAPTER 3. Simple Resistive Circuits
3–22 [c]
Find equivalent resistance on the right side Rr = 5.2 +
(12)(5 + 3) = 10 Ω (12 + 3 + 5)
Find voltage bottom to top across Rr (10)(3) = 30 V Find the equivalent resistance on the left side Rl = 6 +
(40)(45 + 15) = 30 Ω (40 + 45 + 15)
The current in the 6 Ω is 30 = 1 A left to right i6 Ω = 30 Use current division to find io
40 io = (1) = 0.4 A 40 + 15 + 45 P 3.24
[a] v20k =
bottom to top
20 (45) = 36 V 20 + 5
v90k =
90 (45) = 27 V 90 + 60
vx = v20k − v90k = 36 − 27 = 9 V [b] v20k =
20 (Vs ) = 0.8Vs 25
v90k =
90 (Vs ) = 0.6Vs 150
vx = 0.8Vs − 0.6Vs = 0.2Vs P 3.25
15075 = 50 Ω The equivalent resistance to the right of the 90 Ω resistor is (50 + 40)(60 + 30) = 45 Ω
Problems
3–23
The voltage drop across this equivalent resistance is 45 (3) = 1 V 90 + 45 Use voltage division to find v1 , which is the voltage drop across the parallel combination whose equivalent resistance is 50 Ω: 50 (1) = 5/9 V 50 + 40
v1 =
Use voltage division to find v2 : 30 (1) = 1/3 V 30 + 60
v2 = P 3.26
i300Ω =
1000 + 200 (15 × 10−3 ) = 10 mA 1000 + 200 + 300 + 300
v300Ω = (300)(10 × 10−3 ) = 3 V i200Ω = i1 kΩ = 15 × 10−3 − i300Ω = 5 mA v1k = (1000)(5 × 10−3 ) = 5 V vo = 3 − 5 = −2 V P 3.27
5 Ω20 Ω = 4 Ω; Therefore, ig = i6Ω =
P 3.28
4 Ω + 6 Ω = 10 Ω;
1040 = 8 Ω;
125 = 12.5 A 8+2
(40)(12.5) = 10 A; 50
io =
(5)(10) =2A 25
[a] Combine resistors in series and parallel to find the equivalent resistance seen by the source. Use this equivalent resistance to find the current through the source, and use current division to find io : 80 + 70 = 150 Ω i24Ω = io =
10015090 = 36 Ω
60 V =1A 60Ω
36 10090150 (1) = = 0.24 A 150 150
36 + 24 = 60 Ω
3–24
CHAPTER 3. Simple Resistive Circuits [b] Use current division to find the current through the 90 Ω resistor from the source current found in part (a), and use the calculated current to find the power in the 90 Ω resistor: 36 10090150 i90Ω = (1) = = 0.4 A 90 90 p90Ω = i290Ω (90) = (0.4)2 (90) = 14.4 W
P 3.29
[a] v9Ω = (1)(9) = 9 V i2Ω = 9/(2 + 1) = 3 A i4Ω = 1 + 3 = 4 A; v25Ω = (4)(4) + 9 = 25 V i25Ω = 25/25 = 1 A; i3Ω = i25Ω + i9Ω + i2Ω = 1 + 1 + 3 = 5 A; v40Ω = v25Ω + v3Ω = 25 + (5)(3) = 40 V i40Ω = 40/40 = 1 A i520Ω = i40Ω + i25Ω + i4Ω = 1 + 1 + 4 = 6 A v520Ω = (4)(6) = 24 V v32Ω = v40Ω + v520Ω = 40 + 24 = 64 V i32Ω = 64/32 = 2 A; i10Ω = i32Ω + i520Ω = 2 + 6 = 8 A vg = 10(8) + v32Ω = 80 + 64 = 144 V. [b] P20Ω =
(v520Ω )2 242 = = 28.8 W 20 20
P 3.30
4010 = 8 Ω
1560 = 12 Ω
Problems
i1 =
(3)(40) = 2 A; (60)
3–25
vx = 8i1 = 16 V
vg = 20i1 = 40 V v60 = vg − vx = 24 V Pdevice = P 3.31
242 162 402 + + = 75.2 W 60 10 40
[a] The model of the ammeter is an ideal ammeter in parallel with a resistor whose resistance is given by 100 µV = 10 Ω. Rs = 10 µA We can calculate the current through the real meter using current division: (10/99) 10 1 (imeas ) = (imeas ) = imeas 10 + (10/99) 990 + 10 100
im = [b] Rs =
100 µV = 10 Ω. 10 µA
im =
(100/999,990) 1 (imeas ) = (imeas ) 10 + (100/999,990) 100,000
[c] Yes P 3.32
Measured value: 6020.1 = 15.056 Ω ig =
50 = 1.9955 A; (15.056 + 10)
imeas = (1.9955)
60 = 1.495 A 80.1
True value: 6020 = 15 Ω ig =
50 50 = = 2.0 A; (15 + 10) 25
% error =
itrue = (2)
1.495 − 1 × 100 = −0.3488% 1.5
60 = 1.5 A 80
3–26 P 3.33
CHAPTER 3. Simple Resistive Circuits Begin by using current division to find the actual value of the current io : itrue =
15 (50 mA) = 12.5 mA 15 + 45
imeas =
15 (50 mA) = 12.48 mA 15 + 45 + 0.1
12.48 − 1 100 = −0.1664% % error = 12.5 P 3.34
For all full-scale readings the total resistance is RV + Rmovement =
full-scale reading 10−3
We can calculate the resistance of the movement as follows: Rmovement =
20 mV = 20 Ω 1 mA
Therefore,
RV = 1000 (full-scale reading) − 20
[a] RV = 1000(50) − 20 = 49, 980 Ω [b] RV = 1000(5) − 20 = 4980 Ω [c] RV = 1000(0.25) − 20 = 230 Ω [d] RV = 1000(0.025) − 20 = 5 Ω P 3.35
[a] vmeas = (50 × 10−3 )[1545(4980 + 20)] = 0.5612 V [b] vtrue = (50 × 10−3 )(1545) = 0.5625 V 0.5612 − 1 × 100 = −0.23% % error = 0.5625
P 3.36
Original meter: Modified meter:
Re = Re =
50 × 10−3 = 0.01 Ω 5 (0.02)(0.01) = 0.00667 Ω 0.03
.·. (Ifs )(0.00667) = 50 × 10−3 .·. Ifs = 7.5 A
Problems P 3.37
At full scale the voltage across the shunt resistor will be 100 mV; therefore the power dissipated will be (100 × 10−3 )2 RA
PA =
Therefore RA ≥
(100 × 10−3 )2 = 40 mΩ 0.25
Otherwise the power dissipated in RA will exceed its power rating of 0.25 W When RA = 40 mΩ, the shunt current will be iA =
100 × 10−3 = 2.5 A 40 × 10−3
The measured current will be imeas = 2.5 + 0.001 = 2.501 A .·. Full-scale reading is for practical purposes is 2.5 A P 3.38
The current in the shunt resistor at full-scale deflection is iA = ifullscale − 20 × 10−6 The voltage across RA at full-scale deflection is always 10 mV, therefore RA =
10 × 10−3 10 = −3 ifullscale − 2 × 10 1000ifs − 0.02
10 = 1 mΩ 10,000 − 0.02 10 [b] RA = = 10 mΩ 1000 − 0.02 10 = 1Ω [c] RA = 100 − 0.02 10 [d] RA = = 125 Ω 0.1 − 0.02 [a] RA =
P 3.39
[a]
3–27
3–28
CHAPTER 3. Simple Resistive Circuits
10 × 103 i1 + 50 × 103 (i1 − iB ) = 12 50 × 103 (i1 − iB ) = 0.4 + 30iB (0.3 × 103 ) .·.
60i1 − 50iB = 12 × 10−3
50i1 − 59iB = 0.4 × 10−3 Calculator solution yields iB = 553.85 µA [b] With the insertion of the ammeter the equations become 60i1 − 50iB = 12 × 10−3
(no change)
50 × 103 (i1 − iB ) = 2 × 103 iB + 0.4 + 30iB (300) 50i1 − 61iB = 0.4 × 10−3 Calculator solution yields iB = 496.6 µA 496.6 − 1 100 = −10.34% [c] % error = 553.85 P 3.40
[a] vmeter = 100 V [b] Rmeter = (100 Ω/V)(100 V) = 10 kΩ 10 k60 k = 8.57 kΩ vmeter =
8.57 k (100) = 36.36 V 23.57 k
[c] 10 k1 k = 6 kΩ vmeter =
6 (100) = 9.09 V 66
Problems
3–29
[d] vmeter a = 100 V vmeter b + vmeter c = 45.45 V No, because of the loading effect of the meter. P 3.41
[a] Since the unknown voltage is greater than either voltmeter’s maximum reading, the only possible way to use the voltmeters would be to connect them in series. [b]
Rm1 = (300)(1000) = 300 kΩ;
Rm2 = (150)(800) = 120 kΩ
.·. Rm1 + Rm2 = 420 kΩ i1 max =
300 × 10−3 = 1 mA; 300
i2 max =
150 × 10−3 = 1.25 mA 120
.·. imax = 1 mA since meters are in series vmax = 10−3 (300 + 120)103 = 420 V Thus the meters can be used to measure the voltage 399 = 0.95 mA [c] im = 420 × 103 vm1 = (0.95)(300) = 285 V P 3.42
vm2 = (0.95)(120) = 114 V
The current in the series-connected voltmeters is im =
288 = 0.96 mA 300
v80 kΩ = (0.96)(80) = 76.8 V Vpower supply = 288 + 115.2 + 76.8 = 480 V P 3.43
Rmeter = Rm + Rmovement =
750 V = 500 kΩ 1.5 mA
vmeas = (25 kΩ125 kΩ500 kΩ)(30 mA) = (20 kΩ)(30 mA) = 600 V vtrue = (25 kΩ125 kΩ)(30 mA) = (20.833 kΩ)(30 mA) = 625 V % error =
600 − 1 100 = −4% 625
3–30 P 3.44
CHAPTER 3. Simple Resistive Circuits Note – the upper terminal of the voltmeter should be labeled 820 V, not 300 V. [a] Rmeter = 360 kΩ + 200 kΩ50 kΩ = 400 kΩ 400600 = 240 kΩ Vmeter =
240 (300) = 240 V 300
[b] What is the percent error in the measured voltage? True value = % error = P 3.45
[a] R1
=
R2
=
R3
=
[b] Let
600 (300) = 272.73 V 660
240 − 1 100 = −12% 272.73
100 V = 50 kΩ 2 mA 10 V = 5 kΩ 2 mA 1V = 500 Ω 2 mA
ia
=
actual current in the movement
id
=
design current in the movement
ia Then % error = − 1 100 id For the 100 V scale: 100 100 = , ia = 50,000 + 25 50,025 ia 50,000 = 0.9995 = id 50,025 For the 10 V scale: ia 5000 = = 0.995 id 5025 For the 1 V scale: 500 ia = 0.9524 = 525 id P 3.46
id =
100 50,000
% error = (0.9995 − 1)100 = −0.05%
% error = (0.995 − 1.0)100 = −0.5%
% error = (0.9524 − 1.0)100 = −4.76%
[a] Rmovement = 50 Ω R1 + Rmovement =
30 = 30 kΩ 1 × 10−3
R2 + R1 + Rmovement =
.·. R1 = 29,950 Ω
150 = 150 kΩ 1 × 10−3
.·. R2 = 120 kΩ
Problems R3 + R2 + R1 + Rmovement = .·. R3 = 150 kΩ
300 = 300 kΩ 1 × 10−3
3–31
CHAPTER 3. Simple Resistive Circuits
3–32 [b]
imove =
288 (1) = 0.96 mA 300
v1 = (0.96 m)(150 k) = 144 V 144 = 0.192 mA 750 k
i1 =
i2 = imove + i1 = 0.96 m + 0.192 m = 1.152 mA vmeas = vx = 144 + 150i2 = 316.8 V [c] v1 = 150 V;
i2 = 1 m + 0.20 m = 1.20 mA
i1 = 150/750,000 = 0.20 mA .·. vmeas = vx = 150 + (150 k)(1.20 m) = 330 V P 3.47
From the problem statement we have Vs (10) (1) Vs in mV; Rs in MΩ 50 = 10 + Rs 48.75 =
Vs (6) 6 + Rs
[a] From Eq (1)
(2) 10 + Rs = 0.2Vs
.·. Rs = 0.2Vs − 10 Substituting into Eq (2) yields 48.75 =
6Vs 0.2Vs − 6
or
Vs = 52 mV
[b] From Eq (1) 50 =
520 10 + Rs
So Rs = 400 kΩ
or
50Rs = 20
Problems P 3.48
3–33
Since the bridge is balanced, we can remove the detector without disturbing the voltages and currents in the circuit.
It follows that i1 =
ig (R2 + Rx ) ig (R2 + Rx ) = R1 + R2 + R3 + Rx R
i2 =
ig (R1 + R3 ) ig (R1 + R3 ) = R1 + R2 + R3 + Rx R
v 3 = R 3 i1 = v x = i 2 R x .·.
R3 ig (R2 + Rx ) Rx ig (R1 + R3 ) = R R
.·. R3 (R2 + Rx ) = Rx (R1 + R3 ) From which Rx = P 3.49
R2 R3 R1
[a]
The condition for a balanced bridge is that the product of the opposite resistors must be equal: (200)(Rx ) = (500)(800)
so
Rx =
(500)(800) = 2000 Ω 200
3–34
CHAPTER 3. Simple Resistive Circuits [b] The source current is the sum of the two branch currents. Each branch current can be determined using Ohm’s law, since the resistors in each branch are in series and the voltage drop across each branch is 6 V: 6V 6V + = 8.4 mA 200 Ω + 800 Ω 500 Ω + 2000 Ω [c] We can use current division to find the current in each branch: is =
ileft =
500 + 2000 (8.4 mA) = 6 mA 500 + 2000 + 200 + 800
iright = 8.4 mA − 6 mA = 2.4 mA Now we can use the formula p = Ri2 to find the power dissipated by each resistor: p200 = (200)(0.006)2 = 7.2 mW p500 = (500)(0.0024)2 = 2.88 mW
p800 = (800)(0.006)2 = 28.8 mW p2000 = (2000)(0.0024)2 = 11.52 mW
Thus, the 800 Ω resistor absorbs the most power; it absorbs 28.8 mW of power. [d] From the analysis in part (c), the 500 Ω resistor absorbs the least power; it absorbs 2.88 mW of power. P 3.50
Redraw the circuit, replacing the detector branch with a short circuit.
6 kΩ30 kΩ = 5 kΩ 12 kΩ20 kΩ = 7.5 kΩ ig =
75 = 6 mA 5000 + 7500
v1 = 6 mA(5000) = 30 V v2 = 6 mA(7500) = 45 V
Problems i1 =
30 V = 5 mA 6000 Ω
i2 =
45 V = 3.75 mA 12,000 Ω
3–35
id = i1 − i2 = 5 mA − 3.75 mA = 1.25 mA P 3.51
Note the bridge structure is balanced, that is 15 × 5 = 25 × 3, hence there is no current in the 5 kΩ resistor. It follows that the equivalent resistance of the circuit is Req = 0.750 + 11.25 = 12 kΩ The source current is 192/12,000 = 16 mA. The current down through the 3 kΩ resistor is i3k = 16
30 = 10 mA 48
.·. p3k = (10 × 10−3 )2 (3 × 103 ) = 300 mW P 3.52
In order that all four decades (1, 10, 100, 1000) that are used to set R3 contribute to the balance of the bridge, the ratio R2 /R1 should be set to 0.001.
P 3.53
Begin by transforming the Y-connected resistors (10 Ω, 40 Ω, 50 Ω) to ∆-connected resistors. Both the Y-connected and ∆-connected resistors are shown below to assist in using Eqs. 3.44 – 3.46:
Now use Eqs. 3.44 – 3.46 to calculate the values of the ∆-connected resistors: R1 =
(40)(10) = 4 Ω; 10 + 40 + 50
R2 =
(50)(10) = 5 Ω; 10 + 40 + 50
The transformed circuit is shown below:
R3 =
(40)(50) = 20 Ω 10 + 40 + 50
3–36
CHAPTER 3. Simple Resistive Circuits
The equivalent resistance seen by the 24 V source can be calculated by making series and parallel combinations of the resistors to the right of the 24 V source: Req = (15 + 5)(4 + 1) + 20 = 205 + 20 = 4 + 20 = 24 Ω Therefore, the current i in the 24 V source is given by i=
24 V =1A 24 Ω
Use current division to calculate the currents i1 and i2 . Note that the current i1 flows in the branch containing the 15 Ω and 5 Ω series connected resistors, while the current i2 flows in the parallel branch that contains the series connection of the 1 Ω and 4 Ω resistors: i1 =
1+4 5 (i) = (1 A) = 0.2 A, 1 + 4 + 15 + 5 25
and
i2 = 1 A − 0.2 A = 0.8 A
Now use KVL and Ohm’s law to calculate v1 . Note that v1 is the sum of the voltage drop across the 4 Ω resistor, 4i2 , and the voltage drop across the 20 Ω resistor, 20i: v1 = 4i2 + 20i = 4(0.8 A) + 20(1 A) = 3.2 + 20 = 23.2 V Finally, use KVL and Ohm’s law to calculate v2 . Note that v2 is the sum of the voltage drop across the 5 Ω resistor, 5i1 , and the voltage drop across the 20 Ω resistor, 20i: v2 = 5i1 + 20i = 5(0.2 A) + 20(1 A) = 1 + 20 = 21 V P 3.54
[a] Calculate the values of the Y-connected resistors that are equivalent to the 10 Ω, 40 Ω, and 50Ω ∆-connected resistors: RX =
(10)(50) = 5 Ω; 10 + 40 + 50
RZ =
(10)(40) = 4Ω 10 + 40 + 50
RY =
(40)(50) = 20 Ω; 10 + 40 + 50
Replacing the R2 —R3 —R4 delta with its equivalent Y gives
Problems
3–37
Now calculate the equivalent resistance Rab by making series and parallel combinations of the resistors: Rab = 13 + 5 + [(4 + 8)(20 + 4)] + 7 = 33 Ω [b] Calculate the values of the ∆-connected resistors that are equivalent to the 8 Ω, 10 Ω, and 40 Ω Y-connected resistors: (10)(40) + (40)(8) + (8)(10) 800 RX = = = 100 Ω 8 8 (10)(40) + (40)(8) + (8)(10) 800 RY = = = 80 Ω 10 10 800 (10)(40) + (40)(8) + (8)(10) = = 20 Ω RZ = 40 40 Replacing the R2 , R4 , R5 wye with its equivalent ∆ gives
Make series and parallel combinations of the resistors to find the equivalent resistance Rab : 100 Ω50 Ω = 33.33 Ω;
80 Ω4 Ω = 3.81 Ω
.·. 10050 + 804 = 33.33 + 3.81 = 37.14 Ω .·. 37.1420 =
(37.14)(20) = 13 Ω 57.14
.·. Rab = 13 + 13 + 7 = 33 Ω
3–38
CHAPTER 3. Simple Resistive Circuits [c] Convert the delta connection R4 —R5 —R6 to its equivalent wye. Convert the wye connection R3 —R4 —R6 to its equivalent delta.
P 3.55
Replace the upper and lower deltas with the equivalent wyes: R1U =
(50)(10) (50)(40) (40)(10) = 5 Ω; R2U = = 20 Ω; R3U = = 4Ω 100 100 100
R1L =
(60)(10) (60)(30) (30)(10) = 6 Ω; R2L = = 18 Ω; R3L = = 3Ω 100 100 100
The resulting circuit is shown below:
Now make series and parallel combinations of the resistors: (4 + 6)(20 + 32 + 20 + 18) = 1090 = 9 Ω Rab = 33 + 5 + 9 + 3 + 40 = 90 Ω P 3.56
18 + 2 = 20 Ω 2080 = 16 Ω 16 + 4 = 20 Ω 2030 = 12 Ω 12 + 8 = 20 Ω 2060 = 15 Ω 15 + 5 = 20 Ω ig =
240 V = 12 A 20 Ω
Problems io =
3–39
60 (12 A) = 9 A 60 + 20
i30Ω =
20 (9 A) = 3.6 A 20 + 30
p30Ω = (30)(3.6)2 = 388.8 W P 3.57
The top of the pyramid can be replaced by a resistor equal to R1 =
(18)(9) = 6Ω 27
The lower left and right deltas can be replaced by wyes. Each resistance in the wye equals 3 Ω. Thus our circuit can be reduced to
Now the 12 Ω in parallel with 6 Ω reduces to 4 Ω. .·. Rab = 3 + 4 = 3 = 10 Ω P 3.58
Note – the top resistor to the right of the 1.5 Ω resistor is 20 Ω. [a] Convert the upper delta to a wye. R1 =
(50)(50) = 12.5 Ω 200
R2 =
(50)(100) = 25 Ω 200
R3 =
(50)(100) = 25 Ω 200
Convert the lower delta to a wye. R4 =
(60)(80) = 24 Ω 200
R5 =
(60)(60) = 18 Ω 200
3–40
CHAPTER 3. Simple Resistive Circuits R6 =
(60)(80) = 24 Ω 200
Now redraw the circuit using the wye equivalents.
Rab = 1.5 + 12.5 + (25 + 71 + 24)(25 + 31 + 24) + 18 = 1.5 + 12.5 + (12085) + 18 = 1.5 + 12.5 + 48 + 18 = 80 Ω [b] When vab = 400 V 400 ig = =5A 80 120 (5) = 3 A io = 120 + 80 p31Ω P 3.59
=
(31)(3)2 = 279 W
[a] After the 20 Ω—80 Ω—40 Ω wye is replaced by its equivalent delta, the circuit reduces to
Problems Now the circuit can be reduced to
Req = 44 + 28092.5 = 113.53 Ω ig = 5/113.53 = 44.04 mA i = (280/372.5)(44) = 33.11 mA v52.5Ω = (52.5)(33.11 m) = 1.74 V io = 1.74/210 = 8.28 mA [b] v40Ω = (40)(33.11 m) = 1.32 V i1 = 1.32/56 = 23.65 mA [c] Now that io and i1 are known return to the original circuit
i80Ω = 44.04 m − 23.65 m = 20.39 mA i20Ω = 23.65 m − 8.28 m = 15.37 mA i2 = i80Ω + i20Ω = 35.76 mA [d] pdel = (5)(44.04 m) = 220.2 mW
3–41
3–42 P 3.60
CHAPTER 3. Simple Resistive Circuits [a] After the 30 Ω—60 Ω—10 Ω delta is replaced by its equivalent wye, the circuit reduces to
Use current division to calculate i1 : 40 22 + 18 (5 A) = (5 A) = 4 A 22 + 18 + 4 + 6 50 [b] Return to the original circuit and write a KVL equation around the upper left loop: i1 =
(22 Ω)i22Ω + v − (4 Ω)(i1 ) = 0 so
v = (4 Ω)(4 A) − (22 Ω)(5 A − 4 A) = −6 V
[c] Write a KCL equation at the lower center node of the original circuit: −6 v =4+ = 3.9 A 60 60 [d] Write a KVL equation around the bottom loop of the original circuit: i2 = i1 +
−v5A + (4 Ω)(4 A) + (10 Ω)(3.9 A) + (1 Ω)(5 A) = 0 So, Thus, P 3.61
v5A = (4)(4) + (10)(3.9) + (1)(5) = 60 V p5A = (5 A)(60 V) = 300 W
Subtracting Eq. 3.42 from Eq. 3.43 gives R1 − R2 = (Rc Rb − Rc Ra )/(Ra + Rb + Rc ). Adding this expression to Eq. 3.41 and solving for R1 gives R1 = Rc Rb /(Ra + Rb + Rc ). To find R2 , subtract Eq. 3.43 from Eq. 3.41 and add this result to Eq. 3.42. To find R3 , subtract Eq. 3.41 from Eq. 3.42 and add this result to Eq. 3.43. Using the hint, Eq. 3.43 becomes R1 + R3 =
Rb [(R2 /R3 )Rb + (R2 /R1 )Rb ] Rb (R1 + R3 )R2 = (R2 /R1 )Rb + Rb + (R2 /R3 )Rb (R1 R2 + R2 R3 + R3 R1 )
Problems
3–43
Solving for Rb gives Rb = (R1 R2 + R2 R3 + R3 R1 )/R2 . To find Ra : First use Eqs. 3.44–3.46 to obtain the ratios (R1 /R3 ) = (Rc /Ra ) or Rc = (R1 /R3 )Ra and (R1 /R2 ) = (Rb /Ra ) or Rb = (R1 /R2 )Ra . Now use these relationships to eliminate Rb and Rc from Eq. 3.42. To find Rc , use Eqs. 3.44–3.46 to obtain the ratios Rb = (R3 /R2 )Rc and Ra = (R3 /R1 )Rc . Now use the relationships to eliminate Rb and Ra from Eq. 3.41. P 3.62
P 3.63
1 R1 = Ra R1 R2 + R2 R3 + R3 R1 1/G1 = (1/G1 )(1/G2 ) + (1/G2 )(1/G3 ) + (1/G3 )(1/G1 ) G2 G3 (1/G1 )(G1 G2 G3 ) = = G1 + G2 + G3 G1 + G2 + G3 Similar manipulations generate the expressions for Gb and Gc . Ga
=
[a] Rab = 2R1 + Therefore Thus
R2 (2R1 + RL ) = RL 2R1 + R2 + RL 2R1 − RL +
R2 (2R1 + RL ) =0 2R1 + R2 + RL
RL2 = 4R12 + 4R1 R2 = 4R1 (R1 + R2 )
When Rab = RL , the current into terminal a of the attenuator will be Using current division, the current in the RL branch will be vi R2 · RL 2R1 + R2 + RL Therefore and
vo =
vi R2 · RL RL 2R1 + R2 + RL
vo R2 = vi 2R1 + R2 + RL
[b] (600)2 = 4(R1 + R2 )R1 9 × 104 = R12 + R1 R2 vo R2 = 0.6 = vi 2R1 + R2 + 600 .·. 1.2R1 + 0.6R2 + 360 = R2 0.4R2 = 1.2R1 + 360 R2 = 3R1 + 900 .·. 9 × 104 = R12 + R1 (3R1 + 900) = 4R12 + 900R1 .·. R12 + 225R1 − 22,500 = 0
vi /RL
3–44
CHAPTER 3. Simple Resistive Circuits R1 = −112.5 ±
(112.5)2 + 22,500 = −112.5 ± 187.5
.·. R1 = 75 Ω .·. R2 = 3(75) + 900 = 1125 Ω P 3.64
[a] After making the Y-to-∆ transformation, the circuit reduces to
Combining the parallel resistors reduces the circuit to
Now note:
0.75R +
2.25RL2 + 3.75RRL 3RRL = 3R + RL 3R + RL
Therefore
Rab
When Rab = RL , we have Therefore
2.25RL2 + 3.75RRL 3R 3R + RL 3R(3R + 5RL ) = = 2 15R + 9RL 2.25RL + 3.75RRL 3R + 3R + RL
RL2 = R2
or
15RRL + 9RL2 = 9R2 + 15RRL RL = R
Problems [b] When R = RL , the circuit reduces to
ii (3RL ) 1 1 vi ii = = , 4.5RL 1.5 1.5 RL vo = 0.5 Therefore vi io =
P 3.65
[a] 3.5(3R − RL ) = 3R + RL 10.5R − 1050 = 3R + 300 7.5R = 1350,
R = 180 Ω
R2 =
2(180)(300)2 = 4500 Ω 3(180)2 − (300)2
vo =
42 vi = = 12 V 3.5 3.5
io =
12 = 40 mA 300
i1 =
30 42 − 12 = = 6.67 mA 4500 4500
ig =
42 = 140 mA 300
[b]
i2 = 140 m − 6.67 m = 133.33 mA i3 = 40 m − 6.67 m = 33.33 mA i4 = 133.33 m − 33.33 m = 100 mA
1 vo = 0.75RL io = vi , 2
3–45
CHAPTER 3. Simple Resistive Circuits
3–46
p4500 top = (6.67 × 10−3 )2 (4500) = 0.2 W p180 left = (133.33 × 10−3 )2 (180) = 3.2 W p180 right = (33.33 × 10−3 )2 (180) = 0.2 W p180 vertical = (100 × 10−3 )2 (180) = 1.8 W p300 load = (40 × 10−3 )2 (300) = 0.48 W The 180 Ω resistor carrying i2 dissipates the most power. [c] p180 left = 3.2 W [d] Two resistors dissipate minimum power – the 4500 Ω and the 180 Ω carrying i3 . [e] Both resistors dissipate 0.2 W or 200 mW. P 3.66
[a]
va =
vin R4 Ro + R4 + ∆R
vb =
R3 vin R2 + R3
vo = va − vb =
R3 R4 vin − vin Ro + R4 + ∆R R2 + R3
When the bridge is balanced, R3 R4 vin = vin Ro + R4 R2 + R3 .·.
R3 R4 = Ro + R4 R2 + R3
Thus,
vo
R4 vin R4 vin − Ro + R4 + ∆R Ro + R4 1 1 − = R4 vin Ro + R4 + ∆R Ro + R4 R4 vin (−∆R) = (Ro + R4 + ∆R)(Ro + R4 ) −(∆R)R4 vin ≈ (Ro + R4 )2 =
Problems [b] ∆R = 0.03Ro Ro =
R2 R4 (1000)(5000) = = 10,000 Ω R3 500
∆R = (0.03)(104 ) = 300 Ω .·. vo ≈ [c] vo
= = =
P 3.67
−300(5000)(6) = −40 mV (15,000)2
−(∆R)R4 vin (Ro + R4 + ∆R)(Ro + R4 ) −300(5000)(6) (15,300)(15,000) −39.2157 mV −(∆R)R4 vin (Ro + R4 )2
[a] approx value = true value =
−(∆R)R4 vin (Ro + R4 + ∆R)(Ro + R4 )
(Ro + R4 + ∆R) approx value = .·. true value (Ro + R4 )
Ro + R4 + ∆R ∆R − 1 × 100 = × 100 .·. % error = Ro + R4 Ro + R4 But Ro =
R2 R4 R3
.·. % error = [b] % error = P 3.68
R3 ∆R R4 (R2 + R3 )
(500)(300) × 100 = 2% (5000)(1500)
∆R(R3 )(100) = 0.5 (R2 + R3 )R4 ∆R(500)(100) = 0.5 (1500)(5000) .·. ∆R = 75 Ω % change =
75 × 100 = 0.75% 10,000
3–47
3–48 P 3.69
CHAPTER 3. Simple Resistive Circuits [a] From Eq 3.64 we have
i1 i2
2
=
R22 R12 (1 + 2σ)2
Substituting into Eq 3.63 yields R2 =
R22 R1 R12 (1 + 2σ)2
Solving for R2 yields R2 = (1 + 2σ)2 R1 [b] From Eq 3.63 we have R2 i1 = ib R1 + R2 + 2Ra But R2 = (1 + 2σ)2 R1 and Ra = σR1 therefore (1 + 2σ)2 R1 (1 + 2σ)2 i1 = = lb R1 + (1 + 2σ)2 R1 + 2σR1 (1 + 2σ) + (1 + 2σ)2 =
1 + 2σ 2(1 + σ)
It follows that
i1 ib
2
=
(1 + 2σ)2 4(1 + σ)2
Substituting into Eq 3.66 gives Rb = P 3.70
(1 + 2σ)2 Ra (1 + 2σ)2 σR1 = 4(1 + σ)2 4(1 + σ)2
From Eq 3.69 R2 R3 i1 = i3 D But D = (R1 + 2Ra )(R2 + 2Rb ) + 2Rb R2 where Ra = σR1 ; R2 = (1 + 2σ)2 R1 and Rb = Therefore D can be written as
(1 + 2σ)2 σR1 4(1 + σ)2
Problems
D
=
= = =
2(1 + 2σ)2 σR1 (R1 + 2σR1 ) (1 + 2σ) R1 + + 2 4(1 + σ)
(1 + 2σ)2 σR1 2 2(1 + 2σ) R1 4(1 + σ)2
σ (1 + 2σ)σ 3 2 (1 + 2σ) R1 1 + + 2(1 + σ)2 2(1 + σ)2 (1 + 2σ)3 R12 {2(1 + σ)2 + σ + (1 + 2σ)σ} 2(1 + σ)2 (1 + 2σ)3 R12 {1 + 3σ + 2σ 2 } (1 + σ)2 2
(1 + 2σ)4 R12 D= (1 + σ) i1 i3
R2 R3 (1 + σ) (1 + 2σ)4 R12 (1 + 2σ)2 R1 R3 (1 + σ) = (1 + 2σ)4 R12 (1 + σ)R3 = (1 + 2σ)2 R1 When this result is substituted into Eq 3.69 we get .·.
R3 =
=
(1 + σ)2 R32 R1 (1 + 2σ)4 R12
Solving for R3 gives R3 = P 3.71
(1 + 2σ)4 R1 (1 + σ)2
From the dimensional specifications, calculate σ and R3 : σ=
y 0.025 = 0.025; = x 1
R3 =
122 Vdc2 = = 1.2 Ω p 120
Calculate R1 from R3 and σ: R1 =
(1 + σ)2 R3 = 1.0372 Ω (1 + 2σ)4
Calculate Ra , Rb , and R2 : Ra = σR1 = 0.0259 Ω
Rb =
(1 + 2σ)2 σR1 = 0.0068 Ω 4(1 + σ)2
3–49
CHAPTER 3. Simple Resistive Circuits
3–50
R2 = (1 + 2σ)2 R1 = 1.1435 Ω Using symmetry, R4 = R2 = 1.1435 Ω
R5 = R1 = 1.0372 Ω
Rc = Rb = 0.0068 Ω
Rd = Ra = 0.0259 Ω
Test the calculations by checking the power dissipated, which should be 120 W/m. Calculate D, then use Eqs. (3.58)-(3.60) to calculate ib , i1 , and i2 : D = (R1 + 2Ra )(R2 + 2Rb ) + 2R2 Rb = 1.2758 ib =
Vdc (R1 + R2 + 2Ra ) = 21 A D
i1 =
Vdc R2 = 10.7561 A D
i2 =
Vdc (R1 + 2Ra ) = 10.2439 A D
It follows that i2b Rb = 3 W and the power dissipation per meter is 3/0.025 = 120 W/m. The value of i21 R1 = 120 W/m. The value of i22 R2 = 120 W/m. Finally, i21 Ra = 3 W/m. P 3.72
From the solution to Problem 3.71 we have ib = 21 A and i3 = 10 A. By symmetry ic = 21 A thus the total current supplied by the 12 V source is 21 + 21 + 10 or 52 A. Therefore the total power delivered by the source is p12 V (del) = (12)(52) = 624 W. We also have from the solution that pa = pb = pc = pd = 3 W. Therefore the total power delivered to the vertical resistors is pV = (8)(3) = 24 W. The total power delivered to the five horizontal resistors is pH = 5(120) = 600 W. .·.
P 3.73
pdiss = pH + pV = 624 W =
pdel
[a] σ = 0.03/1.5 = 0.02 Since the power dissipation is 150 W/m the power dissipated in R3 must be 200(1.5) or 300 W. Therefore 122 = 0.48 Ω 300 From Table 3.1 we have
R3 =
R1 =
(1 + σ)2 R3 = 0.4269 Ω (1 + 2σ)4
Ra = σR1 = 0.0085 Ω R2 = (1 + 2σ)2 R1 = 0.4617 Ω
Problems Rb =
3–51
(1 + 2σ)2 σR1 = 0.0022 Ω 4(1 + σ)2
Therefore R4 = R2 = 0.4617 Ω
R5 = R1 = 0.4269 Ω
Rc = Rb = 0.0022 Ω
Rd = Ra = 0.0085 Ω
[b] D = [0.4269 + 2(0.0085)][0.4617 + 2(0.0022)] + 2(0.4617)(0.0022) = 0.2090 i1 =
Vdc R2 = 26.51 A D
i21 R1 = 300 W or 200 W/m i2 =
R1 + 2Ra Vdc = 25.49 A D
i22 R2 = 300 W or 200 W/m i21 Ra = 6 W or 200 W/m ib =
R1 + R2 + 2Ra Vdc = 52 A D
i2b Rb = 6 W or 200 W/m isource = 52 + 52 +
12 = 129 A 0.48
pdel = 12(129) = 1548 W pH = 5(300) = 1500 W pV = 8(6) = 48 W
pdel =
pdiss = 1548 W
4 Techniques of Circuit Analysis Assessment Problems AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages:
The two node voltage equations are v1 v1 − v2 v1 + + = 0 −15 + 60 15 5 v2 v2 − v1 5+ + = 0 2 5 Place these equations in standard form: 1 1 1 1 + + + v2 − = v1 60 15 5 5 1 1 1 + + v2 = v1 − 5 2 5 Solving, v1 = 60 V and v2 = 10 V; Therefore, i1 = (v1 − v2 )/5 = 10 A
15 −5
[b] p15A = −(15 A)v1 = −(15 A)(60 V) = −900 W = 900 W(delivered) [c] p5A = (5 A)v2 = (5 A)(10 V) = 50 W= −50 W(delivered)
4–1
4–2
CHAPTER 4. Techniques of Circuit Analysis
AP 4.2 Redraw the circuit, choosing the node voltages and reference node as shown:
The two node voltage equations are: v1 v 1 − v 2 + = 0 −4.5 + 1 6+2 v2 − v1 v2 − 30 v2 + + = 0 12 6+2 4 Place equations in standard form: these 1 1 + v2 − = v1 1 + 8 8 1 1 1 1 + v2 = v1 − + + 8 12 8 4
4.5 7.5
Solving, v1 = 6 V v2 = 18 V To find the voltage v, first find the current i through the series-connected 6 Ω and 2 Ω resistors: i=
v1 − v2 6 − 18 = = −1.5 A 6+2 8
Using a KVL equation, calculate v: v = 2i + v2 = 2(−1.5) + 18 = 15 V AP 4.3 [a] Redraw the circuit, choosing the node voltages and reference node as shown:
The node voltage equations are: v1 − 50 v1 v1 − v2 + + − 3i1 6 8 2 v2 v2 − v1 + + 3i1 −5 + 4 2
=
0
=
0
Problems The dependent source requires the following constraint equation: 50 − v1 6 Place these equations in standard form: 1 1 1 1 50 + + v1 + v2 − + i1 (−3) = 6 8 2 2 6 1 1 1 + v1 − + v2 + i1 (3) = 5 2 4 2 1 50 + v2 (0) + i1 (1) = v1 6 6 Solving, v1 = 32 V; v2 = 16 V; i1 = 3 A Using these values to calculate the power associated with each source: i1 =
p50V = −50i1
=
p5A = −5(v2 ) = p3i1 = 3i1 (v2 − v1 ) =
−150 W −80 W −144 W
[b] All three sources are delivering power to the circuit because the power computed in (a) for each of the sources is negative. AP 4.4 Redraw the circuit and label the reference node and the node at which the node voltage equation will be written:
The node voltage equation is vo vo − 10 vo + 20i∆ =0 + + 40 10 20 The constraint equation required by the dependent source is i∆ = i10 Ω + i30 Ω =
10 − vo 10 + 20i∆ + 10 30
Place these equations in standard form:
4–3
4–4
CHAPTER 4. Techniques of Circuit Analysis
1 1 1 vo + + 40 10 20 1 vo 10 Solving,
vo = 24 V
i∆ (1)
+
i∆ 1 −
+
20 30
=
1
=
1+
10 30
i∆ = −3.2 A
AP 4.5 Redraw the circuit identifying the three node voltages and the reference node:
Note that the dependent voltage source and the node voltages v and v2 form a supernode. The v1 node voltage equation is v1 − v v1 + − 4.8 = 0 7.5 2.5 The supernode equation is v v2 v2 − 12 v − v1 + + + =0 2.5 10 2.5 1 The constraint equation due to the dependent source is ix =
v1 7.5
The constraint equation due to the supernode is v + ix = v2 Place this set of equations in standard form: 1 1 1 + v1 + v − + v2 (0) 7.5 2.5 2.5 1 1 1 1 + +1 + v + v2 v1 − 2.5 2.5 10 2.5 1 + v(0) + v2 (0) v1 − 7.5 v1 (0)
+
v(1)
+
v2 (−1)
Solving this set of equations for v gives v = 8 V v1 = 15 V,
v2 = 10 V,
ix = 2 A
+
ix (0) =
4.8
+
ix (0) =
12
+
ix (1) =
0
+
ix (1) =
0
Problems
4–5
AP 4.6 Redraw the circuit identifying the reference node and the two unknown node voltages. Note that the right-most node voltage is the sum of the 60 V source and the dependent source voltage.
The node voltage equation at v1 is v1 − (60 + 6iφ ) v1 − 60 v1 + + =0 2 24 3 The constraint equation due to the dependent source is iφ =
60 + 6iφ − v1 3
Place these two equations in standard form: 1 1 1 + + + iφ (−2) = 30 + 20 v1 2 24 3 1 + iφ (1 − 2) = 20 v1 3 Solving,
v1 = 48 V
iφ = −4 A
AP 4.7 [a] Redraw the circuit identifying the three mesh currents:
The mesh current equations are: −80 + 5(i1 − i2 ) + 26(i1 − i3 )
=
0
30i2 + 90(i2 − i3 ) + 5(i2 − i1 )
=
0
8i3 + 26(i3 − i1 ) + 90(i3 − i2 )
=
0
4–6
CHAPTER 4. Techniques of Circuit Analysis Place these equations in standard form: 31i1 − 5i2 − 26i3
=
80
−5i1 + 125i2 − 90i3
=
0
−26i1 − 90i2 + 124i3 Solving,
=
0
i1 = 5 A;
i2 = 2 A;
i3 = 2.5 A
p80V = −(80)i1 = −(80)(5) = −400 W Therefore the 80 V source is delivering 400 W to the circuit. [b] p8Ω = (8)i23 = 8(2.5)2 = 50 W, so the 8 Ω resistor dissipates 50 W. AP 4.8 [a] b = 8,
n = 6,
b−n+1=3
[b] Redraw the circuit identifying the three mesh currents:
The three mesh-current equations are −25 + 2(i1 − i2 ) + 5(i1 − i3 ) + 10 =
0
−(−3vφ ) + 14i2 + 3(i2 − i3 ) + 2(i2 − i1 )
=
0
1i3 − 10 + 5(i3 − i1 ) + 3(i3 − i2 )
=
0
The dependent source constraint equation is vφ = 3(i3 − i2 ) Place these four equations in standard form: 7i1 − 2i2 − 5i3 + 0vφ
=
15
−2i1 + 19i2 − 3i3 + 3vφ
=
0
−5i1 − 3i2 + 9i3 + 0vφ
=
10
0i1 + 3i2 − 3i3 + 1vφ
=
0
Solving i1 = 4 A;
i2 = −1 A;
i3 = 3 A;
vφ = 12 V
Problems
4–7
pds = −(−3vφ )i2 = 3(12)(−1) = −36 W Thus, the dependent source is delivering 36 W, or absorbing −36 W. AP 4.9 Redraw the circuit identifying the three mesh currents:
The mesh current equations are: −25 + 6(ia − ib ) + 8(ia − ic )
=
0
2ib + 8(ib − ic ) + 6(ib − ia ) =
0
5iφ + 8(ic − ia ) + 8(ic − ib ) =
0
The dependent source constraint equation is iφ = ia . We can substitute this simple expression for iφ into the third mesh equation and place the equations in standard form: 14ia − 6ib − 8ic
=
25
−6ia + 16ib − 8ic
=
0
−3ia − 8ib + 16ic
=
0
Solving, ia = 4 A;
ib = 2.5 A;
ic = 2 A
Thus, vo = 8(ia − ic ) = 8(4 − 2) = 16 V AP 4.10 Redraw the circuit identifying the mesh currents:
4–8
CHAPTER 4. Techniques of Circuit Analysis Since there is a current source on the perimeter of the i3 mesh, we know that i3 = −16 A. The remaining two mesh equations are −30 + 3i1 + 2(i1 − i2 ) + 6i1
=
0
8i2 + 5(i2 + 16) + 4i2 + 2(i2 − i1 )
=
0
Place these equations in standard form: 11i1 − 2i2
=
30
−2i1 + 19i2
=
−80
Solving: i1 = 2 A, i2 = −4 A, i3 = −16 A The current in the 2 Ω resistor is i1 − i2 = 6 A .·. Thus, the 2 Ω resistors dissipates 72 W.
p2 Ω = (6)2 (2) = 72 W
AP 4.11 Redraw the circuit and identify the mesh currents:
There are current sources on the perimeters of both the ib mesh and the ic mesh, so we know that ib = −10 A;
ic =
2vφ 5
The remaining mesh current equation is −75 + 2(ia + 10) + 5(ia − 0.4vφ ) = 0 The dependent source requires the following constraint equation: vφ = 5(ia − ic ) = 5(ia − 0.4vφ ) Place the mesh current equation and the dependent source equation is standard form: 7ia − 2vφ
=
55
5ia − 3vφ
=
0
Solving: ia = 15 A; Thus, ia = 15 A.
ib = −10 A;
ic = 10 A;
vφ = 25 V
Problems
4–9
AP 4.12 Redraw the circuit and identify the mesh currents:
The 2 A current source is shared by the meshes ia and ib . Thus we combine these meshes to form a supermesh and write the following equation: −10 + 2ib + 2(ib − ic ) + 2(ia − ic ) = 0 The other mesh current equation is −6 + 1ic + 2(ic − ia ) + 2(ic − ib ) = 0 The supermesh constraint equation is i a − ib = 2 Place these three equations in standard form: 2ia + 4ib − 4ic
=
10
−2ia − 2ib + 5ic
=
6
ia − ib + 0ic
=
2
Solving, Thus,
ib = 5 A; ic = 6 A ia = 7 A; p1 Ω = i2c (1) = (6)2 (1) = 36 W
AP 4.13 Redraw the circuit and identify the reference node and the node voltage v1 :
The node voltage equation is v1 − 25 v1 − 20 −2+ =0 15 10
CHAPTER 4. Techniques of Circuit Analysis
4–10
Rearranging and solving,
v1
1 1 20 25 + + =2+ 15 10 15 10
.·. v1 = 35 V
p2A = −35(2) = −70 W Thus the 2 A current source delivers 70 W. AP 4.14 Redraw the circuit and identify the mesh currents:
There is a current source on the perimeter of the i3 mesh, so i3 = 4 A. The other two mesh current equations are −128 + 4(i1 − 4) + 6(i1 − i2 ) + 2i1
=
0
30ix + 5i2 + 6(i2 − i1 ) + 3(i2 − 4) =
0
The constraint equation due to the dependent source is i x = i 1 − i3 = i 1 − 4 Substitute the constraint equation into the second mesh equation and place the resulting two mesh equations in standard form: 12i1 − 6i2
=
144
24i1 + 14i2
=
132
Solving, i1 = 9 A;
i2 = −6 A;
i3 = 4 A;
.·. v4A = 3(i3 − i2 ) − 4ix = 10 V p4A = −v4A (4) = −(10)(4) = −40 W Thus, the 2 A current source delivers 40 W.
ix = 9 − 4 = 5 A
Problems
4–11
AP 4.15 [a] Redraw the circuit with a helpful voltage and current labeled:
Transform the 120 V source in series with the 20 Ω resistor into a 6 A source in parallel with the 20 Ω resistor. Also transform the −60 V source in series with the 5 Ω resistor into a −12 A source in parallel with the 5 Ω resistor. The result is the following circuit:
Combine the three current sources into a single current source, using KCL, and combine the 20 Ω, 5 Ω, and 6 Ω resistors in parallel. The resulting circuit is shown on the left. To simplify the circuit further, transform the resulting 30 A source in parallel with the 2.4 Ω resistor into a 72 V source in series with the 2.4 Ω resistor. Combine the 2.4 Ω resistor in series with the 1.6 Ω resisor to get a very simple circuit that still maintains the voltage v. The resulting circuit is on the right.
Use voltage division in the circuit on the right to calculate v as follows: 8 (72) = 48 V 12 [b] Calculate i in the circuit on the right using Ohm’s law: v=
i=
48 v =6A = 8 8
4–12
CHAPTER 4. Techniques of Circuit Analysis Now use i to calculate va in the circuit on the left: va = 6(1.6 + 8) = 57.6 V Returning back to the original circuit, note that the voltage va is also the voltage drop across the series combination of the 120 V source and 20 Ω resistor. Use this fact to calculate the current in the 120 V source, ia : ia =
120 − 57.6 120 − va = = 3.12 A 20 20
p120V = −(120)ia = −(120)(3.12) = −374.40 W Thus, the 120 V source delivers 374.4 W. AP 4.16 To find RTh , replace the 72 V source with a short circuit:
Note that the 5 Ω and 20 Ω resistors are in parallel, with an equivalent resistance of 520 = 4 Ω. The equivalent 4 Ω resistance is in series with the 8 Ω resistor for an equivalent resistance of 4 + 8 = 12 Ω. Finally, the 12 Ω equivalent resistance is in parallel with the 12 Ω resistor, so RTh = 1212 = 6 Ω. Use node voltage analysis to find vTh . Begin by redrawing the circuit and labeling the node voltages:
The node voltage equations are v1 − vTh v1 − 72 v1 + + = 0 5 20 8 vTh − v1 vTh − 72 + = 0 8 12
Problems
4–13
Place these equations in standard form: 1 1 1 1 72 v1 + + + vTh − = 5 20 8 8 5 1 1 1 + v1 − + vTh = 6 8 8 12 Solving, v1 = 60 V and vTh = 64.8 V. Therefore, the Thévenin equivalent circuit is a 64.8 V source in series with a 6 Ω resistor. AP 4.17 We begin by performing a source transformation, turning the parallel combination of the 15 A source and 8 Ω resistor into a series combination of a 120 V source and an 8 Ω resistor, as shown in the figure on the left. Next, combine the 2 Ω, 8 Ω and 10 Ω resistors in series to give an equivalent 20 Ω resistance. Then transform the series combination of the 120 V source and the 20 Ω equivalent resistance into a parallel combination of a 6 A source and a 20 Ω resistor, as shown in the figure on the right.
Finally, combine the 20 Ω and 12 Ω parallel resistors to give RN = 2012 = 7.5 Ω. Thus, the Norton equivalent circuit is the parallel combination of a 6 A source and a 7.5 Ω resistor. AP 4.18 Find the Thévenin equivalent with respect to A, B using source transformations. To begin, convert the series combination of the −36 V source and 12 kΩ resistor into a parallel combination of a −3 mA source and 12 kΩ resistor. The resulting circuit is shown below:
Now combine the two parallel current sources and the two parallel resistors to give a −3 + 18 = 15 mA source in parallel with a 12 k60 k= 10 kΩ resistor. Then transform the 15 mA source in parallel with the 10 kΩ resistor into a 150 V source in series with a 10 kΩ resistor, and combine this 10 kΩ resistor in series with the 15 kΩ resistor. The Thévenin equivalent is thus a 150 V source in series with a 25 kΩ
4–14
CHAPTER 4. Techniques of Circuit Analysis resistor, as seen to the left of the terminals A,B in the circuit below.
Now attach the voltmeter, modeled as a 100 kΩ resistor, to the Thévenin equivalent and use voltage division to calculate the meter reading vAB : vAB =
100,000 (150) = 120 V 125,000
AP 4.19 Begin by calculating the open circuit voltage, which is also vTh , from the circuit below:
Summing the currents away from the node labeled vTh We have vTh vTh − 24 + 4 + 3ix + =0 8 2 Also, using Ohm’s law for the 8 Ω resistor, vTh ix = 8 Substituting the second equation into the first and solving for vTh yields vTh = 8 V. Now calculate RTh . To do this, we use the test source method. Replace the voltage source with a short circuit, the current source with an open circuit, and apply the test voltage vT , as shown in the circuit below:
Problems Write a KCL equation at the middle node: iT = ix + 3ix + vT /2 = 4ix + vT /2 Use Ohm’s law to determine ix as a function of vT : ix = vT /8 Substitute the second equation into the first equation: iT = 4(vT /8) + vT /2 = vT Thus, RTh = vT /iT = 1 Ω The Thévenin equivalent is an 8 V source in series with a 1 Ω resistor. AP 4.20 Begin by calculating the open circuit voltage, which is also vTh , using the node voltage method in the circuit below:
The node voltage equations are v − (vTh + 160i∆ ) v + − 4 = 0, 60 20 vTh vTh vTh + 160i∆ − v + + = 0 40 80 20 The dependent source constraint equation is i∆ =
vTh 40
4–15
4–16
CHAPTER 4. Techniques of Circuit Analysis Substitute the constraint equation into the node voltage equations and put the two equations in standard form: 1 5 1 v + vTh − = 4 + 60 20 20 1 1 1 5 + vTh = 0 v − + + 20 40 80 20 Solving, v = 172.5 V and vTh = 30 V. Now use the test source method to calculate the test current and thus RTh . Replace the current source with a short circuit and apply the test source to get the following circuit:
Write a KCL equation at the rightmost node: iT =
vT vT vT + 160i∆ + + 80 40 80
The dependent source constraint equation is i∆ =
vT 40
Substitute the constraint equation into the KCL equation and simplify the right-hand side: iT =
vT 10
Therefore, RTh =
vT = 10 Ω iT
Thus, the Thévenin equivalent is a 30 V source in series with a 10 Ω resistor. AP 4.21 First find the Thévenin equivalent circuit. To find vTh , create an open circuit between nodes a and b and use the node voltage method with the circuit
Problems
4–17
below:
The node voltage equations are: vTh − (100 + vφ ) vTh − v1 + = 0 4 4 v1 − 100 v1 − 20 v1 − vTh + = 0 + 4 4 4 The dependent source constraint equation is vφ = v1 − 20 Place these three equations in standard form: 1 1 1 1 vTh + + v1 − + vφ − 4 4 4 4 1 1 1 1 + + + v1 + vφ (0) vTh − 4 4 4 4 vTh (0)
+
v1 (1)
+
vφ (−1)
=
25
=
30
=
20
Solving, vTh = 120 V, v1 = 80 V, and vφ = 60 V. Now create a short circuit between nodes a and b and use the mesh current method with the circuit below:
The mesh current equations are −100 + 4(i1 − i2 ) + vφ + 20 = −vφ + 4i2 + 4(i2 − isc ) + 4(i2 − i1 )
0
=
0
−20 − vφ + 4(isc − i2 ) =
0
4–18
CHAPTER 4. Techniques of Circuit Analysis The dependent source constraint equation is vφ = 4(i1 − isc ) Place these four equations in standard form: 4i1 − 4i2 + 0isc + vφ
=
80
−4i1 + 12i2 − 4isc − vφ
=
0
0i1 − 4i2 + 4isc − vφ
=
20
4i1 + 0i2 − 4isc − vφ
=
0
Solving, i1 = 45 A, i2 = 30 A, isc = 40 A, and vφ = 20 V. Thus, RTh =
vTh 120 = 3Ω = isc 40
[a] For maximum power transfer, R = RTh = 3 Ω [b] The Thévenin voltage, vTh = 120 V, splits equally between the Thévenin resistance and the load resistance, so 120 = 60 V vload = 2 Therefore, pmax =
2 vload 602 = 1200 W = Rload 3
AP 4.22 Sustituting the value R = 3 Ω into the circuit and identifying three mesh currents we have the circuit below:
The mesh current equations are: −100 + 4(i1 − i2 ) + vφ + 20 =
0
−vφ + 4i2 + 4(i2 − i3 ) + 4(i2 − i1 )
=
0
−20 − vφ + 4(i3 − i2 ) + 3i3
=
0
Problems
4–19
The dependent source constraint equation is vφ = 4(i1 − i3 ) Place these four equations in standard form: 4i1 − 4i2 + 0i3 + vφ
=
80
−4i1 + 12i2 − 4i3 − vφ
=
0
0i1 − 4i2 + 7i3 − vφ
=
20
4i1 + 0i2 − 4i3 − vφ
=
0
Solving, i1 = 30 A, i2 = 20 A, i3 = 20 A, and vφ = 40 V. [a] p100V = −(100)i1 = −(100)(30) = −3000 W. Thus, the 100 V source is delivering 3000 W. [b] pdepsource = −vφ i2 = −(40)(20) = −800 W. Thus, the dependent source is delivering 800 W. [c] From Assessment Problem 4.21(b), the power delivered to the load resistor is 1200 W, so the load power is (1200/3800)100 = 31.58% of the combined power generated by the 100 V source and the dependent source.
4–20
CHAPTER 4. Techniques of Circuit Analysis
Problems P 4.1
[a] 11 branches, 8 branches with resistors, 2 branches with independent sources, 1 branch with a dependent source [b] The current is unknown in every branch except the one containing the 8 A current source, so the current is unknown in 10 branches. [c] 9 essential branches – R4 − R5 forms an essential branch as does R8 − 10 V. The remaining seven branches are essential branches that contain a single element. [d] The current is known only in the essential branch containing the current source, and is unknown in the remaining 8 essential branches [e] From the figure there are 6 nodes – three identified by rectangular boxes, two identified with single black dots, and one identified by a triangle. [f] There are 4 essential nodes, three identified with rectangular boxes and one identified with a triangle [g] A mesh is like a window pane, and as can be seen from the figure there are 6 window panes or meshes. P 4.2
Problems
4–21
[a] As can be seen from the figure, the circuit has 2 separate parts. [b] There are 5 nodes – the four black dots and the node betweem the voltage source and the resistor R1 . [c] There are 7 branches, each containing one of the seven circuit components. [d] When a conductor joins the lower nodes of the two separate parts, there is now only a single part in the circuit. There would now be 4 nodes, because the two lower nodes are now joined as a single node. The number of branches remains at 7, where each branch contains one of the seven individual circuit components. P 4.3 [a] From Problem 4.1(d) there are 8 essential branches were the current is unknown, so we need 8 simultaneous equations to describe the circuit. [b] From Problem 4.1(f), there are 4 essential nodes, so we can apply KCL at (4 − 1) = 3 of these essential nodes. These would also be a dependent source constraint equation. [c] The remaining 4 equations needed to describe the circuit will be derived from KVL equations. [d] We must avoid using the topmost mesh and the leftmost mesh. Each of these meshes contains a current source, and we have no way of determining the voltage drop across a current source. P 4.4 [a] There are six circuit components, five resistors and the current source. Since the current is known only in the current source, it is unknown in the five resistors. Therefore there are five unknown currents. [b] There are four essential nodes in this circuit, identified by the dark black dots in Fig. P4.4. At three of these nodes you can write KCL equations that will be independent of one another. A KCL equation at the fourth node would be dependent on the first three. Therefore there are three independent KCL equations. [c]
Sum the currents at any three of the four essential nodes a, b, c, and d. Using nodes a, b, and c we get −ig + i1 + i2 = 0
4–22
CHAPTER 4. Techniques of Circuit Analysis −i1 + i4 + i3 = 0 i 5 − i 2 − i3 = 0 [d] There are three meshes in this circuit: one on the left with the components ig , R1 , and R4 ; one on the top right with components R1 , R2 , and R3 ; and one on the bottom right with components R3 , R4 , and R5 . We cannot write a KVL equation for the left mesh because we don’t know the voltage drop across the current source. Therefore, we can write KVL equations for the two meshes on the right, giving a total of two independent KVL equations. [e] Sum the voltages around two independent closed paths, avoiding a path that contains the independent current source since the voltage across the current source is not known. Using the upper and lower meshes formed by the five resistors gives R 1 i1 + R 3 i3 − R 2 i2 = 0 R 3 i3 + R 5 i5 − R 4 i4 = 0
P 4.5
[a] At node 1:
− ig + i1 + i2 = 0
At node 2:
− i2 + i3 + i4 = 0
At node 3:
ig − i1 − i3 − i4 = 0
[b] There are many possible solutions. For example, solve the equation at node 1 for ig : ig = i1 + i2 Substitute this expression for ig into the equation at node 3: (i1 + i2 ) − i1 − i3 − i4 = 0
so
i2 − i3 − i4 = 0
Multiply this last equation by -1 to get the equation at node 2: −(i2 − i3 − i4 ) = −0
so
− i2 + i3 + i4 = 0
Problems
4–23
P 4.6
Note that we have chosen the lower node as the reference node, and that the voltage at the upper node with respect to the reference node is vo . Write a KCL equation (node voltage equation)by summing the currents leaving the upper node: vo vo + 25 + + 0.04 = 0 120 + 5 25 Solve by multiplying both sides of the KCL equation by 125 and collecting the terms involving vo on one side of the equation and the constants on the other side of the equation: vo + 25 + 5vo + 5 = 0 .·. 6vo = −30 so vo = −30/6 = −5 V P 4.7 [a] From the solution to Problem 4.6 we know vo = −5 V; therefore p40mA = (−5)(0.04) = −0.2 W The power developed by the 40 mA source is 200 mW [b] The current into the negative terminal of the 25 V source in the figure of Problem 4.6 is ig = (−5 + 25)/125 = 160 mA The power in the 25 V source is p25V = −(25)(0.16) = −4 W The power developed by the 25 V source is 4 W =
(0.16)2 (5) = 128 mW
p120Ω
=
(0.16)2 (120) = 3.072 W
p25Ω
=
(−5)2 /25 = 1 W
[c] p5Ω
pdis = 0.128 + 3.072 + 1 = 4.2 W pdev = 0.2 + 4 = 4.2 W (checks!)
P 4.8
[a] The node voltage equation is: vo + 25 vo + + 0.04 = 0 125 25
4–24
CHAPTER 4. Techniques of Circuit Analysis Solving, vo + 25 + 5vo + 5 = 0
.·.
6vo = −30
so
vo = −5 V
[b] Let vx = voltage drop across 40 mA source: vx = vo − (100)(0.04) = −5 − 4 = −9 V p40mA = (−9)(0.04) = −360 mW The power developed by the 40 mA source is 360 mW [c] Let ig = current into negative terminal of 25 V source: ig = (−5 + 25)/125 = 160 mA p25V = −(25)(0.16) = −4 W The power developed by the 25 V source is 4 W =
(0.16)2 (5) = 128 mW
p120Ω
=
(0.16)2 (120) = 3.072 W
p25Ω
=
(−5)2 /25 = 1 W
p100Ω
=
(0.04)2 (100) = 160 mW
[d] p5Ω
pdis = 0.128 + 3.072 + 1 + 0.160 = 4.36 W pdev = 0.360 + 4 = 4.36 W (checks!)
[e] vo is independent of any finite resistance connected in series with the 40 mA current source P 4.9
The two node voltage equations are: v1 − v2 v1 + = 0 −6 + 40 8 v2 v2 v2 − v1 + + +1 = 0 8 80 120 Place these equations in standard form: 1 1 1 v1 + + v2 − = 40 8 8 1 1 1 1 + + + v2 = v1 − 8 8 80 120 Solving, v1 = 120 V and v2 = 96 V.
6 −1
Check this result by calculating the power associated with each component:
Problems Component
Power Delivered (W)
6A
−(6 A)(120 V) = −720
Power Absorbed (W)
40 Ω
1202 = 360 40
8Ω
(120 − 96)2 = 72 8
80 Ω
962 = 115.2 80
120 Ω
962 = 76.8 120
1A
(96 V)(1 A) = 96
Total
−720
720
P 4.10 [a]
The two node voltage equations are: v1 v1 − 44 v1 − v2 + + = 0 6 4 1 v 2 v2 − v1 v2 + 2 + + = 0 3 1 2 Place these equations in standard form: 1 1 44 v1 + +1 + v2 (−1) = 6 4 4 1 1 2 +1+ v1 (−1) + v2 = − 3 2 2 v2 = 6 V Solving, v1 = 12 V; Now calculate the branch currents from the node voltage values: 44 − 12 =8A ia = 4 12 =2A ib = 6 12 − 6 =6A ic = 1 6 =2A id = 3 6+2 =4A ie = 2
4–25
CHAPTER 4. Techniques of Circuit Analysis
4–26
[b] psources = p44V + p2V = −(44)ia − (2)ie = −(44)(8) − (2)(4) = −352 − 8 = −360 W Thus, the power developed in the circuit is 360 W. Note that the resistors cannot develop power! P 4.11 [a]
v1 − 110 v1 − v2 v1 − v3 + + =0 so 11v1 − 2v2 − v3 = 880 2 8 16 v2 − v1 v2 v2 − v3 + =0 so −3v1 + 12v2 − v3 = 0 + 8 3 24 v3 + 110 v3 − v2 v3 − v1 + + =0 so −3v1 − 2v2 + 29v3 = −2640 2 24 16 Solving, v1 = 74.64 V; v2 = 11.79 V; v3 = −82.5 V 110 − v1 = 17.68 A 2 v2 = 3.93 A i2 = 3 v3 + 110 = 13.75 A i3 = 2
Thus, i1 =
[b]
v1 − v2 = 7.86 A 8 v2 − v3 i5 = = 3.93 A 24 v1 − v3 i6 = = 9.82 A 16 i4 =
Pdev = 110i1 + 110i3 = 3457.14 W
Pdis = i21 (2) + i22 (3) + i23 (2) + i24 (8) + i25 (24) + i26 (16) = 3457.14 W
P 4.12
The two node voltage equations are: v1 − v2 v1 − 150 v1 + + = 0 20 80 40 v2 v2 − v1 − 11.25 + = 0 40 4
Problems Place these equations in standard form: 1 1 1 1 + v1 + v2 (− ) + 20 80 40 40 1 1 1 v1 − + v2 + 40 40 4 Solving, v1 = 100 V; v2 = 50 V
=
150 20
=
11.25
4–27
P 4.13
At vo : Solving,
−2 +
vo − 45 vo + =0 50 4+1
vo = 50 V
p2A = −(50)(2) = −100 W Thus, the 2 A current source delivers 100 W, or the current source extracts −100 W from the circuit. P 4.14
The three node voltage equations are: v1 − v2 v1 − 40 v1 + + = 0 4 40 2 v 2 − v1 v2 − v3 − 28 = 0 + 2 4 v3 v3 − v2 + + 28 = 0 2 4 Place these equations in standard form: 1 1 1 1 + + v1 + v2 − + 4 40 2 2 1 1 1 + v1 − + v2 + 2 2 4 1 + v2 − + v1 (0) 4
v3 (0)
=
40 4
=
28
=
−28
1 v3 − 4 1 1 + v3 2 4
4–28
CHAPTER 4. Techniques of Circuit Analysis Solving, v1 = 60 V;
v2 = 73 V;
v3 = −13 V.
p28A = −va (28 A) = −(v2 − v3 )(28 A) = −(73 + 13)(28) = −2408 W The 28 A source delivers 2408 W. P 4.15
The node voltage equations are: v1 + 40 v1 v1 − v2 + = 0 +5+ 12 25 20 v2 − v1 v2 − v3 + − 7.5 − 5 = 0 20 40 v3 − v2 v3 + + 7.5 = 0 40 40 Place these equations in standard form: 1 1 1 1 v1 + v2 − + + + 12 25 20 20 1 1 1 + v2 + + v1 − 20 20 40 1 + v2 − + v1 (0) 40 v2 = 132 V; v3 Solving, v1 = −10 V; Find the power: i40V
=
(−10 + 40)/12 = 2.5 A
p40V
=
−(2.5)(40) = −100 W
p5A
=
(5)(−10 − 132) = −710 W
p7.5A
=
(7.5)(−84 − 132) = −1620 W
p12Ω
=
(−10 + 40)2 /12 = 75 W
p25Ω
=
(−10)2 /25 = 4 W
v3 (0)
(abs)
(del) (del)
−
=
12.5
=
−7.5
1 v3 − 40 1 1 v3 + 40 40 = −84 V.
(del)
(abs)
40 −5 12
=
Problems p20Ω
=
(132 + 10)2 /20 = 1008.2 W
(abs)
p40Ω
=
(132 + 84)2 /40 = 1166.4 W
(abs)
p40Ω
=
(−84)2 /40 = 176.4 W
P 4.16 [a]
(abs)
pdiss = 75 + 4 + 1008.2 + 1166.4 + 176.4 = 2430 W pdev = 100 + 710 + 1620 W = 2430 W
(CHECKS)
vo − vn vo − v 1 v o − v 2 vo − v 3 + + + ··· + =0 R R R R .·. nvo = v1 + v2 + v3 + · · · + vn .·. vo =
1 1 n [v1 + v2 + v3 + · · · + vn ] = vk n n k=1
1 [b] vo = (120 + 60 − 30) = 50 V 3 P 4.17 [a]
The node voltage equation is: vo − 6.25i∆ vo − 45 vo + + =0 100 5 25 The dependent source constraint equation is: −0.45 +
45 − vo 25 Place these equations in standard form: 1 1 1 6.25 + + vo + i∆ − = 100 5 25 5 1 vo + i∆ (1) = 25 Solving, vo = 15 V; i∆ = 1.2 A 15 − 7.5 vo − 6.25i∆ = = 1.5 A [b] ids = 5 5 pds = [6.25(1.2)](1.5) = 11.25 W Thus, the dependent source absorbs 11.25 W i∆ =
45 + 0.45 25 45 25
4–29
CHAPTER 4. Techniques of Circuit Analysis
4–30
[c] p450mA = −(0.45)(15) = −6.75 W p45V = −(1.2)(45) = −54 W pdev = 6.75 + 54 = 60.75 W Thus the independent sources develop 60.75 W Also, pdis = pds + p100Ω + p5Ω + p25Ω = 11.25 + (15)2 /100 + (1.5)2 (5) + (1.2)2 (25) = 11.25 + 2.25 + 11.25 + 36 = 60.75 W (checks!) P 4.18 [a]
The node voltage equations are: v1 v1 − v2 = 0 −5io + + 20 5 v2 v2 − v3 v2 − v1 + + = 0 5 40 10 v3 − v2 v3 − 11.5io v3 − 96 = 0 + + 10 5 4 The dependent source constraint equation is: io = v2 /40 Place these equations in standard form: 1 1 1 + v1 + v2 − + v3 (0) + io (−5) = 0 20 5 5 1 1 1 1 1 + + + v2 + v3 − + io (0) = 0 v1 − 5 5 40 10 10 1 1 1 1 11.5 96 + + + v3 + io − = v1 (0) + v2 − 10 10 5 4 5 4 1 v1 (0) + v2 − + v3 (0) + io (1) = 0 40 v2 = 120 V; v3 = 78 V; io = 3 A Solving, v1 = 156 V;
Problems
4–31
[b] Calculate the power: pcccs
=
−[5(3)](156) = −2340 W
p20Ω
=
(156)2 /20 = 1216.8 W
p5Ω
=
(156 − 120)2 /5 = 259.2 W
p40Ω
=
(120)2 /40 = 360 W
p10Ω
=
(120 − 78)2 /10 = 176.4 W
p5Ω
=
(78 − 11.5 · 3)2 /5 = 378.45 W
p4Ω
=
(78 − 96)2 /4 = 81 W
p96V
=
[(78 − 96)/4](96) = −432 W
pccvs
=
[(78 − 3 · 11.5)/5](11.5 · 3) = 300.15 W
pdev = 2340 + 432 = 2772 W pdis = 1216.8 + 259.2 + 360 + 176.4 + 378.45 + 81 + 300.15 = 2772 W (checks) Thus, the circuit dissipates 2772 W
P 4.19
The node voltage equation is vo vo − 160 vo − 150iσ + + =0 10 100 30 + 20 The dependent source constraint equation is: vo iσ = − 100 Place these equations in standard form: 1 1 1 150 160 + + + iσ − = vo 10 100 50 50 10 1 vo + iσ (1) = 0 100 Solving, vo = 100 V; iσ = −1 A Now find the power: 160 − 100 −1=5A io = 10 pds = [150(−1)](5) = −750 W. Thus, the dependent source delivers 750 W
CHAPTER 4. Techniques of Circuit Analysis
4–32 P 4.20 [a]
The node voltage equations are: v1 v1 − v2 v1 + + = 0 −25 + 40 160 10 v2 − v1 v2 v2 − 84i∆ + + = 0 10 20 8 The dependent source constraint equation is: i∆ = v1 /160 Place in standard these three equations form: 1 1 1 1 + v1 + v2 − + i∆ (0) = 25 + 40 160 10 10 1 1 1 1 84 + + + v2 + i∆ − = 0 v1 − 10 10 20 8 8 1 + v2 (0) + i∆ (1) = 0 v1 − 160 v2 = 212 V; i∆ = 2.2 A Solving, v1 = 352 V; Now calculate the power. Only the two sources can develop power, so focus on the sources: p25A = −(352)(25) = −8800 W idep source
=
(v2 − 84i∆ )/8 = (212 − 84 · 2.2)/8 = 3.4 A
pdep source
=
(84 · 2.2)(3.4) = 628.32 W
Thus, only the current source develops power, so the total power developed in the circuit is 8800 W [b] The dependent source and all of the resistors dissipate the power developed by the current source. Check that the power developed equals the power dissipated: p40Ω
=
(352)2 /40 = 3097.6 W
p160Ω
=
(352)2 /160 = 774.4 W
p10Ω
=
(352 − 212)2 /10 = 1960 W
p20Ω
=
(212)2 /20 = 2247.2 W
p8Ω
=
(212 − 84 · 2.2)2 /8 = 92.48 W
pdiss = 628.32 + 3097.6 + 774.4 + 1960 + 2247.2 + 92.48 = 8800 W so the power balances.
Problems P 4.21
The two node voltage equations are: v1 − v2 v1 − 40 v1 + + = 0 5 50 10 v2 v2 − 40 v 2 − v1 − 10 + + = 0 10 40 8 Place these equations in standard form: 1 1 1 1 + + + v2 − v1 5 50 10 10 1 1 1 1 v1 − + v2 + + 10 10 40 8
=
40 5
=
10 +
40 8
v2 = 80 V. Solving, v1 = 50 V; Thus, vo = v1 − 40 = 50 − 40 = 10 V. POWER CHECK: ig = (50 − 40)/5 + (80 − 40)/8 = 7 A p40V
=
(40)(7) = 280 W
p5Ω
=
(50 − 40)2 /5 = 20 W
p8Ω
=
(80 − 40)2 /8 = 200 W
(abs)
p10Ω
=
(80 − 50)2 /10 = 90 W
(abs)
p50Ω
=
502 /50 = 50 W
p40Ω
=
802 /40 = 160 W
p10A
=
−(80)(10) = −800 W
(abs) (abs)
(abs) (abs) (del)
pabs = 280 + 20 + 200 + 90 + 50 + 160 = 800 W =
pdel
4–33
CHAPTER 4. Techniques of Circuit Analysis
4–34 P 4.22
The node voltage equations are: v1 v1 − 2.26 v1 − v2 + + = 20 50 25 v2 v2 − 2.26 v2 − v1 + + = 40 50 100
0 0
Place these equations in standard form: 1 1 1 1 + v1 + v2 − + 20 50 25 50 1 1 1 1 v1 − + v2 + + 50 40 50 100 Solving, Thus,
v1 = 1.3 V; io =
v2 = 1.5 V.
v1 − v2 1.3 − 1.5 = = −4 mA 50 50
P 4.23 [a]
The node voltage equations are:
= =
2.26 20 2.26 40
Problems v2 − 230 v2 − v4 v2 − v3 + + = 0 1 1 1 v 3 − v2 v3 − v5 v3 + + = 0 1 1 1 v4 − 230 v4 − v2 v4 − v5 + + = 0 5+1 1 2 v5 v 5 − v4 v5 − v3 + + = 0 2 1 5+1 Place these equations in standard form: v2 (1 + 1 + 1) + v3 (−1) + v4 (−1) + v5 (0)
=
230
= 0 v2 (−1) + v3 (1 + 1 + 1) + v4 (0) + v5 (−1) 1 1 1 230 +1+ v2 (−1) + v3 (0) + v4 + v5 − = 6 2 2 6 1 1 1 +1+ v2 (0) + v3 (−1) + v4 − + v5 = 0 2 2 6 Solving, v2 = 150 V; v3 = 80 V; v4 = 140 V; v5 = 90 V Find the power dissipated by the 2 Ω resistor: 140 − 90 v4 − v5 = = 25 A i2Ω = 2 2 p2Ω = (25)2 (2) = 1250 W [b] Find the power developed by the 230 V source: v2 − 230 v4 − 230 + = −80 − 15 = −95 A i230V = 1 6 p230V = (230)(−95) = −21,850 W, so the source supplies 21,850 W Check:
Pdis
=
(80)2 (1) + (15)2 (1) + (15)2 (5) + (70)2 (1) + (10)2 (1) +(25)2 (2) + (10)2 (1) + (80)2 (1) + (15)2 (5) + (15)2 (1)
=
21,850 W(checks)
4–35
CHAPTER 4. Techniques of Circuit Analysis
4–36 P 4.24 [a]
There is only one node voltage equation: va va − 80 va + 30 + + + 0.01 = 0 5000 500 1000 Solving, va + 30 + 10va + 5va − 400 + 50 = 0 .·. va = 20 V Calculate the currents: i1 = (−30 − 20)/5000 = −10 mA i2
=
20/500 = 40 mA
i4
=
80/4000 = 20 mA
i5
=
(80 − 20)/1000 = 60 mA
so
16va = 320
i3 + i4 + i5 − 10 mA = 0 so i3 = 0.01 − 0.02 − 0.06 = −0.07 = −70 mA =
(30)(−0.01) = −0.3 W
p10mA
=
(20 − 80)(0.01) = −0.6 W
p80V
=
(80)(−0.07) = −5.6 W
p5k
=
(−0.01)2 (5000) = 0.5 W
p500Ω
=
(0.04)2 (500) = 0.8 W
p1k
=
(80 − 20)2 /(1000) = 3.6 W
p4k
=
(80)2 /(4000) = 1.6 W
[b] p30V
pabs = 0.5 + 0.8 + 3.6 + 1.6 = 6.5 W pdel = 0.3 + 0.6 + 5.6 = 6.5 W (checks!)
Problems
4–37
P 4.25
The two node voltage equations are: vb vb − vc + = 0 7+ 3 1 vc − vb vc − 4 −2vx + + = 0 1 2 The constraint equation for the dependent source is: vx = vc − 4 Place these equations in standard form: 1 + vc (−1) + vx (0) = −7 +1 vb 3 1 4 + vc 1 + + vx (−2) = vb (−1) 2 2 vb (0)
+
vc (1)
Solving, vo = vb = 1.5 V
+
vx (−1) =
4
Also, vc = 9 V and vx = 5 V.
P 4.26
This circuit has a supernode includes the nodes v1 , v2 and the 25 V source. The supernode equation is 2+
v2 v1 v2 + =0 + 50 150 20 + 55
The supernode constraint equation is v2 + 25 = v1 Place these two equations in standard form:
CHAPTER 4. Techniques of Circuit Analysis
4–38
v1
1 50
v1 (1)
+ v2
1 1 + 150 75
+ v2 (−1)
= −2 =
25
Solving, v1 = −37.5 V and v2 = −62.5 V. p25V = (25)i25 i25 = −2 A − i50 = −2 A − Thus,
−37.5 v1 =2A− = −2 A + 0.75 A = −1.25 A 50 50
p25V = (25)(−1.25) = −31.25 W
The 25 V source delivers 31.25 W. P 4.27
The supernode equation is: v1 − 4v∆ v1 − 4v∆ v1 − 100 v1 + + + =0 10 60 20 30 The constraint equation for the dependent source is: 4v∆ = v1 − v∆ Place these equations in standard form: 1 4 1 1 1 4 + v∆ − − = v1 + + + 10 60 20 30 20 30 v1 (1)
+ v∆ (−5)
=
100 10 0
v∆ = 15 V Solving, v1 = 75 V; Thus, vo = 100 − v1 = 25 V P 4.28 Calculate currents and voltages needed to calculate the power for the various components: 81.6 − 108 v4 − v3 = = −3.3 A = iφ 8 8 40 40 iφ = (−3.3) = −44 V 3 3 40 v1 = v4 + iφ = 81.6 − 44 = 37.6 V 3
Problems .·.
v3 + v∆
=
120
1.75v∆
=
(1.75)(12) = 21 A
i120V
=
iccvs
=
4–39
v∆ = 120 − 108 = 12 V
37.6 − 120 108 − 120 v1 − 120 v3 − 120 + = + = −26.6 A 4 2 4 2 −37.6 120 − 37.6 0 − v1 v2 − v1 + = + = 18.72 A 20 4 20 4
Now calculate the power associated with each circuit element: p20Ω
=
(37.6)2 /20 = 70.688 W
p4Ω
=
(37.6 − 120)2 /4 = 1697.44 W
p120V
=
(120)(−26.6) = −3192 W
p2Ω
=
(12)2 /2 = 72 W
p40Ω
=
(108)2 /40 = 291.6 W
p8Ω
=
(108 − 81.6)2 /8 = 87.12 W
p80Ω
=
(81.6)2 /80 = 83.232 W
pvccs
=
(81.6)[1.75(12)] = 1713.6 W
pccvs
=
(18.72)(−44) = −823.68 W
pabs =
Now
pdel = 4015.6 W
sum the powers: ptotal = 70.688 + 1697.44 − 3192 + 72 + 291.6 + 87.12 +83.232 + 1712.6 − 823.68 = 0 W Thus, the power balances and the staff analyst has correctly calculated the voltage values
P 4.29
The supernode equation is: v1 − 60 v3 − v2 v3 v1 + + + − 0.625v∆ = 0 100 10 20 400 The node voltage equation at v2 is:
4–40
CHAPTER 4. Techniques of Circuit Analysis v2 − 60 v2 v2 − v3 + + =0 5 200 20 The supernode constraint equation is: v3 − v1 = 175iφ The two dependent source constraint equations are: v∆ = v2 − 60 iφ = −v2 /200 Place form: the four equations above in standard 1 1 1 1 1 + + v1 + v2 − + v3 + iφ (0) + v∆ (−0.625) = 100 10 20 400 20 1 1 1 1 + v1 (0) + v2 + v3 − + iφ (0) + v∆ (0) = + 5 200 20 20
60 10 60 5
v1 (1) + v2 (0) + v3 (−1) + iφ (175) + v∆ (0)
=
0
v1 (0) + v2 (1) + v3 (0) + iφ (0) + v∆ (−1)
=
60
=
0
v1 (0) + v2
1 + v3 (0) + iφ (1) + v∆ (0) 200
Solving, v1 = −60.75 V v2 = 30 V; v3 = −87 V; iφ = −0.15 A; Calculate the power for the 60 V source: v1 − 60 v2 − 60 + i60V = 10 5 −60.75 − 60 30 − 60 + = −18.075 A = 10 5 =
p60V
v∆ = −30 V
(60)(−18.075) = −1084.5 W
Thus, the 60 V source delivers 1084.5 W P 4.30 From Eq. 4.16, From Eq. 4.17, From Eq. 4.19, iB
iB = vc /(1 + β)RE iB = (vb − Vo )/(1 + β)RE
=
VCC (1 + β)RE R2 + Vo R1 R2 1 − Vo (1 + β)RE R1 R2 + (1 + β)RE (R1 + R2 )
=
[VCC R2 /(R1 + R2 )] − Vo VCC R2 − Vo (R1 + R2 ) = R1 R2 + (1 + β)RE (R1 + R2 ) [R1 R2 /(R1 + R2 )] + (1 + β)RE
Problems P 4.31 [a]
The mesh current equations are: −60 + 4i1 + 10(i1 − i2 ) + 1i1
=
0
20 + 3i2 + 10(i2 − i1 ) + 2i2
=
0
Place the equations in standard form: i1 (4 + 10 + 1) + i2 (−10) =
60
i1 (−10) + i2 (3 + 10 + 2) =
−20
Solving, i1 = 5.6 A; i2 = 2.4 A Now solve for the requested currents: ib = i1 − i2 = 3.2 A; ia = i1 = 5.6 A;
ic = −i2 = −2.4 A
[b] If the polarity of the 60 V source is reversed, we have the following mesh current equations in standard form: i1 (4 + 10 + 1) + i2 (−10) =
−60
i1 (−10) + i2 (3 + 10 + 2) =
−20
Solving, i1 = −8.8 A; i2 = −7.2 A Now solve for the requested currents: ib = i1 − i2 = −1.6 A; ia = i1 = −8.8 A; P 4.32 [a]
The mesh current equations are:
ic = −i2 = 7.2 A
4–41
4–42
CHAPTER 4. Techniques of Circuit Analysis −230 + 1(i1 − i2 ) + 2(i1 − i3 ) + 115 + 4i1
=
0
6i2 + 3(i2 − i3 ) + 1(i2 − i1 )
=
0
=
0
460 + 5i3 − 115 + 2(i3 − i1 ) + 3(i3 − i2 ) Place these equations in standard form: i1 (1 + 2 + 4) + i2 (−1) + i3 (−2) =
115
i1 (−1) + i2 (6 + 3 + 1) + i3 (−3) =
0
i1 (−2) + i2 (−3) + i3 (5 + 2 + 3) = −345 i2 = −10.6 A; i3 = −36.8 A Solving, i1 = 4.4 A; The only components that can develop power in the circuit are the sources: p230V
=
−(230)(4.4) = −1012 W
p115V
=
−(115)(−36.8 − 4.4) = 4738 W
p460V = (460)(−36.8) = −16,928 W pdev = 1012 + 16,928 = 17940 W .·. [b] From part (a) we know that the 115 V source is dissipating power; compute the power dissipated by the resistors: p1Ω
=
(1)(4.4 + 10.6)2 = 225 W
p4Ω
=
(4)(4.4)2 = 77.44 W
p6Ω
=
(6)(−10.6)2 = 674.16 W
p2Ω
=
(2)(4.4 + 36.8)2 = 3394.88 W
p3Ω
=
(3)(−10.6 + 36.8)2 = 2059.32 W
p5Ω = (5)(−36.8)2 = 6771.2 W pdis = 4738 + 225 + 77.44 + 674.16 + 3394.88 + 2059.32 + 6771.2 .·. = 17940 W (checks!) P 4.33
The mesh current equations are:
Problems −135 + 3(i1 − i2 ) + 20(i1 − i3 ) + 2i1
=
0
5i2 + 4(i2 − i3 ) + 3(i2 − i1 )
=
0
10iσ + 1i3 + 20(i3 − i1 ) + 4(i3 − i2 )
=
0
The dependent source constraint equation is: i σ = i 2 − i1 Place these equations in standard form: i1 (3 + 20 + 2) + i2 (−3) + i3 (−20) + iσ (0)
=
135
i1 (−3) + i2 (5 + 4 + 3) + i3 (−4) + iσ (0)
=
0
i1 (−20) + i2 (−4) + i3 (1 + 20 + 4) + iσ (10) =
0
i1 (1) + i2 (−1) + i3 (0) + iσ (1)
0
=
Solving, i1 = 64.8 A, i2 = 39 A; i3 = 68.4 A; Calculate the power: p20Ω = 20(68.4 − 64.8)2 = 259.2 W Thus the 20 Ω resistor dissipates 259.2 W. P 4.34
The mesh current equations: −132 + 1i1 + 3(i1 − i3 ) + 2(i1 − i2 )
=
0
−7iφ + 2(i2 − i1 ) + 10(i2 − i3 )
=
0
5i3 + 10(i3 − i2 ) + 3(i3 − i1 )
=
0
The dependent source constraint equation: i φ = i 2 − i3 Place these equations in standard form: i1 (1 + 3 + 2) + i2 (−2) + i3 (−3) + iφ (0)
=
132
i1 (−2) + i2 (10 + 2) + i3 (−10) + iφ (−7)
=
0
i1 (−3) + i2 (−10) + i3 (5 + 10 + 3) + iφ (0) =
0
i1 (0) + i2 (−1) + i3 (1) + iφ (1)
0
=
iσ = −25.8 A
4–43
CHAPTER 4. Techniques of Circuit Analysis
4–44
Solving,
i1 = 48 A;
i2 = 36 A;
i3 = 28 A;
iφ = 8 A
Solve for the power: pdep source = −7(iφ )i2 = −7(8)(36) = −2016 W Thus, the dependent source is developing 2016 W. P 4.35
The mesh current equations: 53(i2 − i3 ) + 5(i1 − i3 ) + 3(i1 − i2 )
=
0
30 + 3(i2 − i1 ) + 20(i2 − i3 ) + 7i2
=
0
−30 + 2i3 + 20(i3 − i2 ) + 5(i3 − i1 )
=
0
i1 (5 + 3) + i2 (53 − 3) + i3 (−53 − 5) =
0
i1 (−3) + i2 (3 + 20 + 7) + i3 (−20)
=
−30
i1 (−5) + i2 (−20) + i3 (2 + 20 + 5)
=
30
Place these equations in standard form:
Solving,
i1 = 186 A;
i2 = 81.6 A;
i3 = 96 A
Calculate the power: p30V(left)
=
(30)(81.6) = 2448 W
p30V(right)
=
−(30)(96) = −2880 W
pdep source
=
53(81.6 − 96)(186) = −141,955.2 W
p3Ω
=
(3)(186 − 81.6)2 = 32,698.08 W
p5Ω
=
(5)(186 − 96)2 = 40,500 W
p20Ω
=
(20)(81.6 − 96)2 = 4147.2 W
p7Ω
=
(7)(81.6)2 = 46,609.92 W
p2Ω
=
(2)(96)2 = 18,432 W
pdev = 2880 + 141,955.2 = 144,835.2 W
Problems
pdis = 2448 + 32,698.08 + 40,500 + 4147.2 + 46,609.92 + 18,432 = 144,835.2 W(checks)
Thus the dependent source develops 141,955.2 W. P 4.36 [a]
10 = 18i1 − 16i2 0 = −16i1 + 28i2 + 4i∆ 4 = 8i∆ Solving, i1 = 1 A;
i2 = 0.5 A;
i∆ = 0.5 A
v0 = 16(i1 − i2 ) = 16(0.5) = 8 V [b] p4i∆ = 4i∆ i2 = (4)(0.5)(0.5) = 1 W (abs) .·. p4i∆ (deliver) = −1 W P 4.37
600 = 25.6i1 − 16i2 − 5.6i3 −424 = −16i1 + 20i2 − 0.8i3 30 = i3 Solving, i1 = 35 A;
i2 = 8 A;
i3 = 30 A
4–45
CHAPTER 4. Techniques of Circuit Analysis
4–46
[a] v30A = 0.8(i2 − i3 ) + 5.6(i1 − i3 ) = 0.8(8 − 30) + 5.6(35 − 30) = 10.4 V p30A = 30v30A = 30(10.4) = 312 W (abs) Therefore, the 30 A source delivers −312 W. [b] p600V = −600(35) = −21,000 W(del) p424V = 424(8) = 3392 W(abs) Therefore, the total power delivered is 21,000 W =
(35)2 (4) = 4900 W
p3.2Ω
=
(8)2 (3.2) = 204.8 W
p16Ω
=
(35 − 8)2 (16) = 11,664 W
p5.6Ω
=
(35 − 30)2 (5.6) = 140 W
p0.8Ω
=
(−30 + 8)2 (0.8) = 387.2 W
[c] p4Ω
presistors = 17,296 W pabs = 17,296 + 312 + 3392 = 21,000 W (CHECKS)
P 4.38 [a]
The mesh current equation for the right mesh is: 5400(i1 − 0.005) + 3700i1 − 150(0.005 − i1 ) = 0 .·. i1 = 3 mA Solving, 9250i1 = 27.75 Then, i∆ = 0.005 − i1 = 0.005 − 0.003 = 0.002 = 2 mA [b] vo = (0.005)(10,000) + (0.002)(5400) = 60.8 V p5mA = −(60.8)(0.005) = −304 mW Thus, the 5 mA source delivers 304 mW
Problems [c] 150i∆ = 150(0.002) = 0.3 V pdep source = 150i∆ i1 = −(0.3)(0.003) = −0.9 mW The dependent source delivers 0.9 mW. P 4.39
Mesh equations: 7i1 + 1(i1 − i3 ) + 2(i1 − i2 )
=
0
−125 + 2(i2 − i1 ) + 3(i2 − i3 ) + 75 =
0
Constraint equations: v∆ = 2(i1 − i2 ) i3 = −0.5v∆ ; Place these equations in standard form: i1 (7 + 1 + 2) + i2 (−2) + i3 (−1) + v∆ (0) =
0
i1 (−2) + i2 (2 + 3) + i3 (−3) + v∆ (0)
=
50
i1 (0) + i2 (0) + i3 (1) + v∆ (0.5)
=
0
i1 (2) + i2 (−2) + i3 (0) + v∆ (−1)
=
0
Solving, i1 = 6 A; i2 = 22 A; i3 = 16 A; v∆ = −32 V Solve the outer loop KVL equation to find vcs : −125 + 7i1 + vcs = 0; .·. vcs = 125 − 7(6) = 83 V Calculate the power: p125V
=
−(125)(22) = −2750 W
p75V
=
(75)(22 − 16) = 450 W
pdep source
=
−(83)[0.5(−32)] = 1328 W
Thus, the total power developed is 2750 W. CHECK: p7Ω = (6)2 (7) = 252 W p2Ω
=
(22 − 6)2 (2) = 512 W
p3Ω
=
(22 − 16)2 (3) = 108 W
p1Ω
=
(16 − 6)2 (1) = 100 W
4–47
CHAPTER 4. Techniques of Circuit Analysis
4–48 .·.
pabs = 450 + 1328 + 252 + 512 + 108 + 100 = 2750 W (checks!)
P 4.40
Since the bottom left mesh current value is known, we need only two mesh current equations: 1i1 + 4(i1 − i2 ) + 5(i1 − 20)
=
0
6.5i1 + 20(i2 − 20) + 4(i2 − i1 )
=
0
Place these equations in standard form: i1 (1 + 4 + 5) + i2 (−4)
=
100
i1 (6.5 − 4) + i2 (20 + 4) =
400
i2 = 15 A Solving, i1 = 16 A; Find v: −v + 5(20 − i1 ) + 20(20 − i2 ) = 0 .·. Calculate the power: p20A
=
−(120)(20) = −2400 W
pdep source
=
[6.5(16)](15) = 1560 W
p1Ω
=
1(16)2 = 256 W
p5Ω
=
5(20 − 16)2 = 80 W
p4Ω
=
4(16 − 15)2 = 4 W
p20Ω
=
20(20 − 15)2 = 500 W
v = 5(4) + 20(5) = 120 V
pdev = 2400 W pdis = 1560 + 256 + 80 + 4 + 500 = 2400 W (checks)
The power developed by the 20 A source is 2400 W
Problems
4–49
P 4.41 [a]
The mesh current equations are: −20 + 1(i1 − i3 ) + 25(i1 − i2 ) + 4i1
=
0
80 + 3i2 + 25(i2 − i1 ) + 2(i2 − i3 )
=
0
The constraint equation is: i3 = 45i∆ = 45(i1 − i2 ) Place these equations in standard form: i1 (1 + 25 + 4) + i2 (−25) + i3 (−1) =
20
i1 (−25) + i2 (3 + 25 + 2) + i3 (−2) =
−80
i1 (−45) + i2 (45) + i3 (1)
0
=
Solving, i1 = 8 A; i2 = 7 A; i3 = 45 A Find the power in the 2 Ω resistor: p2Ω = 2(i2 − i3 )2 = 2(−38)2 = 2888 W The 2 Ω resistor dissipates 2888 W. [b] Find the power developed by the sources: vo + 80 + 3(7) + 4(8) − 20 = 0 .·.
vo = 20 − 80 − 21 − 32 = −113 V
pdep source
=
(−113)[45(8 − 7)] = −5085 W
p80V
=
(80)(7) = 560 W
p20V
=
−(20)(8) = −160 W
pdev = 5085 + 160 = 5245 W
The percent of the power developed that is deliverd to the 2 Ω resistor is: 2888 × 100 = 55.06% 5245
CHAPTER 4. Techniques of Circuit Analysis
4–50 P 4.42 [a]
The mesh current equations are: 75 + 6i1 + 12(i1 − i2 ) − 7i∆
=
0
15i2 + 60(i2 − i3 ) + 7i∆ + 12(i2 − i1 )
=
0
The two constraint equations are: i∆
=
−i2
i3
=
1.6v∆ = 1.6(6i1 ) = 9.6i1
Place these equations in standard form: i1 (6 + 12) + i2 (−12) + i3 (0) + i∆ (−7)
=
−75
i1 (−12) + i2 (15 + 60 + 12) + i3 (−60) + i∆ (7) =
0
i1 (0) + i2 (1) + i3 (0) + i∆ (1)
=
0
i1 (9.6) + i2 (0) + i3 (−1) + i∆ (0)
=
0
Solving, i1 = 4 A; i2 = 29.4 A; i3 = 38.4 A; Calculate the power associated with the three sources: v
=
60(i2 − i3 ) = −540 V
v∆
=
6i1 = 6(4) = 24 V
p75V
=
(75)(4) = 300 W
pCCVS
=
−7(−29.4)(4 − 29.4) = −5227.32 W
pVCCS
=
(−540)[1.6(24)] = −20,736 W
i∆ = −29.4 A
The two dependent sources are generating a total of 5227.32 + 20,736 = 25,963.32 W. [b] Find the power dissipated. Remember that the 75 V source is generating 300 W, as calculated in part (a): p6Ω
=
(6)(4)2 = 96 W
p12Ω
=
(12)(4 − 29.4)2 = 7741.92 W
p15Ω
=
(15)(29.4)2 = 12,965.4 W
p60Ω
=
(60)(29.4 − 38.4)2 = 4860 W
pdis = 300 + 96 + 7741.92 + 12,965.4 + 4860 = 25,963.32 W(checks)
Problems Thus the power dissipated in the circuit is 25,963.32 W. P 4.43
The supermesh equation is: −20 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0 The supermesh constraint equation is : i 1 − i2 = 6 Place these equations in standard form: i1 (4 + 1) + i2 (9 + 6) =
20 + 90
i1 (1) + i2 (−1)
6
=
Solving, i1 = 10 A; Now find the power:
i2 = 4 A
p4Ω
=
102 (4) = 400 W
p1Ω
=
102 (1) = 100 W
p9Ω
=
42 (9) = 144 W
p6Ω
=
42 (6) = 96 W
p20V
=
−(20)(10) = −200 W
v6A
=
9i2 − 90 + 6i2 = (9)(4) − 90 + (6)(4) = −30 V
p6A
=
(−30)(6) = −180 W
p90V
=
−(90)(4) = −360 W
In summary:
pdev = 200 + 180 + 360 = 740 W pdiss = 400 + 100 + 144 + 96 = 740 W Thus the power dissipated in the circuit is 740 W
4–51
CHAPTER 4. Techniques of Circuit Analysis
4–52 P 4.44
The supermesh equation is: −120 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0 The supermesh constraint equation is : i 1 − i2 = 6 Place these equations in standard form: i1 (4 + 1) + i2 (9 + 6) =
120 + 90
i1 (1) + i2 (−1)
6
=
Solving, i1 = 15 A; Now find the power:
i2 = 9 A
p4Ω
=
152 (4) = 900 W
p1Ω
=
152 (1) = 225 W
p9Ω
=
92 (9) = 729 W
p6Ω
=
92 (6) = 486 W
p120V
=
−(120)(15) = −1800 W
vo
=
9i2 − 90 + 6i2 = 9(9) − 90 + 6(9) = 45 V
p6A
=
(45)(6) = 270 W
p90V
=
−(90)(9) = −810 W
In summary:
pdev = 900 + 225 + 729 + 486 + 270 = 2610 W now dissipating power!) pdiss = 1800 + 810 = 2610 W Thus the power dissipated in the circuit is 2610 W P 4.45 [a]
(note that the 6 A source is
Problems
4–53
The supermesh equation is: −60 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0 The supermesh constraint equation is : i 1 − i2 = 6 Place these equations in standard form: i1 (4 + 1) + i2 (9 + 6) =
60 + 90
i1 (1) + i2 (−1)
6
=
Solving, i1 = 12 A; Now find the power:
i2 = 6 A
p4Ω
=
122 (4) = 576 W
p1Ω
=
122 (1) = 144 W
p9Ω
=
62 (9) = 324 W
p6Ω
=
62 (6) = 216 W
p60V
=
−(60)(20) = −720 W
vo
=
9i2 − 90 + 6i2 = 9(6) − 90 + 6(6) = 0 V (the 6 A source acts like a short circuit carrying 6 A of current)
p6A
=
(0)(6) = 0 W
p90V = −(90)(6) = −540 W In summary: pdev = 576 + 144 + 324 + 216 = 1260 W (note that the power of the 6 A source is zero) pdiss = 720 + 540 = 1260 W Thus the power dissipated in the circuit is 1260 W [b]
Now there is no longer a supermesh. The two simple mesh current equations are: −60 + 4i1 + 1i1 = 0 −90 + 6i2 + 9i2
=
0
Since these equations are uncoupled, each can be solved separately:
CHAPTER 4. Techniques of Circuit Analysis
4–54
.·.
i1 = 60/5 = 12 A
15i2 = 90 .·.
i2 = 90/15 = 6 A
5i1 = 60
Since the currents are the same as in part (a), the power will be the same as calculated in part (a). Thus, the power dissipated in the circuit is again 1260 W. [c] As noted in part (a), the 6 A source has zero voltage drop, so is equivalent to a short circuit (which has no voltage drop by definition) carrying 6 A of current, as in the circuit of part (b). P 4.46 [a]
The i1 mesh current equation: −100 + 5(i1 − i2 ) + 10(i1 − i3 ) + 2i1 = 0 The i2 — i3 supermesh equationa: 2i2 + 20i3 + 10(i3 − i1 ) + 5(i2 − i1 ) = 0 The supermesh constraint: i3 − i2 = 1.2ib = 1.2i1 Place these equations in standard form: i1 (5 + 10 + 2) + i2 (−5) + i3 (−10)
=
100
i1 (−10 − 5) + i2 (2 + 5) + i3 (20 + 10) =
0
i1 (1.2) + i2 (1) + i3 (−1)
0
=
Solving, i1 = 7.4 A; i2 = −4.2 A; Solve for the requested currents: ia
=
i2 = −4.2 A
ib
=
i1 = 7.4 A
ic
=
i3 = 4.68 A
id
=
i1 − i2 = 11.6 A
ie
=
i1 − i3 = 2.72 A
[b] Find vcs : 2i2 + vcs + 5(i2 − i1 ) = 0
.·.
i3 = 4.68 A
vcs = −2(−4.2) − 5(−4.2 − 7.4) = 66.4 V
Problems Calculate the power: p100V
=
−(100)(7.4) = −740 W
pdep source
=
−(66.4)[1.2(7.4)] = −589.632 W
p2Ω
=
2(−4.2)2 = 35.28 W
p5Ω
=
5(7.4 + 4.2)2 = 672.8 W
p2Ω
=
2(7.4)2 = 109.52 W
p10Ω
=
10(7.4 − 4.68)2 = 73.984 W
p20Ω
=
20(4.68)2 = 438.048 W
pdev = 740 + 589.632 = 1329.632 W pdis = 35.28 + 672.8 + 109.52 + 73.984 + 438.048 = 1329.632 W
P 4.47 [a]
The i2 mesh current equation: −4id + 10(i2 − i4 ) + 5(i2 − i3 ) = 0 The i3 — i4 supermesh equation: 40(i3 − 19) + 5(i3 − i2 ) + 10(i4 − i2 ) − 240 = 0 The supermesh constraint equation: i4 − i3 = 2ib = 2(i2 − i3 ) Place the equations in standard form: i2 (10 + 5) + i3 (−5) + i4 (−10 − 4) =
0
i2 (−5 − 10) + i3 (40 + 5) + i4 (10)
=
240 + (40)(19)
i2 (2) + i3 (−1) + i4 (−1)
=
0
Solving, i2 = 18 A; i3 = 26 A; Solve for the requested currents:
i4 = 10 A
4–55
CHAPTER 4. Techniques of Circuit Analysis
4–56
ia = 19 − i3 = 19 − 26 = −7 A ib = i2 − i3 = 18 − 26 = −8 A ic = i2 − i4 = 18 − 10 = 8 A id = i4 = 10 A ie = i2 = 18 A [b] Find the power in the circuit: va
=
40ia = 40(−7) = −280 V
vb
=
−10ic − 240 = −10(8) − 240 = −320 V
p19A
=
−(−280)(19) = 5320 W
pCCCS
=
−(−320)(2)(−8) = −5120 W
pCCVS
=
−(4)(10)(18) = −720 W
p240V
=
−(240)(10) = −2400 W
p40Ω
=
(40)(−7)2 = 1960 W
p5Ω
=
(5)(−8)2 = 320 W
p10Ω
=
(10)(8)2 = 640 W
pdev = 5120 + 720 + 2400 = 8240 W pdis = 5320 + 1960 + 320 + 640 = 8240 W(checks)
P 4.48 [a]
125 =
10i1 − 0.4i2 − 9.4i3
125 =
−0.4i1 + 20i2 − 19.4i3
0
−9.4i1 − 19.4i2 + 50i3
=
Solving, i1 = 23.93 A;
i2 = 17.79 A;
i3 = 11.40 A
Problems v1
=
9.4(i1 − i3 ) = 117.76 V
v2
=
19.4(i2 − i3 ) = 123.90 V
v3
=
21.2i3 = 241.66 V
[b] pR1
=
(i1 − i3 )2 (9.4) = 1475.22 W
pR2
=
(i2 − i3 )2 (19.4) = 791.29 W
pR3
=
i23 (21.2) = 2754.64 W
[c]
4–57
pdev = 125(i1 + i2 ) = 5213.99 W
pload = 5021.15 W
% delivered =
5021.15 × 100 = 96.3% 5213.99
[d]
250 = 29.2i1 − 28.8i2 0 = −28.8i1 + 50i2 Solving, i1 = 19.82 A;
i2 = 11.42 A
i1 − i2
=
8.41 A
v1
=
(8.41)(9.4) = 79.01 V
v2
=
8.41(19.4) = 163.06 V
Note v1 is low and v2 is high. Therefore, loads designed for 125 V would not function properly, and could be damaged.
4–58
CHAPTER 4. Techniques of Circuit Analysis
P 4.49
125 = (R1 + 0.6)ia − 0.4ib − R1 ic 125 = −0.4ia + (R2 + 0.6)ib − R2 ic 0 = −R1 ia − R2 ib + (R1 + R2 + 21.2)ic
∆=
(R + 0.6) 1 −0.4 −R1
−0.4 (R2 + 0.6) −R2
−R1 −R2 (R1 + R2 + 21.2)
When R1 = R2 , ∆ reduces to ∆ = 21.6R12 + 25.84R1 + 4.24. Na
Nb
ia =
=
125 125 0
=
125 [2R1 R2 + R1 + 22.2R2 + 21.2]
=
(R + 0.6) 1 −0.4 −R1
=
125 [2R1 R2 + 22.2R1 + R2 + 21.2]
Na , ∆
−0.4 −R1 (R2 + 0.6) −R2 −R2 (R1 + R2 + 21.2)
ib =
ineutral = ia − ib =
125 −R1 125 −R2 0 (R1 + R2 + 21.2)
Nb ∆ 125[(R1 − R2 ) + 22.2(R2 − R1 )] N a − Nb = ∆ ∆
Problems
4–59
Now note that when R1 = R2 , ineutral reduces to ineutral =
0 =0 ∆
P 4.50
The mesh current equations: −240 + 12i1 + 20(i1 − i2 )
=
0
20(i2 − i1 ) + 15(i2 + 4) + 50(i2 + idc ) + 40i2
=
0
Place these equations in standard form: i1 (12 + 20) + i2 (−20) + idc (0)
=
240
i1 (−20) + i2 (20 + 15 + 50 + 40) + idc (50) =
−60
But if the power associated with the 4 A source is zero, the voltage drop across the source must be zero. This means that the voltage drop across the 15 Ω resistor is also zero, so the 15 Ω resistor is effectively removed from the circuit. Once this happens, i2 = −4 A. Substitute this value into the first equation and solve for i1 : 32i1 − 20(−4) = 240 .·. 32i1 = 160 so i1 = 5 A Now substitute this value for i1 into the second equation and solve for idc : −20(5) + 125(−4) + 50idc = −60 so 50idc = −60 + 100 + 500 = 540 .·. idc = 540/50 = 10.8 A
CHAPTER 4. Techniques of Circuit Analysis
4–60 P 4.51 [a]
Write the mesh current equations. Note that if io = 0, then i1 = 0: −23 + 5(−i2 ) + 10(−i3 ) + 46
=
0
30i2 + 15(i2 − i3 ) + 5i2
=
0
Vdc + 25i3 − 46 + 10i3 + 15(i3 − i2 )
=
0
i2 (−5) + i3 (−10) + Vdc (0)
=
−23
i2 (30 + 15 + 5) + i3 (−15) + Vdc (0)
=
0
Place the equations in standard form:
i2 (−15) + i3 (25 + 10 + 15) + Vdc (1) =
46
Solving, i2 = 0.6 A; i3 = 2 A; Vdc = −45 V Thus, the value of Vdc required to make io = 0 is −45 V. [b] Calculate the power: p23V
=
−(23)(0) = 0 W
p46V
=
−(46)(2) = −92 W
pVdc
=
(−45)(2) = −90 W
p30Ω
=
(30)(0.6)2 = 10.8 W
p5Ω
=
(5)(0.6)2 = 1.8 W
p15Ω
=
(15)(2 − 0.6)2 = 29.4 W
p10Ω
=
(10)(2)2 = 40 W
p20Ω
=
(20)(0)2 = 0 W
p25Ω
=
(25)(2)2 = 100 W
pdev = 92 + 90 = 182 W pdis = 10.8 + 1.8 + 29.4 + 40 + 0 + 100 = 182 W(checks)
Problems
4–61
P 4.52 [a] There are three unknown node voltages and only two unknown mesh currents. Use the mesh current method to minimize the number of simultaneous equations. [b]
The mesh current equations: 2i1 + 10(i1 − i2 ) + 8(i1 − 4) =
0
4i2 + 1(i2 − 4) + 10(i2 − i1 ) = 0 Place the equations in standard form: i1 (2 + 10 + 8) + i2 (−10) =
32
i1 (−10) + i2 (4 + 1 + 10) = 4 i2 = 2 A Solving, i1 = 2.6 A; Find the power in the 10 Ω resistor: i10Ω = i1 − i2 = 0.6 A p10Ω = (0.6)2 (10) = 3.6 W [c] No, the voltage across the 4 A current source is readily available from the mesh currents, and solving two simultaneous mesh-current equations is less work than solving three node voltage equations. [d] vg = 2i1 + 4i2 = 2(2.6) + 4(2) = 13.2 V p4A = −(13.2)(4) = −52.8 W Thus the 4 A source develops 52.8 W. P 4.53 [a] There are three unknown node voltages and three unknown mesh currents, so the number of simultaneous equations required is the same for both methods. The node voltage method has the advantage of having to solve the three simultaneous equations for one unknown voltage provided the connection at either the top or bottom of the circuit is used as the reference node. Therefore recommend the node voltage method. [b]
4–62
CHAPTER 4. Techniques of Circuit Analysis The node voltage equations are: v1 v1 − v2 v1 − v3 + + = 0 1 8 10 v2 − v1 v2 − v3 v2 + + = 0 −4 + 20 8 2 v3 − v1 v3 − v2 v3 + + = 0 10 2 4 Put the equations in standard form: 1 1 1 1 v1 1 + + + v2 − = 0 + v3 − 8 10 8 10 1 1 1 1 1 + + + v2 = 4 v1 − + v3 − 8 20 8 2 2 1 1 1 1 1 + + v2 − + v3 = 0 v1 − + 10 2 2 10 4 Solving, v1 = 1.72 V; v2 = 11.33 V; v3 = 6.87 V p4A = −(11.33)(4) = −45.32 W Therefore, the 4 A source is developing 45.32 W
P 4.54 [a] The node voltage method requires summing the currents at two supernodes in terms of four node voltages and using two constraint equations to reduce the system of equations to two unknowns. If the connection at the bottom of the circuit is used as the reference node, then the voltages controlling the dependent sources are node voltages. This makes it easy to formulate the constraint equations. The current in the 20 V source is obtained by summing the currents at either terminal of the source. The mesh current method requires summing the voltages around the two meshes not containing current sources in terms of four mesh currents. In addition the voltages controlling the dependent sources must be expressed in terms of the mesh currents. Thus the constraint equations are more complicated, and the reduction to two equations and two unknowns involves more algebraic manipulation. The current in the 20 V source is found by subtracting two mesh currents. Because the constraint equations are easier to formulate in the node voltage method, it is the preferred approach.
Problems [b]
Node voltage equations: v2 v1 + − 0.2 + 3 × 10−3 v3 = 0 100 250 v4 v3 + − 3 × 10−3 v3 + 0.2 = 0 500 200 Constraints: v2 − v1 = 20;
v4 − v3 = 0.4vα ; vα = v2
Solving, v2 = 44 V io = 0.2 − 44/250 = 24 mA p20V = 20io = 480 mW (abs) P 4.55 [a] Apply source transformations to both current sources to get
io =
−(5.4 + 0.6) = −1 mA 2700 + 2300 + 1000
[b]
The node voltage equations:
4–63
4–64
CHAPTER 4. Techniques of Circuit Analysis v1 − v2 v1 + = 0 2700 2300 v2 − v1 v2 + + 0.6 × 10−3 = 0 1000 2300 Place form: these equations in standard 1 1 1 + + v2 − = v1 2700 2300 2300 1 1 1 + v1 − + v2 = 2300 1000 2300 Solving, v1 = 2.7 V; v2 = 0.4 V −2 × 10−3 +
.·. io = P 4.56 [a]
v2 − v1 = −1 mA 2300
2 × 10−3 −0.6 × 10−3
Problems
io = −135/30,000 = −4.5 mA [b]
va
=
ia
=
ib
=
vb
=
ig
=
p120V
=
Check: p8.4mA
(7500)(−0.0045) = −33.75 V −33.75 va = = −0.375 mA 90,000 90,000 −8.4 × 10−3 + 0.375 × 10−3 + 4.5 × 10−3 = −3.525 mA (6000)(3.525 × 10−3 ) − 33.75 = −12.6 V −12.6 − 120 = −3.315 mA 40,000 (120)(−3.315 × 10−3 ) = −397.8 mW =
(−33.75)(8.4 × 10−3 ) = −283.5 mW
Pdev
=
397.8 + 283.5 = 681.3 mW
Pdis
=
= P 4.57 [a]
(−12.6)2 (−33.75)2 + 60,000 90,000 −3 2 +(6000)(−3.525 × 10 ) + (7500)(−4.5 × 10−3 )2 (40,000)(−3.315 × 10−3 )2 +
681.3 mW
4–65
CHAPTER 4. Techniques of Circuit Analysis
4–66
.·. vo =
250 (480) = 400 V; 300
io =
400 = 1.6 A 250
[b]
p520V = −(520)(3.6) = −1872 W Therefore, the 520 V source is developing 1872 kW. [c] v = −(16)(1) − 40(2.6) = −120 V p1A = (−120)(1) = −120 W Therefore the 1 A source is developing 120 W. [d] Calculate the power dissipated by the resistors: p16Ω
=
(16)(1)2 = 16 W
p260Ω
=
(260)(2)2 = 1040 W
p40Ω
=
(40)(2.6)2 = 270.4 W
p4Ω
=
(4)(1.6)2 = 10.24 W
p250Ω
=
(250)(1.6)2 = 640 W
p6Ω
=
(6)(1.6)2 = 15.36 W
pdev = 120 + 1872 = 1992 W pdev = 16 + 1040 + 270.4 + 10.24 + 640 + 15.36 = 1992 W (CHECKS)
Problems
4–67
P 4.58 [a] Applying a source transformation to each current source yields
Now combine the 12 V and 5 V sources into a single voltage source and the 6 Ω, 6 Ω and 5 Ω resistors into a single resistor to get
Now use a source transformation on each voltage source, thus
which can be reduced to
.·. io =
8.5 (−1) = −0.85 A 10
[b]
The mesh current equations are:
4–68
CHAPTER 4. Techniques of Circuit Analysis 6(ia − 2) + 6ia + 5(ia − 1) + 17(ia − io ) − 34 =
0
1.5io + 34 + 17(io − ia )
0
=
Put these equations in standard form: ia (6 + 6 + 5 + 17) + io (−17) =
12 + 5 + 34
ia (−17) + io (1.5 + 17)
−34
Solving, P 4.59 VTh =
ia = 1.075 A;
=
io = −0.85 A
30 (80) = 60 V 30 + 10
RTh = 1030 + 2.5 = 10 Ω
P 4.60
Write and solve the node voltage equation at v1 : v1 − 60 v1 + −4=0 10 40 4v1 − 240 + v1 − 160 = 0 .·. v1 = 400/5 = 80 V Calculate VTh : VTh = v1 + (8)(4) = 80 + 32 = 112 V Calculate RTh by removing the independent sources and making series and parallel combinations of the resistors: RTh = 8 + 4010 = 8 + 8 = 16 Ω
Problems
4–69
P 4.61 After making a source transformation the circuit becomes
The mesh current equations are: −500 + 8(i1 − i2 ) + 12i1
=
0
−300 + 30i2 + 5.2i2 + 8(i2 − i1 )
=
0
Put the equations in standard form: i1 (8 + 12) + i2 (−8)
=
500
i1 (−8) + i2 (30 + 5.2 + 8) =
300
i2 = 12.5 A Solving, i1 = 30 A; VTh = 5.2i2 + 12i1 = 425 V RTh = (812 + 5.2)30 = 7.5 Ω
P 4.62 First we make the observation that the 10 mA current source and the 10 kΩ resistor will have no influence on the behavior of the circuit with respect to the terminals a,b. This follows because they are in parallel with an ideal voltage source. Hence our circuit can be simplified to
CHAPTER 4. Techniques of Circuit Analysis
4–70
or
Therefore the Norton equivalent is determined by adding the current sources and combining the resistors in parallel:
P 4.63 [a] First, find the Thévenin equivalent with respect to a,b using a succession of source transformations.
Problems
.·. VTh = 54 V
vmeas =
4–71
RTh = 4.5 kΩ
85.5 (54) = 51.3 V 90
51.3 − 54 [b] %error = × 100 = −5% 54 P 4.64 [a] Open circuit:
The node voltage equations are: v1 − 17.4 v1 + 0.1 = 0 + 40 15 v2 v2 − 17.4 = 0 −0.1 + + 14 26 The above equations are decoupled, so just solve the second equation for v2 and use v2 to solve for voc : −36.4 + 26v2 + 14v2 − 243.6 = 0 .·. v2 = 280/40 = 7 V 10 (7) = 5 V voc = 10 + 4 Short circuit:
CHAPTER 4. Techniques of Circuit Analysis
4–72
Write a node voltage equation at v2 : v2 − 17.4 v2 + =0 −0.1 + 26 4 Solving, −5.2 + 2v2 − 34.8 + 13v2 = 0 .·. Calculate the short circuit current: isc = (40/15)/4 = 2/3 A Therefore, RTh = 5/(2/3) = 7.5 Ω
[b]
RTh = 10(26 + 4) = 7.5 Ω (CHECKS)
v2 = 40/15 V
Problems
4–73
P 4.65
OPEN CIRCUIT Use Ohm’s law to solve for v2 on the right hand side of the circuit: v2 = −80ib (50,000) = −40 × 105 ib Use this value of v2 to express the value of the dependent voltage source in terms of ib : 4 × 10−5 v2 = 4 × 10−5 (−40 × 105 ib ) = −160ib Write the mesh current equation for the ib mesh: 1310ib − 160ib + 100(ib − 500 × 10−6 ) = 0 Solving, 1250ib = 0.05 .·. ib = 0.05/1250 = 40 µ A Thus, VTh = v2 = −40 × 105 ib = −40 × 105 (40 × 10−6 ) = −160 V SHORT CIRCUIT v2 = 0;
isc = −80ib
Calculate ib using current division on the left hand side of the circuit: ib =
100 500 × 10−6 = 35.461 µ A 100 + 1310
Calculate the short circuit current from the right hand side of the circuit: isc = −80(35.461 × 10−6 ) = −2.8369 × 10−3 mA Calculate RTh from the short circuit current and open circuit voltage: RTh =
−160 = 56.4 kΩ −2.8369 × 10−3
4–74
CHAPTER 4. Techniques of Circuit Analysis
P 4.66
12.72 = VTh − 2RTh
12 = VTh − 20RTh Solving the above equations for VTh and RTh yields VTh = 12.8 V, .·. IN = 320 A,
RTh = 40 mΩ RN = 40 mΩ
P 4.67 First, find the Thévenin equivalent with respect to Ro .
Problems Ro
io
vo
Ro
io
vo
0
12
0
20
4
80
2
10 20
30
3
90
6
7.5 45
40
2.4
96
10
6 60
50
2 100
15 4.8 72
70
1.5 105
4–75
P 4.68
The node voltage equations are: v1 − 90 v1 v1 − v2 + + 15,000 10,000 4000 v2 − v3 v2 v2 − v1 + + − 19i∆ 4000 40,000 5000 v3 v3 − v2 + + 19i∆ 5000 89,000
=
0
=
0
=
0
The dependent source constraint equation is: v1 − v2 i∆ = 4000 Substitute the constraint equation into the node voltage equations and put the three remaining equations in standard form: 1 1 1 1 90 v1 + + + v2 − + v3 (0) = 15,000 10,000 4000 4000 15,000
v1
19 1 1 19 1 1 1 − + + + − + v2 + v3 − =0 4000 4000 4000 40,000 5000 4000 5000
19 1 19 1 1 − + + v2 − + v3 4000 5000 4000 5000 89,000
v1
Solving, v1 = 32.75 V; VTh = v3 = −19.8 V
v2 = 30.58 V;
=0
v3 = −19.8 V
4–76
CHAPTER 4. Techniques of Circuit Analysis
The mesh current equations are: −90 + 15,000i1 + 10,000(i1 − i∆ )
=
0
4000i∆ + 40,000(i∆ − isc ) + 10,000(i∆ − i1 )
=
0
40,000(isc − i∆ ) + 5000(isc + 19i∆ )
=
0
Put these equations in standard form: i1 (25,000) + i∆ (−10,000) + isc (0)
=
+90
i1 (−10,000) + i∆ (54,000) + isc (−40,000) =
0
i1 (0) + i∆ (55,000) + isc (45,000)
0
=
Solving, i1 = 3745.62 µA; i∆ = 364.04 µ A; isc = −444.94 µ A RTh = −19.8/ − 444.94 × 10−6 = 44.5 kΩ
isc = −444.94 µ A
P 4.69 [a] Use source transformations to simplify the left side of the circuit.
ib =
7.7 − 5.5 = 0.1 mA 22,000
Problems Let
4–77
Ro = Rmeter 1.3 kΩ = 5.5/4.4 × 10−3 = 1250 Ω
(Rmeter )(1300) = 1250; .·. Rmeter + 1300
Rmeter =
(1250)(1300) = 32.5 kΩ 50
[b] Actual value of ve : 7.7 = 97.22 µ A ib = 22,000 + 44(1300) ve = 44ib (1300) = 5.56 V % error =
5.5 − 5.56 × 100 = −1.10% 5.56
P 4.70 [a] Find the Thévenin equivalent with respect to the terminals of the ammeter. This is most easily done by first finding the Thévenin with respect to the terminals of the 4.8 Ω resistor. Thévenin voltage: note iφ is zero.
VTh − 24 VTh VTh VTh + + + =0 2 100 25 20 50VTh + VTh + 4VTh + 5VTh = 50(24) Short-circuit current:
isc = 12 + 2isc , RTh =
.·. isc = −12 A
20 = −1.67 Ω −12
.·.
VTh = 50(24)/60 = 20 V
4–78
CHAPTER 4. Techniques of Circuit Analysis
Rtotal =
20 = 3.333 Ω 6
Rmeter = 3.333 − 3.133 = 0.20 Ω [b] Actual current:
iactual =
20 = 6.383 A 3.133
% error =
6 − 6.383 × 100 = −6% 6.383
P 4.71
i1 = 100/20,000 = 5 mA 100 = VTh − 0.005RTh ,
VTh = 100 + 0.005RTh
Problems
i2 = 200/50,000 = 4 mA 200 = VTh − 0.004RTh ,
VTh = 200 + 0.004RTh
.·. 100 + 0.005RTh = 200 + 0.004RTh
so
RTh = 100 kΩ
VTh = 100 + 500 = 600 V
P 4.72
Use voltage division to calculate v1 and v2 : 501 (5) = 4.168053 V v1 = 501 + 100 5000 v2 = (5) = 4.1666667 V 5000 + 1000 Now calculate VTh : VTh = v1 − v2 = 4.168053 − 4.1666667 = 1.3866 mV Calculate RTh by removing the voltage source and creating series and parallel combinations of the resisitors:
4–79
CHAPTER 4. Techniques of Circuit Analysis
4–80
(100)(501) (1000)(5000) + = 916.69 Ω RTh = 100501 + 10005000 = 601 6000 The resulting Thévenin equivalent circuit is shown below:
Use KVL to calculate igal : igal =
1.3866 × 10−3 = 1.43 µA 916.69 + 50
P 4.73 VTh = 0, since circuit contains no independent sources.
iT =
vT − v1 vT − 40i∆ + 20 60
v1 − 40i∆ v1 v1 − vT + + =0 16 80 20 .·. 10v1 − 200i∆ = 4vT .·. 12.5v1 = 4vT ;
60 vT = 18.75 Ω = iT 3.2
RTh = 18.75 Ω
−v1 , 80
v1 = 0.32vT
60iT = 4vT − 2.5v1 = 3.2vT .·.
i∆ =
200i∆ = −2.5v1
Problems
4–81
P 4.74 VTh = 0 since there are no independent sources in the circuit. To find RTh , apply a 1 A test source and calculate the voltage drop across the test source. Use the mesh current method.
The mesh current equations for the two meshes on the left: −10ix + 5(ix − iy ) + 40ix
=
0
10ix + 20(iy − 1) + 10iy + 5(iy − ix )
=
0
Place these equations in standard form: ix (−10 + 5 + 40) + iy (−5)
=
ix (10 − 5) + iy (20 + 10 + 5) =
0 20
Solving, ix = 80 mA; iy = 560 mA Find the voltage drop across the 1 A source: vT = 20(1 − 0.56) = 8.8 V .·. RTh = vT /1 A = 8.8/1 = 8.8 Ω
P 4.75 We begin by finding the Thévenin equivalent with respect to Ro . After making a couple of source transformations the circuit simplifies to
i∆ =
160 − 30i∆ ; 50
i∆ = 2 A
4–82
CHAPTER 4. Techniques of Circuit Analysis VTh = 20i∆ + 30i∆ = 50i∆ = 100 V Using the test-source method to find the Thévenin resistance gives
iT =
vT vT − 30(−vT /30) + 30 20
1 4 2 1 iT + = = = vT 30 10 30 15 RTh =
vT 15 = = 7.5 Ω iT 2
Thus our problem is reduced to analyzing the circuit shown below.
100 p= 7.5 + Ro
2
Ro = 250
104 Ro = 250 Ro2 + 15Ro + 56.25 104 Ro = Ro2 + 15Ro + 56.25 250 40Ro = Ro2 + 15Ro + 56.25 Ro2 − 25Ro + 56.25 = 0 √ Ro = 12.5 ± 156.25 − 56.25 = 12.5 ± 10 Ro = 22.5 Ω Ro = 2.5 Ω
Problems P 4.76 [a] Find the Thévenin equivalent with respect to the terminals of RL . Open circuit voltage:
The mesh current equations are: −240 + 3(i1 − i2 ) + 20(i1 − i3 ) + 2i1
=
0
2i2 + 4(i2 − i3 ) + 3(i2 − i1 )
=
0
10iβ + 1i3 + 20(i3 − i1 ) + 4(i3 − i2 )
=
0
The dependent source constraint equation is: i β = i 2 − i1 Place these equations in standard form: i1 (3 + 20 + 2) + i2 (−3) + i3 (−20) + iβ (0)
=
240
i1 (−3) + i2 (2 + 4 + 3) + i3 (−4) + iβ (0)
=
0
i1 (−20) + i2 (−4) + i3 (4 + 1 + 20) + iβ (10) =
0
i1 (1) + i2 (−1) + i3 (0) + iβ (1)
0
Solving, i1 = 99.6 A; i2 = 78 A; VTh = 20(i1 − i3 ) = −24 V Short-circuit current:
The mesh current equations are:
=
i3 = 100.8 A;
iβ = −21.6 A
4–83
CHAPTER 4. Techniques of Circuit Analysis
4–84
−240 + 3(i1 − i2 ) + 2i1
=
0
2i2 + 4(i2 − i3 ) + 3(i2 − i1 ) =
0
10iβ + 1i3 + 4(i3 − i2 )
0
=
The dependent source constraint equation is: i β = i 2 − i1 Place these equations in standard form: i1 (3 + 2) + i2 (−3) + i3 (0) + iβ (0)
=
240
i1 (−3) + i2 (2 + 4 + 3) + i3 (−4) + iβ (0) =
0
i1 (0) + i2 (−4) + i3 (4 + 1) + iβ (10)
=
0
i1 (1) + i2 (−1) + i3 (0) + iβ (1)
=
0
Solving,
i1 = 92 A;
isc = i1 − i3 = −4 A;
i2 = 73.33 A; i3 = 96 A; iβ = −18.67 A VTh −24 = 6Ω RTh = = isc −4
RL = RTh = 6 Ω [b] pmax =
122 = 24 W 6
P 4.77 [a]
v v − 12 v − 10 + + =0 12,000 20,000 12,500 Solving, v10k =
v = 7.03125 V
10,000 (7.03125) = 5.625 V 12,500
.·. VTh = v − 10 = −4.375 V
Problems
4–85
RTh = [(12,00020,000) + 2500] = 5 kΩ Ro = RTh = 5 kΩ [b]
pmax = (−437.5 × 10−6 )2 (5000) = 957.03 µ W P 4.78 Write KCL equations at each of the labeled nodes, place them in standard form, and solve:
v1 v1 − v2 + =0 4000 8000
At v1 :
− 3 × 10−3 +
At v2 :
v2 − v1 v2 − 10 v2 − v3 + + =0 8000 20,000 2500
At v3 :
v3 v3 − 10 v3 − v2 + + =0 2500 10,000 5000
CHAPTER 4. Techniques of Circuit Analysis
4–86
Standard form:
1 1 1 + + v2 − + v3 (0) = 0.003 4000 8000 8000
1 1 1 1 1 10 + + − + v2 + v3 − = 8000 8000 20,000 2500 2500 20,000
v1
v1
v1 (0) + v2
1 1 1 1 + + − + v3 2500 2500 10,000 5000
=
10 5000
Calculator solution: v1 = 10.890625 V
v2 = 8.671875 V
v3 = 7.8125 V
Calculate currents: i2 =
10 − v2 = 66.40625 µ A 20,000
i3 =
10 − v3 = 437.5 µ A 5000
Calculate power delivered by the sources: p3mA = (3 × 10−3 )v1 = (3 × 10−3 )(10.890625) = 32.671875 mW p10Vmiddle = i2 (10) = (66.40625 × 10−6 )(10) = 0.6640625 mW p10Vtop = i3 (10) = (437.5 × 10−6 )(10) = 4.375 mW pdeliveredtotal = 32.671875 + 0.6640625 + 4.375 = 37.7109375 mW Calculate power absorbed by the 5 kΩ resistor and the percentage power: p5k = i23 (5000) = (437.5 × 10−6 )2 (5000) = 0.95703125 mW % delivered to Ro :
0.95793125 (100) = 2.54% 37.7109375
P 4.79 [a] From the solution of Problem 4.68 we have RTh = 44.5 kΩ and VTh = −19.8 V. Therefore Ro = RTh = 44.5 kΩ [b] p =
(−9.9)2 = 2.2 mW 44,500
Problems
4–87
[c]
The node voltage equations are: v1 v1 − v2 v1 − 90 + + = 0 15,000 10,000 4000 v2 v2 − v3 v2 − v1 + + − 19i∆ = 0 4000 40,000 5000 v3 v3 v3 − v2 + + 19i∆ + = 0 5000 89,000 44,500 The dependent source constraint equation is: v1 − v2 i∆ = 4000 Place these equations in standard form: 1 1 90 1 1 v1 + v2 − + v3 (0) + i∆ (0) = + + 15,000 10,000 4000 4000 15,000
v1
v1 (0) + v2
1 1 1 1 1 − + v2 + v3 − + i∆ (−19) = 0 + + 4000 4000 40,000 5000 5000
1 1 1 1 + + − + v3 + i∆ (19) = 0 5000 5000 89,000 44,500
1 1 + v2 − + v3 (0) + i∆ (−1) = 0 4000 4000 Solving, v1 = 33.2818 V; v2 = 31.4697 V; v3 = −9.9 V; Calculate the power: 90 + 33.2818 = 3.78 mA ig = 15,000
v1
p90V = −(90)(3.78 × 10−3 ) = −340.31 mW pdep source = (v3 − v2 )(19i∆ ) = −356.07 mW pdev = 340.31 + 356.07 = 696.38 mW % delivered =
2.2 × 10−3 × 100 = 0.316% 696.38 × 10−3
P 4.80 [a] From the solution to Problem 4.67 we have
i∆ = 453 µA
CHAPTER 4. Techniques of Circuit Analysis
4–88
Ro (Ω)
Po (W)
Ro (Ω)
0
0
20 320.00
2
200.00
30 270.00
6
337.50
40 230.40
10
360.00
50 200.00
15
345.60
70 157.50
Po (W)
[b]
[c] Ro = 10 Ω,
Po (max) = 360 W
P 4.81 Find the Thévenin equivalent with respect to the terminals of Ro . Open circuit voltage:
(440 − 220) = 5ia − 2ib − 3ic 0 = −2ia + 10ib − ic ic = 0.5v∆ ;
v∆ = 2(ia − ib );
i c = i a − ib
Problems Solving, ia = 96.8 A; ib = 26.4 A; ic = 70.4 A; v∆ = 140.8 V .·. VTh = 7ib = 184.8 V Short circuit current:
440 − 220 = 5ia − 2isc − 3ic 0 = −2ia + 3isc − 1ic ic = 0.5v∆ ;
v∆ = 2(ia − isc )
Solving, isc = 60 A;
ia = 80 A;
.·.
ic = ia − isc
ic = 20 A;
RTh = VTh /isc = 184.8/60 = 3.08 Ω Ro = 3.08 Ω p Ro =
(92.4)2 = 2772 W 3.08
With Ro equal to 3.08 Ω the circuit becomes
v∆ = 40 V
4–89
CHAPTER 4. Techniques of Circuit Analysis
4–90
220 = 5i1 − 3(0.5)(2)(i1 − i3 ) − 2i3 = 2i1 + i3 .·.
2i1 = 220 − i3 = 220 − 43.2 = 176.8
v∆ = 2(i1 − i3 ) = 90.4 V i2 = 0.5v∆ = 45.2 A Thus we have
vc = 220 + 3(43.2) − 2 = 347.6 V
.·. i1 = 88.4 A
Problems
4–91
Therefore, the only source developing power is the 440 V source. p440V = −(440)(88.4) = −38,896 W % delivered =
Power delivered is 38,896 W
2772 (100) = 7.13% 38,896
P 4.82 [a] We begin by finding the Thévenin equivalent with respect to the terminals of Ro . Open circuit voltage
The mesh current equations are: −100 + 4(i1 − i2 ) + 80(i1 − i3 ) + 16i1
=
0
124i∆ + 8(i2 − i3 ) + 4(i2 − i1 )
=
0
50 + 12i3 + 80(i3 − i1 ) + 8(i3 − i2 )
=
0
The constraint equation is: i∆ = i3 − i1 Place these equations in standard form: i1 (4 + 80 + 16) + i2 (−4) + i3 (−80) + i∆ (0) =
100
i1 (−4) + i2 (8 + 4) + i3 (−8) + i∆ (124)
0
=
i1 (−80) + i2 (−8) + i3 (12 + 80 + 8) + i∆ (0) =
−50
i1 (1) + i2 (0) + i3 (−1) + i∆ (1)
0
Solving, i1 = 4.7 A; i2 = 10.5 A; Also, VTh = vab = −80i∆ = 48 V Now find the short-circuit current.
=
i3 = 4.1 A;
i∆ = −0.6 A
4–92
CHAPTER 4. Techniques of Circuit Analysis
Note with the short circuit from a to b that i∆ is zero, hence 124i∆ is also zero. The mesh currents are: −100 + 4(i1 − i2 ) + 16i1 = 0 8(i2 − i3 ) + 4(i2 − i1 )
=
0
50 + 12i3 + 8(i3 − i2 )
=
0
Place these equations in standard form: i1 (4 + 16) + i2 (−4) + i3 (0)
=
100
i1 (−4) + i2 (8 + 4) + i3 (−8) =
0
i1 (0) + i2 (−8) + i3 (12 + 8)
−50
=
Solving, i1 = 5 A; i2 = 0 A; Then, isc = i1 − i3 = 7.5 A RTh = 48/7.5 = 6.4 Ω
For maximum power transfer [b] pmax =
i3 = −2.5 A
Ro = RTh = 6.4 Ω
24 = 90 W 6.4 2
P 4.83 From the solution of Problem 4.82 we know that when Ro is 6.4 Ω, the voltage across Ro is 24 V, positive at the upper terminal. Therefore our problem reduces to the analysis of the following circuit. In constructing the circuit we have used the fact that i∆ is −0.3 A, and hence 124i∆ is −37.2 V.
Problems
Using the node voltage method to find v1 and v2 yields 4.05 +
24 − v1 24 − v2 =0 + 4 8
2v1 + v2 = 104.4;
v1 + 37.2 = v2
Solving, v1 = 22.4 V; v2 = 59.6 V. It follows that 22.4 − 100 = −4.85 A = ig1 16 59.6 − 50 = 0.8 A ig2 = 12 59.6 − 24 i2 = = 4.45 A 8 ids
=
−4.45 − 0.8 = −5.25 A
p100V
=
100ig1 = −485 W
p50V
=
50ig2 = 40 W
pds
=
37.2ids = −195.3 W
.·.
pdev = 485 + 195.3 = 680.3 W
.·. % delivered =
90 (100) = 13.23% 680.3
.·. 13.23% of developed power is delivered to load
4–93
4–94
CHAPTER 4. Techniques of Circuit Analysis
P 4.84 [a] Open circuit voltage
Node voltage equations: v1 − 60 v1 − 4i∆ v1 − v2 + + 2 5 4 v2 − v1 + 2v∆ 4 Constraint equations:
=
0
=
0
60 − v1 v1 − v2 i∆ = 4 Place the equations in standard form: 1 1 1 1 4 + + + v2 − + i∆ − + v∆ (0) = 30 v1 2 5 4 4 5 1 1 + v2 + i∆ (0) + v∆ (2) = 0 v1 − 4 4 v∆
=
v1 (1) + v2 (0) + i∆ (0) + v∆ (1)
=
60
v1 (1) + v2 (−1) + i∆ (−4) + v∆ (0)
=
0
Solving, v1 = 20 V; Short circuit current:
v2 = −300 V;
The node voltage equation: v1 − 60 v1 − 4i∆ v1 + + =0 2 5 4 The constraint equation:
i∆ = 80 A;
v∆ = 40 V
Problems i∆ = v1 /4 Place instandard these equations form: 1 1 1 4 + + v1 + i∆ − = 30 2 5 4 5 1 + i∆ (−1) = 0 v1 4 Solving, v1 = 40 V; i∆ = 10 A Then, v∆ = 60 − 40 = 20 V and isc = i∆ − 2v∆ = 10 − 40 = −30 A Thus, RTh = −300/ − 30 = 10 Ω [b]
pmax =
(150)2 = 2250 W 10
[c]
The node voltage equation: va − 60 va − 4i∆ va + 150 =0 + + 2 5 4 The constraint equation is: va + 150 i∆ = 4 Place the equations in standard form: 1 1 1 150 4 va + i∆ − = 30 − + + 2 5 4 4 5 1 150 va − + i∆ (1) = 4 4 i∆ = 45 A Solving, va = 30 V; Calculate the power:
4–95
4–96
CHAPTER 4. Techniques of Circuit Analysis i60V
=
va − 60 = −15 A 2
p60V
=
(60)(−15) = −900 W
iccvs
=
va − 4i∆ = −30 A 5
pccvs
=
4(45)(−30) = −5400 W
pvccs
=
(−150)[2(30)] = −9000 W
pdev = 900 + 5400 + 9000 = 15,300 W
% delivered =
2250 × 100 = 14.7% 15,3000
P 4.85 [a] First find the Thévenin equivalent with respect to Ro . Open circuit voltage: iφ = 0; 50iφ = 0
v1 − 280 v1 − 280 v1 v1 + + + + 0.5125v∆ = 0 100 10 25 400 v∆ =
(280 − v1 ) 5 = 56 − 0.2v1 25
v1 = 210 V;
v∆ = 14 V
VTh = 280 − v∆ = 280 − 14 = 266 V Short circuit current
Problems
v1 − 280 v2 v2 v1 + + + + 0.5125(280) = 0 100 10 20 400 v∆ = 280 V v2 + 50iφ = v1 iφ =
280 v2 + = 56 + 0.05v2 5 20
v2 = −968 V;
v1 = −588 V
iφ = isc = 56 + 0.05(−968) = 7.6 A RTh = VTh /isc = 266/7.6 = 35 Ω .·. Ro = 35 Ω [b]
pmax = (133)2 /35 = 505.4 W
4–97
CHAPTER 4. Techniques of Circuit Analysis
4–98 [c]
v1 − 280 v2 − 133 v2 v1 + + + + 0.5125(280 − 133) = 0 100 10 20 400 v2 + 50iφ = v1 ;
iφ = 133/35 = 3.8 A
Therefore, v1 = −189 V and v2 = −379 V; thus, 280 − 133 280 + 189 + = 76.30 A 5 10
ig =
p280V (dev) = (280)(76.3) = 21,364 W P 4.86 [a] Since 0 ≤ Ro < ∞ maximum power will be delivered to the 8 Ω resistor when Ro = 0. 242 = 72 W [b] P = 8 P 4.87 [a] 110 V source acting alone:
Re = i =
10(14) 35 = Ω 24 6
132 110 = A 5 + 35/6 13
35 v = 6
132 770 V = 13 13
4 A source acting alone:
Problems
5 Ω10 Ω = 50/15 = 10/3 Ω 10/3 + 2 = 16/3 Ω 16/312 = 48/13 Ω Hence our circuit reduces to:
It follows that va = 4(48/13) = (192/13) V and 5 −va (10/3) = − va = −(120/13) V v = (16/3) 8
.·.
v = v + v =
vo2 = 250 W [b] p = 10
770 120 − = 50 V 13 13
4–99
4–100
CHAPTER 4. Techniques of Circuit Analysis
P 4.88 70-V source acting alone:
v = 70 − 4ib is =
v vb + = ia + ib 2 10
70 = 20ia + vb ia = .·.
70 − vb 20 ib =
v 70 − vb 11 vb v + − = vb + − 3.5 2 10 20 20 10
v = vb + 2ib .·. vb = v − 2ib .·. ib =
11 v (v − 2ib ) + − 3.5 20 10
13 70 v − .·. v = 70 − 4 42 42 50-V source acting alone:
v = −4ib
or
or
ib =
v =
13 70 v − 42 42
1610 3220 = V 94 47
Problems v = vb + 2ib v = −50 + 10id .·. id =
v + 50 10
is =
vb v + 50 + 2 10
ib =
11 vb v v v + 50 v + 50 + is = b + b + = vb + 20 20 2 10 20 10
vb = v − 2ib .·. ib =
11 v + 50 (v − 2ib ) + 20 10
13 100 v + v = −4 42 42
Thus, Hence,
v = v + v =
or
v = −
200 V 47
1410 1610 200 = = 30 V − 47 47 47
20 (10) = 5 V 20 + 5 + 15
20 V source acting alone:
vo2 =
ib =
P 4.89 10 V source acting alone:
vo1 =
13 100 v + 42 42
or
13.333 (20) = 5 V 13.333 + 10 + 30
4–101
4–102
CHAPTER 4. Techniques of Circuit Analysis
6 A current source acting alone:
Node voltage equations: v1 − v2 v1 + −6 15 5 v2 v2 − v3 v2 − v1 + + 5 40 10 v3 − v2 v3 + +6 10 30
=
0
=
0
=
0
In standard form: 1 1 1 + + v2 − + v3 (0) v1 15 5 5 1 1 1 1 1 + + + v2 + v3 − v1 − 5 5 40 10 10 1 1 1 + + v3 v1 (0) + v2 − 10 10 30 Solving, Note that Finally,
=
6
=
0
=
−6
v1 = 22.5 V; v2 = 0 V; v3 = −45 V vo3 = v2 = 0 V vo = vo1 + vo2 + vo3 = 5 + 5 + 0 = 10 V
P 4.90 Voltage source acting alone:
vo1 vo1 − 25 vo1 − 25 + − 2.2 =0 4000 20,000 4000
Problems Simplifying
5vo1 − 125 + vo1 − 11vo1 + 275 = 0
.·. vo1 = 30 V Current source acting alone:
vo2 vo2 vo2 + 0.005 − 2.2 =0 + 4000 20,000 4000 Simplifying
5vo2 + vo2 + 100 − 11vo2 = 0
.·. vo2 = 20 V vo = vo1 + vo2 = 30 + 20 = 50 V P 4.91 Voltage source acting alone:
io1 =
−135 = −2.25 A 40 + 10025
vo1 =
60 (−135) = −90 V 90
Current source acting alone:
4–103
CHAPTER 4. Techniques of Circuit Analysis
4–104
v1 v1 + + 18 = 0 30 60 −18 +
.·.
v1 = −360 V;
vo2 = 360 V
v2 − v3 v2 + =0 80 20
v3 − v2 v3 v3 + + =0 20 25 40 .·.
v2 = 441.6 V;
v3 = 192 V;
io2 = 192/40 = 4.8 A
.·. vo = vo1 + vo2 = −90 + 360 = 270 V io = io1 + io2 = −2.25 + 4.8 = 2.55 A P 4.92 6 A source:
30 Ω5 Ω60 Ω = 4 Ω .·. io1 =
20 (6) = 4.8 A 20 + 5
10 A source:
Problems
io2 =
4 (10) = 1.6 A 25
75 V source:
io3 = −
4 (15) = −2.4 A 25
io = io1 + io2 + io3 = 4.8 + 1.6 − 2.4 = 4 A P 4.93 [a] By hypothesis io + io = 3.5 mA.
i o =
2000 (−0.005) = −1.25 mA; 8000
.·. io = 3.5 − 1.25 = 2.25 mA
[b] With all three sources in the circuit write a single node voltage equation. vb vb − 8 + + 0.005 − 0.010 = 0 2000 6000
4–105
4–106
CHAPTER 4. Techniques of Circuit Analysis .·. vb = 13.5 V 13.5 vb = = 2.25 mA 6000 6000
io = P 4.94 [a]
voc = VTh = 75 V; Therefore [b] iL =
RTh =
vo VTh − vo = RL RTh
Therefore
RTh
iL =
60 = 3 A; 20
iL =
75 − 60 15 = RTh RTh
15 = 5Ω 3
VTh VTh − vo = = − 1 RL vo /RL vo
P 4.95 [a]
v v − v2 v − v1 + + =0 2xr R 2r(L − x)
1 1 v2 1 v1 v + + + = 2xr R 2r(L − x) 2xr 2r(L − x) v=
v1 RL + xR(v2 − v1 ) RL + 2rLx − 2rx2
[b] Let D = RL + 2rLx − 2rx2 dv (RL + 2rLx − 2rx2 )R(v2 − v1 ) − [v1 RL + xR(v2 − v1 )]2r(L − 2x) = dx D2 dv = 0 when numerator is zero. dx The numerator simplifies to x2 +
RL(v2 − v1 ) − 2rv1 L2 2L − v1 x+ =0 (v2 − v1 ) 2r(v2 − v1 )
Problems Solving for the roots of the quadratic yields x=
L −v1 ± v2 − v1
L [c] x = −v1 ± v2 − v1 v2 = 1200 V,
v1 v2 −
R (v2 − v1 )2 2rL
R (v1 − v2 )2 v1 v2 − 2rL
v1 = 1000 V,
r = 5 × 10−5 Ω/m;
4–107
L = 16 km
R = 3.9 Ω
16,000 L = 80; = v2 − v1 1200 − 1000
v1 v2 = 1.2 × 106
R 3.9(−200)2 (v1 − v2 )2 = = 0.975 × 105 2rL (10 × 10−5 )(16 × 103 ) √ x = 80{−1000 ± 1.2 × 106 − 0.0975 × 106 } = 80{−1000 ± 1050} = 80(50) = 4000 m [d] vmin
=
v1 RL + R(v2 − v1 )x RL + 2rLx − 2rx2
=
(1000)(3.9)(16 × 103 ) + 3.9(200)(4000) (3.9)(16,000) + 10 × 10−5 (16,000)(4000) − 10 × 10−5 (16 × 106 )
=
975 V
P 4.96 [a] In studying the circuit in Fig. P4.96 we note it contains six meshes and six essential nodes. Further study shows that by replacing the parallel resistors with their equivalent values the circuit reduces to four meshes and four essential nodes as shown in the following diagram. The node Voltage approach will require solving three node Voltage equations along with equations involving v∆ and iβ . The mesh-current approach will require writing one supermesh equation plus three constraint equations involving the three current sources. Thus at the outset we know the supermesh equation can be reduced to a single unknown current. Since we are interested in the power developed by the 1 V source, we will retain the mesh current ib and eliminate the mesh currents ia , ic And id . The supermesh is denoted by the dashed line in the following figure.
4–108
CHAPTER 4. Techniques of Circuit Analysis
[b] Summing the voltages around the supermesh yields 4 −9iβ + ia + 0.75ib + 1 + 5ib + 7(ic − id ) + 8ic = 0 3 Note that iβ = ib And multiply the equation by 12: −108ib + 16ia + 9ib + 12 + 60ib + 84(ic − id ) + 96ic = 0 or 16ia − 39ib + 180ic − 84id = −12 Now note: ib − ic = 3iβ = 3ib ;
.·. ic = −2ib
whence 16ia − 39ib − 360ib − 84id = −12 Now use the constraint that ia − ic = −2 ia = −2 + ic = −2 − 2ib Therefore −32 − 32ib − 399ib − 84id = −12 −431ib − 84id = 20 Now use the constraint −4 ia = 8ia = −16 − 16ib id = −6v∆ = −6 3
Problems
4–109
Therefore −431ib − 84(−16 − 16ib ) = 20 or 913ib = −1324 .·. ib ≈ −1.45 A p1V = 1ib ∼ = −1.45 W;
.·. p1V (developed) ∼ = 1.45 W
P 4.97
vB − 3vx vB − vE + − 0.1 = 0 4 7 vE vE − 3vx vE − vB At node E: + + +5=0 6 5 7 vD + 13v∆ vD =0 At node D: − 5 + 0.1 + 3 2 Constraint: v∆ = vB − vE Constraint: vx = −v∆ + 5i∆ − 0.9 Constraint: i∆ = (3vx − vB )/4 In standard form: 1 1 1 3 vB + vD (0) + vE − + v∆ (0) + vx − + i∆ (0) = 0.1 + 4 7 7 4 1 1 13 + + vE (0) + v∆ + vx (0) + i∆ (0) = 4.9 vB (0) + vD 2 3 3 1 1 1 1 3 + + + vD (0) + vE + v∆ (0) + vx − + i∆ (0) = −5 vB − 7 6 5 7 5 B–C supernode:
vB (−1) + vD (0) + vE (1) + v∆ (1) + vx (0) + i∆ (0)
=
0
vB (0) + vD (0) + vE (0) + v∆ (1) + vx (1) + i∆ (−5)
=
−0.9
vB (1) + vD (0) + vE (0) + v∆ (0) + vx (−3) + i∆ (4)
=
0
4–110
CHAPTER 4. Techniques of Circuit Analysis
vB = −11.17 V; vD = −20.95 V; vE = −16.33 V; v∆ = 5.16 V; vx = −2.87 V; i∆ = 0.64 A p5A = (vE − vD )(5) = 23.1 W The 5 A source absorbs 23.1 W Solving,
P 4.98
The mesh equations are: −125 + 0.15ia + 18.4(ia − ic ) + 0.25(ia − ib )
=
0
−125 + 0.25(ib − ia ) + 38.4(ib − id ) + 0.15ib
=
0
0.15ic + 18.4(ic − ie ) + 0.25(ic − id ) + 18.4(ic − ia )
=
0
0.15id + 38.4(id − ib ) + 0.25(id − ic ) + 38.4(id − ie )
=
0
11.6ie + 38.4(ie − id ) + 18.4(ie − ic )
=
0
Place these equations in standard form: ia (18.8) + ib (−0.25) + ic (−18.4) + id (0) + ie (0)
=
125
ia (−0.25) + ib (38.8) + ic (0) + id (−38.4) + ie (0)
=
125
ia (−18.4) + ib (0) + ic (37.2) + id (−0.25) + ie (−18.4) =
0
ia (0) + ib (−38.4) + ic (−0.25) + id (77.2) + ie (−38.4) =
0
ia (0) + ib (0) + ic (−18.4) + id (−38.4) + ie (68.4)
0
Solving, ia = 32.77 A; ib = 26.46 A; Find the requested voltages: v1 = 18.4(ic − ie ) = 113.90 V v2 = 38.4(id − ie ) = 120.19 V v3 = 11.6ie = 233.62 V
ic = 26.33 A;
=
id = 23.27 A;
ie = 20.14 A
Problems P 4.99
100 = 6ia − 1ib + 0ic − 2id − 2ie + 0if − 1ig 0 = −1ia + 4ib − 2ic + 0id + 0ie + 0if + 0ig 0 = 0ia − 2ib + 13ic − 3id + 0ie + 0if + 0ig 0 = −2ia + 0ib − 3ic + 9id − 4ie + 0if + 0ig 0 = −2ia + 0ib + 0ic − 4id + 9ie − 3if + 0ig 0 = 0ia + 0ib + 0ic + 0id − 3ie + 13if − 2ig 0 = −1ia + 0ib + 0ic + 0id + 0ie − 2if + 4ig A computer solution yields ia = 30 A;
ie = 15 A;
ib = 10 A;
if = 5 A;
ic = 5 A;
ig = 10 A;
id = 15 A .·. i = id − ie = 0 A CHECK: p1T = p1B = (ib )2 = (ig )2 = 100 W p1L = (ia − ib )2 = (ia − ig )2 = 400 W p2C = 2(ib − ic )2 = (ig − if )2 = 50 W p3 = 3(ic − id )2 = 3(ie − if )2 = 300 W p4 = 4(id − ie )2 = 0 W p8 = 8(ic )2 = 8(if )2 = 200 W p2L = 2(ia − id )2 = 2(ia − ie )2 = 450 W
4–111
CHAPTER 4. Techniques of Circuit Analysis
4–112
P 4.100
pabs
=
100 + 400 + 50 + 200 + 300 + 450 + 0 + 450 + 300+ 200 + 50 + 400 + 100 = 3000 W
pgen
=
100ia = 100(30) = 3000 W (CHECKS)
−R1 [R2 (R3 + R4 ) + R3 R4 ] dv1 = dIg1 (R1 + R2 )(R3 + R4 ) + R3 R4 dv1 R1 R3 R4 = dIg2 (R1 + R2 )(R3 + R4 ) + R3 R4 dv2 −R1 R3 R4 + dIg1 (R1 + R2 )(R3 + R4 ) + R3 R4 dv2 R3 R4 (R1 + R2 ) = dIg2 (R1 + R2 )(R3 + R4 ) + R3 R4
P 4.101 From the solution to Problem 4.100 we have dv1 175 −25[5(125) + 3750] =− V/A = dIg1 30(125) + 3750 12 and dv2 −(25)(50)(75) = −12.5 V/A = dIg1 30(125) + 3750 By hypothesis, ∆Ig1 = 11 − 12 = −1 A .·. ∆v1 = (−
175 175 )(−1) = = 14.5833 V 12 12
Thus, v1 = 25 + 14.5833 = 39.5833 V Also, ∆v2 = (−12.5)(−1) = 12.5 V Thus, v2 = 90 + 12.5 = 102.5 V The PSpice solution is v1 = 39.5830 V and v2 = 102.5000 V These values are in agreement with our predicted values.
Problems P 4.102 From the solution to Problem 4.100 we have (25)(50)(75) dv1 = = 12.5 V/A dIg2 30(125) + 3750 and (50)(75)(30) dv2 = 15 V/A = dIg2 30(125) + 3750 By hypothesis, ∆Ig2 = 17 − 16 = 1 A .·. ∆v1 = (12.5)(1) = 12.5 V Thus, v1 = 25 + 12.5 = 37.5 V Also, ∆v2 = (15)(1) = 15 V Thus, v2 = 90 + 15 = 105 V The PSpice solution is v1 = 37.5 V and v2 = 105 V These values are in agreement with our predicted values. P 4.103 From the solutions to Problems 4.100 — 4.102 we have 175 dv1 V/A; =− dIg1 12
dv1 = 12.5 V/A dIg2
dv2 = −12.5 V/A; dIg1
dv2 = 15 V/A dIg2
By hypothesis, ∆Ig1 = 11 − 12 = −1 A ∆Ig2 = 17 − 16 = 1 A Therefore, ∆v1 =
175 + 12.5 = 27.0833 V 12
4–113
4–114
CHAPTER 4. Techniques of Circuit Analysis
∆v2 = 12.5 + 15 = 27.5 V Hence v1 = 25 + 27.0833 = 52.0833 V v2 = 90 + 27.5 = 117.5 V The PSpice solution is v1 = 52.0830 V and v2 = 117.5 V These values are in agreement with our predicted values. P 4.104 By hypothesis, ∆R1 = 27.5 − 25 = 2.5 Ω ∆R2 = 4.5 − 5 = −0.5 Ω ∆R3 = 55 − 50 = 5 Ω ∆R4 = 67.5 − 75 = −7.5 Ω So ∆v1 = 0.5833(2.5) − 5.417(−0.5) + 0.45(5) + 0.2(−7.5) = 4.9168 V .·. v1 = 25 + 4.9168 = 29.9168 V ∆v2 = 0.5(2.5) + 6.5(−0.5) + 0.54(5) + 0.24(−7.5) = −1.1 V .·. v2 = 90 − 1.1 = 88.9 V The PSpice solution is v1 = 29.6710 V and v2 = 88.5260 V Note our predicted values are within a fraction of a volt of the actual values.
The Operational Amplifier
5
Assessment Problems AP 5.1 [a] This is an inverting amplifier, so vo = (−Rf /Ri )vs = (−80/16)vs , vs ( V)
0.4
2.0
so
vo = −5vs
3.5 −0.6 −1.6 −2.4
vo ( V) −2.0 −10.0 −15.0 3.0 8.0 10.0 Two of the vs values, 3.5 V and −2.4 V, cause the op amp to saturate. [b] Use the negative power supply value to determine the largest input voltage: −15 = −5vs ,
vs = 3 V
Use the positive power supply value to determine the smallest input voltage: 10 = −5vs , Therefore
vs = −2 V − 2 ≤ vs ≤ 3 V
AP 5.2 From Assessment Problem 5.1 vo = (−Rf /Ri )vs = (−Rx /16,000)vs = (−Rx /16,000)(−0.640) = 0.64Rx /16,000 = 4×10−5 Rx Use the negative power supply value to determine one limit on the value of Rx : 4×10−5 Rx = −15
so
Rx = −15/4×10−5 = −375 kΩ
Since we cannot have negative resistor values, the lower limit for Rx is 0. Now use the positive power supply value to determine the upper limit on the value of Rx : 4×10−5 Rx = 10
so
Rx = 10/4×10−5 = 250 kΩ
Therefore, 0 ≤ Rx ≤ 250 kΩ 5–1
5–2
CHAPTER 5. The Operational Amplifier
AP 5.3 [a] This is an inverting summing amplifier so vo = (−Rf /Ra )va + (−Rf /Rb )vb = −(250/5)va − (250/25)vb = −50va − 10vb Substituting the values for va and vb : vo = −50(0.1) − 10(0.25) = −5 − 2.5 = −7.5 V [b] Substitute the value for vb into the equation for vo from part (a) and use the negative power supply value: vo = −50va − 10(0.25) = −50va − 2.5 = −10 V Therefore 50va = 7.5,
so
va = 0.15 V
[c] Substitute the value for va into the equation for vo from part (a) and use the negative power supply value: vo = −50(0.10) − 10vb = −5 − 10vb = −10 V; Therefore 10vb = 5,
so
vb = 0.5 V
[d] The effect of reversing polarity is to change the sign on the vb term in each equation from negative to positive. Repeat part (a): vo = −50va + 10vb = −5 + 2.5 = −2.5 V Repeat part (b): vo = −50va + 2.5 = −10 V;
50va = 12.5,
va = 0.25 V
Repeat part (c): vo = −5 + 10vb = 15 V;
10vb = 20;
vb = 2.0 V
AP 5.4 [a] Write a node voltage equation at vn ; remember that for an ideal op amp, the current into the op amp at the inputs is zero: vn − vo vn + =0 4500 63,000 Solve for vo in terms of vn by multiplying both sides by 63,000 and collecting terms: 14vn + vn − vo = 0
so
vo = 15vn
Now use voltage division to calculate vp . We can use voltage division because the op amp is ideal, so no current flows into the non-inverting input terminal and the 400 mV divides between the 15 kΩ resistor and the Rx resistor: vp =
Rx (0.400) 15,000 + Rx
Problems
5–3
Now substitute the value Rx = 60 kΩ: vp =
60,000 (0.400) = 0.32 V 15,000 + 60,000
Finally, remember that for an ideal op amp, vn = vp , so substitute the value of vp into the equation for v0 vo = 15vn = 15vp = 15(0.32) = 4.8 V [b] Substitute the expression for vp into the equation for vo and set the resulting equation equal to the positive power supply value:
0.4Rx vo = 15 15,000 + Rx
=5
15(0.4Rx ) = 5(15,000 + Rx ) so Rx = 75 kΩ AP 5.5 [a] Since this is a difference amplifier, we can use the expression for the output voltage in terms of the input voltages and the resistor values given in Eq. 5.22: vo =
20(60) 50 vb − va 10(24) 10
Simplify this expression and substitute in the value for vb : vo = 5(vb − va ) = 20 − 5va Set this expression for vo to the positive power supply value: 20 − 5va = 10 V
so va = 2 V
Now set the expression for vo to the negative power supply value: 20 − 5va = −10 V
so
va = 6 V
Therefore 2 ≤ va ≤ 6 V [b] Begin as before by substituting the appropriate values into Eq. 5.22: vo =
8(60) vb − 5va = 4vb − 5va 10(12)
Now substitute the value for vb : vo = 4(4) − 5va = 16 − 5va Set this expression for vo to the positive power supply value: 16 − 5va = 10 V
so va = 1.2 V
Now set the expression for vo to the negative power supply value: 16 − 5va = −10 V
so
va = 5.2 V
Therefore 1.2 ≤ va ≤ 5.2 V
5–4
CHAPTER 5. The Operational Amplifier
AP 5.6 [a] Replace the op amp with the more realistic model of the op amp from Fig. 5.15:
Write the node voltage equation at the left hand node: vn − vg vn − vo vn + + =0 500,000 5000 100,000 Multiply both sides by 500,000 and simplify: vn + 100vn − 100vg + 5vn − 5v0 = 0 so
21.2vn − vo = 20vg
Write the node voltage equation at the right hand node: vo − 300,000(−vn ) vo − vn =0 + 5000 100,000 Multiply through by 100,000 and simplify: 20vo + 6 × 106 vn + vo − vn = 0 so 6 × 106 vn + 21vo = 0 Use Cramer’s method to solve for vo : ∆=
21.2 6 × 106
−1 21
= 6,000,445.2
No =
21.2 6 × 106
vo =
No = −19.9985vg ; ∆
20vg 0
= −120 × 106 vg
so
vo = −19.9985 vg
[b] Use Cramer’s method again to solve for vn : N1 =
vn =
20vg 0
−1
21
= 420vg
N1 = 6.9995 × 10−5 vg ∆
vg = 1 V,
vn = 69.995 µ V
Problems
5–5
[c] The resistance seen at the input to the op amp is the ratio of the input voltage to the input current, so calculate the input current as a function of the input voltage: vg − 6.9995 × 10−5 vg vg − vn = 5000 5000 Solve for the ratio of vg to ig to get the input resistance: ig =
Rg =
vg 5000 = = 5000.35 Ω ig 1 − 6.9995 × 10−5
[d] This is a simple inverting amplifier configuration, so the voltage gain is the ratio of the feedback resistance to the input resistance: vo 100,000 = −20 =− vg 5000 Since this is now an ideal op amp, the voltage difference between the two input terminals is zero; since vp = 0, vn = 0 Since there is no current into the inputs of an ideal op amp, the resistance seen by the input voltage source is the input resistance: Rg = 5000 Ω
5–6
CHAPTER 5. The Operational Amplifier
Problems P 5.1
[a] The five terminals of the op amp are identified as follows:
[b] The input resistance of an ideal op amp is infinite, which constrains the value of the input currents to 0. Thus, in = 0 A. [c] The open-loop voltage gain of an ideal op amp is infinite, which constrains the difference between the voltage at the two input terminals to 0. Thus, (vp − vn ) = 0. [d] Write a node voltage equation at vn : vn − 2.5 vn − vo + =0 10,000 40,000 But vp = 0 and vn = vp = 0. Thus, −2.5 vo − = 0 so 10,000 40,000 P 5.2
vb − va vb − vo + = 0, 20 100
vo = −10 V
therefore
vo = 6vb − 5va
[a] va = 4 V,
vb = 0 V,
vo = −15 V
[b] va = 2 V,
vb = 0 V,
vo = −10 V
[c] va = 2 V,
vb = 1 V,
vo = −4 V
[d] va = 1 V,
vb = 2 V,
vo = 7 V
[e] If vb = 1.6 V,
(sat)
vo = 9.6 − 5va = ±15
.·. −1.08 ≤ va ≤ 4.92 V P 5.3
vo = −(0.5 × 10−3 )(10 × 103 ) = −5 V .·. io =
P 5.4
−5 = −1 mA 5000
Since the current into the inverting input terminal of an ideal op-amp is zero, the voltage across the 2.2 MΩ resistor is (2.2 × 106 )(3.5 × 10−6 ) or 7.7 V. Therefore the voltmeter reads 7.7 V.
Problems
P 5.5
[a] ia =
5–7
25 × 10−3 = 5 µA 5000
va = −50 × 103 ia = −250 mV [b]
va va − vo va + + =0 50,000 10,000 40,000 .·. 4va + 20va + 5va − 5vo = 0 .·. vo = 29va /5 = −1.45 V
[c] ia = 5 µA va − vo −vo + = 78.33 µ A [d] io = 30,000 40,000 P 5.6
[a] The gain of an inverting amplifier is the ratio of the feedback resistor to the input resistor. If the gain of the inverting amplifier is to be 6, the feedback resistor must be 6 times as large as the input resistor. There are many possible designs that use only 20 kΩ resistors. We present two here. Use a single 20 kΩ resistor as the input resistor, and use six 20 kΩ resistors in series as the feedback resistor to give a total of 120 kΩ.
Alternately, Use a single 20 kΩ resistor as the feedback resistor and use six 20 kΩ resistors in parallel as the input resistor to give a total of 3.33 kΩ.
[b] To amplify a 3 V signal without saturating the op amp, the power supply voltages must be greater than or equal to the product of the input voltage and the amplifier gain. Thus, the power supplies should have a magnitude of (3)(6) = 18 V. P 5.7
[a] The circuit shown is a non-inverting amplifier.
5–8
CHAPTER 5. The Operational Amplifier [b] We assume the op amp to be ideal, so vn = vp = 3V. Write a KCL equation at vn : 3 − vo 3 + =0 40,000 80,000 Solving, vo = 9 V.
P 5.8
vp =
18 (12) = 9 V = vn 24
vn − 24 vn − vo + =0 30 20 vo = (45 − 48)/3 = −1.0 V iL =
vo 1 × 10−3 = − × 10−3 = −200 × 10−6 5 5
iL = −200 µA P 5.9
[a] First, note that vn = vp = 2.5 V Let vo1 equal the voltage output of the op-amp. Then 2.5 − vg 2.5 − vo1 + = 0, 5000 10,000
.·. vo1 = 7.5 − 2vg
Also note that vo1 − 2.5 = vo ,
[b] Yes, the circuit designer is correct!
.·. vo = 5 − 2vg
Problems P 5.10
[a] Let v∆ be the voltage from the potentiometer contact to ground. Then 0 − vg 0 − v∆ + =0 2000 50,000 .·. v∆ = −25(40 × 10−3 ) = −1 V
−25vg − v∆ = 0,
v∆ − vo v∆ − 0 v∆ + + =0 αR∆ 50,000 (1 − α)R∆ v∆ − vo v∆ + 2v∆ + =0 α 1−α 1 1 vo v∆ +2+ = α 1−α 1−α
(1 − α) .·. vo = −1 1 + 2(1 − α) + α When
α = 0.2,
When
α = 1,
vo = −1(1 + 1.6 + 4) = −6.6 V vo = −1(1 + 0 + 0) = −1 V
.·. −6.6 V ≤ vo ≤ −1 V
(1 − α) = −7 [b] −1 1 + 2(1 − α) + α α + 2α(1 − α) + (1 − α) = 7α α + 2α − 2α2 + 1 − α = 7α .·. 2α2 + 5α − 1 = 0 P 5.11
so
α∼ = 0.186
[a] Replace the combination of vg , 1.6 kΩ, and the 6.4 kΩ resistors with its Thévenin equivalent.
−[12 + σ50] (0.2) 1.28 At saturation vo = −5 V; therefore
Then
−
vo =
12 + σ50 (0.2) = −5, 1.28
or
σ = 0.4
Thus for 0 ≤ σ < 0.40 the operational amplifier will not saturate.
5–9
CHAPTER 5. The Operational Amplifier
5–10
[b] When Also
σ = 0.272,
−(12 + 13.6) (0.2) = −4 V 1.28
vo vo + + io = 0 10 25.6
.·. io = − P 5.12
vo =
vo 4 4 vo − = + mA = 556.25 µA 10 25.6 10 25.6
[a]
vn − va vn − vo + =0 R R 2vn − va = vo va va − vn va − vo + =0 + Ra R R va va
vo 1 2 vn = + − Ra R R R
R 2+ − vn = vo Ra
vn = vp = va + vg .·. 2vn − va = 2va + 2vg − va = va + 2vg .·. va − vo = −2vg 2va + va
(1)
R − va − vg = vo Ra
.·. va 1 +
R − vo = vg Ra
(2)
Now combining equations (1) and (2) yields −va
R = −3vg Ra
Problems or
va = 3vg
Hence ia = [b] At saturation
Ra R va 3vg = Ra R
.·. va 1 +
Q.E.D.
Vo = ± Vcc
.·. va = ± Vcc − 2vg and
5–11
R Ra
(3)
= ± Vcc + vg
(4)
Dividing Eq (4) by Eq (3) gives
P 5.13
1+
± Vcc + vg R = Ra ± Vcc − 2vg
.·.
± Vcc + vg 3vg R = −1= Ra ± Vcc − 2vg ± Vcc − 2vg
or
Ra =
(± Vcc − 2vg ) R 3vg
Q.E.D.
[a] Assume the op-amp is operating within its linear range, then iL =
8 = 2 mA 4000
For RL = 4 kΩ
vo = (4 + 4)(2) = 16 V
Now since vo < 20 V our assumption of linear operation is correct, therefore iL = 2 mA [b] 20 = 2(4 + RL );
RL = 6 kΩ
[c] As long as the op-amp is operating in its linear region iL is independent of RL . From (b) we found the op-amp is operating in its linear region as long as RL ≤ 6 kΩ. Therefore when RL = 16 kΩ the op-amp is saturated. We can estimate the value of iL by assuming ip = in iL . Then iL = 20/(4,0000 + 16,000) = 1 mA. To justify neglecting the current into the op-amp assume the drop across the 50 kΩ resistor is negligible, and the input resistance to the op-amp is at least 500 kΩ. Then ip = in = (8 − 4)/(500 × 103 ) = 8 µA. But 8 µA 1 mA, hence our assumption is reasonable.
CHAPTER 5. The Operational Amplifier
5–12 [d]
P 5.14
[a] Let vo1 = output voltage of the amplifier on the left. Let vo2 = output voltage of the amplifier on the right. Then vo1 = ia =
−47 (1) = −4.7 V; 10
−220 (−0.15) = 1.0 V 33
vo2 − vo1 = 5.7 mA 1000
[b] ia = 0 when vo1 = v02 Thus −47 (vL ) = 1 10 vL = − P 5.15
vo2 =
so from (a)
vo2 = 1 V
10 = −212.77 mV 47
(60 × 10−3 )2 = 6 µW (600) 600 (60 × 10−3 ) = 1.2 mV = 30,000
[a] p600Ω = [b] v600Ω
p600Ω =
(1.2 × 10−3 )2 = 2.4 nW (600)
pa 6 × 10−6 = = 2500 pb 2.4 × 10−9 [d] Yes, the operational amplifier serves several useful purposes: [c]
Problems
5–13
• First, it enables the source to control 2, 500 times as much power delivered to the load resistor. When a small amount of power controls a larger amount of power, we refer to it as power amplification. • Second, it allows the full source voltage to appear across the load resistor, no matter what the source resistance. This is the voltage follower function of the operational amplifier. • Third, it allows the load resistor voltage (and thus its current) to be set without drawing any current from the input voltage source. This is the current amplification function of the circuit. P 5.16
[a] This circuit is an example of an inverting summing amplifier. 220 220 220 [b] vo = − va − vc = −8 + 15 − 11 = −4 V vb − 33 22 80 [c] vo = −19 − 10vb = ±6 .·. vb = −1.3 V
when
vo = −6 V;
vb = −2.5 V
when
vo = 6 V
.·. −2.5 V ≤ vb ≤ −1.3 V P 5.17
[a] Write a KCL equation at the inverting input to the op amp: vd − v a v d − v b vd − vc vd vd − vo + + + + =0 40,000 22,000 100,000 352,000 220,000 Multiply through by 220,000, plug in the values of input voltages, and rearrange to solve for vo :
vo = 220,000
−1 −5 4 + + 40,000 22,000 100,000
8 8 + + 352,000 220,000
= 14 V
[b] Write a KCL equation at the inverting input to the op amp. Use the given values of input voltages in the equation: 8 8 − vo 8 − va 8−9 8 − 13 + + =0 + + 40,000 22,000 100,000 352,000 220,000 Simplify and solve for vo : 44 − 5.5va − 10 − 11 + 5 + 8 − vo = 0 so
vo = 36 − 5.5va
Set vo to the positive power supply voltage and solve for va : 36 − 5.5va = 15
.·.
va = 3.818 V
Set vo to the negative power supply voltage and solve for va : 36 − 5.5va = −15
.·.
va = 9.273 V
CHAPTER 5. The Operational Amplifier
5–14
Therefore, 3.818 V ≤ va ≤ 9.273 V P 5.18
[a]
8−4 8−9 8 − 13 8 8 − v0 =0 + + + + 40,000 22,000 100,000 352,000 Rf 8 − vo = −2.7272 × 10−5 Rf For
vo = 15 V,
For
vo = −15 V,
so
Rf =
8 − vo −2.727 × 10−5
Rf = 256.7 kΩ Rf < 0
so this solution is not possible.
15 − 8 15 + = −1.527 mA [b] io = −(if + i10k ) = − 3 256.7 × 10 10,000 P 5.19
We want the following expression for the output voltage: vo = −(2va + 4vb + 6vc + 8vd ) This is an inverting summing amplifier, so each input voltage is amplified by a gain that is the ratio of the feedback resistance to the resistance in the forward path for the input voltage: vo = −
48 48 48 48 va + vb + vc + vd Ra Rb Rc Rd
Solve for each input resistance value to yield the desired gain: .·. Ra = 48,000/2 = 24 kΩ Rc = 48,000/6 = 8 kΩ Rb = 48,000/4 = 12 kΩ
Rd = 48,000/8 = 6 kΩ
The final circuit is shown here:
P 5.20
[a] vp = vs ,
vn =
Therefore vo =
R1 v o , R1 + R2
vn = vp
R1 + R2 R2 vs = 1 + vs R1 R1
Problems
5–15
[b] vo = vs [c] Because vo = vs , thus the output voltage follows the signal voltage.
P 5.21
Rf Rf Rf (0.15) + (0.1) + (0.25) vo = − 3000 5000 25,000 −6 = −8 × 10−5 Rf ;
P 5.22
.·. 0 ≤ Rf ≤ 75 kΩ
Rf = 75 kΩ;
[a] This circuit is an example of the non-inverting amplifier. [b] Use voltage division to calculate vp : vp =
10,000 vs vs = 10,000 + 30,000 4
Write a KCL equation at vn = vp = vs /4: vs /4 vs /4 − vo + =0 4000 28,000 Solving, vo = 7vs /4 + vs /4 = 2vs [c] 2vs = 8
vs = 4 V
so
2vs = −12
so
vs = −6 V
Thus, −6 V ≤ vs ≤ 4 V. P 5.23
[a] vp = vn = .·.
68,000 vg = 0.85vg 80,000
0.85vg − vo 0.85vg + = 0; 30,000 63,000
.·. vo = 2.635vg = 2.635(4),
vo = 10.54 V
[b] vo = 2.635vg = ±12 vg = ±4.554 V, [c]
−4.554 ≤ vg ≤ 4.554 V
0.85vg − vo 0.85vg + =0 30,000 Rf
0.85Rf + 0.85 vg = vo = ±12 30,000
.·. 1.7×10−3 Rf + 51 = ±360; P 5.24
1.7×10−3 Rf = 360 − 51;
[a] This circuit is an example of a non-inverting summing amplifier.
Rf = 181.76 kΩ
CHAPTER 5. The Operational Amplifier
5–16
[b] Write a KCL equation at vp and solve for vp in terms of vs : vp − 6 vp − vs + =0 15,000 30,000 2vp − 2vs + vp − 6 = 0
so
vp = 2vs /3 + 2
Now write a KCL equation at vn and solve for vo : vn − vo vn + =0 20,000 60,000
so
vo = 4vn
Since we assume the op amp is ideal, vn = vp . Thus, vo = 4(2vs /3 + 2) = 8vs /3 + 8 [c] 8vs /3 + 8 = 16 8vs /3 + 8 = −12
vs = 3 V
so so
vs = −7.5 V
Thus, −7.5 V ≤ vs ≤ 3 V. P 5.25
[a] The circuit is a non-inverting summing amplifier. vp − vb vp − va + =0 [b] 3.3 × 103 4.7 × 103 .·. vp = 0.5875va + 0.4125vb vn vn − vo =0 + 10,000 100,000 .·. vo = 11vn = 11vp = 6.4625va + 4.5375vb = 8.03 V [c] vp = vn =
vo = 730 mV 11
ia =
va − vp = −100 µA 3.3 × 103
ib =
vb − vp = 100 µA 4.7 × 103
[d] 6.4625 for va 4.5375 for vb P 5.26
[a]
vp − va vp − vb vp − vc vp + + + =0 Ra Rb Rc Rg .·. vp =
Rb Rc Rg Ra Rc Rg Ra Rb Rg va + vb + vc D D D
where D = Rb Rc Rg + Ra Rc Rg + Ra Rb Rg + Ra Rb Rc
Problems
5–17
vn − vo vn + =0 Rs Rf vn
1 1 + Rs Rf
=
vo Rf
Rf vn = kvn .·. vo = 1 + Rs
Rf where k = 1 + Rs vp = vn
.·. vo = kvp or kRg Rb Rc kRg Ra Rc kRg Ra Rb va + vb + vc D D D
vo =
kRg Rb Rc =3 D
.·.
Rb = 1.5 Ra
kRg Ra Rc =2 D
.·.
Rc =2 Rb
kRg Ra Rb =1 D
.·.
Rc =3 Ra
Since
Ra = 2 kΩ
Rb = 3 kΩ
Rc = 6 kΩ
.·. D = [(3)(6)(4) + (2)(6)(4) + (2)(3)(4) + (2)(3)(6)] × 109 = 180 × 109 k(4)(3)(6) × 109 =3 180 × 109 k=
540 × 109 = 7.5 72 × 109
.·. 7.5 = 1 +
Rf Rs
Rf = 6.5 Rs Rf = (6.5)(12,000) = 78 kΩ [b] vo = 3(0.8) + 2(1.5) + 2.10 = 7.5 V vn = vp = ia =
7.5 = 1.0 V 7.5
−0.2 0.8 − 1 = = −0.1 mA = −100 µA 2000 2000
CHAPTER 5. The Operational Amplifier
5–18
P 5.27
[a]
ib =
1.5 − 1.0 0.5 = = 166.67 µA 3000 3000
ic =
1.1 2.10 − 1.0 = = 183.33 µA 6000 6000
ig =
1 = 250 µA 4000
is =
1 vn = = 83.33 µA 12,000 12,000
vp − va vp − vb vp − vc + + =0 Ra Rb Rc .·. vp =
Rb Rc Ra Rc Ra Rb va + vb + vc D D D
where D = Rb Rc + Ra Rc + Ra Rb vn − vo vn + =0 10,000 Rf
Rf + 1 vn = vo 10,000
Let
Rf +1=k 10,000
vo = kvn = kvp .·. vo = .·.
kRb Rc kRa Rc kRa Rb va + vb + vc D D D Rc .·. =5 Ra
kRb Rc =5 D
kRa Rc =4 D kRa Rb =1 D
.·.
Rc =4 Rb
.·. Rc = 5Ra = 5 kΩ Rb = Rc /4 = 1.25 kΩ .·. D = (1.25)(5) + (1)(5) + (1.25)(1) = 12.5 × 106 .·. k = .·.
(5)(12.5) × 106 5D = = 10 Rb Rc (1.25)(5) × 106
Rf + 1 = 10, 10,000
Rf = 90 kΩ
Problems [b] vo = 5(0.5) + 4(1) + 1.5 = 8 V vn = vo /10 = 0.8 V = vp
P 5.28
ia =
0.5 − 0.8 va − vp = = −300 µA 1000 1000
ib =
1 − 0.8 vb − vp = = 160 µA 1250 1250
ic =
1.5 − 0.8 vc − vp = = 140 µA 5000 5000
[a] Assume va is acting alone. Replacing vb with a short circuit yields vp = 0, therefore vn = 0 and we have 0 − va 0 − vo + + in = 0, Ra Rb Therefore va vo =− , Rb Ra
vo = −
in = 0
Rb va Ra
Assume vb is acting alone. Replace va with a short circuit. Now v b Rd Rc + Rd
vp = vn =
vn vn − vo + + in = 0, Ra Rb
1 1 + Ra Rb
vo = vo = [b]
Rd Ra
Rd v vb − o = 0 Rc + Rd Rb
Rb +1 Ra
vo
+
vo
Ra + Rb Rc + Rd
Rd Rd vb = Rc + Rd Ra
Rd = Ra
=
Ra + Rb Rc + Rd
Eq. (5.22) reduces to
Ra + Rb vb Rc + Rd
Ra + Rb Rb vb − va Rc + Rd Ra
Rb , Ra
Rd Ra = Rb Rc , Rd When Ra
in = 0
therefore
therefore
=
Rd (Ra + Rb ) = Rb (Rc + Rd )
Ra Rc = Rb Rd
Rb Ra
vo =
Rb Rb Rb vb − va = (vb − va ). Ra Ra Ra
5–19
CHAPTER 5. The Operational Amplifier
5–20 P 5.29
Use voltage division to find vp : vp =
2000 (5) = 1 V 2000 + 8000
Write a KCL equation at vn and solve it for vo : vn − va vn − vo + =0 5000 Rf
so
Rf Rf + 1 vn − va = vo 5000 5000
Since the op amp is ideal, vn = vp = 1V, so vo =
Rf Rf +1 − va 5000 5000
To satisfy the equation,
Rf +1 =5 5000
and
Rf =4 5000
Thus, Rf = 20 kΩ. P 5.30
[a]
v p − v c vp − v d vp + =0 + 9,000 24,000 72,000 .·. vp = (2/3)vc + 0.25vd = vn vn − vo vn − va vn − vb + =0 + 12,000 18,000 144,000 .·. vo
=
21vn − 12va − 8vb
=
21[(2/3)vc + 0.25vd ] − 12va − 8vb
=
21(0.4 + 0.2) − 12(0.5) − 8(0.3) = 4.2 V
Problems [b] vo = 14vc + 4.2 − 6 − 2.4 ±15 = 14vc − 4.2 .·. 14vc = ±15 + 4.2 .·. vc = 1.371 V
and
vc = −0.771 V
.·. −771 ≤ vc ≤ 1371 mV P 5.31
[a] vo =
Rd (Ra + Rb ) Rb 47(110) vb − (0.80) − 10(0.67) va = Ra (Rc + Rd ) Ra 10(80)
vo = 5.17 − 6.70 = −1.53 V (800)(47) = 470 mV 80
[b] vn = vp = ia =
(670 − 470)10−3 = 20 µA 10 × 103
Ra =
va 670 × 10−3 = = 33.5 kΩ ia 20 × 10−6
[c] Rin b = Rc + Rd = 80 kΩ P 5.32
vp =
v b Rb = vn Ra + Rb
vn − va vn − vo =0 + 4700 Rf vn .·.
Rf v a Rf +1 − = vo 4700 4700
Rf Rb Rf +1 va = vo vb − 4700 Ra + Rb 4700
.·.
Rf = 10; 4700
.·.
Rf + 1 = 11 4700
.·. 11
Rb Ra + Rb
Rf = 47 kΩ
= 10
11Rb = 10Rb + 10Ra
Rb = 10Ra
5–21
CHAPTER 5. The Operational Amplifier
5–22
Ra + Rb = 220 kΩ 11Ra = 220 kΩ Ra = 20 kΩ Rb = 220 − 20 = 200 kΩ P 5.33
v p = v n = R b ib Rb ib − 3000ia Rb ib − vo + =0 3000 Rf
Rb Rb vo + i b − ia = 3000 Rf Rf
Rb Rf + R b ib − R f i a vo = 3000 .·. Rf = 2000 Ω (2/3)Rb + Rb = 2000 .·. Rb = 1200 Ω P 5.34
vo =
Rd (Ra + Rb ) Rb vb − va Ra (Rc + Rd ) Ra
By hypothesis: Rb /Ra = 4;
Rc + Rd = 470 kΩ;
Rd (Ra + 4Ra ) =3 .·. Ra 470,000
Rd = 282 kΩ;
so
Rd (Ra + Rb ) =3 Ra (Rc + Rd )
Rc = 188 kΩ
Also, when vo = 0 we have vn vn − va + =0 Ra Rb
Ra .·. vn 1 + Rb ia =
= va ;
va − 0.8va va = 0.2 ; Ra Ra
.·. Ra = 4.4 kΩ;
vn = 0.8va Rin =
Rb = 17.6 kΩ
va = 5Ra = 22 kΩ ia
Problems P 5.35
[a] vn = vp = αvg vn − vg vn − vo + =0 R1 Rf (vn − vg )
α
vo
Rf + vn − vo = 0 R1
vo
α
vo
α
=
(αvg − vg )4 + αvg
=
[(α − 1)4 + α]vg
=
(5α − 4)vg
=
(5α − 4)(2) = 10α − 8
vo
0.0 −8 V
0.4 −4 V
0.8 0 V
0.1 −7 V
0.5 −3 V
0.9 1 V
0.2 −6 V
0.6 −2 V
1.0 2 V
0.3 −5 V
0.7 −1 V
[b] Rearranging the equation for vo from (a) gives
Rf Rf + 1 vg α + − vg vo = R1 R1 Therefore, slope =
Rf + 1 vg ; R1
intercept = −
[c] Using the equations from (b),
Rf + 1 vg ; −6 = R1
Solving, vg = −2 V; P 5.36
vp =
Rf =2 R1
1500 (−18) = −3 V = vn 9000
18 − 3 −3 − vo =0 + 1600 Rf
Rf 4=− vg R1
Rf vg R1
5–23
CHAPTER 5. The Operational Amplifier
5–24
.·. vo =
15 Rf − 3 1600
vo = 9 V;
Rf = 1280 Ω
vo = −9 V; But
Rf ≥ 0,
Rf = −640 Ω .·. Rf = 1280 Ω
(24)(26) + (25)(25) = 24.98 (2)(1)(25) (1)(24) − 25(1) [b] Acm = = −0.04 1(25) 24.98 = 624.50 [c] CMRR = 0.04
P 5.37
[a] Adm =
P 5.38
Acm =
(20)(50) − (50)Rx 20(50 + Rx )
Adm =
50(20 + 50) + 50(50 + Rx ) 2(20)(50 + Rx )
Adm Rx + 120 = Acm 2(20 − Rx ) .·.
Rx + 120 = ±1000 for the limits on the value of Rx 2(20 − Rx )
If we use +1000
Rx = 19.93 kΩ
If we use −1000
Rx = 20.07 kΩ
19.93 kΩ ≤ Rx ≤ 20.07 kΩ P 5.39
[a] Replace the op amp with the model from Fig. 5.15:
Problems
5–25
Write two node voltage equations, one at the left node, the other at the right node: vn − vo vn vn − vg + + =0 5000 100,000 500,000 vo + 3 × 105 vn vo − vn vo + + =0 5000 100,000 1000 Simplify and place in standard form: 106vn − 5vo = 100vg (6×106 − 1)vn + 121vo = 0 Let vg = 1 V and solve the two simultaneous equations: vo = −19.9915 V;
vn = 403.2 µV
[b] From the solution in part (a), vn = 403.2 µV. [c] ig =
vg − 403.2 × 10−6 vg vg − vn = 5000 5000
Rg =
vg 5000 = = 5002.02 Ω ig 1 − 403.2 × 10−6
[d] For an ideal op amp, the voltage gain is the ratio between the feedback resistor and the input resistor: vo 100,000 =− = −20 vg 5000 For an ideal op amp, the difference between the voltages at the input terminals is zero, and the input resistance of the op amp is infinite. Therefore, vn = vp = 0 V; P 5.40
Rg = 5000 Ω
Note – the load resistor should have the value 4 kΩ. [a] Replace the op amp with the model shown in Fig. 5.15. The node voltage equation at the inverting input: vn − vo vn − vg vn + =0 + 80,000 40,000 500,000 Simplify: 12.5vn + vn − vg + 6.25vn − 6.25vo = 0 The node voltage equation at the op amp output: vo vo − 20,000(vp − vn ) vo − vn + + =0 4000 5000 80,000
5–26
CHAPTER 5. The Operational Amplifier Simplify: 20vo + 16vo − 320,000(vp − vn ) + vo − vn = 0 From the input, vp − vn = 0.8(vg − vn ) Substituting into the equation written at the output, 20vo + 16vo − 256,000(vg − vn ) + vo − vn = 0 Now let vg = 1 V; plug this value into both the input and output equations and simplify into two simultaneous equations: 19.75vn − 6.25vo = 1 255,999vn + 37vo = 256,000 These equations are in standard form, so solve them to yield vo = 2.9986 V; vn = 999.571 mV Thus, vo 2.9986 = 2.9986 = vg 1 [b] From part (a), vn = 999.571 mV. Use this value to solve for vp : vp = 0.8(1 − vn ) + vn = 999.914 mV [c] vp − vn = 343.6 µ V 1 − 999.914 × 10−3 vg − vp = = 859 pA [d] ig = 100,000 100,000 [e] For an ideal op amp, vn = vp = vg , so the KVL equation at the inverting node is vg − vo vo + =0 40,000 80,000 Then, vo = 3vg
so
vo =3 vg
Also, vn = vp = 1 V;
vp − vn = 0 V;
ig = 0 A
Problems P 5.41
[a]
vn − vTh vn − 0.88 vn + =0 + 1600 500,000 24,000 vTh + 105 vn vTh − vn + =0 2000 24,000 Solving, vTh = −13.198 V Short-circuit current calculation:
vn vn − 0.88 vn − 0 + =0 + 500,000 1600 24,000 .·. vn = 0.823 V isc =
105 vn − vn = −41.13 A 24,000 2000
RTh =
vTh = 320.90 mΩ isc
5–27
CHAPTER 5. The Operational Amplifier
5–28
[b] The output resistance of the inverting amplifier is the same as the Thévenin resistance, i.e., Ro = RTh = 320.90 mΩ [c]
330 vo = (−13.198) = −13.185 V 330.32
vn + 13.185 vn − 0.88 vn + + =0 1600 500,000 24,000 .·. vn = 941.92 µV ig =
0.88 − 941.92 × 10−6 = 549.41 µA 1600
Rg = P 5.42
[a] vTh =
0.88 (1600) = 1601.7 Ω 0.88 − 941.92 × 10−6 −24 (0.88) = −13.2 V 1.6
RTh = 0, since op-amp is ideal
Problems [b] Ro = RTh = 0 Ω [c] Rg = 1.6 kΩ P 5.43
since
vn = 0
[a]
vn − vo vn − vg + =0 15,000 135,000 .·. vo = 10vn − 9vg Also
vo = A(vp − vn ) = −Avn
.·. vn =
−vo A
.·. vo 1 + vo =
10 = −9vg A
−9A vg (10 + A)
−9(90)(0.4) = −3.24 V (10 + 90) [c] vo = −9(0.4) = −3.60 V −9(0.4)A [d] −3.42 = 10 + A
[b] vo =
.·. A = 190 P 5.44
From Eq. 5.57,
1 1 1 vref + + = vn R + ∆R R + ∆R R − ∆R Rf
−
vo Rf
Substituting Eq. 5.59 for vp = vn : vref
1 R+∆R
+
1 R−∆R
vref
= 1 R + ∆R (R − ∆R) R+∆R +
+
1 Rf
1 R−∆R
+
1 Rf
−
vo Rf
5–29
5–30
CHAPTER 5. The Operational Amplifier Rearranging,
1 1 vo − = vref Rf R − ∆R R + ∆R
Thus, vo = vref
2∆R Rf 2 R − ∆R2
P 5.45
i1 =
15 − 10 = 1 mA 5000
i2 + i1 + 0 = 10 mA;
i2 = 9 mA
vo2 = 10 + (400)(9) × 10−3 = 13.6 V i3 =
15 − 13.6 = 0.7 mA 2000
i4 = i3 + i1 = 1.7 mA vo1 = 15 + 1.7(0.5) = 15.85 V
Problems P 5.46
vp =
5.6 vg = 0.7vg = 7 sin(π/3)t V 8.0
vn − vo vn + =0 15,000 75,000 6vn = vo ;
vn = vp
.·. vo = 42 sin(π/3)t V vo = 0
0≤t≤∞
t≤0
At saturation 42 sin .·.
π t = ±21; 3
π π t= , 3 6
t = 0.50 s,
P 5.47
5π , 6 2.50 s,
π sin t = ±0.5 3 7π , 6 3.50 s,
11π , 6
etc.
5.50 s,
etc.
It follows directly from the circuit that vo = −16vg From the plot of vg we have vg = 0, t < 0 vg
=
t
0 ≤ t ≤ 0.5
vg
=
−t + 1
0.5 ≤ t ≤ 1.5
vg
=
t−2
1.5 ≤ t ≤ 2.5
vg
=
−t + 3
2.5 ≤ t ≤ 3.5
t−4
3.5 ≤ t ≤ 4.5,
vg = Therefore
etc.
5–31
5–32
CHAPTER 5. The Operational Amplifier vo
=
−16t
0 ≤ t ≤ 0.5
vo
=
16t − 16
0.5 ≤ t ≤ 1.5
vo
=
−16t + 32
1.5 ≤ t ≤ 2.5
vo
=
16t − 48
2.5 ≤ t ≤ 3.5
vo = −16t + 64 3.5 ≤ t ≤ 4.5, etc. These expressions for vo are valid as long as the op amp is not saturated. Since the peak values of vo are ±5, the output is clipped at ±5. The plot is shown below.
P 5.48
[a] Use Eq. 5.61 to solve for Rf ; note that since we are using 1% strain gages, ∆ = 0.01: vo R (5)(120) Rf = = = 2 kΩ 2∆vref (2)(0.01)(15) [b] Now solve for ∆ given vo = 50 mV: ∆=
vo R (0.05)(120) = 100 × 10−6 = 2Rf vref 2(2000)(15)
The change in strain gage resistance that corresponds to a 50 mV change in output voltage is thus ∆R = ∆R = (100 × 10−6 )(120) = 12 mΩ
Problems P 5.49
[a]
Let R1 = R + ∆R vp vp − vin vp + + =0 Rf R R1 .·. vp
1 1 1 vin + + = Rf R R1 R1
.·. vp =
RRf vin = vn RR1 + Rf R1 + Rf R
vn vn − vin vn − vo + + =0 R R Rf
vn
1 1 1 vo vin + + − = R R Rf Rf R
vo R + 2Rf vin = − .·. vn RRf R Rf
vin R + 2Rf RRf vin vo − = .·. Rf RRf [RR1 + Rf R1 + Rf R] R
1 R + 2Rf vo − = vin .·. Rf RR1 + Rf R1 + Rf R R .·. vo =
[R2 + 2RRf − R1 (R + Rf ) − RRf ]Rf vin R[R1 (R + Rf ) + RRf ]
Now substitute R1 = R + ∆R and get vo =
−∆R(R + Rf )Rf vin R[(R + ∆R)(R + Rf ) + RRf ]
If ∆R R (R + Rf )Rf (−∆R)vin vo ≈ R2 (R + 2Rf ) [b] vo ≈
47 × 104 (48 × 104 )(−95)15 ≈ −3.384 V 108 (95 × 104 )
5–33
5–34
P 5.50
CHAPTER 5. The Operational Amplifier [c] vo =
−95(48 × 104 )(47 × 104 )15 = −3.368 V 104 [(1.0095)104 (48 × 104 ) + 47 × 108 ]
[a] vo ≈
(R + Rf )Rf (−∆R)vin R2 (R + 2Rf )
vo = .·.
(R + Rf )(−∆R)Rf vin R[(R + ∆R)(R + Rf ) + RRf ]
R[(R + ∆R)(R + Rf ) + RRf ] approx value = true value R2 (R + 2Rf )
R[(R + ∆R)(R + Rf ) + RRf ] − R2 (R + 2Rf ) .·. Error = R2 (R + 2Rf ) = .·. % error = [b] % error = P 5.51
1=
∆R(R + Rf ) × 100 R(R + 2Rf )
95(48 × 104 ) × 100 = 0.48% 104 (95 × 104 )
∆R(48 × 104 ) × 100 104 (95 × 104 )
.·. ∆R =
9500 = 197.91667 Ω 48
.·. % change in R = P 5.52
∆R (R + Rf ) R (R + 2Rf )
197.19667 × 100 ≈ 1.98% 104
[a] It follows directly from the solution to Problem 5.49 that vo =
[R2 + 2RRf − R1 (R + Rf ) − RRf ]Rf vin R[R1 (R + Rf ) + RRf ]
Now R1 = R − ∆R. Substituting into the expression gives vo =
(R + Rf )Rf (∆R)vin R[(R − ∆R)(R + Rf ) + RRf ]
Now let ∆R R and get vo ≈
(R + Rf )Rf ∆Rvin R2 (R + 2Rf )
Problems [b] It follows directly from the solution to Problem 5.49 that .·.
approx value R[(R − ∆R)(R + Rf ) + RRf ] = true value R2 (R + 2Rf )
.·. Error =
(R − ∆R)(R + Rf ) + RRf − R(R + 2Rf ) R(R + 2Rf )
= .·. % error =
−∆R(R + Rf ) R(R + 2Rf ) −∆R(R + Rf ) × 100 R(R + 2Rf )
[c] R − ∆R = 9810 Ω .·. vo ≈ [d] % error =
.·. ∆R = 10,000 − 9810 = 190 Ω
(48 × 104 )(47 × 104 )(190)(15) ≈ 6.768 V 108 (95 × 104 ) −190(48 × 104 )(100) = −0.96% 104 (95 × 104 )
5–35
6 Inductance, Capacitance, and Mutual Inductance
Assessment Problems AP 6.1 [a] ig = 8e−300t − 8e−1200t A v=L
dig = −9.6e−300t + 38.4e−1200t V, dt
t > 0+
v(0+ ) = −9.6 + 38.4 = 28.8 V [b] v = 0 when
38.4e−1200t = 9.6e−300t
or
t = (ln 4)/900 = 1.54 ms
[c] p = vi = 384e−1500t − 76.8e−600t − 307.2e−2400t W dp = 0 when e1800t − 12.5e900t + 16 = 0 [d] dt Let x = e900t
and solve the quadratic x2 − 12.5x + 16 = 0
x = 1.45,
t=
x = 11.05,
ln 1.45 = 411.05 µs 900
t=
ln 11.05 = 2.67 ms 900
p is maximum at t = 411.05 µs [e] pmax = 384e−1.5(0.41105) − 76.8e−0.6(0.41105) − 307.2e−2.4(0.41105) = 32.72 W [f] imax = 8[e−0.3(1.54) − e−1.2(1.54) ] = 3.78 A wmax = (1/2)(4 × 10−3 )(3.78)2 = 28.6 mJ [g] W is max when i is max, i is max when di/dt is zero. When di/dt = 0, v = 0, therefore t = 1.54 ms.
6–1
6–2
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
AP 6.2 [a] i = C
d dv = 24 × 10−6 [e−15,000t sin 30,000t] dt dt
= [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A,
π ms = −31.66 mA, [b] i 80
i(0+ ) = 0.72 A
π ms = 20.505 V, v 80
p = vi = −649.23 mW [c] w =
1 Cv 2 = 126.13 µJ 2
1 t i dx + v(0− ) AP 6.3 [a] v = C 0−
=
t 1 3 cos 50,000x dx = 100 sin 50,000t V 0.6 × 10−6 0−
[b] p(t) = vi = [300 cos 50,000t] sin 50,000t = 150 sin 100,000t W, [c] w(max) =
p(max) = 150 W
1 2 Cvmax = 0.30(100)2 = 3000 µJ = 3 mJ 2
60(240) = 48 mH 300 [b] i(0+ ) = 3 + −5 = −2 A 125 t [c] i = (−0.03e−5x ) dx − 2 = 0.125e−5t − 2.125 A 6 0+ 50 t (−0.03e−5x ) dx + 3 = 0.1e−5t + 2.9 A [d] i1 = + 3 0
AP 6.4 [a] Leq =
i2 =
25 t (−0.03e−5x ) dx − 5 = 0.025e−5t − 5.025 A 6 0+
i1 + i2 = i AP 6.5 v1 = 0.5 × 106
t 0+
v2 = 0.125 × 106
240 × 10−6 e−10x dx − 10 = −12e−10t + 2 V
t 0+
240 × 10−6 e−10x dx − 5 = −3e−10t − 2 V
v1 (∞) = 2 V,
v2 (∞) = −2 V
1 1 W = (2)(4) + (8)(4) × 10−6 = 20 µJ 2 2
Problems AP 6.6 [a] Summing the voltages around mesh 1 yields di1 d(i2 + ig ) +8 + 20(i1 − i2 ) + 5(i1 + ig ) = 0 dt dt or 4
di1 di2 dig 4 + 25i1 + 8 − 20i2 = − 5ig + 8 dt dt dt
Summing the voltages around mesh 2 yields 16
di1 d(i2 + ig ) +8 + 20(i2 − i1 ) + 780i2 = 0 dt dt
or di1 di2 dig − 20i1 + 16 + 800i2 = −16 8 dt dt dt [b] From the solutions given in part (b) i1 (0) = −0.4 − 11.6 + 12 = 0;
i2 (0) = −0.01 − 0.99 + 1 = 0
These values agree with zero initial energy in the circuit. At infinity, i1 (∞) = −0.4A;
i2 (∞) = −0.01A
When t = ∞ the circuit reduces to
.·. i1 (∞) = −
7.8 7.8 + = −0.4A; 20 780
From the solutions for i1 and i2 we have di1 = 46.40e−4t − 60e−5t dt di2 = 3.96e−4t − 5e−5t dt Also,
dig = 7.84e−4t dt
Thus di1 = 185.60e−4t − 240e−5t 4 dt
i2 (∞) = −
7.8 = −0.01A 780
6–3
6–4
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 25i1 = −10 − 290e−4t + 300e−5t 8
di2 = 31.68e−4t − 40e−5t dt
20i2 = −0.20 − 19.80e−4t + 20e−5t 5ig = 9.8 − 9.8e−4t 8
dig = 62.72e−4t dt
Test: 185.60e−4t − 240e−5t − 10 − 290e−4t + 300e−5t + 31.68e−4t − 40e−5t +0.20 + 19.80e−4t − 20e−5t = −[9.8 − 9.8e−4t + 62.72e−4t ] ?
−9.8 + (300 − 240 − 40 − 20)e−5t +(185.60 − 290 + 31.68 + 19.80)e−4t = −(9.8 + 52.92e−4t ) ?
−9.8 + 0e−5t + (237.08 − 290)e−4t = −9.8 − 52.92e−4t ?
−9.8 − 52.92e−4t = −9.8 − 52.92e−4t
(OK)
Also, 8
di1 = 371.20e−4t − 480e−5t dt
20i1 = −8 − 232e−4t + 240e−5t 16
di2 = 63.36e−4t − 80e−5t dt
800i2 = −8 − 792e−4t + 800e−5t 16
dig = 125.44e−4t dt
Test: 371.20e−4t − 480e−5t + 8 + 232e−4t − 240e−5t + 63.36e−4t − 80e−5t −8 − 792e−4t + 800e−5t = −125.44e−4t ?
(8 − 8) + (800 − 480 − 240 − 80)e−5t +(371.20 + 232 + 63.36 − 792)e−4t = −125.44e−4t ?
(800 − 800)e−5t + (666.56 − 792)e−4t = −125.44e−4t ?
−125.44e−4t = −125.44e−4t
(OK)
Problems
Problems P 6.1
[a] i =
0
t0
d [8e−20t + 17] dt
= 0.32e−20t mA, t>0 −20t CHECK: i1 + i2 = 1.92e mA = io P 6.29
[ 12 (8 × 10−6 )(−5)2 + 12 (10 × 10−6 )(25)2 + 12 (2 × 10−6 )(25)2 ]
[a] w(0) =
3850 µJ
= [b]
vo (∞) =
−17 V
v1 (∞) =
17 V [ 12 (8 × 10−6 )(−17)2 + 12 (12 × 10−6 )(17)2 ]
w(∞) =
2890 µJ
= [c] w =
∞ 0
(20e−20t )(1.92 × 10−3 e−20t ) dt = 960 µJ
CHECK: 3850 − 2890 = 960 µJ 960 × 100 = 24.9% [d] % delivered = 3850 [e] w(40 ms) =
0
(20e−20t )(1.92 × 10−3 e−20t ) dt
=
e−40t 0.04 0.0384 −40 0
=
960 × 10−6 (1 − e−1.6 ) = 766.2 µJ
% delivered = P 6.30
0.04
766.2 (100) = 79.8% 960
From Figure 6.17(a) we have 1 t 1 t v= i + v1 (0) + i dx + v2 (0) + · · · C1 0 C2 0
t 1 1 + + ··· i dx + v1 (0) + v2 (0) + · · · v= C1 C2 0
Problems
Therefore P 6.31
1 1 1 = + + ··· , Ceq C1 C2
6–25
veq (0) = v1 (0) + v2 (0) + · · ·
From Fig. 6.18(a) i = C1
dv dv dv + C2 + · · · = [C1 + C2 + · · ·] dt dt dt
Therefore Ceq = C1 + C2 + · · ·. Because the capacitors are in parallel, the initial voltage on every capacitor must be the same. This initial voltage would appear on Ceq . dio dt
P 6.32
=
5{e−2000t [−8000 sin 4000t + 4000 cos 4000t] −2000e−2000t [2 cos 4000t + sin 4000t]}
dio + (0 ) = 5[1(4000) + (−2000)(2)] = 0 dt dio v2 (0+ ) = 10 × 10−3 (0+ ) = 0 dt + + v1 (0 ) = 40io (0 ) + v2 (0+ ) = 40(10) + 0 = 400V P 6.33
vc
vL
vo
P 6.34
=
1 − 0.625 × 10−6
t 0
−16,000x
1.5e
dx −
t 0
0.5e
=
150(e−16,000t − 1) − 200(e−4000t − 1) − 50
=
150e−16,000t − 200e−4000t V dio 25 × 10−3 dt
=
−4000x
dx − 50
=
25 × 10−3 (−24,000e−16,000t + 2000e−4000t )
=
−600e−16,000t + 50e−4000t V
=
vc − vL
=
(150e−16,000t − 200e−4000t ) − (−600e−16,000t + 50e−4000t )
=
750e−16,000t − 250e−4000t V,
[a] −2
di2 dig + 16 + 32i2 = 0 dt dt
16
di2 dig + 32i2 = 2 dt dt
[b] i2 = e−t − e−2t A di2 = −e−t + 2e−2t A/s dt
t>0
6–26
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance ig = 8 − 8e−t A dig = 8e−t A/s dt .·. −16e−t + 32e−2t + 32e−t − 32e−2t = 16e−t [c] v1
di2 dig −2 dt dt
=
4
=
4(8e−t ) − 2(−e−t + 2e−2t )
=
34e−t − 4e−2t V,
t>0
Also [d] v1 (0) = 34 − 4 = 30 V; di2 dig v1 (0) = 4 (0) − 2 (0) dt dt = 4(8) − 2(−1 + 2) = 32 − 2 = 30 V Yes, the initial value of v1 is consistent with known circuit behavior. P 6.35
[a] Yes, vo = 20(i2 − i1 ) + 60i2 [b] vo
=
20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t )+ 60(1 − 52e−5t + 51e−4t )
=
20(−3 − 116e−5t + 119e−4t ) + 60 − 3120e−5t + 3060e−4t
vo
=
−5440e−5t + 5440e−4t V
[c] vo
=
L2
=
d d (15 + 36e−5t − 51e−4t ) + 8 (4 + 64e−5t − 68e−4t ) dt dt −5t −4t −5t −2880e + 3264e − 2560e + 2176e−4t
vo
=
−5440e−5t + 5440e−4t V
[a] vg
=
5(ig − i1 ) + 20(i2 − i1 ) + 60i2
=
5(16 − 16e−5t − 4 − 64e−5t + 68e−4t )+
=
P 6.36
d di1 (ig − i2 ) + M dt dt
16
20(1 − 52e−5t + 51e−4t − 4 − 64e−5t + 68e−4t )+ 60(1 − 52e−5t + 51e−4t ) =
60 + 5780e−4t − 5840e−5t V
[b] vg (0) = 60 + 5780 − 5840 = 0 V
Problems [c] pdev
=
v g ig
=
960 + 92,480e−4t − 94,400e−5t − 92,480e−9t +
6–27
93,440e−10t W [d] pdev (∞) = 960 W [e] i1 (∞) = 4 A;
i2 (∞) = 1 A;
ig (∞) = 16 A;
p5Ω = (16 − 4)2 (5) = 720 W p20Ω = 32 (20) = 180 W p60Ω = 12 (60) = 60 W
.·. P 6.37
pabs = 720 + 180 + 60 = 960 W
pdev =
pabs = 960 W
[a] Rearrange by organizing the equations by di1 /dt, i1 , di2 /dt, i2 and transfer the ig terms to the right hand side of the equations. We get 4
di1 di2 dig + 25i1 − 8 − 20i2 = 5ig − 8 dt dt dt
−8
di2 dig di1 + 80i2 = 16 − 20i1 + 16 dt dt dt
[b] From the given solutions we have di1 = −320e−5t + 272e−4t dt di2 = 260e−5t − 204e−4t dt Thus, 4
di1 = −1280e−5t + 1088e−4t dt
25i1 = 100 + 1600e−5t − 1700e−4t 8
di2 = 2080e−5t − 1632e−4t dt
20i2 = 20 − 1040e−5t + 1020e−4t 5ig = 80 − 80e−5t 8
dig = 640e−5t dt
6–28
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance Thus, −1280e−5t + 1088e−4t + 100 + 1600e−5t − 1700e−4t − 2080e−5t +1632e−4t − 20 + 1040e−5t − 1020e−4t = 80 − 80e−5t − 640e−5t ?
80 + (1088 − 1700 + 1632 − 1020)e−4t +(1600 − 1280 − 2080 + 1040)e−5t = 80 − 720e−5t ?
80 + (2720 − 2720)e−4t + (2640 − 3360)e−5t = 80 − 720e−5t 8
(OK)
di1 = −2560e−5t + 2176e−4t dt
20i1 = 80 + 1280e−5t − 1360e−4t 16
di2 = 4160e−5t − 3264e−4t dt
80i2 = 80 − 4160e−5t + 4080e−4t 16
dig = 1280e−5t dt
2560e−5t − 2176e−4t − 80 − 1280e−5t + 1360e−4t + 4160e−5t − 3264e−4t +80 − 4160e−5t + 4080e−4t = 1280e−5t ?
(−80 + 80) + (2560 − 1280 + 4160 − 4160)e−5t +(1360 − 2176 − 3264 + 4080)e−4t = 1280e−5t ?
0 + 1280e−5t + 0e−4t = 1280e−5t
P 6.38
[a] L2 = N1 = N2 [b] P1 =
P1 =
=
L1 = L2
(0.09)2 = 50 mH (0.75)2 (0.288)
288 = 2.4 50
L1 0.288 = = 0.2 × 10−6 Wb/A 2 N1 (1200)2
P2 = P 6.39
M2 k 2 L1
(OK)
L2 0.05 = = 0.2 × 10−6 Wb/A 2 N2 (500)2
L1 = 2 nWb/A; N12
P12 = P21 =
P2 =
L2 = 2 nWb/A; N22
M = 1.2 nWb/A N1 N2
P11 = P1 − P21 = 0.8 nWb/A
M = k L1 L2 = 180 µH
Problems P 6.40
6–29
7.2 M = √ = 0.8 [a] k = √ L1 L2 81 √ [b] M = 81 = 9 mH N1 2 N12 P1 L1 = 2 = [c] L2 N2 P2 N2 .·.
N1 N2
2
=
27 =9 3
N1 =3 N2 P 6.41
√ [a] M = k L1 L2 = 0.8 324 = 14.4 mH
P1 =
L1 36 × 10−3 = = 900 nWb/A N12 (200)2
dφ11 P11 = = 0.1; dφ21 P21
P21 = 10P11
P1 = P11 + P21 = 11P11 P11 =
1 P1 = 81.82 nWb/A 11
P21 = 10P11 = 818.18 nWb/A N2 =
M 14.4 × 10−3 = = 88 turns N1 P21 (200)(818.18 × 10−9 )
L2 9 × 10−3 = = 1162.19 nWb/A N22 (88)2 [c] P11 = 81.82 nWb/A [see part (a)] φ22 P22 [d] = φ12 P12
[b] P2 =
P12 = P21 = 818.18 nWb/A P22 = P2 − P12 = 1162.19 × 10−9 − 818.18 × 10−9 = 344.01 nWb/A 344.01 φ22 = 0.4205 = φ12 818.18 P 6.42
[a] Dot terminal 1; the flux is up in coil 1-2, and down in coil 3-4. Assign the current into terminal 4; the flux is down in coil 3-4. Therefore, dot terminal 4. Hence, 1 and 4 or 2 and 3. [b] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4. Assign the current into terminal 4; the flux is right-to-left in coil 3-4. Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3.
6–30
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance [c] Dot terminal 2; the flux is up in coil 1-2, and right-to-left in coil 3-4. Assign the current into terminal 4; the flux is right-to-left in coil 3-4. Therefore, dot terminal 4. Hence, 2 and 4 or 1 and 3. [d] Dot terminal 1; the flux is down in coil 1-2, and down in coil 3-4. Assign the current into terminal 4; the flux is down in coil 3-4. Therefore, dot terminal 4. Hence, 1 and 4 or 2 and 3.
P 6.43
[a]
1 P11 P22 P11 = 1+ 1+ = 1+ 2 k P12 P12 P21 Therefore P12 P21 k2 = (P21 + P11 )(P12 + P22 )
1+
P22 P12
Now note that φ1 = φ11 + φ21 = P11 N1 i1 + P21 N1 i1 = N1 i1 (P11 + P21 ) and similarly φ2 = N2 i2 (P22 + P12 ) It follows that (P11 + P21 ) = and
φ1 N 1 i1
(P22 + P12 ) =
φ2 N 2 i2
Therefore φ12 φ21 (φ12 /N2 i2 )(φ21 /N1 i1 ) k2 = = (φ1 /N1 i1 )(φ2 /N2 i2 ) φ1 φ2 or k=
φ21 φ12
φ1
φ2
[b] The fractions (φ21 /φ1 ) and (φ12 /φ2 ) are by definition less than 1.0, therefore k < 1. P 6.44
[a] vab = L1
di di di di di + L2 + M + M = (L1 + L2 + 2M ) dt dt dt dt dt
It follows that [b] vab = L1
Lab = (L1 + L2 + 2M )
di di di di di − M + L2 − M = (L1 + L2 − 2M ) dt dt dt dt dt
Therefore Lab = (L1 + L2 − 2M )
Problems P 6.45
6–31
When the switch is opened the induced voltage is negative at the dotted terminal. Since the voltmeter kicks upscale, the induced voltage across the voltmeter must be positive at its positive terminal. Therefore, the voltage is negative at the negative terminal of the voltmeter. Thus, the lower terminal of the unmarked coil has the same instantaneous polarity as the dotted terminal. Therefore, place a dot on the lower terminal of the unmarked coil.
P 6.46
[a] vab = L1 0 = L1
di2 d(i1 − i2 ) +M dt dt
di2 d(i1 − i2 ) d(i2 − i1 ) di2 −M +M + L2 dt dt dt dt
Collecting coefficients of [di1 /dt] and [di2 /dt], the two mesh-current equations become di1 di2 + (M − L1 ) vab = L1 dt dt and di1 di2 0 = (M − L1 ) + (L1 + L2 − 2M ) dt dt Solving for [di1 /dt] gives di1 L1 + L2 − 2M vab = dt L1 L2 − M 2 from which we have
vab =
L1 L2 − M 2 L1 + L2 − 2M
.·. Lab =
di1 dt
L1 L2 − M 2 L1 + L2 − 2M
[b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses, therefore Lab = P 6.47
L1 L2 − M 2 L1 + L2 + 2M
[a] W = (0.5)L1 i21 + (0.5)L2 i22 + M i1 i2
M = 0.85 (18)(32) = 20.4 mH W = [9(36) + 16(81) + 20.4(54)] = 2721.6 mJ [b] W = [324 + 1296 + 1101.6] = 2721.6 mJ [c] W = [324 + 1296 − 1101.6] = 518.4 mJ [d] W = [324 + 1296 − 1101.6] = 518.4 mJ
6–32
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P 6.48
[a] M = 1.0 (18)(32) = 24 mH,
i1 = 6 A
Therefore 16i22 + 144i2 + 324 = 0,
9 ± Therefore i2 = − 2
2
9 2
i22 + 9i2 + 20.25 = 0
− 20.25 = −4.5 ±
√
0
Therefore i2 = −4.5 A [b] No, setting W equal to a negative value will make the quantity under the square root sign negative. P 6.49
When the button is not pressed we have
C2
dv d = C1 (vs − v) dt dt
or (C1 + C2 )
dvs dv = C1 dt dt
dvs dv C1 = dt (C1 + C2 ) dt Assuming C1 = C2 = C dvs dv = 0.5 dt dt or v = 0.5vs (t) + v(0) When the button is pressed we have
Problems
C1
dv dv d(v − vs ) + C3 + C2 =0 dt dt dt
.·.
C2 dv dvs = dt C1 + C2 + C3 dt
Assuming C1 = C2 = C3 = C 1 dvs dv = dt 3 dt 1 v = vs (t) + v(0) 3 Therefore interchanging the fixed capacitor and the button has no effect on the change in v(t). P 6.50
With no finger touching and equal 10 pF capacitors v(t) =
10 (vs (t)) + 0 = 0.5vs (t) 20
With a finger touching Let
Ce = equivalent capacitance of person touching lamp
Ce =
(10)(100) = 9.091 pF 110
Then
C + Ce = 10 + 9.091 = 19.091 pF
.·. v(t) =
10 vs = 0.344vs 29.091
.·. ∆v(t) = (0.5 − 0.344)vs = 0.156vs
6–33
6–34 P 6.51
CHAPTER 6. Inductance, Capacitance, and Mutual Inductance With no finger on the button the circuit is
C1
dv d (v − vs ) + C2 (v + vs ) = 0 dt dt
when
C1 = C2 = C
(2C)
dv =0 dt
With a finger on the button
C1
d(v − vs ) d(v + vs ) dv + C2 + C3 =0 dt dt dt
(C1 + C2 + C3 )
when
dvs dvs dv + C2 − C1 =0 dt dt dt
C1 = C2 = C3 = C
(3C)
dv =0 dt
.·. there is no change in the output voltage of this circuit.
7 Response of First-Order RL and RC Circuits
Assessment Problems AP 7.1 [a] The circuit for t < 0 is shown below. Note that the inductor behaves like a short circuit, effectively eliminating the 2 Ω resistor from the circuit.
First combine the 30 Ω and 6 Ω resistors in parallel: 306 = 5 Ω Use voltage division to find the voltage drop across the parallel resistors: 5 (120) = 75 V v= 5+3 Now find the current using Ohm’s law: 75 v i(0− ) = − = − = −12.5 A 6 6 1 1 [b] w(0) = Li2 (0) = (8 × 10−3 )(12.5)2 = 625 mJ 2 2 [c] To find the time constant, we need to find the equivalent resistance seen by the inductor for t > 0. When the switch opens, only the 2 Ω resistor remains connected to the inductor. Thus, L 8 × 10−3 τ= = = 4 ms R 2 [d] i(t) = i(0− )et/τ = −12.5e−t/0.004 = −12.5e−250t A, t≥0 [e] i(5 ms) = −12.5e−250(0.005) = −12.5e−1.25 = −3.58 A 7–1
7–2
CHAPTER 7. Response of First-Order RL and RC Circuits So w (5 ms) = 12 Li2 (5 ms) = 12 (8) × 10−3 (3.58)2 = 51.3 mJ w (dis) = 625 −51.3 =573.7 mJ 573.7 % dissipated = 100 = 91.8% 625
AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor:
Using current division, 10 (6.4) = 4 A i(0− ) = 10 + 6 Now use the circuit for t > 0 to find the equivalent resistance seen by the inductor, and use this value to find the time constant:
L 0.32 = 0.1 s = Req 3.2 Use the initial inductor current and the time constant to find the current in the inductor: i(t) = i(0− )e−t/τ = 4e−t/0.1 = 4e−10t A, t ≥ 0 Use current division to find the current in the 10 Ω resistor: 4 4 (−i) = (−4e−10t ) = −0.8e−10t A, t ≥ 0+ io (t) = 4 + 10 + 6 20 Req = 4(6 + 10) = 3.2 Ω,
.·.
τ=
Finally, use Ohm’s law to find the voltage drop across the 10 Ω resistor: vo (t) = 10io = 10(−0.8e−10t ) = −8e−10t V, t ≥ 0+ [b] The initial energy stored in the inductor is 1 1 w(0) = Li2 (0− ) = (0.32)(4)2 = 2.56 J 2 2 Find the energy dissipated in the 4 Ω resistor by integrating the power over all time: di t ≥ 0+ v4Ω (t) = L = 0.32(−10)(4e−10t ) = −12.8e−10t V, dt
Problems p4Ω (t) = w4Ω (t) =
2 v4Ω = 40.96e−20t W, 4
∞ 0
7–3
t ≥ 0+
40.96e−20t dt = 2.048 J
Find the percentage of the initial energy in the inductor dissipated in the 4 Ω resistor:
2.048 % dissipated = 100 = 80% 2.56 AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves like an open circuit.
Find the voltage drop across the open circuit by finding the voltage drop across the 50 kΩ resistor. First use current division to find the current through the 50 kΩ resistor: 80 × 103 (7.5 × 10−3 ) = 4 mA 80 × 103 + 20 × 103 + 50 × 103 Use Ohm’s law to find the voltage drop: v(0− ) = (50 × 103 )i50k = (50 × 103 )(0.004) = 200 V i50k =
[b] To find the time constant, we need to find the equivalent resistance seen by the capacitor for t > 0. When the switch opens, only the 50 kΩ resistor remains connected to the capacitor. Thus, τ = RC = (50 × 103 )(0.4 × 10−6 ) = 20 ms [c] v(t) = v(0− )e−t/τ = 200e−t/0.02 = 200e−50t V, t ≥ 0 1 1 [d] w(0) = Cv 2 = (0.4 × 10−6 )(200)2 = 8 mJ 2 2 1 1 2 [e] w(t) = Cv (t) = (0.4 × 10−6 )(200e−50t )2 = 8e−100t mJ 2 2 The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains: 8 × 10−3 e−100t = 2 × 10−3 ,
e100t = 4,
t = (ln 4)/100 = 13.86 ms
AP 7.4 [a] This circuit is actually two RC circuits in series, and the requested voltage, vo , is the sum of the voltage drops for the two RC circuits. The circuit for t < 0 is shown below:
7–4
CHAPTER 7. Response of First-Order RL and RC Circuits
Find the current in the loop and use it to find the initial voltage drops across the two RC circuits: 15 = 0.2 mA, v5 (0− ) = 4 V, i= v1 (0− ) = 8 V 75,000 There are two time constants in the circuit, one for each RC subcircuit. τ5 is the time constant for the 5 µF – 20 kΩ subcircuit, and τ1 is the time constant for the 1 µF – 40 kΩ subcircuit: τ5 = (20 × 103 )(5 × 10−6 ) = 100 ms; τ1 = (40 × 103 )(1 × 10−6 ) = 40 ms Therefore, v5 (t) = v5 (0− )e−t/τ5 = 4e−t/0.1 = 4e−10t V, t ≥ 0 v1 (t) = v1 (0− )e−t/τ1 = 8e−t/0.04 = 8e−25t V, t ≥ 0 Finally, vo (t) = v1 (t) + v5 (t) = [8e−25t + 4e−10t ] V, t≥0 [b] Find the value of the voltage at 60 ms for each subcircuit and use the voltage to find the energy at 60 ms: v5 (60 ms) = 4e−10(0.06) ∼ v1 (60 ms) = 8e−25(0.06) ∼ = 1.79 V, = 2.20 V w1 (60 ms) = 12 Cv12 (60 ms) = 12 (1 × 10−6 )(1.79)2 ∼ = 1.59 µJ w5 (60 ms) = 12 Cv52 (60 ms) = 12 (5 × 10−6 )(2.20)2 ∼ = 12.05 µJ w(60 ms) = 1.59 + 12.05 = 13.64 µJ Find the initial energy from the initial voltage: w(0) = w1 (0) + w2 (0) = 12 (1 × 10−6 )(8)2 + 12 (5 × 10−6 )(4)2 = 72 µJ Now calculate the energy dissipated at 60 ms and compare it to the initial energy: wdiss = w(0) − w(60 ms) = 72 − 13.64 = 58.36 µJ % dissipated = (58.36 × 10−6 /72 × 10−6 )(100) = 81.05 % AP 7.5 [a] Use the circuit at t < 0, shown below, to calculate the initial current in the inductor:
Problems
7–5
i(0− ) = 24/2 = 12 A = i(0+ ) Note that i(0− ) = i(0+ ) because the current in an inductor is continuous. [b] Use the circuit at t = 0+ , shown below, to calculate the voltage drop across the inductor at 0+ . Note that this is the same as the voltage drop across the 10 Ω resistor, which has current from two sources — 8 A from the current source and 12 A from the initial current through the inductor.
v(0+ ) = −10(8 + 12) = −200 V [c] To calculate the time constant we need the equivalent resistance seen by the inductor for t > 0. Only the 10 Ω resistor is connected to the inductor for t > 0. Thus, τ = L/R = (200 × 10−3 /10) = 20 ms [d] To find i(t), we need to find the final value of the current in the inductor. When the switch has been in position a for a long time, the circuit reduces to the one below:
Note that the inductor behaves as a short circuit and all of the current from the 8 A source flows through the short circuit. Thus, if = −8 A Now, i(t) = if + [i(0+ ) − if ]e−t/τ = −8 + [12 − (−8)]e−t/0.02 = −8 + 20e−50t A, t ≥ 0 [e] To find v(t), use the relationship between voltage and current for an inductor: v(t) = L
di(t) = (200 × 10−3 )(−50)(20e−50t ) = −200e−50t V, dt
t ≥ 0+
7–6
CHAPTER 7. Response of First-Order RL and RC Circuits
AP 7.6 [a]
From Example 7.6, vo (t) = −60 + 90e−100t V Write a KVL equation at the top node and use it to find the relationship between vo and vA : vA vA + 75 vA − vo + + =0 8000 160,000 40,000 20vA − 20vo + vA + 4vA + 300 = 0 25vA = 20vo − 300 vA = 0.8vo − 12 Use the above equation for vA in terms of vo to find the expression for vA : vA (t) = 0.8(−60 + 90e−100t ) − 12 = −60 + 72e−100t V,
t ≥ 0+
[b] t ≥ 0+ , since there is no requirement that the voltage be continuous in a resistor. AP 7.7 [a] Use the circuit shown below, for t < 0, to calculate the initial voltage drop across the capacitor:
i=
40 × 103 (10 × 10−3 ) = 3.2 mA 125 × 103
vc (0− ) = (3.2 × 10−3 )(25 × 103 ) = 80 V
so
vc (0+ ) = 80 V
Now use the next circuit, valid for 0 ≤ t ≤ 10 ms, to calculate vc (t) for that interval:
Problems For
7–7
0 ≤ t ≤ 100 ms:
τ = RC = (25 × 103 )(1 × 10−6 ) = 25 ms vc (t) = vc (0− )et/τ = 80e−40t V,
0 ≤ t ≤ 10 ms
[b] Calculate the starting capacitor voltage in the interval t ≥ 10 ms, using the capacitor voltage from the previous interval: vc (0.01) = 80e−40(0.01) = 53.63 V Now use the next circuit, valid for t ≥ 10 ms, to calculate vc (t) for that interval:
For
t ≥ 10 ms :
Req = 25 kΩ100 kΩ = 20 kΩ τ = Req C = (20 × 103 )(1 × 10−6 ) = 0.02 s Therefore vc (t) = vc (0.01+ )e−(t−0.01)/τ = 53.63e−50(t−0.01) V,
t ≥ 0.01 s
[c] To calculate the energy dissipated in the 25 kΩ resistor, integrate the power absorbed by the resistor over all time. Use the expression p = v 2 /R to calculate the power absorbed by the resistor. w25 k =
0.01 0
∞ [80e−40t ]2 [53.63e−50(t−0.01) ]2 dt + dt = 2.91 mJ 25,000 25,000 0.01
[d] Repeat the process in part (c), but recognize that the voltage across this resistor is non-zero only for the second interval: w100 kΩ =
∞ 0.01
[53.63e−50(t−0.01) ]2 dt = 0.29 mJ 100,000
We can check our answers by calculating the initial energy stored in the capacitor. All of this energy must eventually be dissipated by the 25 kΩ resistor and the 100 kΩ resistor. Check:
wstored = (1/2)(1 × 10−6 )(80)2 = 3.2 mJ wdiss = 2.91 + 0.29 = 3.2 mJ
AP 7.8 [a] Note – the 30 Ω resistor should be a 3 Ω resistor; the resistor in parallel with the 8 A current source should be 9 Ω. Prior to switch a closing at t = 0, there are no sources connected to the inductor; thus, i(0− ) = 0. At the instant A is closed, i(0+ ) = 0.
7–8
CHAPTER 7. Response of First-Order RL and RC Circuits For 0 ≤ t ≤ 1 s,
The equivalent resistance seen by the 10 V source is 2 + (30.8). The current leaving the 10 V source is 10 = 3.8 A 2 + (30.8) The final current in the inductor, which is equal to the current in the 0.8 Ω resistor is 3 (3.8) = 3 A i(∞) = 3 + 0.8 The resistance seen by the inductor is calculated to find the time constant: 0.8 + (23) = 2 Ω
τ=
2 L = = 1s R 2
Therefore, i = i(∞) + [i(0+ ) − i(∞)]e−t/τ = 3 − 3e−t A,
0 ≤ t ≤ 1s
For part (b) we need the value of i(t) at t = 1 s: i(1) = 3 − 3e−1 = 1.896 A . [b] For t > 1 s
Use current division to find the final value of the current: 9 (−8) = −4.8 A i= 9+6 The equivalent resistance seen by the inductor is used to calculate the time constant: 2 L = = 0.8 s 3(9 + 6) = 2.5 Ω τ= R 2.5
Problems
7–9
Therefore, i = i(∞) + [i(1+ ) − i(∞)]e−(t−1)/τ = −4.8 + 6.696e−1.25(t−1) A,
t≥1s
AP 7.9 0 ≤ t ≤ 32 ms:
32×10−3 1 32×10−3 1 1 vo = − −10 dt + 0 = − (−10t) =− (−320 × 10−3 ) RCf 0 RCf RCf 0
RCf = (200 × 103 )(0.2 × 10−6 ) = 40 × 10−3
so
1 = 25 RCf
vo = −25(−320 × 10−3 ) = 8 V t ≥ 32 ms:
t 1 t 1 1 vo = − 5 dy + 8 = − (5y) +8 = − 5(t − 32 × 10−3 ) + 8 −3 −3 RCf 32×10 RCf RCf 32×10
RCf = (250 × 103 )(0.2 × 10−6 ) = 50 × 10−3
so
1 = 20 RCf
vo = −20(5)(t − 32 × 10−3 ) + 8 = −100t + 11.2 The output will saturate at the negative power supply value: −15 = −100t + 11.2
.·.
t = 262 ms
7–10
CHAPTER 7. Response of First-Order RL and RC Circuits
AP 7.10 [a] Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: vp = Vf + [Vo − Vf ]e−t/τ = −2 + (0 + 2)e−t/τ τ = (160 × 103 )(10 × 10−9 ) = 10−3 ; vp = −2 + 2e−625t V;
1/τ = 625
vn = vp
Write a KVL equation at the inverting input, and use it to determine vo : vn − vo vn + =0 10,000 40,000 .·. vo = 5vn = 5vp = −10 + 10e−625t V The output will saturate at the negative power supply value: −10 + 10e−625t = −5;
e−625t = 1/2;
t = ln 2/625 = 1.11 ms
[b] Use RC circuit analysis to determine the expression for the voltage at the non-inverting input: vp = Vf + [Vo − Vf ]e−t/τ = −2 + (1 + 2)e−625t = −2 + 3e−625t V The analysis for vo is the same as in part (a): vo = 5vp = −10 + 15e−625t V The output will saturate at the negative power supply value: −10 + 15e−625t = −5;
e−625t = 1/3;
t = ln 3/625 = 1.76 ms
Problems
7–11
Problems P 7.1
[a] t < 0
2 kΩ6 kΩ = 1.5kΩ Find the current from the voltage source by combining the resistors in series and parallel and using Ohm’s law: ig (0− ) =
40 = 20 mA (1500 + 500)
Find the branch currents using current division: i1 (0− ) =
2000 (0.02) = 5 mA 8000
i2 (0− ) =
6000 (0.02) = 15 mA 8000
[b] The current in an inductor is continuous. Therefore, i1 (0+ ) = i1 (0− ) = 5 mA i2 (0+ ) = −i1 (0+ ) = −5 mA [c] τ =
(when switch is open)
0.4 × 10−3 L = = 5 × 10−5 s; R 8 × 103
i1 (t) = i1 (0+ )e−t/τ = 5e−20,000t mA, [d] i2 (t) = −i1 (t)
when
1 = 20,000 τ t≥0
t ≥ 0+
.·. i2 (t) = −5e−20,000t mA,
t ≥ 0+
[e] The current in a resistor can change instantaneously. The switching operation forces i2 (0− ) to equal 15 mA and i2 (0+ ) = −5 mA. P 7.2
[a] i(0) = 60 V/(10 Ω + 5 Ω) = 4 A 4 L = = 80 ms [b] τ = R 45 + 5
7–12
CHAPTER 7. Response of First-Order RL and RC Circuits [c] i = 4e−t/0.08 = 4e−12.5t A,
t≥0
v1 = −45i = −180e−12.5t V v2 = L
t ≥ 0+
di = (4)(−12.5)(4e−12.5t ) = −200e−12.5t V dt
t ≥ 0+
[d] pdiss = i2 (45) = 720e−25t W wdiss =
t 0
720e−25x dx = 720
e−25x t = 28.8 − 28.8e−25t J −25 0
wdiss (40 ms) = 28.8 − 28.8e−1 = 18.205 J 1 w(0) = (4)(4)2 = 32 J 2 18.205 (100) = 56.89% % dissipated = 32 P 7.3
[a] io (0− ) = 0
since the switch is open for t < 0.
[b] For t = 0− the circuit is:
120 Ω60 Ω = 40 Ω .·. ig = iL (0− ) =
12 = 0.24 A = 240 mA 10 + 40
120 ig = 160 mA 180
[c] For t = 0+ the circuit is:
120 Ω40 Ω = 30 Ω
Problems .·. ig =
7–13
12 = 0.30 A = 300 mA 10 + 30
120 300 = 225 mA ia = 160 .·. io (0+ ) = 225 − 160 = 65 mA [d] iL (0+ ) = iL (0− ) = 160 mA [e] io (∞) = ia = 225 mA since the switch short circuits the branch containing the 20 Ω [f] iL (∞) = 0, resistor and the 100 mH inductor. L 1 100 × 10−3 [g] τ = = 5 ms; = 200 = R 20 τ .·. iL = 0 + (160 − 0)e−200t = 160e−200t mA, [h] vL (0− ) = 0
t≥0
since for t < 0 the current in the inductor is constant
[i] Refer to the circuit at t = 0+ and note: 20(0.16) + vL (0+ ) = 0; [j] vL (∞) = 0,
.·. vL (0+ ) = −3.2 V
since the current in the inductor is a constant at t = ∞.
[k] vL (t) = 0 + (−3.2 − 0)e−200t = −3.2e−200t V, [l] io = ia − iL = 225 − 160e−200t mA, P 7.4
t ≥ 0+
400e−5t v =R= = 40 Ω i 10e−5t 1 [b] τ = = 200 ms 5 L = 200 × 10−3 [c] τ = R [a]
L = (200 × 10−3 )(40) = 8 H 1 1 [d] w(0) = L[i(0)]2 = (8)(10)2 = 400 J 2 2 [e] wdiss =
t 0
4000e−10x dx = 400 − 400e−10t
0.8w(0) = (0.8)(400) = 320 J 400 − 400e−10t = 320 Solving, t = 160.9 ms.
.·. e10t = 5
t ≥ 0+
7–14 P 7.5
CHAPTER 7. Response of First-Order RL and RC Circuits [a] iL (0) =
12 = 2A 6
io (0+ ) =
12 − 2 = 6 − 2 = 4A 2
io (∞) =
12 = 6A 2
[b] iL = 2e−t/τ ;
τ=
L 1 = s R 4
iL = 2e−4t A io = 6 − iL = 6 − 2e−4t A,
t ≥ 0+
[c] 6 − 2e−4t = 5 1 = 2e−4t .·. t = 173.3 ms
e6t = 2 P 7.6
1 w(0) = (30 × 10−3 )(32 ) = 135 mJ 2 1 w(0) = 27 mJ 5 iR = 3e−t/τ pdiss = i2R R = 9Re−2t/τ wdiss =
t 0
wdiss = 9R
R(9)e−2x/τ dx e−2x/τ to = −4.5τ R(e−2to /τ − 1) = 4.5L(1 − e−2to /τ ) −2/τ 0
4.5L = (4.5)(30) × 10−3 = 0.135; 1 − e−2to /τ = e2to /τ = 1.25; R=
to = 15 µs
1 5 2to R 2to = = ln 1.25 τ L
L ln 1.25 30 × 10−3 ln 1.25 = = 223.14 Ω 2to 30 × 10−6
Problems P 7.7
1 [a] w(0) = LIg2 2
wdiss
to
e−2t/τ to (−2/τ ) 0 0 1 1 = Ig2 Rτ (1 − e−2to /τ ) = Ig2 L(1 − e−2to /τ ) 2 2 = σw(0)
wdiss =
Ig2 Re−2t/τ dt = Ig2 R
1 2 1 2 LIg (1 − e−2to /τ ) = τ LI 2 2 g 1 1 − e−2to /τ = σ; e2to /τ = (1 − σ)
.·.
1 2to ; = ln τ (1 − σ) R=
R(2to ) = ln[1/(1 − σ)] L
L ln[1/(1 − σ)] 2to
(30 × 10−3 ) ln[1/0.8] 30 × 10−6 R = 223.14 Ω
[b] R =
P 7.8
[a] t < 0
iL (0− ) =
150 (12) = 10 A 180
t≥0
τ=
1.6 × 10−3 = 200 × 10−6 ; 8
io = −10e−5000t A
t≥0
1/τ = 5000
7–15
7–16
CHAPTER 7. Response of First-Order RL and RC Circuits 1 [b] wdel = (1.6 × 10−3 )(10)2 = 80 mJ 2 [c] 0.95wdel = 76 mJ .·. 76 × 10−3 =
to 0
8(100e−10,000t ) dt
to −3 −3 −10,000t −3 −10,000to · . . 76 × 10 = −80 × 10 e ) = 80 × 10 (1 − e 0
.·. e−10,000to = 4 × 10−3 .·. P 7.9
so
to = 552.1 µs
552.1 × 10−6 to = = 2.76 τ 200 × 10−6
so
to ≈ 2.76τ
For t < 0+
ig =
−48 = −6.5 A 6 + (181.5)
iL (0− ) =
18 (−6.5) = −6 A = iL (0+ ) 18 + 1.5
For t > 0
iL (t) = iL (0+ )e−t/τ A, τ=
t≥0
0.5 L = = 0.0125 s; R 10 + 12.45 + (5426)
iL (t) = −6e−80t A, io (t) =
1 = 80 τ
t≥0
54 54 (−iL (t)) = (6e−80t ) = 4.05e−80t V, 80 80
t ≥ 0+
Problems P 7.10
From the solution to Problem 7.9, i54Ω =
26 (−iL ) = −1.95e−80t A 80
P54Ω = 54(i54Ω )2 = 205.335e−160t W
wdiss =
0.0125 0
205.335e−160t dt
205.335 −160t 0.0125 e = −160 0 −2 = 1.28(1 − e ) = 1.11 J 1 wstored = (0.5)(−6)2 = 9 mJ. 2 % diss = P 7.11
1.11 × 100 = 12.3% 9
[a] t < 0 :
iL (0− ) = iL (0+ ) =
i∆ =
70 (11.84) = 11.2 A 70 + 4
70 iT = 0.4375iT 160
vT = 30i∆ + iT
(90)(70) (90)(70) = 30(0.4375)iT + iT = 52.5iT 160 160
vT = RTh = 52.5 Ω iT
7–17
7–18
CHAPTER 7. Response of First-Order RL and RC Circuits
τ=
20 × 10−3 L = = .·. R 52.5
iL = 11.2e−2625t A, [b] vL = L
1 = 2625 τ
t≥0
diL = 20 × 10−3 (−2625)(11.2e−2625t ) = −588e−2625t V, dt
[c]
vL = 30i∆ + 90i∆ = 120i∆ i∆ = P 7.12
vL = −4.9e−2625t A 120
t ≥ 0+
1 w(0) = (20 × 10−3 )(11.2)2 = 1254.4 mJ 2 p30i∆ = −30i∆ iL = −30(−4.9e−2625t )(11.2e−2625t ) = 1646.4e−5250t W w30i∆ =
∞ 0
1646.4e
% dissipated = P 7.13
−5250t
e−5250t ∞ dt = 1646.4 = 313.6mJ −5250 0
313.6 (100) = 25% 1254.4
t0
Find Thévenin resistance seen by inductor
iT = 4vT ;
τ=
5 × 10−3 L = = 20 ms; R 0.25
io = 4e−50t A, vo = L P 7.14
vT 1 = RTh = = 0.25 Ω iT 4 1/τ = 50
t≥0
dio = (5 × 10−3 )(−200e−50t ) = −e−50t V, dt
t < 0:
iL (0+ ) = 8 A
t ≥ 0+
7–19
7–20
CHAPTER 7. Response of First-Order RL and RC Circuits t > 0:
Re = τ=
(10)(40) + 10 = 18 Ω 50 1 = 250 τ
L 0.072 = 4 ms; = Re 18
.·. iL = 8e−250t A .·. vo = −10iL − 0.072 = 64e−250t A P 7.15
t ≥ 0+
1 w(0) = (72 × 10−3 )(8)2 = 2304 mJ 2 p40Ω = w40Ω =
642 −500t vo2 = e = 102.4e−500t W 40 40 ∞
%diss = P 7.16
diL = −80e−250t + 144e−250t dt
0
102.4e−500t dt = 204.8 mJ
204.8 (100) = 8.89% 2304
[a] vo (t) = vo (0+ )e−t/τ .·. vo (0+ )e−1×10 .·. e1×10
−3 /τ
−3 /τ
= 0.5vo (0+ )
=2
.·. τ =
1 × 10−3 L = R ln 2
.·. L =
10 × 10−3 = 14.43 mH ln 2
Problems [b] vo (0+ ) = −10iL (0+ ) = −10(1/10)30 × 10−3 = −30 mV .·.
vo = −0.03e−t/τ V,
p10Ω =
t ≥ 0+
vo2 = 9 × 10−5 e−2t/τ 10
w10Ω (1 ms) =
10−3 0+
9 × 10−5 e−2t/τ dt
= 4.5τ × 10−5 (1 − e−2(0.001)/τ ) τ=
1 1000 ln 2
.·.
w10Ω (1 ms) = 48.69 nJ
1 1 wL (0) = Li2L (0) = (14.43 × 10−3 )(3 × 10−3 )2 = 64.92 nJ 2 2 %dissipated in 1 ms = P 7.17
48.69 (100) = 75% 64.92
[a] t < 0 :
t = 0+ :
33 = iab + 9 + 15, [b] At
iab = 9 A,
t = ∞:
iab = 165/5 = 33 A,
t=∞
t = 0+
7–21
7–22
CHAPTER 7. Response of First-Order RL and RC Circuits
[c] i1 (0) = 9,
τ1 =
i2 (0) = 15,
12.5 × 10−3 = 2.5 ms 5 3.75 × 10−3 1.25 ms 3
τ2 =
i1 (t) = 9e−400t A,
t≥0
i2 (t) = 15e−800t A,
t≥0
iab = 33 − 9e−400t − 15e−800t A,
t ≥ 0+
33 − 9e−400t − 15e−800t = 19 14 = 9e−400t + 15e−800t Let
x = e−400t
.·.
x2 = e−800t
Substituting, 15x2 + 9x − 14 = 0 so .·. P 7.18
t=
x = 0.7116 = e−400t
[ln(1/0.7116)] = 850.6 µs 400
[a] t < 0
1 kΩ4 kΩ = 0.8 kΩ 20 kΩ80 kΩ = 16 kΩ (105 × 10−3 )(0.8 × 103 ) = 84 V
Problems
iL (0− ) =
84 = 5 mA 16,800
t>0
τ=
6 L = × 10−3 = 250 µs; R 24
iL (t) = 5e−4000t mA,
1 = 4000 τ
t≥0
p4k = 25 × 10−6 e−8000t (4000) = 0.10e−8000t W wdiss =
t 0
0.10e−8000x dx = 12.5 × 10−6 [1 − e−8000t ] J
1 w(0) = (6)(25 × 10−6 ) = 75 µJ 2 0.10w(0) = 7.5 µJ 12.5(1 − e−8000t ) = 7.5; t=
.·. e8000t = 2.5
ln 2.5 = 114.54 µs 8000
[b] wdiss (total) = 75(1 − e−8000t ) µJ wdiss (114.54 µs) = 45 µJ % = (45/75)(100) = 60%
7–23
7–24 P 7.19
CHAPTER 7. Response of First-Order RL and RC Circuits [a] t < 0:
t = 0+ :
t > 0:
iR = −2e−t/τ A;
τ=
5 L = = 666.67 ms R 7.5
iR = −2e−1.5t A
vR = (7.5)(−2e−1.5t ) = −15e−1.5t V
.·.
1 = 1.5 τ
Problems v1 = 1.25[(−1.5)(−2e−1.5t )] = 3.75e−1.5t V, vo = −v1 − vR = 11.25e−1.5t V [b] io = P 7.20
t ≥ 0+
1 t 11.25e−1.5x dx + 0 = 1.25 − 1.25e−1.5t A 6 0
t≥0
[a] From the solution to Problem 7.19, iR = −2e−1.5t A pR = (−2e−1.5t )2 (7.5) = 30e−3t W wdiss =
∞ 0
30e−3t dt
e−3t ∞ = 10 J = 30 −3 0
1 1 [b] wtrapped = (10)(−1.25)2 + (6)(1.25)2 = 12.5 J 2 2 1 CHECK: w(0) = 2 (1.25)(2)2 + 12 (10)(2)2 = 22.5 J .·. w(0) = wdiss + wtrapped P 7.21
[a] v1 (0− ) = v1 (0+ ) = 40 V
v2 (0+ ) = 0
Ceq = (1)(4)/5 = 0.8 µF
τ = (25 × 103 )(0.8 × 10−6 ) = 20ms; i=
40 −50t e = 1.6e−50t mA, 25,000
1 = 50 τ t ≥ 0+
−1 t v1 = −6 1.6 × 10−3 e−50x dx + 40 = 32e−50t + 8 V, t≥0 10 0 t 1 1.6 × 10−3 e−50x dx + 0 = −8e−50t + 8 V, t≥0 v2 = 4 × 10−6 0
7–25
7–26
CHAPTER 7. Response of First-Order RL and RC Circuits 1 [b] w(0) = (10−6 )(40)2 = 800 µJ 2 1 1 [c] wtrapped = (10−6 )(8)2 + (4 × 10−6 )(8)2 = 160 µJ. 2 2 The energy dissipated by the 25 kΩ resistor is equal to the energy dissipated by the two capacitors; it is easier to calculate the energy dissipated by the capacitors (final voltage on the equivalent capacitor is zero): 1 wdiss = (0.8 × 10−6 )(40)2 = 640 µJ. 2 Check: wtrapped + wdiss = 160 + 640 = 800 µJ;
P 7.22
w(0) = 800 µJ.
[a] Calculate the initial voltage drop across the capacitor: v(0) = (2.7 k3.3 k)(40 mA) = (1485)(40 × 10−3 ) = 59.4 V The equivalent resistance seen by the capacitor is Re = 3 k(2.4 k + 3.6 k) = 3 k6 k = 2 kΩ τ = Re C = (2000)(0.5) × 10−6 = 1000 µs; v = v(0)e−t/τ = 59.4e−1000t V io =
1 = 1000 τ
t≥0
v = 9.9e−1000t mA, 2.4 k + 3.6 k
t ≥ 0+
1 [b] w(0) = (0.5 × 10−6 )(59.4)2 = 882.09 µJ 2 i3k
59.4e−1000t = 19.8e−1000t mA = 3000
p3k = [(19.8 × 10−3 )e−1000t ]2 (3000) = 1.176e−2000t −6 e−2000x 500×10 1.176 −1 (e − 1) = 371.72 µJ w3k (500 µs) = 1.176 = −2000 0 −2000
%= P 7.23
371.72 × 100 = 42.14% 882.09
v = 4 kΩ i 1 1 1 [b] = = 25; C= = 10 µF τ RC (25)(4 × 103 ) 1 = 40 ms [c] τ = 25 1 [d] w(0) = (10 × 10−6 )(48)2 = 11.52 mJ 2 [a] R =
Problems [e] wdiss (60 ms) =
0.06 2 v
R
0
dt =
0.06 0
(48e−25t )2 dt (4 × 103 )
e−50t 0.06 = 0.576 = −5.74 × 10−4 + 0.01152 = 10.95 mJ −50 0
P 7.24
[a] t < 0:
i1 (0− ) = i2 (0− ) =
3V = 100 mA 30 Ω
[b] t > 0:
i1 (0+ ) =
0.2 = 100 mA 2
i2 (0+ ) =
−0.2 = −25 mA 8
[c] Capacitor voltage cannot change instantaneously, therefore, i1 (0− ) = i1 (0+ ) = 100 mA [d] Switching can cause an instantaneous change in the current in a resistive branch. In this circuit i2 (0− ) = 100 mA [e] vc = 0.2e−t/τ V,
and t≥0
i2 (0+ ) = −25 mA Re = 2||(5 + 3) = 1.6 Ω
τ = 1.6(2 × 10−6 ) = 3.2 × 10−6 s vc = 0.2e−312,500t V,
t≥0
vc = 0.1e−312,500t A, t≥0 2 −vc = −25e−312,500t mA, [f] i2 = t ≥ 0+ 8 i1 =
7–27
7–28 P 7.25
CHAPTER 7. Response of First-Order RL and RC Circuits [a] t < 0:
Re = 12 k||68 k = 10.2 kΩ vo (0) =
10,200 (−120) = −102 V 10,200 + 1800
t > 0:
1 = 25 τ
τ = [(10/3) × 10−6 )(12,000) = 40 ms; vo = −102e−25t V, p=
t≥0
vo2 = 867 × 10−3 e−50t W 12,000
wdiss =
12×10−3 0
867 × 10−3 e−50t dt
= 17.34 × 10−3 (1 − e−50(12×10
1 [b] w(0) = 2
−3 )
) = 7.82 mJ
10 (102)2 × 10−6 = 17.34 mJ 3
0.75w(0) = 13 mJ to 0
867 × 10−3 e−50x dx = 13 × 10−3
.·. 1 − e−50to = 0.75;
e50to = 4;
so
to = 27.73 ms
Problems P 7.26
[a]
vT = 20 × 103 (iT + αv∆ ) + 5 × 103 iT v∆ = 5 × 103 iT vT = 25 × 103 iT + 20 × 103 α(5 × 103 iT ) RTh = 25,000 + 100 × 106 α τ = RTh C = 40 × 10−3 = RTh (0.8 × 10−6 ) RTh = 50 kΩ = 25,000 + 100 × 106 α α=
25,000 = 2.5 × 10−4 A/V 100 × 106
[b] vo (0) = (−5 × 10−3 )(3600) = −18 V t > 0:
vo = −18e−25t V,
t≥0
v∆ − vo v∆ + + 2.5 × 10−4 v∆ = 0 5000 20,000
t 0:
τ = 2 ms;
1/τ = 500
vo = 10e−500t V,
t≥0
io = e−500t mA,
t ≥ 0+
−500t
i24kΩ = e
16 = 0.4e−500t mA, 40
t ≥ 0+
p24kΩ = (0.16 × 10−6 e−1000t )(24,000) = 3.84e−1000t mW w24kΩ =
∞ 0
3.84 × 10−3 e−1000t dt = −3.84 × 10−6 (0 − 1) = 3.84 µJ
7–31
7–32
CHAPTER 7. Response of First-Order RL and RC Circuits 1 1 w(0) = (0.25 × 10−6 )(40)2 + (1 × 10−6 )(50)2 = 1.45 mJ 2 2 % diss (24 kΩ) =
3.84 × 10−6 × 100 = 0.26% 1.45 × 10−3
[b] p400Ω = 400(1 × 10−3 e−500t )2 = 0.4 × 10−3 e−1000t w400Ω =
∞ 0
p400 dt = 0.40 µJ
% diss (400Ω) = i16kΩ = e−500t
0.4 × 10−6 × 100 = 0.03% 1.45 × 10−3
24 = 0.6e−500t mA, 40
t ≥ 0+
p16kΩ = (0.6 × 10−3 e−500t )2 (16,000) = 5.76 × 10−3 e−1000t W w16kΩ =
∞ 0
5.76 × 10−3 e−1000t dt = 5.76 µJ
% diss (16kΩ) = 0.4% [c]
wdiss = 3.84 + 5.76 + 0.4 = 10 µJ
wtrapped = w(0) − % trapped =
wdiss = 1.45 × 10−3 − 10 × 10−6 = 1.44 mJ
1.44 × 100 = 99.31% 1.45
Check: 0.26 + 0.03 + 0.4 + 99.31 = 100% P 7.30
[a] Ce =
(2 + 1)6 = 2 µF 2+1+6
vo (0) = −5 + 30 = 25 V τ = (2 × 10−6 )(250 × 103 ) = 0.5 s;
vo = 25e−2t V,
t > 0+
1 =2 τ
Problems
7–33
1 1 [b] wo = (3 × 10−6 )(30)2 + (6 × 10−6 )(5)2 = 1425 µJ 2 2 1 wdiss = (2 × 10−6 )(25)2 = 625 µJ 2 % diss = [c] io =
1425 − 625 × 100 = 56.14% 1425
vo = 100e−2t µA −3 250 × 10
t t 1 −6 −2x 100 × 10 e dx − 5 = −16.67 e−2x dx − 5 v1 = − −6 6 × 10 0 0 e−2x t = −16.67 −5 = 8.33e−2t − 13.33 V t≥0 −2 0
[d] v1 + v2 = vo v2 = vo − v1 = 25e−2t − 8.33e−2t + 13.33 = 16.67e−2t + 13.33 V
t≥0
1 1 [e] wtrapped = (6 × 10−6 )(13.33)2 + (3 × 10−6 )(13.33)2 = 800 µJ 2 2 wdiss + wtrapped = 625 + 800 = 1425 µJ P 7.31
(check)
[a] At t = 0− the voltage on each capacitor will be 150 V(5 × 30), positive at the upper terminal. Hence at t ≥ 0+ we have
.·. isd (0+ ) = 5 +
150 150 + = 1055 A 0.2 0.5
At t = ∞, both capacitors will have completely discharged. .·. isd (∞) = 5 A [b] isd (t) = 5 + i1 (t) + i2 (t) τ1 = 0.2(10−6 ) = 0.2 µs τ2 = 0.5(100 × 10−6 ) = 50 µs
7–34
CHAPTER 7. Response of First-Order RL and RC Circuits .·. i1 (t) = 750e−5×10 t A, 6
i2 (t) = 300e−20,000t A,
t ≥ 0+ t≥0
.·. isd = 5 + 750e−5×10 t + 300e−20,000t mA, 6
P 7.32
t ≥ 0+
[a] t < 0:
io (0− ) =
6000 (40 m) = 24 mA 6000 + 4000
vo (0− ) = (3000)(24 m) = 72 V i2 (0− ) = 40 − 24 = 16 mA v2 (0− ) = (6000)(16 m) = 96 V t>0
τ = RC = (1000)(0.2 × 10−6 ) = 200 µs;
io (t) = [b]
24 e−t/τ = 24e−5000t mA, 1 × 103
1 = 5000 τ
t ≥ 0+
Problems
t 1 24 × 10−3 e−5000x dx + 72 −6 0.6 × 10 0 e−5000x t (40,000) +72 −5000 0 −8e−5000t + 8 + 72 [−8e−5000t + 80] V, t≥0
vo = = = vo =
[c] wtrapped = (1/2)(0.3 × 10−6 )(80)2 + (1/2)(0.6 × 10−6 )(80)2 wtrapped = 2880 µJ. Check: 1 wdiss = (0.2 × 10−6 )(24)2 = 57.6 µJ 2 1 1 w(0) = (0.3 × 10−6 )(96)2 + (0.6 × 10−6 )(72)2 = 2937.6 µJ. 2 2 wtrapped + wdiss = w(0) 2880 + 57.6 = 2937.6 P 7.33
OK.
[a] t < 0
iL (0− ) = −5 A t>0
iL (∞) = τ=
40 − 80 = −2 A 4 + 16
4 × 10−3 L = = 200 µs; R 4 + 16
1 = 5000 τ
7–35
7–36
CHAPTER 7. Response of First-Order RL and RC Circuits iL = iL (∞) + [iL (0+ ) − iL (∞)]e−t/τ = −2 + (−5 + 2)e−5000t = −2 − 3e−5000t A,
t≥0
vo = 16iL + 80 = 16(−2 − 3e−5000t ) + 80 = 48 − 48e−5000t V, [b] vL = L
diL = 4 × 10−3 (−5000)[−3e−5000t ] = 60e−5000t V, dt
t ≥ 0+
t ≥ 0+
vL (0+ ) = 60 V From part (a) Check:
vo (0+ ) = 0 V
at t = 0+ the circuit is:
vL (0+ ) = 40 + (5 A)(4 Ω) = 60 V, P 7.34
[a] t < 0
KVL equation at the top node: vo vo vo + + 50 = 8 40 10 Multiply by 40 and solve: 2000 = (5 + 1 + 4)vo ; .·. io (0− ) = t>0
vo = 200 V
vo = 200/10 = 20 A 10
vo (0+ ) = 80 − (16 Ω)(5 A) = 0 V
Problems Use voltage division to find the Thévenin voltage: 40 (800) = 200 V 40 + 120 Remove the voltage source and make series and parallel combinations of resistors to find the equivalent resistance:
VTh = vo =
RTh = 10 + 12040 = 10 + 30 = 40 Ω The simplified circuit is:
40 × 10−3 L = = 1 ms; R 40 200 = 5A io (∞) = 40
1 = 1000 τ
τ=
.·. io = io (∞) + [io (0+ ) − io (∞)]e−t/τ = 5 + (20 − 5)e−1000t = 5 + 15e−1000t A, [b] vo
vo P 7.35
t≥0
=
dio dt 10(5 + 15e−1000t ) + 0.04(−1000)(15e−1000t )
=
50 + 150e−1000t − 600e−1000t
=
50 − 450e−1000t V,
=
10io + L
t ≥ 0+
After making a Thévenin equivalent we have
For t < 0, the 15 Ω resistor is bypassed: io (0− ) = io (0+ ) = 50/5 = 10 A
7–37
7–38
CHAPTER 7. Response of First-Order RL and RC Circuits τ=
16 × 10−3 L = = 8 × 10−4 ; R 5 + 15
i(∞) =
1 = 1250 τ
V 50 = = 2.5 A Req 5 + 15
io = io (∞) + [io (0+ ) − io (∞)]e−t/τ = 2.5 + (10 − 2.5)e−1250t = 2.5 + 7.5e−1250t A, t ≥ 0 vo = L P 7.36
dio = 16 × 10−3 (−1250)(7.5e−1250t ) = −150e−1250t V, dt
[a] vo (0+ ) = −Ig R2 ;
τ=
t ≥ 0+
L R1 + R2
vo (∞) = 0 vo (t) = −Ig R2 e−[(R1 +R2 )/L]t V, [b] vo = −(10)(15)e−
(5+15) t 0.016
t ≥ 0+
= −150e−1250t V,
t ≥ 0+
[c] vo (0+ ) → ∞, and the duration of vo (t) → zero L τ= [d] vsw = R2 io ; R1 + R2 io (0+ ) = Ig ; Therefore
Therefore [e] |vsw (0+ )| → ∞;
io (∞) = Ig
R1 R1 + R2
io (t) =
Ig R1 R1 +R2
io (t) =
R1 Ig (R1 +R2 )
vsw =
+ Ig − +
R1 Ig (1+R1 /R2 )
Ig R1 R1 +R2
e−[(R1 +R2 )/L]t
R2 Ig e−[(R1 +R2 )/L]t (R1 +R2 )
+
R2 Ig e−[(R1 +R2 )/L]t , (1+R1 /R2 )
t ≥ 0+
duration → 0
P 7.37
Opening the inductive circuit causes a very large voltage to be induced across the inductor L. This voltage also appears across the switch (part [e] of Problem 7.36) causing the switch to arc over. At the same time, the large voltage across L damages the meter movement.
P 7.38
[a] From Eqs. (7.35) and (7.42)
Vs Vs −(R/L)t i= e + Io − R R v = (Vs − Io R)e−(R/L)t .·.
Vs = 4; R
Io −
Vs =4 R
Problems R = 40 L
Vs − Io R = −80; .·. Io = 4 +
Vs = 8A R
Now since Vs = 4R we have 4R − 8R = −80; Vs = 80 V;
R = 20 Ω
L=
[b] i = 4 + 4e−40t ;
R = 0.5 H 40
i2 = 16 + 32e−40t + 16e−80t
1 1 w = Li2 = (0.5)[16 + 32e−40t + 16e−80t ] = 4 + 8e−40t + 4e−80t 2 2 .·. 4 + 8e−40t + 4e−80t = 9 or
e−80t + 2e−40t − 1.25 = 0
Let x = e−40t : x2 + 2x − 1.25 = 0;
Solving,
x = 0.5;
x = −2.5
But x ≥ 0 for all t. Thus, e−40t = 0.5; P 7.39
e40t = 2;
t = 25 ln 2 = 17.33 ms
For t < 0
vx vx − 480 − 0.8vφ + =0 15 21 vφ =
vx − 480 21
vx − 480 vx − 480 vx − 0.8 + 15 21 21
vs − 480 vx + 0.2 = = 21vx + 3(vx − 480) = 0 15 21
7–39
7–40
CHAPTER 7. Response of First-Order RL and RC Circuits .·.
24vx = 1440 so
io (0− ) =
vx = 60 V
vx =4A 15
t>0
Find Thévenin equivalent with respect to a, b
VTh − 320 VTh − 320 − 0.8 =0 5 5
vT vT = (iT + 0.8vφ )(5) = iT + 0.8 5
VTh = 320 V
(5)
Problems vT = 5iT + 0.8vT
.·. 0.2vT = 5iT
vT = RTh = 25 Ω iT
io (∞) = 320/40 = 8 A τ=
80 × 10−3 = 2 ms; 40
1/τ = 500
io = 8 + (4 − 8)e−500t = 8 − 4e−500t A, P 7.40
t > 0;
t≥0
calculate vo (0+ )
va − vo (0+ ) va + = 20 × 10−3 15 5 .·. va = 0.75vo (0+ ) + 75 × 10−3 15 × 10−3 +
vo (0+ ) − va vo (0+ ) + − 9i∆ + 50 × 10−3 = 0 5 8
13vo (0+ ) − 8va − 360i∆ = −2600 × 10−3 i∆ =
vo (0+ ) − 9i∆ + 50 × 10−3 8
.·. i∆ =
vo (0+ ) + 5 × 10−3 80
.·. 360i∆ = 4.5vo (0+ ) + 1800 × 10−3
7–41
7–42
CHAPTER 7. Response of First-Order RL and RC Circuits 8va = 6vo (0+ ) + 600 × 10−3 .·. 13vo (0+ ) − 6vo (0+ ) − 600 × 10−3 − 4.5vo (0+ )− 1800 × 10−3 = −2600 × 10−3 2.5vo (0+ ) = −200 × 10−3 ;
vo (0+ ) = −80 mV
vo (∞) = 0 Find the Thévenin resistance seen by the 4 mH inductor:
iT =
vT vT + − 9i∆ 20 8
i∆ =
vT − 9i∆ 8
iT =
vT 10vT 9vT + − 20 80 80
.·. 10i∆ =
vT ; 8
i∆ =
vT 80
1 1 iT 1 5 + = = = S vT 20 80 80 16 .·. RTh = 16Ω τ=
4 × 10−3 = 0.25 ms; 16
1/τ = 4000
.·. vo = 0 + (−80 − 0)e−4000t = −80e−4000t mV,
t ≥ 0+
Problems P 7.41
[a]
1t v Vs + v dx = R L 0 R v 1 dv + =0 R dt L dv R + v=0 dt L [b]
dv R =− v dt L dv R dt = − v dt dt L dv R = − dt v L
.·.
v(t) v(0+ )
dy R t dx =− y L 0+
v(t)
ln y
v(0+ )
=−
R t L
v(t) R =− t ln + v(0 ) L −(R/L)t
v(t) = v(0 )e +
;
t>0
τ=
1 40
Vs − Io R = Vs − Io R v(0 ) = R
.·. v(t) = (Vs − Io R)e−(R/L)t P 7.42
+
7–43
7–44
CHAPTER 7. Response of First-Order RL and RC Circuits io = 5e−40t A,
t≥0
vo = 40io = 200e−40t V, 200e−40t = 100; .·. t = P 7.43
t > 0+
e40t = 2
1 ln 2 = 17.33 ms 40
1 1 [a] wdiss = Le i2 (0) = (1)(5)2 = 12.5 J 2 2 1 t (200)e−40x dx − 5 [b] i3H = 3 0 = 1.67(1 − e−40t ) − 5 = −1.67e−40t − 3.33 A i1.5H =
1 t (200)e−40x dx + 0 1.5 0
= −3.33e−40t + 3.33 A 1 wtrapped = (4.5)(3.33)2 = 25 J 2 1 [c] w(0) = (3)(5)2 = 37.5 J 2 P 7.44
[a] t < 0
t>0
iL (0− ) = iL (0+ ) = 25 mA;
τ=
24 × 10−3 = 0.2 ms; 120
1 = 5000 τ
iL (∞) = −50 mA iL = −50 + (25 + 50)e−5000t = −50 + 75e−5000t mA, vo = −120[75 × 10−3 e−5000t ] = −9e−5000t V,
t ≥ 0+
t≥0
Problems t 1 [b] i1 = −9e−5000x dx + 10 × 10−3 = (30e−5000t − 20) mA, −3 60 × 10 0 t 1 −9e−5000x dx + 15 × 10−3 = (45e−5000t − 30) mA, [c] i2 = 40 × 10−3 0
P 7.45
t≥0 t≥0
[a] Let v be the voltage drop across the parallel branches, positive at the top node, then v 1 t 1 t −Ig + + v dx + v dx = 0 R g L1 0 L2 0 1 1 t v + + v dx = Ig Rg L1 L2 0
v 1 t + v dx = Ig R g Le 0 v 1 dv + =0 Rg dt Le dv Rg + v=0 dt Le
Therefore v = Ig Rg e−t/τ ; τ = Le /Rg Thus 1 t Ig Rg e−x/τ t Ig Le i1 = Ig Rg e−x/τ dx = = (1 − e−t/τ ) L1 0 L1 (−1/τ ) 0 L1 i1 =
Ig L2 (1 − e−t/τ ) and L 1 + L2
[b] i1 (∞) = P 7.46
7–45
L2 Ig ; L1 + L2
i2 =
i2 (∞) =
Ig L1 (1 − e−t/τ ) L1 + L2
L1 Ig L1 + L2
For t < 0, i80mH (0) = 50 V/10 Ω = 5 A For t > 0, after making a Thévenin equivalent we have
Vs Vs −t/τ + Io − i= e R R
7–46
CHAPTER 7. Response of First-Order RL and RC Circuits R 8 1 = = = 80 τ L 100 × 10−3 Io = 5 A;
If =
−80 Vs = = −10 A R 8
i = −10 + (5 + 10)e−80t = −10 + 15e−80t A, vo = 0.08 P 7.47
t≥0
di = 0.08(−1200e−80t ) = −96e−80t V, dt
For t < 0
Simplify the circuit: 80/10,000 = 8 mA,
10 kΩ40 kΩ24 kΩ = 6 kΩ
8 mA − 3 mA = 5 mA 5 mA × 6 kΩ = 30 V Thus, for t < 0
.·. vo (0− ) = vo (0+ ) = 30 V t>0
t > 0+
Problems Simplify the circuit: 8 mA + 2 mA = 10 mA 10 k40 k24 k = 6 kΩ (10 mA)(6 kΩ) = 60 V Thus, for t > 0
vo (∞) = −60 V τ = RC = (10 k)(0.05 µ) = 0.5 ms;
1 = 2000 τ
vo = vo (∞) + [vo (0+ ) − vo (∞)]e−t/τ = −60 + [30 − (−60)]e−2000t = −60 + 90e−2000t V P 7.48
t≥0
[a] Simplify the circuit for t > 0 using source transformation:
Since there is no source connected to the capacitor for t < 0 vo (0− ) = vo (0+ ) = 0 V From the simplified circuit, vo (∞) = 60 V τ = RC = (20 × 103 )(0.5 × 10−6 ) = 10 ms
1/τ = 100
vo = vo (∞) + [vo (0+ ) − vo (∞)]e−t/τ = (60 − 60e−100t ) V,
t≥0
7–47
7–48
CHAPTER 7. Response of First-Order RL and RC Circuits [b] ic = C
dvo dt
ic = 0.5 × 10−6 (−100)(−60e−100t ) = 3e−100t mA v1 = 8000ic + vo = (8000)(3 × 10−3 )e−100t + (60 − 60e−100t ) = 60 − 36e−100t V io =
v1 = 1 − 0.6e−100t mA, 60 × 103
t ≥ 0+
[c] i1 (t) = io + ic = 1 + 2.4e−100t mA t ≥ 0+ v1 [d] i2 (t) = = 4 − 2.4e−100t mA t ≥ 0+ 3 15 × 10 + [e] i1 (0 ) = 1 + 2.4 = 3.4 mA At t = 0+ : Re = 15 k60 k8 k = 4800 Ω v1 (0+ ) = (5 × 10−3 )(4800) = 24 V i1 (0+ ) = P 7.49
v1 (0+ ) v1 (0+ ) + = 0.4 m + 3 m = 3.4mA 60,000 8000 −t/RC
[a] v = Is R + (Vo − Is R)e .·. Is R = 40,
(checks)
Vo −t/RC i = Is − e R
Vo − Is R = −24
.·. Vo = 16 V Is −
Vo = 3 × 10−3 ; R
Is −
.·. Is − 0.4Is = 3 × 10−3 ; R=
16 = 3 × 10−3 ; R
R=
40 Is
Is = 5 mA
40 × 103 = 8 kΩ 5
1 = 2500; RC
C=
1 10−3 = = 50 nF; 2500R 20 × 103
τ = RC =
[b] v(∞) = 40 V 1 w(∞) = (50 × 10−9 )(1600) = 40 µJ 2 0.81w(∞) = 32.4 µJ v 2 (to ) =
32.4 × 10−6 = 1296; 25 × 10−9
40 − 24e−2500to = 36;
v(to ) = 36 V
e2500to = 6;
.·. to = 716.70 µs
1 = 400 µs 2500
Problems P 7.50
[a] For t > 0:
τ = RC = 250 × 103 × 8 × 10−9 = 2 ms; vo = 50e−500t V, [b] io =
P 7.52
t ≥ 0+
t −1 −6 × 200 × 10 e−500x dx + 50 = 10e−500t + 40 V, 40 × 10−9 0
1 1 [a] w = Ceq vo2 = (8 × 10−9 )(502 ) = 10 µJ 2 2 1 [b] wtrapped = (40)2 (50 × 10−9 ) = 40 µJ 2 1 [c] w(0) = (40 × 10−9 )(502 ) = 50 µJ 2 For t > 0 VTh = (−25)(16,000)ib = −400 × 103 ib ib =
1 = 500 τ
50e−500t vo = = 200e−500t µA 250,000 250,000
v1 = P 7.51
7–49
33,000 (120 × 10−6 ) = 49.5 µA 80,000
VTh = −400 × 103 (49.5 × 10−6 ) = −19.8 V RTh = 16 kΩ
vo (∞) = −19.8 V;
vo (0+ ) = 0
τ = (16, 000)(0.25 × 10−6 ) = 4 ms;
1/τ = 250
t≥0
7–50
CHAPTER 7. Response of First-Order RL and RC Circuits vo = −19.8 + 19.8e−250t V,
t≥0
1 w(t) = (0.25 × 10−6 )vo2 = w(∞)(1 − e−250t )2 J 2 (1 − e−250t )2 =
0.36w(∞) = 0.36 w(∞)
1 − e−250t = 0.6 e−250t = 0.4 P 7.53
.·.
t = 3.67 ms
[a]
io (0+ ) =
−36 = −7.2 mA 5000
[b] io (∞) = 0 [c] τ = RC = (5000)(0.8 × 10−6 ) = 4 ms [d] io = 0 + (−7.2)e−250t = −7.2e−250t mA,
t ≥ 0+
[e] vo = −[36 + 1800(−7.2 × 10−3 e−250t )] = −36 + 12.96e−250t V, P 7.54
[a] vo (0− ) = vo (0+ ) = 120 V
vo (∞) = −150 V;
τ = 2 ms;
1 = 500 τ
vo = −150 + (120 − (−150))e−500t vo = −150 + 270e−500t V,
t≥0
[b] io = −0.04 × 10−6 (−500)[270e−500t ] = 5.4e−500t mA,
t ≥ 0+
t ≥ 0+
Problems
7–51
[c] vg = vo − 12.5 × 103 io = −150 + 202.5e−500t V [d] vg (0+ ) = −150 + 202.5 = 52.5 V Checks: vg (0+ ) = io (0+ )[37.5 × 103 ] − 150 = 202.5 − 150 = 52.5 V vg = −3 + 4.05e−500t mA i50k = 50k vg = −1 + 1.35e−500t mA i150k = 150k (ok) -io + i50k + i150k + 4 = 0 P 7.55
For t < 0, t > 0:
vo (0) = (−3 m)(15 k) = −45 V
VTh = −20 × 103 i∆ +
−75 10 (75) = −20 × 103 + 15 = 45 V 50 50 × 103
vT = −20 × 103 i∆ + 8 × 103 iT = −20 × 103 (0.2)iT + 8 × 103 iT = 4 × 103 iT RTh =
vT = 4 kΩ iT
t>0
vo = 45 + (−45 − 45)e−t/τ
7–52
CHAPTER 7. Response of First-Order RL and RC Circuits
1 τ = RC = (4000) × 10−6 = 250 µs; 16 vo = 45 − 90e−4000t V, P 7.56
vo (0) = 45 V;
1 = 4000 τ
t≥0
vo (∞) = −45 V
RTh = 20 kΩ τ = (20 × 103 )
1 × 10−6 = 1.25 × 10−3 ; 16
v = −45 + (45 + 45)e−800t = −45 + 90e−800t V, P 7.57
1 = 800 τ t≥0
t < 0; io (0− ) =
20 (10 × 10−3 ) = 2 mA; 100
vo (0− ) = (2 × 10−3 )(50,000) = 100 V
t = ∞:
−3
io (∞) = −5 × 10
20 = −1 mA; 100
RTh = 50 kΩ50 kΩ = 25 kΩ;
C = 16 nF 1 = 2500 τ
τ = (25,000)(16 × 10−9 ) = 0.4 ms; .·. vo (t) = −50 + 150e−2500t V,
ic = C
dvo = −6e−2500t mA, dt
vo (∞) = io (∞)(50,000) = −50 V
t≥0
t ≥ 0+
Problems i50k =
vo = −1 + 3e−2500t mA, 50,000
t ≥ 0+
io = ic + i50k = −(1 + 3e−2500t ) mA, P 7.58
t ≥ 0+
[a] Let i be the current inthe clockwise direction around the circuit. Then 1 t 1 t Vg = iRg + i dx + i dx C1 0 C2 0
1 1 t 1t iRg + + i dx = iRg + i dx C1 C2 Ce 0 0
=
Now differentiate the equation 0 = Rg
i di + dt Ce
Therefore i =
1 di + i=0 dt Rg Ce
Vg −t/τ Vg −t/Rg Ce e = e ; Rg Rg
τ = Rg Ce t
v1 (t) =
1 t Vg −x/τ Vg e−x/τ e dx = C1 0 Rg Rg C1 −1/τ
v1 (t) =
Vg C2 (1 − e−t/τ ); C1 + C2
τ = Rg Ce
v2 (t) =
Vg C1 (1 − e−t/τ ); C1 + C2
τ = Rg Ce
[b] v1 (∞) = P 7.59
or
C2 Vg ; C1 + C2
v2 (∞) =
[a]
1 t Is R = Ri + i dx + Vo C 0+ 0=R .·.
i di + +0 dt C
di i + =0 dt RC
0
=−
C1 Vg C1 + C2
Vg Ce −t/τ (e − 1) C1
7–53
7–54
CHAPTER 7. Response of First-Order RL and RC Circuits [b]
i di =− ; dt RC i(t) i(0+ )
ln
di dt =− i RC
1 t dy =− dx y RC 0+
−t i(t) = + i(0 ) RC
i(t) = i(0+ )e−t/RC ;
i(0+ ) =
Is R − Vo Vo = Is − R R
Vo −t/RC e .·. i(t) = Is − R P 7.60
[a] t < 0
t>0
vo (0− ) = vo (0+ ) = 40 V vo (∞) = 80 V τ = (0.16 × 10−6 )(6.25 × 103 ) = 1 ms; vo = 80 − 40e−1000t V, [b] io = −C
t≥0
dvo = −0.16 × 10−6 [40,000e−1000t ] dt
= −6.4e−1000t mA;
t ≥ 0+
1/τ = 1000
Problems t −1 [c] v1 = −6.4 × 10−3 e−1000x dx + 32 −6 0.2 × 10 0
= 64 − 32e−1000t V,
t≥0
t −1 −6.4 × 10−3 e−1000x dx + 8 [d] v2 = 0.8 × 10−6 0
= 16 − 8e−1000t V,
t≥0
1 1 [e] wtrapped = (0.2 × 10−6 )(64)2 + (0.8 × 10−6 )(16)2 = 512 µJ. 2 2 P 7.61
[a] vc (0+ ) = 50 V [b] Use voltage division to find the final value of voltage: 20 (−30) = −24 V 20 + 5 [c] Find the Thévenin equivalent with respect to the terminals of the capacitor: vc (∞) =
VTh = −24 V,
RTh = 205 = 4 Ω,
Therefore τ = Req C = 4(25 × 10−9 ) = 0.1 µs The simplified circuit for t > 0 is:
−24 − 50 = −18.5 A 4 [e] vc = vc (∞) + [vc (0+ ) − vc (∞)]e−t/τ [d] i(0+ ) =
7
= −24 + [50 − (−24)]e−t/τ = −24 + 74e−10 t V, [f] i = C P 7.62
dvc 7 7 = (25 × 10−9 )(−107 )(74e−10 t ) = −18.5e−10 t A, dt
[a] Use voltage division to find the initial value of the voltage: 9k (120) = 90 V 9k + 3k [b] Use Ohm’s law to find the final value of voltage: vc (0+ ) = v9k =
vc (∞) = v40k = −(1.5 × 10−3 )(40 × 103 ) = −60 V
t≥0 t ≥ 0+
7–55
7–56
CHAPTER 7. Response of First-Order RL and RC Circuits [c] Find the Thévenin equivalent with respect to the terminals of the capacitor: VTh = −60 V,
RTh = 10 k + 40 k = 50 kΩ
τ = RTh C = 1 ms = 1000 µs [d] vc = vc (∞) + [vc (0+ ) − vc (∞)]e−t/τ = −60 + (90 + 60)e−1000t = −60 + 150e−1000t V,
t≥0
We want vc = −60 + 150e−1000t = 0: Therefore t = P 7.63
ln(150/60) = 916.3 µs 1000
[a] For t < 0, calculate the Thévenin equivalent for the circuit to the left and right of the 400-mH inductor. We get
i(0− ) =
−60 − 200 = −13 mA 15 k + 5 k
i(0− ) = i(0+ ) = −13 mA [b] For t > 0, the circuit reduces to
Therefore i(∞) = −60/5,000 = −12 mA 400 × 10−3 L = = 80 µs R 5000 [d] i(t) = i(∞) + [i(0+ ) − i(∞)]e−t/τ [c] τ =
= −12 + [−13 + 12]e−12,500t = −12 − e−12,500t mA, P 7.64
[a] From Example 7.10, Leq = τ=
36 − 16 5 L1 L2 − M 2 = = H L1 + L2 − 2M 20 − 8 3
(5/3) 1 Leq = = R (50/3) 10
t≥0
Problems io =
100 100 −10t = 6 − 6e−10t A − e (50/3) (50/3)
t≥0
50 50 io = 100 − (6 − 6e−10t ) = 100e−10t V, 3 3 di2 di1 +4 [c] vo = 2 dt dt
[b] vo = 100 −
io = i1 + i2 di1 di2 dio = + dt dt dt dio di1 di1 di2 = − = 60e−10t − dt dt dt dt
.·. 100e
−10t
.·.
di1 di1 =2 + 4 60e−10t − dt dt
di1 = 70e−10t dt
di1 = 70e−10t dt i1 0
dx = 70
t 0
e−10y dy
e−10y t · = 7 − 7e−10t A, . . i1 = 70 −10 0
[d] i2
[e] vo
=
i o − i1
=
6 − 6e−10t − 7 + 7e−10t
=
−1 + e−10t A,
=
L2
=
18(−10e−10t ) + 4(70e−10t )
= Also, vo
t≥0
t≥0
di1 di2 +M dt dt
100e−10t V,
t ≥ 0+
(checks)
di2 di1 +M dt dt
=
L1
=
2(70e−10t ) + 4(−10e−10t )
=
100e−10t V,
t ≥ 0+
CHECKS
t ≥ 0+
7–57
7–58
CHAPTER 7. Response of First-Order RL and RC Circuits i1 (0) = 7 − 7 = 0; agrees with initial conditions; i2 (0) = −1 + 1 = 0; agrees with initial conditions; The final values of io , i1 , and i2 can be checked via the conservation of Wb-turns: io (∞)Leq = 6 × (5/3) = 10 Wb-turns i1 (∞)L1 + i2 (∞)M = 7(2) − 1(4) = 10 Wb-turns i2 (∞)L2 + i1 (∞)M = −1(18) + 7(4) = 10 Wb-turns Thus our solutions make sense in terms of known circuit behavior.
P 7.65
[a] Leq = τ=
(3)(15) = 2.5 H 3 + 15 Leq 1 2.5 = s = R 7.5 3
io (0) = 0;
io (∞) =
120 = 16 A 7.5
.·. io = 16 − 16e−3t A,
t≥0
vo = 120 − 7.5io = 120e−3t V,
t ≥ 0+
1 t 40 40 −3t i1 = 120e−3x dx = − e A, 3 0 3 3 i 2 = i o − i1 =
8 8 −3t − e A, 3 3
t≥0
t≥0
[b] io (0) = i1 (0) = i2 (0) = 0, consistent with initial conditions. vo (0+ ) = 120 V, consistent with io (0) = 0. vo = 3
di1 = 120e−3t V, dt
t ≥ 0+
or di2 = 120e−3t V, t ≥ 0+ dt The voltage solution is consistent with the current solutions. vo = 15
λ1 = 3i1 = 40 − 40e−3t Wb-turns λ2 = 15i2 = 40 − 40e−3t Wb-turns .·. λ1 = λ2 as it must, since vo =
dλ1 dλ2 = dt dt
λ1 (∞) = λ2 (∞) = 40 Wb-turns
Problems λ1 (∞) = 3i1 (∞) = 3(40/3) = 40 Wb-turns λ2 (∞) = 15i2 (∞) = 15(8/3) = 40 Wb-turns .·. i1 (∞) and i2 (∞) are consistent with λ1 (∞) and λ2 (∞). P 7.66
[a] From Example 7.10, Leq = τ=
50 − 25 L1 L2 − M 2 = = 1H L1 + L2 + 2M 15 + 10
L 1 = ; R 20
1 = 20 τ
.·. io (t) = 4 − 4e−20t A,
t≥0
[b] vo = 80 − 20io = 80 − 80 + 80e−20t = 80e−20t V, di2 di1 −5 = 80e−20t V [c] vo = 5 dt dt
t ≥ 0+
io = i1 + i2 di1 di2 dio = + = 80e−20t A/s dt dt dt di1 di2 = 80e−20t − dt dt
.·.
.·. 80e−20t = 5 .·. 10 t1 0
di1 di1 − 400e−20t + 5 dt dt
di1 = 480e−20t ; dt
dx =
t 0
di1 = 48e−20t dt
48e−20y dy
48 −20y t −20t e i1 = A, = 2.4 − 2.4e −20 0
t≥0
[d] i2 = io − i1 = 4 − 4e−20t − 2.4 + 2.4e−20t = 1.6 − 1.6e−20t A,
t≥0
[e] io (0) = i1 (0) = i2 (0) = 0, consistent with zero initial stored energy. vo = Leq
dio = 1(80)e−20t = 80e−20t V, dt
t ≥ 0+ (checks)
Also, vo = 5
di2 di1 −5 = 80e−20t V, dt dt
t ≥ 0+ (checks)
7–59
7–60
CHAPTER 7. Response of First-Order RL and RC Circuits vo = 10
di1 di2 −5 = 80e−20t V, dt dt
t ≥ 0+ (checks)
vo (0+ ) = 80 V, which agrees with io (0+ ) = 0 A io (∞) = 4 A;
io (∞)Leq = (4)(1) = 4 Wb-turns
i1 (∞)L1 + i2 (∞)M = (2.4)(5) + (1.6)(−5) = 4 Wb-turns (ok) i2 (∞)L2 + i1 (∞)M = (1.6)(10) + (2.4)(−5) = 4 Wb-turns (ok) Therefore, the final values of io , i1 , and i2 are consistent with conservation of flux linkage. Hence, the answers make sense in terms of known circuit behavior. P 7.67
[a] Leq = 5 + 10 − 2.5(2) = 10 H τ=
10 1 L = = ; R 40 4
i = 2 − 2e−4t A,
1 =4 τ t≥0
di di di1 − 2.5 = 2.5 = 2.5(8e−4t ) = 20e−4t V, t ≥ 0+ dt dt dt di di di1 − 2.5 = 7.5 = 7.5(8e−4t ) = 60e−4t V, t ≥ 0+ [c] v2 (t) = 10 dt dt dt [d] i(0) = 2 − 2 = 0, which agrees with initial conditions.
[b] v1 (t) = 5
80 = 40i1 + v1 + v2 = 40(2 − 2e−4t ) + 20e−4t + 60e−4t = 80 V Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, the answers make sense in terms of known circuit behavior. P 7.68
[a] Leq = 5 + 10 + 2.5(2) = 20 H τ=
20 1 L = = ; R 40 2
i = 2 − 2e−2t A,
1 =2 τ t≥0
di di di1 + 2.5 = 7.5 = 7.5(4e−2t ) = 30e−2t V, t ≥ 0+ dt dt dt di di di1 + 2.5 = 12.5 = 12.5(4e−2t ) = 50e−2t V, t ≥ 0+ [c] v2 (t) = 10 dt dt dt [d] i(0) = 0, which agrees with initial conditions.
[b] v1 (t) = 5
80 = 40i1 + v1 + v2 = 40(2 − 2e−2t ) + 30e−2t + 50e−2t = 80 V Therefore, Kirchhoff’s voltage law is satisfied for all values of t ≥ 0. Thus, the answers make sense in terms of known circuit behavior.
Problems P 7.69
Use voltage division to find the initial voltage: vo (0) =
60 (50) = 30 V 40 + 60
Use Ohm’s law to find the final value of voltage: vo (∞) = (−5 mA)(20 kΩ) = −100 V τ = RC = (20 × 103 )(250 × 10−9 ) = 5 ms;
1 = 200 τ
vo = vo (∞) + [vo (0+ ) − vo (∞)]e−t/τ = −100 + (30 + 100)e−200t = −100 + 130e−200t V, P 7.70
[a] t < 0:
Using Ohm’s law, ig =
800 = 12.5 A 40 + 6040
Using current division, 60 (12.5) = 7.5 A = i(0+ ) 60 + 40 [b] 0 ≤ t ≤ 1 ms: i(0− ) =
i = i(0+ )e−t/τ = 7.5e−t/τ 1 R 40 + 12060 = 1000 = = τ L 80 × 10−3 i = 7.5e−1000t i(200µs) = 7.5e−10
3 (200×10−6 )
= 7.5e−0.2 = 6.14 A
t≥0
7–61
7–62
CHAPTER 7. Response of First-Order RL and RC Circuits [c] i(1ms) = 7.5e−1 = 2.7591 A 1 ms ≤ t < ∞
R 40 1 = = = 500 τ L 80 × 10−3 i = i(1 ms)e−(t−1 ms)/τ = 2.7591e−500(t−0.001) A i(6ms) = 2.7591e−500(0.005) = 2.7591e−2.5 = 226.48 mA [d] 0 ≤ t ≤ 1 ms: i = 7.5e−1000t v=L
di = (80 × 10−3 )(−1000)(7.5e−1000t ) = −600e−1000t V dt
v(1− ms) = −600e−1 = −220.73 V [e] 1 ms ≤ t ≤ ∞: i = 2.7591e−500(t−0.001) v=L
di = (80 × 10−3 )(−500)(2.591e−500(t−0.001) ) dt
= −110.4e−500(t−0.001) V v(1+ ms) = −110.4 V P 7.71
Note that for t > 0, vo = (4/6)vc , where vc is the voltage across the 0.5 µF capacitor. Thus we will find vc first. t 20+ ms
vo (25ms) = −8.26e−100(0.025−0.02) = −5 V P 7.74
From the solution to Problem 7.73, the initial energy is 1 w(0) = (50 mH)(10 A)2 = 2.5 J 2 0.04w(0) = 0.1 J .·.
1 (50 × 10−3 )i2L = 0.1 so 2
iL = 2 A
Again, from the solution to Problem 7.73, t must be between 10 ms and 20 ms since i(10 ms) = 3.68 A
and
i(20 ms) = 1.65 A
For 10 ms ≤ t ≤ 20 ms: i = 3.68e−80(t−0.01) = 2 e80(t−0.01) = P 7.75
3.68 2
so
t − 0.01 = 0.0076
.·.
t = 17.6 ms
0 ≤ t ≤ 10 µs:
τ = RC = (4 × 103 )(20 × 10−9 ) = 80 µs;
1/τ = 12,500
Problems vo (0) = 0 V;
vo (∞) = −20 V
vo = −20 + 20e−12,500t V
0 ≤ t ≤ 10 µs
10 µs ≤ t ≤ ∞:
t = ∞:
i=
−50 V = −2.5 mA 20 kΩ
vo (∞) = (−2.5 × 10−3 )(16,000) + 30 = −10 V vo (10 µs) = −20 + 20−0.125 = −2.35 V vo = −10 + (−2.35 + 10)e−(t − 10×10
−6 )/τ
RTh = 4 kΩ16 kΩ = 3.2 kΩ τ = (3200)(20 × 10−9 ) = 64 µs; vo = −10 + 7.65e−15,625(t − 10×10 P 7.76
1/τ = 15,625
−6 )
10 µs ≤ t ≤ ∞
0 ≤ t ≤ 200 µs;
Re = 150100 = 60 kΩ;
10 × 10−9 (60,000) = 200 µs τ= 3
7–67
7–68
CHAPTER 7. Response of First-Order RL and RC Circuits vc = 300e−5000t V vc (200 µs) = 300e−1 = 110.36 V 200 µs ≤ t ≤ ∞:
Re = 3060 + 12040 = 20 + 30 = 50 kΩ τ=
10 × 10−9 (50,000) = 166.67 µs; 3
1 = 6000 τ
vc = 110.36e−6000(t − 200 µs) V vc (300 µs) = 110.36e−6000(100 µs) = 60.57 V io (300 µs) = i1 =
60.57 = 1.21 mA 50,000
60 2 io = io ; 90 3
i2 =
40 1 i o = io 160 4
2 1 5 5 isw = i1 − i2 = io − io = io = (1.21 × 10−3 ) = 0.50 mA 3 4 12 12 P 7.77
t < 0:
vc (0− ) = (20 × 10−3 )(500) = 10 V = vc (0+ )
Problems 0 ≤ t ≤ 50 ms:
τ = ∞;
1/τ = 0;
vo = 10e−0 = 10 V
50 ms ≤ t ≤ ∞:
τ = (6.25 k)(0.16 µ) = 1 ms;
1/τ = 1000;
Summary: vo = 10 V,
0 ≤ t ≤ 50 ms
vo = 10e−1000(t − 0.05) V, P 7.78
50 ms ≤ t ≤ ∞
t < 0:
iL (0− ) = 10 V/5 Ω = 2 A = iL (0+ )
vo = 10e−1000(t − 0.05) V
7–69
7–70
CHAPTER 7. Response of First-Order RL and RC Circuits 0 ≤ t ≤ 5:
τ = 5/0 = ∞ iL (t) = 2e−t/∞ = 2e−0 = 2 iL (t) = 2 A,
0≤t≤5s
5 ≤ t ≤ ∞:
τ=
5 = 5 s; 1
1/τ = 0.2
iL (t) = 2e−0.2(t −5) A, P 7.79
t≥5s
[a] 0 ≤ t ≤ 2.5 ms vo (0+ ) = 80 V;
vo (∞) = 0
L = 2 ms; R
1/τ = 500
τ=
vo (t) = 80e−500t V,
0+ ≤ t ≤ 2.5 ms
vo (2.5− ms) = 80e−1.25 = 22.92 V io (2.5− ms) =
(80 − 22.92) = 2.85 A 20
vo (2.5+ ms) = −20(2.85) = −57.08 V vo (∞) = 0;
τ = 2 ms;
vo = −57.08e−500(t − 0.0025) V
1/τ = 500 2.5+ ms ≤ t ≤ ∞
Problems [b]
[c] vo (5 ms) = −16.35 V io = P 7.80
+16.35 = 817.68 mA 20
[a] io (0) = 0;
io (∞) = 25 mA
R 2000 1 = = × 103 = 8000 τ L 250 io = (25 − 25e−8000t ) mA, vo = 0.25
0 ≤ t ≤ 75 µs
dio = 50e−8000t V, dt
0+ ≤ t ≤ 75− µs
75+ µs ≤ t ≤ ∞: io (75µs) = 25 − 25e−0.6 = 11.28 mA; io = 11.28e−8000(t−75×10 vo = (0.25)
−6 )
io (∞) = 0
mA
dio = −22.56e−8000(t−75µs) dt
.·. t < 0 :
vo
=
0
0+ ≤ t ≤ 75− µs :
vo
=
50e−8000t V
75+ µs ≤ t ≤ ∞ :
vo
=
−22.56e−8000(t−75µs)
[b] vo (75− µs) = 50e−0.6 = 27.44 V vo (75+ µs) = −22.56 V [c] io (75− µs) = io (75+ µs) = 11.28 mA
7–71
7–72 P 7.81
CHAPTER 7. Response of First-Order RL and RC Circuits [a] 0 ≤ t < 1 ms: vc (0+ ) = 0;
vc (∞) = 50 V;
RC = 400 × 103 (0.01 × 10−6 ) = 4 ms;
1/RC = 250
vc = 50 − 50e−250t vo = 50 − 50 + 50e−250t = 50e−250t V,
0 ≤ t ≤ 1 ms
1 ms < t ≤ ∞: vc (1 ms) = 50 − 50e−0.25 = 11.06 V vc (∞) = 0 V τ = 4 ms;
1/τ = 250
vc = 11.06e−250(t − 0.001) V vo = −vc = −11.06e−250(t − 0.001) V,
1 ms < t ≤ ∞
[b]
P 7.82
[a] t < 0; vo = 0 0 ≤ t ≤ 4 ms: τ = (200 × 103 )(0.025 × 10−6 ) = 5 ms; vo = 100 − 100e−200t V,
0 ≤ t ≤ 4 ms
vo (4 ms) = 100(1 − e−0.8 ) = 55.07 V 4 ms ≤ t ≤ 8 ms: vo = −100 + 155.07e−200(t−0.004) V vo (8 ms) = −100 + 155.07e−0.8 = −30.32 V 8 ms ≤ t ≤ ∞: vo = −30.32e−200(t−0.008) V
1/τ = 200
Problems [b]
[c] t ≤ 0 : vo = 0 0 ≤ t ≤ 4 ms: τ = (50 × 103 )(0.025 × 10−6 ) = 1.25 ms vo = 100 − 100e−800t V,
1/τ = 800
0 ≤ t ≤ 4 ms
vo (4 ms) = 100 − 100e−3.2 = 95.92 V 4 ms ≤ t ≤ 8 ms: vo = −100 + 195.92e−800(t−0.004) V,
4 ms ≤ t ≤ 8 ms
vo (8 ms) = −100 + 195.92e−3.2 = −92.01 V 8 ms ≤ t ≤ ∞: vo = −92.01e−800(t−0.008) V,
8 ms ≤ t ≤ ∞
7–73
7–74 P 7.83
CHAPTER 7. Response of First-Order RL and RC Circuits [a] τ = RC = (20,000)(0.2 × 10−6 ) = 4 ms; io = v o = 0 io (0+ ) = 20
1/τ = 250
t0
−160 × 10−3 +
20 + iC (0+ ) + 0 = 0; 125
.·. iC (0+ ) = 0
1 1 = 800 rad/s = 2RC 2(125)(5 × 10−6 ) ωo2 =
1 1 = = 64 × 104 −3 LC (312.5 × 10 )(5 × 10−6 )
.·. α2 = ωo2
critically damped
Problems [a] vo = Vf + D1 te−800t + D2 e−800t Vf = 0 dvo (0) = −800D2 + D1 = 0 dt vo (0+ ) = 20 = D2 D1 = 800D2 = 16,000 V/s .·. vo = 16,000te−800t + 20e−800t V,
t ≥ 0+
[b] iL = If + D3 te−800t + D4 e−800t iL (0+ ) = 0;
20 diL (0+ ) = = 64 A/s dt 312.5 × 10−3
If = 160 mA;
.·. 0 = 160 + D4 ;
D4 = −160 mA;
−800D4 + D3 = 64;
D3 = −64 A/s
.·. iL = 160 − 64,000te−800t − 160e−800t mA P 8.33
[a] wL =
∞ 0
pdt =
∞ 0
t≥0
vo iL dt
vo = 16,000te−800t + 20e−800t V iL = 0.16 − 64te−800t − 0.16e−800t A p = 3.2e−800t + 2560te−800t − 3840te−1600t −1,024,000t2 e−1600t − 3.2e−1600t W ∞
wL = 3.2
−800t
e
0
∞
dt + 2560
∞
−1,024,000
2 −1600t
te
0
−800t
te
0
∞
dt − 3480
∞
dt − 3.2
∞
e−1600t dt
0
∞
2560 −800t e−800t + e (−800t − 1) = 3.2 2 −800 0 (800) 0 ∞
3840 −1600t − e (−1600t − 1) 2 (1600) 0
∞
1,024,000 −1600t 2 2 − e (1600 t + 3200t + 2) (−1600)3 0 ∞
e−1600t − 3.2 (−1600) 0
0
te−1600t dt
8–25
8–26
CHAPTER 8. Natural and Step Responses of RLC Circuits All the upper limits evaluate to zero hence 3.2 3840 (1,024,000)(2) 3.2 2560 − − − + = 4 mJ 2 2 3 800 800 1600 1600 1600 Note this value corresponds to the final energy stored in the inductor, i.e.
wL =
1 wL (∞) = (312.5 × 10−3 )(0.16)2 = 4 mJ. 2 [b] v = 16,000te−800t + 20e−800t V v = 128te−800t + 0.16e−800t A 125
iR =
pR = viR = 2,048,000t2 e−1600t + 5120te−1600t + 3.2e−1600t wR =
∞ 0
pR dt ∞
= 2,048,000 =
0
∞
t2 e−1600t dt + 5120
0
∞
te−1600t dt + 3.2
2,048,000e−1600t [16002 t2 + 3200t + 2] −16003
∞ 0
e−1600t dt
0
+
∞ 3.2e−1600t ∞ 5120e−1600t (−1600t − 1) + 16002 (−1600) 0 0
Since all the upper limits evaluate to zero we have 2,048,000(2) 5120 3.2 = 5 mJ + + 16003 16002 1600 [c] 160 = iR + iC + iL (mA) wR =
iR + iL = 160 + 64,000te−800t mA .·. iC = 160 − (iR + iL ) = −64,000te−800t mA = −64te−800t A pC = viC = [16,000te−800t + 20e−800t ][−64te−800t ] = −1,024,000t2 e−1600t − 1280e−1600t wC = −1,024,000
∞ 0
t2 e−1600t dt − 1280
∞
te−1600t dt
0
∞ 1280e−1600t ∞ −1,024,000e−1600t 2 2 − [1600 t + 3200t + 2] (−1600t − 1) wC = −16003 16002 0 0
Since all upper limits evaluate to zero we have wC =
−1,024,000(2) 1280(1) − = −1 mJ 16003 16002
Problems
8–27
Note this 1 mJ corresponds to the initial energy stored in the capacitor, i.e., 1 wC (0) = (5 × 10−6 )(20)2 = 1 mJ. 2 Thus wC (∞) = 0 mJ which agrees with the final value of v = 0. [d] is = 160 mA ps (del) = 160v mW = 0.16[16,000te−800t + 20e−800t ] = 3.2e−800t + 2560te−800t W ∞
ws = 3.2
0
e−800t dt +
∞ 0
2560te−800t dt
∞ 3.2e−800t ∞ 2560e−800t (−800t − 1) = + −800 0 8002 0
= [e] wL = 4 mJ
(absorbed)
wR = 5 mJ
(absorbed)
wC = 1 mJ
(delivered)
wS = 8 mJ
(delivered)
P 8.34
2560 3.2 + = 8 mJ 800 800
wdel = wabs = 9 mJ.
vC (0+ ) =
3.75 × 103 (150) = 50 V 11.25 × 103
iL (0+ ) = 100 mA; α=
iL (∞) =
150 = 20 mA 7500
1 1 = = 800 2RC 2(2500)(0.25 × 10−6 )
ωo2 =
1 1 = = 106 LC (4)(0.25 × 10−6 )
α2 = 64 × 104 ;
α2 < ωo2 ;
.·.
underdamped
√ s1,2 = −800 ± j 8002 − 106 = −800 ± j600 rad/s
8–28
CHAPTER 8. Natural and Step Responses of RLC Circuits iL
=
If + B1 e−αt cos ωd t + B2 e−αt sin ωd t
=
20 + B1 e−800t cos 600t + B2 e−800t sin 600t
iL (0) = 20 × 10−3 + B1 ;
B1 = 100 m − 20 m = 80 mA
50 diL (0) = 600B2 − 800B1 = = 12.5 dt 4 .·. 600B2 = 800(80 × 10−3 ) + 12.5;
B2 = 127.5 mA
.·. iL = 20 + 80e−800t cos 600t + 127.5e−800t sin 600t mA, P 8.35
[a] 2α = 5000;
α = 2500 rad/s
α2 − ωo2 = 1500;
α=
R = 2500; 2L
ωo2 =
1 = 4 × 106 ; LC
ωo2 = 4 × 106 ;
ωo = 2000 rad/s
R = 5000L L=
109 = 5H 4 × 106 (50)
R = 25,000 Ω [b] i(0) = 0 L
t≥0
di(0) = vc (0); dt
.·. vc2 (0) = 3600;
1 (50) × 10−9 vc2 (0) = 90 × 10−6 2 vc (0) = 60 V
60 di(0) = = 12 A/s dt 5 [c] i(t) = A1 e−1000t + A2 e−4000t i(0) = A1 + A2 = 0 di(0) = −1000A1 − 4000A2 = 12 dt Solving, .·. A1 = 4 mA;
A2 = −4 mA
i(t) = 4e−1000t − 4e−4000t mA
t≥0
Problems [d]
di(t) = −4e−1000t + 16e−4000t dt di = 0 when 16e−4000t = 4e−1000t dt or e3000t = 4 .·. t =
ln 4 µs = 462.10 µs 3000
[e] imax = 4e−0.4621 − 4e−1.8484 = 1.89 mA di [f] vL (t) = 5 = [−20e−1000t + 80e−4000t ] V, dt P 8.36
α = 2000 rad/s;
ωd = 1500 rad/s
ωo2 − α2 = 225 × 104 ; α=
R = 2000; 2L
t ≥ 0+
ωo2 = 625 × 104 ;
wo = 2500 rad/s
R = 4000L
1 = 625 × 104 ; LC
L=
1 (625 ×
104 )(80
× 10−9 )
= 2H
.·. R = 8 kΩ i(0+ ) = B1 = 7.5 mA;
at t = 0+
60 + vL (0+ ) − 30 = 0;
.·.
vL (0+ ) = −30 V
di(0+ ) −30 = = −15 A/s dt 2 di(0+ ) · = 1500B2 − 2000B1 = −15 .. dt .·. 1500B2 = 2000(7.5 × 10−3 ) − 15; .·. i = 7.5e−2000t sin 1500t mA,
t≥0
.·. B2 = 0 A
8–29
8–30 P 8.37
CHAPTER 8. Natural and Step Responses of RLC Circuits From Prob. 8.36 we know vc will be of the form vc = B3 e−2000t cos 1500t + B4 e−2000t sin 1500t From Prob. 8.36 we have vc (0) = −30 V = B3 and iC (0) dvc (0) 7.5 × 10−3 = = 93.75 × 103 = dt C 80 × 10−9 dvc (0) = 1500B4 − 2000B3 = 93,750 dt .·. 1500B4 = 2000(−30) + 93,750;
B4 = 22.5 V
vc (t) = −30e−2000t cos 1500t + 22.5e−2000t sin 1500t V P 8.38
[a] ωo2 = α=
t≥0
1 1 = = 25 × 106 −3 LC (80 × 10 )(0.5 × 10−6 ) R = ωo = 5000 rad/s 2L
.·. R = (5000)(2)L = 800 Ω [b] i(0) = iL (0) = 30 mA vc (0) = 800i(0) + 80 × 10−3
di(0) dt
20 − 800(30 × 10−3 ) di(0) = 80 × 10−3 dt .·.
di(0) = −50 A/s dt
[c] vC = D1 te−5000t + D2 e−5000t vC (0) = D2 = 20 V −iL (0) iC (0) dvC (0) = D1 − 5000D2 = = dt C C D1 − 100,000 = −
30 × 10−3 = −60,000 0.5 × 10−6
vC = 40,000te−5000t + 20e−5000t V,
.·.
t≥0
D1 = 40,000 V/s
Problems P 8.39
[a] For t > 0:
Since i(0− ) = i(0+ ) = 0 va (0+ ) = 72 V [b] va = 5000i +
t 1 i dx + 72 0.1 × 10−6 0
di dva = 5000 + 10 × 106 i dt dt di(0+ ) di(0+ ) dva (0+ ) = 5000 + 10 × 106 i(0+ ) = 5000 dt dt dt −L
di(0+ ) = 72 dt
72 di(0+ ) =− = −28.8 A/s dt 2.5 dva (0+ ) · = −144,000 V/s .. dt [c] α =
R 12,500 = = 2500 rad/s 2L 2(2.5) 1 1 = = 4 × 106 LC (2.5)(0.1 × 10−6 ) √ = −2500 ± 25002 − 4 × 106 = −2500 ± 1500 rad/s
ωo2 = s1,2
Overdamped: va = A1 e−1000t + A2 e−4000t va (0) = 72 = A1 + A2 dva (0) = −144,000 = −1000A1 − 4000A2 dt Solving,
A1 = 48;
A2 = 24
va = 48e−1000t + 24e−4000t V,
t ≥ 0+
8–31
8–32 P 8.40
CHAPTER 8. Natural and Step Responses of RLC Circuits ωo2 =
α=
1 1 = 25 = LC (10)(4 × 10−3 )
R 80 = = 4; 2L 2(10)
α2 < ωo2
.·.
α2 = 16
underdamped
√ s1,2 = −4 ± j 9 = −4 ± j3 rad/s i = B1 e−4t cos 3t + B2 e−4t sin 3t i(0) = B1 = −240/100 = −2.4 A di (0) = 3B2 − 4B1 = 0 dt .·. B2 = −3.2 A i = −2.4e−4t cos 3t − 3.2 sin 3t A, P 8.41
t≥0
t < 0:
i(0) =
240 240 = = 6A 8 + 3070 + 11 40
vo (0) = 240 − 8(6) −
70 (6)(20) = 108 V 100
Problems t > 0:
α=
20 R = = 10, 2L 2(1)
ωo2 =
α2 = 100
1 1 = 200 = LC (1)(5 × 10−3 )
ωo2 > α2
underdamped
s1,2 = −100 ±
√
100 − 200 = −10 ± j10 rad/s
vo = B1 e−10t cos 10t + B2 e−10t sin 10t vo (0) = B1 = 108 V C
dvo −6 = = −1200 V/s dt 5 × 10−3
dvo (0) = −6, dt
dvo (0) = −10B1 + 10B2 = −1200 dt 10B2 = −1200 + 10B1 = −1200 + 1080; .·. vo = 108e−10t cos 10t − 12e−10t sin 10t V, P 8.42
B2 = −120/10 = −12 V t≥0
[a] t < 0: 80 = 100 mA; vo = 500io = (500)(0.01) = 50 V 800 t > 0: R 500 α= = = 105 rad/s −3 2L 2(2.5 × 10 ) io =
ωo2 =
1 1 = = 100 × 108 LC (2.5 × 10−3 )(40 × 10−9 )
α2 = ωo2
.·.
critically damped
8–33
8–34
CHAPTER 8. Natural and Step Responses of RLC Circuits .·. io (t) = D1 te−10 t + D2 e−10 5
5t
io (0) = D2 = 100 mA dio (0) = −αD2 + D1 = 0 dt .·.
D1 = 105 (100 × 10−3 ) = 10,000 5
5
io (t) = 10,000te−10 t + 0.1e−10 t A, 5
[b] vo (t) = D3 te−10 t + D4 e−10
t≥0
5t
vo (0) = D4 = 50 C
dvo (0) = −0.1 dt
dvo −0.1 = −25 × 105 V/s = −αD4 + D3 (0) = dt 40 × 10−9 .·.
D3 = 105 (50) − 25 × 105 = 25 × 105 5
5
vo (t) = 25 × 105 te−10 t + 50e−10 t V, P 8.43
α=
t≥0
8000 R = = 4000 rad/s 2L 2(1)
ωo2 =
1 1 = = 20 × 106 LC (1)(50 × 10−9 )
s1,2 = −4000 ±
√
40002 − 20 × 106 = −4000 ± j2000 rad/s
vo = Vf + B1 e−4000t cos 2000t + B2 e−4000t sin 2000t vo (0) = 0 = Vf + B1 vo (∞) = 80 V;
.·. B1 = −80 V
dvo (0) = 0 = 2000B2 − 4000B1 dt .·. 2000B2 = 4000(−80)
.·.
B2 = −160 V
vo = 80 − 80e−4000t cos 2000t − 160e−4000t sin 2000t V,
t≥0
Problems P 8.44
t < 0:
iL (0) =
−150 = −5 A 30
vC (0) = 18iL (0) = −90 V t > 0:
α=
10 R = = 50 rad/s 2L 2(0.1)
ωo2 =
1 1 = = 5000 LC (0.1)(2 × 10−3 )
ωo > α2
.·.
s1,2 = −50 ±
underdamped √
502 − 5000 = −50 ± j50
vc = 60 + B1 e−50t cos 50t + B2 e−50t sin 50t vc (0) = −90 = 60 + B1 C
dvc (0) = −5; dt
.·.
B1 = −150
−5 dvc (0) = = −2500 dt 2 × 10−3
dvc (0) = −50B1 + 50B2 = −2500 dt
.·.
B2 = −200
vc = 60 − 150e−50t cos 50t − 200e−50t sin 50t V,
t≥0
8–35
8–36 P 8.45
CHAPTER 8. Natural and Step Responses of RLC Circuits iC (0) = 0; α=
vo (0) = 50 V
8000 R = = 25,000 rad/s 2L 2(160 × 10−3 )
ωo2 =
1 1 = = 625 × 106 −3 LC (160 × 10 )(10 × 10−9 )
.·. α2 = ωo2 ;
critical damping
vo (t) = Vf + D1 te−25,000t + D2 e−25,000t Vf = 250 V vo (0) = 250 + D2 = 50;
D2 = −200 V
dvo (0) = −25,000D2 + D1 = 0 dt D1 = 25,000D2 − 5 × 106 V/s vo = 250 − 5 × 106 te−25,000t − 200e−25,000t V, P 8.46
t≥0
[a] t < 0:
io (0− ) =
120 = 4 mA 30,000
vC (0− ) = 80 − (10,000)(0.004) = 40 V t = 0+ :
5 kΩ20 kΩ = 4 kΩ .·. vo (0+ ) = −(0.004)(4000) + 40 = 40 − 16 = 24 V
Problems [b] vo (t) = vc − 4000io dvo + dvc + dio (0 ) = (0 ) − 4000 (0+ ) dt dt dt dvc + −4 × 10−3 (0 ) = = −2048 V/s dt (125/64) × 10−6 −vL (0+ ) + vo (0+ ) + 40 = 0 vL = 64 V 64 dio + = 12.8 A/s (0 ) = dt 5 dvo + (0 ) = −2048 − 4000(12.8) = −53,248 V/s dt [c] ωo2 = α=
1 1 = = 10.24 × 104 LC (5)[(125/64) × 10−6 ] 4000 R = = 400 rad/s; 2L 2(5)
α2 > ωo2
α2 = 16 × 104
overdamped
s1,2 = −400 ± 240 rad/s vo (t) = Vf + A1 e−160t + A2 e−640t Vf = vo (∞) = −40 V −40 + A1 + A2 = 24 −160A1 − 640A2 = −53,248 Solving,
A1 = −25.6;
A2 = 89.6
.·. vo (t) = −40 − 25.6e−160t + 89.6e−640t V, P 8.47
t ≥ 0+
[a] vc = Vf + [B1 cos ωd t + B2 sin ωd t] e−αt dvc = [(ωd B2 − αB1 ) cos ωd t − (αB2 + ωd B1 ) sin ωd t]e−αt dt Since the initial stored energy is zero, vc (0 ) = 0 and +
It follows that
dvc (0+ ) =0 dt
B1 = −Vf
and
B2 =
αB1 ωd
8–37
8–38
CHAPTER 8. Natural and Step Responses of RLC Circuits When these values are substituted into the expression for [dvc /dt], we get dvc = dt But
α2 + ωd Vf e−αt sin ωd t ωd
Vf = V
dvc = dt
Therefore [b]
α2 α2 + ωd2 ω2 + ωd = = o ωd ωd ωd
and
dvc = 0 when dt
ωo2 V e−αt sin ωd t ωd
sin ωd t = 0,
ωd t = nπ
or
where n = 0, 1, 2, 3, . . . Therefore t = [c] When
tn = and
nπ ωd
nπ , ωd
cos ωd tn = cos nπ = (−1)n
sin ωd t = sin nπ = 0
Therefore vc (tn ) = V [1 − (−1)n e−αnπ/ωd ] [d] It follows from [c] that v(t1 ) = V + V e−(απ/ωd ) Therefore But
e−(απ/ωd ) vc (t1 ) − V = −(3απ/ω ) = e(2απ/ωd ) d vc (t3 ) − V e
2π = t3 − t1 = Td , ωd
P 8.48
vc (t1 ) − V 1 ln α= Td vc (t3 ) − V α=
vc (t3 ) = V + V e−(3απ/ωd )
and
thus
α=
1 [vc (t1 ) − V ] ln Td [vc (t3 ) − V ]
;
7000 63.84 ln ≈ 1000; 2π 26.02
Td = t3 − t1 =
ωd =
2π 3π π − = ms 7 7 7
2π = 7000 rad/s Td
ωo2 = ωd2 + α2 = 49 × 106 + 106 = 50 × 106 L=
1 = 200 mH; (50 × 106 )(0.1 × 10−6 )
R = 2αL = 400 Ω
Problems P 8.49
[a] Let i be the current in the direction of the voltage drop vo (t). Then by hypothesis i = if + B1 e−αt cos ωd t + B2 e−αt sin ωd t if = i(∞) = 0,
i(0) =
Vg = B1 R
Therefore i = B1 e−αt cos ωd t + B2 e−αt sin ωd t L
di(0) = 0, dt
therefore
di(0) =0 dt
di = [(ωd B2 − αB1 ) cos ωd t − (αB2 + ωd B1 ) sin ωd t] e−αt dt Therefore ωd B2 − αB1 = 0; Therefore
B2 =
α α Vg B1 = ωd ωd R
α2 Vg ωd Vg di vo = L = − L + sin ωd t e−αt dt ωd R R
LVg =− R Vg L =− R vo = − [b]
α2 + ωd sin ωd t e−αt ωd
α2 + ωd2 −αt e sin ωd t ωd
Vg −αt e sin ωd t V, RCωd
t ≥ 0+
Vg dvo =− {ωd cos ωd t − α sin ωd t}e−αt dt ωd RC dvo = 0 when dt
tan ωd t =
ωd α
Therefore ωd t = tan−1 (ωd /α)
ωd 1 tan−1 t= ωd α P 8.50
8–39
(smallest t)
[a] From Problem 8.49 we have vo =
−Vg −αt e sin ωd t RCωd
α=
R 4800 = 37,500 rad/s = 2L 2(64 × 10−3 )
ωo2 =
1 1 = = 3906.25 × 106 LC (64 × 10−3 )(4 × 10−9 )
8–40
CHAPTER 8. Natural and Step Responses of RLC Circuits ωd =
ωo2 − α2 = 50 krad/s
−Vg −(−72) = 75 = RCωd (4800)(4 × 10−9 )(50 × 103 ) .·. vo = 75e−37,500t sin 50,000t V, [b] From Problem 8.49
ωd 1 tan−1 td = ωd α
t≥0
50,000 1 tan−1 = 50,000 37,500
td = 18.55 µs [c] vmax = 75e−0.0375(18.55) sin[(0.05)(18.55)] = 29.93 V [d] R = 480 Ω;
α = 3750 rad/s
ωd = 62,387.4 rad/s vo = 601.08e−3750t sin 62,387.4t V,
t≥0
td = 24.22 µs vmax = 547.92 V P 8.51
[a]
1 d2 vo = vg dt2 R1 C1 R2 C2 1 10−6 = = 250 R1 C1 R2 C2 (100)(400)(0.5)(0.2) × 10−6 × 10−6 d2 vo = 250vg .·. dt2 0 ≤ t ≤ 0.5− : vg = 80 mV d2 vo = 20 dt2 Let g(t) g(0)
g(t) =
dvo , dt
dx = 20
dg = 20 or dt
t
dy 0
g(t) − g(0) = 20t, g(t) =
then
dvo = 20t dt
g(0) =
dvo (0) = 0 dt
dg = 20 dt
Problems dvo = 20t dt vo (t) vo (0)
dx = 20
t
vo (t) − vo (0) = 10t2 ,
x dx; 0
vo (t) = 10t2 V,
0 ≤ t ≤ 0.5−
1 dvo1 =− vg = −20vg = −1.6 dt R1 C1 dvo1 = −1.6 dt vo1 (t) vo1 (0)
dx = −1.6
t
dy 0
vo1 (t) − vo1 (0) = −1.6t,
vo1 (0) = 0
0 ≤ t ≤ 0.5−
vo1 (t) = −1.6t V, 0.5+ ≤ t ≤ tsat : d2 vo = −10, dt2
let
dg(t) = −10; dt g(t) g(0.5+ )
g(t) =
dvo dt
dg(t) = −10 dt
dx = −10
t
dy 0.5
g(t) − g(0.5+ ) = −10(t − 0.5) = −10t + 5 g(0.5+ ) = C
dvo (0.5+ ) dt
0 − vo1 (0.5+ ) dvo (0.5+ ) = dt 400 × 103
vo1 (0.5+ ) = vo (0.5− ) = −1.6(0.5) = −0.80 V .·. C
0.80 dvo (0.5+ ) = = 2 µA dt 0.4 × 106
2 × 10−6 dvo + (0.5 ) = = 10 V/s dt 0.2 × 10−6 .·. g(t) = −10t + 5 + 10 = −10t + 15 = .·. dvo = −10t dt + 15 dt vo (t) vo (0.5+ )
dx =
t 0.5+
−10y dy +
t 0.5+
15 dy
dvo dt
vo (0) = 0
8–41
8–42
CHAPTER 8. Natural and Step Responses of RLC Circuits vo (t) − vo (0.5 ) = −5y +
2
t
0.5
+ 15y
t
0.5
vo (t) = vo (0.5+ ) − 5t2 + 1.25 + 15t − 7.5 vo (0.5+ ) = vo (0.5− ) = 2.5 V .·. vo (t) = −5t2 + 15t − 3.75 V, dvo1 = −20(−0.04) = 0.8, dt dvo1 = 0.8 dt;
0.5+ ≤ t ≤ tsat
0.5+ ≤ t ≤ tsat
vo1 (t) vo1 (0.5+ )
dx = 0.8
0.5+
dy
vo1 (0.5+ ) = vo1 (0.5− ) = −0.8 V
vo1 (t) − vo1 (0.5+ ) = 0.8t − 0.4; .·. vo1 (t) = 0.8t − 1.2 V,
t
0.5+ ≤ t ≤ tsat
Summary: 0 ≤ t ≤ 0.5− s :
vo1 = −1.6t V,
0.5+ s ≤ t ≤ tsat :
vo = 10t2 V
vo1 = 0.8t − 1.2 V,
vo = −5t2 + 15t − 3.75 V
[b] −12.5 = −5t2sat + 15tsat − 3.75 .·. 5t2sat − 15tsat − 8.75 = 0 Solving,
tsat = 3.5 sec
vo1 (tsat ) = 0.8(3.5) − 1.2 = 1.6 V P 8.52
τ1 = (106 )(0.5 × 10−6 ) = 0.50 s 1 = 2; τ1 .·.
τ2 = (5 × 106 )(0.2 × 10−6 ) = 1 s;
dvo d2 vo + 2vo = 20 +3 2 dt dt
s2 + 3s + 2 = 0 (s + 1)(s + 2) = 0;
s1 = −1,
vo = Vf + A1 e−t + A2 e−2t ; vo = 10 + A1 e−t + A2 e−2t
s2 = −2
Vf =
20 = 10 V 2
.·.
1 =1 τ2
Problems vo (0) = 0 = 10 + A1 + A2 ; .·. A1 = −20,
dvo (0) = 0 = −A1 − 2A2 dt
A2 = 10 V
vo (t) = 10 − 20e−t + 10e−2t V, dvo1 + 2vo1 = −1.6; dt
0 ≤ t ≤ 0.5 s
.·. vo1 = −0.8 + 0.8e−2t V,
vo (0.5) = 10 − 20e−0.5 + 10e−1 = 1.55 V vo1 (0.5) = −0.8 + 0.8e−1 = −0.51 V At
iC =
C
t = 0.5 s
0 + 0.51 1.55 − 0 − = 0.954 µA 400 × 103 5 × 106
dvo = 0.954 µA; dt
0.954 dvo = = 4.773 V/s dt 0.2
t ≥ 0.5 s d2 vo dvo + 2vo = −10 +3 2 dt dt vo (∞) = −5 .·. vo = −5 + A1 e−(t−0.5) + A2 e−2(t−0.5) 1.55 = −5 + A1 + A2
0 ≤ t ≤ 0.5 s
8–43
8–44
CHAPTER 8. Natural and Step Responses of RLC Circuits dvo (0.5) = 4.773 = −A1 − 2A2 dt .·. A1 + A2 = 6.55;
−A1 − 2A2 = 4.773
Solving, A1 = 17.87 V;
A2 = −11.32 V
.·. vo = −5 + 17.87e−(t−0.5) − 11.32e−2(t−0.5) V,
t ≥ 0.5 s
dvo1 + 2vo1 = 0.8 dt .·. vo1 = 0.4 + (−0.51 − 0.4)e−2(t−0.5) = 0.4 − 0.91e−2(t−0.5) V,
t ≥ 0.5 s
P 8.53
At t = 0 the voltage across each capacitor is zero. It follows that since the operational amplifiers are ideal, the current in the 500 kΩ is zero. Therefore there cannot be an instantaneous change in the current in the 1 µF capacitor. Since the capacitor current equals C(dvo /dt), the derivative must be zero.
P 8.54
[a] From Example 8.13 therefore
dg(t) = 2, dt
g(t) − g(0) = 2t;
iR =
d2 vo =2 dt2 g(t) =
dvo dt
g(t) = 2t + g(0);
5 dvo (0) × 10−3 = 1 µA = iC = −C 500 dt
−1 × 10−6 dvo (0) = = −1 = g(0) dt 1 × 10−6 dvo = 2t − 1 dt dvo = 2t dt − dt vo − vo (0) = t2 − t; vo = t2 − t + 8,
vo (0) = 8 V 0 ≤ t ≤ tsat
g(0) =
dvo (0) dt
Problems
8–45
[b] t2 − t + 8 = 9 t2 − t − 1 = 0 √ t = (1/2) ± ( 5/2) ∼ = 1.62 s,
tsat ∼ = 1.62 s
(Negative value has no physical significance.) P 8.55
Part (1) — Example 8.14, with R1 and R2 removed: [a] Ra = 100 kΩ;
1 d2 vo = 2 dt Ra C1
C1 = 0.1 µF;
Rb = 25 kΩ;
1 vg ; Rb C2
1 = 100 Ra C1
vg = 250 × 10−3 ;
therefore
dvo (0) , our solution is dt The second op-amp will saturate when or
tsat =
1 = 40 Rb C2
d2 vo = 1000 dt2
[b] Since vo (0) = 0 = vo = 6 V,
C2 = 1 µF
vo = 500t2 V
6/500 ∼ = 0.1095 s
dvo1 1 vg = −25 =− dt Ra C1 [d] Since vo1 (0) = 0, vo1 = −25t V [c]
At
t = 0.1095 s,
vo1 ∼ = −2.74 V
Therefore the second amplifier saturates before the first amplifier saturates. Our expressions are valid for 0 ≤ t ≤ 0.1095 s. Once the second op-amp saturates, our linear model is no longer valid. Part (2) — Example 8.14 with vo1 (0) = −2 V and vo (0) = 4 V: [a] Initial conditions will not change the differential equation; hence the equation is the same as Example 8.14. [b] vo = 5 + A1 e−10t + A2 e−20t (from Example 8.14) vo (0) = 4 = 5 + A1 + A2
8–46
CHAPTER 8. Natural and Step Responses of RLC Circuits 2 4 + iC (0+ ) − =0 100 25 iC (0+ ) =
dvo (0+ ) 4 mA = C 100 dt
0.04 × 10−3 dvo (0+ ) = = 40 V/s dt 10−6 dvo = −10A1 e−10t − 20A2 e−20t dt dvo + (0 ) = −10A1 − 20A2 = 40 dt Therefore −A1 − 2A2 = 4 and Thus, A1 = 2 and A2 = −3 vo = 5 + 2e−10t − 3e−20t V,
A1 + A2 = −1
t≥0
[c] Same as Example 8.14: dvo1 + 20vo1 = −25 dt [d] From Example 8.14: vo1 (∞) = −1.25 V;
v1 (0) = −2 V
(given)
Therefore vo1 = −1.25 + (−2 + 1.25)e−20t = −1.25 − 0.75e−20t V, P 8.56
t≥0
[a]
2C
dva va − vg va + + =0 dt R R
(1) Therefore
va vg dva + = ; dt RC 2RC
(2) Therefore
va dvb + = 0, dt RC
0 − va d(0 − vb ) +C =0 R dt
va = −RC
dvb dt
Problems dvb d(vb − vo ) 2vb +C +C =0 R dt dt dvb vb 1 dvo + = dt RC 2 dt
(3) Therefore
From (2) we have
d2 vb dva = −RC 2 dt dt
and
va = −RC
dvb dt
When these are substituted into (1) we get (4) − RC
vg d2 vb dvb = − 2 dt dt 2RC
Now differentiate (3) to get (5)
1 d2 vo 1 dvb d2 vb = + dt2 RC dt 2 dt2
But from (4) we have (6)
vg 1 dvb d2 vb =− 2 2 + 2 dt RC dt 2R C
Now substitute (6) into (5) vg d2 vo =− 2 2 2 dt R C d2 vo vg = 2 2 2 dt R C The two equations are the same except for a reversal in algebraic sign.
[b] When R1 C1 = R2 C2 = RC :
[c] Two integrations of the input signal with one operational amplifier. P 8.57
[a] f (t) = =
inertial force + frictional force + spring force M [d2 x/dt2 ] + D[dx/dt] + Kx
D f d2 x − = [b] 2 dt M M Given
vA =
d2 x , dt2
dx K − x dt M then
1 t d2 x 1 dx dy = − vB = − 2 R1 C1 0 dy R1 C1 dt
vC = −
1 t 1 vB dy = x R2 C2 0 R1 R2 C1 C2
vD = −
R3 R3 dx · vB = R4 R4 R1 C1 dt
8–47
8–48
CHAPTER 8. Natural and Step Responses of RLC Circuits
R5 + R6 R5 + R6 1 vC = · ·x vE = R6 R6 R1 R2 C1 C2 vF =
−R8 f (t), R7
Therefore
vA = −(vD + vE + vF )
R8 R3 dx R5 + R6 d2 x − = f (t) − x dt2 R7 R4 R1 C1 dt R6 R1 R2 C1 C2
Therefore M =
R7 , R8
D=
R3 R7 R8 R4 R1 C1
and
K=
R7 (R5 + R6 ) R8 R6 R1 R2 C1 C2
Box Number Function
P 8.58
1
inverting and scaling
2
inverting and scaling
3
integrating and scaling
4
integrating and scaling
5
inverting and scaling
6
noninverting and scaling
[a] Given that the current response is underdamped we know i will be of the form i = If + [B1 cos ωd t + B2 sin ωd t]e−αt where and
α= ωd =
R 2L
ωo2 − α2 =
1 − α2 LC
The capacitor will force the final value of i to be zero, therefore If = 0. By hypothesis i(0+ ) = Vdc /R therefore B1 = Vdc /R. At t = 0+ the voltage across the primary winding is zero hence di(0+ )/dt = 0. From our equation for i we have di = [(ωd B2 − αB1 ) cos ωd t − (ωd B1 + αB2 ) sin ωd t]e−αt dt Hence di(0+ ) = ωd B2 − αB1 = 0 dt Thus α αVdc B2 = B1 = ωd ωd R It follows directly that α Vdc cos ωd t + sin ωd t e−αt i= R ωd
Problems [b] Since ωd B1 − αB1 = 0 it follows that di = −(ωd B1 + αB2 )e−αt sin ωd t dt αB2 =
But
α2 Vdc ωd R
and
ωd B1 =
ωd Vdc R
Therefore
αB2
Vdc ωd2 + α2 ωd Vdc α2 Vdc + = = R ωd R R ωd
ωd B1
+
But
ωd2 + α2 = ωo2 =
1 LC
Hence Vdc ωd RLC
ωd B1 + αB2 = Now since v1 = −L
v1 = L
di dt
we get
Vdc Vdc −αt e−αt sin ωd t = − e sin ωd t ωd RLC ωd RC
[c] vc = Vdc − iR − L
di dt
iR = Vdc cos ωd t +
vc = Vdc − Vdc
α sin ωd t e−αt ωd
Vdc −αt α e sin ωd t cos ωd t + sin ωd t e−αt + ωd ωd RC
= Vdc − Vdc e−αt cos ωd t +
−αt
= Vdc 1 − e
αVdc −αt Vdc − e sin ωd t ωd RC ωd
1 cos ωd t + ωd
= Vdc [1 − e−αt cos ωd t + Ke−αt sin ωd t] P 8.59
vsp = Vdc
1 − α e−αt sin ωd t RC
a e−αt sin ωd t 1− ωd RC
−aVdc d −αt dvsp = [e sin ωd t] dt ωd RC dt =
−aVdc [−αe−αt sin ωd t + ωd cos ωd te−αt ] ωd RC
=
aVdc e−αt [α sin ωd t − ωd cos ωd t] ωd RC
8–49
8–50
CHAPTER 8. Natural and Step Responses of RLC Circuits dvsp = 0 when dt or
tan ωd t =
α sin ωd t = ωd cos ωd t
ωd ; α
ωd t = tan−1
ωd 1 tan−1 .·. tmax = ωd α
ωd α
Note that because tan θ is periodic, i.e., tan θ = tan(θ ± nπ), where n is an integer, there are an infinite number of solutions for t where dvsp /dt = 0, that is t=
tan−1 (ωd /α) ± nπ ωd
Because of e−αt in the expression for vsp and knowing t ≥ 0 we know vsp will be maximum when t has its smallest positive value. Hence tmax = P 8.60
tan−1 (ωd /α) . ωd
[a] vc = Vdc [1 − e−αt cos ωd t + Ke−αt sin ωd t] dvc d = Vdc [1 + e−αt (K sin ωd t − cos ωd t)] dt dt = Vdc {(−αe−αt )(K sin ωd t − cos ωd t)+ e−αt [ωd K cos ωd t + ωd sin ωd t]} = Vdc e−αt [(ωd − αK) sin ωd t + (α + ωd K) cos ωd t] dvc = 0 when dt or
tan ωd t =
(ωd − αK) sin ωd t = −(α + ωd K) cos ωd t
α + ωd K αK − ωd
α + ωd K .·. ωd t ± nπ = tan−1 αK − ωd
α + ωd K 1 tc = tan−1 ± nπ ωd αK − ωd α=
4 × 103 R = = 666.67 rad/s 2L 6
ωd =
109 − (666.67)2 = 28,859.81 rad/s 1.2
Problems 1 K= ωd
8–51
1 − α = 21.63 RC
1 −1 1 tan (−43.29) + nπ = {−1.55 + nπ} ωd ωd
tc =
The smallest positive value of t occurs when n = 1, therefore tc max = 55.23 µs [b] vc (tc max ) = 12[1 − e−αtc max cos ωd tc max + Ke−αtc max sin ωd tc max ] = 262.42 V [c] From the text example the voltage across the spark plug reaches its maximum value in 53.63 µs. If the spark plug does not fire the capacitor voltage peaks in 55.23 µs. When vsp is maximum the voltage across the capacitor is 262.15 V. If the spark plug does not fire the capacitor voltage reaches 262.42 V. P 8.61
1 1 [a] w = L[i(0+ )]2 = (5)(16) × 10−3 = 40 mJ 2 2 3 R 3 × 10 = 300 rad/s [b] α = = 2L 10
ωd =
109 − (300)2 = 28,282.68 rad/s 1.25
1 106 4 × 106 = = RC 0.75 3
tmax
ωd 1 = tan−1 ωd α
vsp (tmax ) = 12 −
= 55.16 µs
12(50)(4 × 106 ) −αtmax e sin ωd tmax = −27,808.04 V 3(28,282.68)
[c] vc (tmax ) = 12[1 − e−αtmax cos ωd tmax + Ke−αtmax sin ωd tmax ]
1 1 − α = 47.13 K= ωd RC vc (tmax ) = 568.15 V
Sinusoidal Steady State Analysis
Assessment Problems AP 9.1 [a] V = 170/−40◦ V [b] 10 sin(1000t + 20◦ ) = 10 cos(1000t − 70◦ ) .·.
I = 10/−70◦ A
[c] I = 5/36.87◦ + 10/−53.13◦ = 4 + j3 + 6 − j8 = 10 − j5 = 11.18/−26.57◦ A [d] sin(20,000πt + 30◦ ) = cos(20,000πt − 60◦ ) Thus, V = 300/45◦ − 100/−60◦ = 212.13 + j212.13 − (50 − j86.60) = 162.13 + j298.73 = 339.90/61.51◦ mV AP 9.2 [a] v = 18.6 cos(ωt − 54◦ ) V [b] I = 20/45◦ − 50/− 30◦ = 14.14 + j14.14 − 43.3 + j25 = −29.16 + j39.14 = 48.81/126.68◦ Therefore i = 48.81 cos(ωt + 126.68◦ ) mA [c] V = 20 + j80 − 30/15◦ = 20 + j80 − 28.98 − j7.76 = −8.98 + j72.24 = 72.79/97.08◦ v = 72.79 cos(ωt + 97.08◦ ) V AP 9.3 [a] ωL = (104 )(20 × 10−3 ) = 200 Ω [b] ZL = jωL = j200 Ω 9–1
9
9–2
CHAPTER 9. Sinusoidal Steady State Analysis [c] VL = IZL = (10/30◦ )(200/90◦ ) × 10−3 = 2/120◦ V [d] vL = 2 cos(10,000t + 120◦ ) V −1 −1 = = −50 Ω ωC 4000(5 × 10−6 ) [b] ZC = jXC = −j50 Ω 30/25◦ V = = 0.6/115◦ A [c] I = ZC 50/−90◦ [d] i = 0.6 cos(4000t + 115◦ ) A
AP 9.4 [a] XC =
AP 9.5 I1 = 100/25◦ = 90.63 + j42.26 I2 = 100/145◦ = −81.92 + j57.36 I3 = 100/−95◦ = −8.72 − j99.62 I4 = −(I1 + I2 + I3 ) = (0 + j0) A, AP 9.6 [a] I =
therefore
i4 = 0 A
125/−60◦ 125 /(−60 − θZ )◦ = |Z|/θz |Z|
But −60 − θZ = −105◦
.·. θZ = 45◦
Z = 90 + j160 + jXC .·. XC = −70 Ω; .·. C = [b] I =
XC = −
1 = −70 ωC
1 = 2.86 µF (70)(5000)
125/−60◦ Vs = = 0.982/−105◦ A; Z (90 + j90)
.·. |I| = 0.982 A
AP 9.7 [a]
ω = 2000 rad/s ωL = 10 Ω,
−1 = −20 Ω ωC
Zxy = 20j10 + 5 + j20 =
20(j10) + 5 − j20 (20 + j10)
= 4 + j8 + 5 − j20 = (9 − j12) Ω
Problems [b] ωL = 40 Ω, Zxy
−1 = −5 Ω ωC
(20)(j40) = 5 − j5 + 20j40 = 5 − j5 + 20 + j40
9–3
= 5 − j5 + 16 + j8 = (21 + j3) Ω
[c] Zxy
20(jωL) j106 = + 5− 20 + jωL 25ω
20ω 2 L2 j400ωL j106 + + 5 − 400 + ω 2 L2 400 + ω 2 L2 25ω The impedance will be purely resistive when the j terms cancel, i.e., =
106 400ωL = 400 + ω 2 L2 25ω Solving for ω yields ω = 4000 rad/s. [d] Zxy =
20ω 2 L2 + 5 = 10 + 5 = 15 Ω 400 + ω 2 L2
AP 9.8 The frequency 4000 rad/s was found to give Zxy = 15 Ω in Assessment Problem 9.7. Thus, V = 150/0◦ ,
Is =
V 150/0◦ = 10/0◦ A = Zxy 15
Using current division, IL =
20 (10) = 5 − j5 = 7.07/−45◦ A 20 + j20
iL = 7.07 cos(4000t − 45◦ ) A,
Im = 7.07 A
AP 9.9 After replacing the delta made up of the 50 Ω, 40 Ω, and 10 Ω resistors with its equivalent wye, the circuit becomes
9–4
CHAPTER 9. Sinusoidal Steady State Analysis The circuit is further simplified by combining the parallel branches, (20 + j40)(5 − j15) = (12 − j16) Ω Therefore I =
136/0◦ = 4/28.07◦ A 14 + 12 − j16 + 4
AP 9.10 V1 = 240/53.13◦ = 144 + j192 V V2 = 96/−90◦ = −j96 V jωL = j(4000)(15 × 10−3 ) = j60 Ω 6 × 106 1 = −j = −j60 Ω jωC (4000)(25) Perform source transformations: 144 + j192 V1 = = 3.2 − j2.4 A j60 j60 V2 96 = −j4.8 A = −j 20 20
Combine the parallel impedances: Y =
1 1 1 j5 1 1 + + + = = j60 30 −j60 20 j60 12
Z=
1 = 12 Ω Y
Vo = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48/36.87◦ V vo = 48 cos(4000t + 36.87◦ ) V
Problems
9–5
AP 9.11 Use the lower node as the reference node. Let V1 = node voltage across the 20 Ω resistor and VTh = node voltage across the capacitor. Writing the node voltage equations gives us V1 V1 − 10Ix − 2/45◦ + = 0 and 20 j10
VTh =
−j10 (10Ix ) 10 − j10
We also have Ix =
V1 20
Solving these equations for VTh gives VTh = 10/45◦ V. To find the Thévenin impedance, we remove the independent current source and apply a test voltage source at the terminals a, b. Thus
It follows from the circuit that 10Ix = (20 + j10)Ix Therefore Ix = 0 and
ZTh =
VT , IT
IT =
VT VT + −j10 10
therefore ZTh = (5 − j5) Ω
AP 9.12 The phasor domain circuit is as shown in the following diagram:
9–6
CHAPTER 9. Sinusoidal Steady State Analysis The node voltage equation is V V V V − 100/−90◦ −10 + + + + =0 5 −j(20/9) j5 20 Therefore V = 10 − j30 = 31.62/−71.57◦ Therefore v = 31.62 cos(50,000t − 71.57◦ ) V
AP 9.13 Let Ia , Ib , and Ic be the three clockwise mesh currents going from left to right. Summing the voltages around meshes a and b gives 33.8 = (1 + j2)Ia + (3 − j5)(Ia − Ib ) and 0 = (3 − j5)(Ib − Ia ) + 2(Ib − Ic ). But Vx = −j5(Ia − Ib ), therefore Ic = −0.75[−j5(Ia − Ib )]. Solving for I = Ia = 29 + j2 = 29.07/3.95◦ A. √ ωM = 80 Ω AP 9.14 [a] M = 0.4 0.0625 = 0.1 H, Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) Ω Therefore |Z22 | = 500 Ω,
80 Zr = 500 [b] I1 =
2
∗ Z22 = (400 − j300) Ω
(400 − j300) = (10.24 − j7.68) Ω
245.20 = 0.50/− 53.13◦ A 184 + 100 + j400 + Zr
i1 = 0.5 cos(800t − 53.13◦ ) A
jωM j80 I1 = (0.5/− 53.13◦ ) = 0.08/0◦ A [c] I2 = Z22 500/36.87◦ i2 = 80 cos 800t mA
Problems Vs 25 × 103 /0◦ = Z1 + Z2 /a2 1500 + j6000 + (25)2 (4 − j14.4)
AP 9.15 I1 =
= 4 + j3 = 5/36.87◦ A V1 = Vs − Z1 I1 = 25,000/0◦ − (4 + j3)(1500 + j6000) = 37,000 − j28,500 V2 = −
I2 = Also,
1 V1 = −1480 + j1140 = 1868.15/142.39◦ V 25
V2 1868.15/142.39◦ = 125/− 143.13◦ A = Z2 4 − j14.4 I2 = −25I1
9–7
9–8
CHAPTER 9. Sinusoidal Steady State Analysis
Problems P 9.1
[a] ω = 2πf = 3769.91 rad/s,
f=
ω = 600 Hz 2π
[b] T = 1/f = 1.67 ms [c] Vm = 10 V [d] v(0) = 10 cos(−53.13◦ ) = 6 V −53.13◦ (2π) [e] φ = −53.13◦ ; φ= = −0.9273 rad 360◦ [f] V = 0 when 3769.91t − 53.13◦ = 90◦ . Now resolve the units: 143.13◦ = 2.498 rad, (3769.91 rad/s)t = (180◦ /π)
t = 662.64 µs
[g] (dv/dt) = (−10)3769.91 sin(3769.91t − 53.13◦ ) (dv/dt) = 0 when
3769.91t − 53.13◦ = 0◦
53.13◦ 3769.91t = = 0.9273 rad 57.3◦ /rad
or
Therefore t = 245.97 µs
P 9.2
Vrms = T /2 0
1 T /2 2 2 2π t dt Vm sin T 0 T
Vm2
sin
2
2π Vm2 T /2 4π V 2T t dt = 1 − cos t dt = m T 2 0 T 4
Therefore Vrms = P 9.3
1 Vm2 T Vm = T 4 2
[a] 40 V [b] 2πf = 100π;
f = 50Hz
[c] ω = 100π = 314.159 rad/s 2π π [d] θ(rad) = (60◦ ) = = 1.05 rad ◦ 360 3 ◦ [e] θ = 60 1 1 = 20 ms [f] T = = f 50 [g] v = −40 when π 100πt + = π; .·. t = 6.67 ms 3
Problems
0.01 π + [h] v = 40 cos 100π t − 3 3
= 40 cos[100πt − (π/3) + (π/3)] = 40 cos 100πt V [i] 100π(t − to ) + (π/3) = 100πt − (π/2) .·. 100πto =
5π ; 6
to = 8.33 ms
[j] 100π(t + to ) + (π/3) = 100πt + 2π .·. 100πto =
5π ; 3
to = 16.67 ms
16.67 ms to the left P 9.4
[a] Left as φ becomes more positive [b] Left P 9.5
[a] By hypothesis v = 80 cos(ωt + θ) dv = −80ω sin(ωt + θ) dt .·. 80ω = 80,000; ω = 1000 rad/s [b] f =
ω = 159.155 Hz; 2π
T =
1 = 6.28 ms f
−2π/3 = −0.3333, .·. θ = −90 − (−0.3333)(360) = 30◦ 6.28 .·. v = 80 cos(1000t + 30◦ ) V
9–9
9–10
P 9.6
CHAPTER 9. Sinusoidal Steady State Analysis [a]
T = 8 + 2 = 10 ms; 2 f=
T = 20 ms
1 1 = = 50Hz T 20 × 10−3
[b] v = Vm sin(ωt + θ) ω = 2πf = 100π rad/s 100π(−2 × 10−3 ) + θ = 0;
.·. θ =
π rad = 36◦ 5
v = Vm sin[100πt + 36◦ ] 80.9 = Vm sin 36◦ ;
Vm = 137.64 V
v = 137.64 sin[100πt + 36◦ ] = 137.64 cos[100πt − 54◦ ] V P 9.7
u= = =
to +T to
Vm2
Vm2 cos2 (ωt + φ) dt
to +T 1
Vm2
2 V2 = m 2 Vm2 = 2
2
t
o to +T to
+
1 cos(2ωt + 2φ) dt 2
dt +
to +T to
cos(2ωt + 2φ) dt
P 9.8
1 T+ sin(2ωt + 2φ) |ttoo +T 2ω 1 T+ [sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)] 2ω T 1 2 2 T (0) = Vm = Vm + 2 2ω 2 √ √ Vm = 2Vrms = 2(120) = 169.71 V
P 9.9
[a] The numerical values of the terms in Eq. 9.8 are Vm = 20, R/L = 1066.67, √ R2 + ω 2 L2 = 100
ωL = 60
φ = 25◦ ,
θ = 36.87◦
θ = tan−1 60/80,
Substitute these values into Equation 9.9:
i = −195.72e−1066.67t + 200 cos(800t − 11.87◦ ) mA,
t≥0
[b] Transient component = −195.72e−1066.67t mA Steady-state component = 200 cos(800t − 11.87◦ ) mA [c] By direct substitution into Eq 9.9 in part (a), i(1.875 ms) = 28.39 mA [d] 200 mA,
800 rad/s,
−11.87◦
Problems [e] The current lags the voltage by 36.87◦ . P 9.10
[a] From Eq. 9.9 we have L
Vm R cos(φ − θ) −(R/L)t ωLVm sin(ωt + φ − θ) di √ = √ 2 e − dt R + ω 2 L2 R 2 + ω 2 L2
−Vm R cos(φ − θ)e−(R/L)t Vm R cos(ωt + φ − θ) √ √ Ri = + R 2 + ω 2 L2 R 2 + ω 2 L2
R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ) di √ L + Ri = Vm dt R 2 + ω 2 L2
But R ωL √ = cos θ and = sin θ R 2 + ω 2 L2 R 2 + ω 2 L2 Therefore the right-hand side reduces to √
Vm cos(ωt + φ) At t = 0, Eq. 9.9 reduces to −Vm cos(φ − θ) Vm cos(φ − θ) + √ 2 =0 i(0) = √ 2 R − ω 2 L2 R + ω 2 L2 Vm cos(ωt + φ − θ) [b] iss = √ 2 R + ω 2 L2 Therefore diss −ωLVm L =√ 2 sin(ωt + φ − θ) dt R + ω 2 L2 and Vm R Riss = √ 2 cos(ωt + φ − θ) R + ω 2 L2
R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ) diss √ L + Riss = Vm dt R 2 + ω 2 L2 = Vm cos(ωt + φ) P 9.11
[a] Y = 50/60◦ + 100/− 30◦ = 111.8/− 3.43◦ y = 111.8 cos(500t − 3.43◦ ) [b] Y = 200/50◦ − 100/60◦ = 102.99/40.29◦ y = 102.99 cos(377t + 40.29◦ ) [c] Y = 80/30◦ − 100/− 225◦ + 50/− 90◦ = 161.59/− 29.96◦ y = 161.59 cos(100t − 29.96◦ )
9–11
9–12
CHAPTER 9. Sinusoidal Steady State Analysis [d] Y = 250/0◦ + 250/120◦ + 250/− 120◦ = 0 y=0
P 9.12
[a] 1000Hz [b] θv = 0◦ 200 200/0◦ = /− 90◦ = 25/− 90◦ ; [c] I = jωL ωL 200 200 = 25; ωL = = 8Ω [d] ωL 25 8 [e] L = = 1.27 mH 2π(1000) [f] ZL = jωL = j8 Ω
P 9.13
θi = −90◦
[a] ω = 2πf = 314,159.27 rad/s 10 × 10−3 /0◦ V = jωC(10 × 10−3 )/0◦ = 10 × 10−3 ωC/90◦ = [b] I = ZC 1/jωC .·. θi = 90◦ [c] 628.32 × 10−6 = 10 × 10−3 ωC 1 10 × 10−3 = = 15.92 Ω, ωC 628.32 × 10−6 [d] C =
.·. XC = −15.92 Ω
1 1 = 15.92(ω) (15.92)(100π × 103 )
C = 0.2 µF
−1 [e] Zc = j ωC P 9.14
= −j15.92 Ω
[a] jωL = j(2 × 104 )(300 × 10−6 ) = j6 Ω 1 1 = −j10 Ω; = −j 4 jωC (2 × 10 )(5 × 10−6 )
Ig = 922/30◦ A
Problems [b] Vo = 922/30◦ Ze Ze =
1 ; Ye
1 1 1 +j + 10 10 8 + j6
Ye =
Ye = 0.18 + j0.04 S Ze =
1 = 5.42/− 12.53◦ Ω 0.18 + j0.04
Vo = (922/30◦ )(5.42/− 12.53◦ ) = 5000.25/17.47◦ V [c] vo = 5000.25 cos(2 × 104 t + 17.47◦ ) V P 9.15
[a] ZL = j(8000)(5 × 10−3 ) = j40 Ω ZC =
−j = −j100 Ω (8000)(1.25 × 10−6 )
600/20◦ = 8.32/76.31◦ A 40 + j40 − j100 [c] i = 8.32 cos(8000t + 76.31◦ ) A [b] I =
P 9.16
Z = 4 + j(50)(0.24) − j
Io =
1 = 4 + j4 = 5.66/45◦ Ω (50)(0.0025)
0.1/− 90◦ V = = 17.68/− 135◦ mA Z 5.66/45◦
io (t) = 17.68 cos(50t − 135◦ ) mA P 9.17
[a] Y =
1 1 1 + + 3 + j4 16 − j12 −j4
= 0.12 − j0.16 + 0.04 + j0.03 + j0.25 = 0.16 + j0.12 = 200/36.87◦ mS [b] G = 160 mS [c] B = 120 mS
9–13
9–14
CHAPTER 9. Sinusoidal Steady State Analysis [d] I = 8/0◦ A, IC =
V=
8 I = = 40/−36.87◦ V Y 0.2/36.87◦
V 40/−36.87◦ = = 10/53.13◦ A ◦ ZC 4/−90
iC = 10 cos(ωt + 53.13◦ ) A, P 9.18
Im = 10 A
ZL = j(2000)(60 × 10−3 ) = j120 Ω ZC =
−j = −j40 Ω (2000)(12.5 × 10−6 )
Construct the phasor domain equivalent circuit:
Using current division: I=
(120 − j40) (0.5) = 0.25 − j0.25 A 120 − j40 + 40 + j120
Vo = j120I = 30 + j30 = 42.43/45◦ V vo = 42.43 cos(2000t + 45◦ ) V P 9.19
[a] Vg = 300/78◦ ; .·. Z =
Ig = 6/33◦
300/78◦ Vg = = 50/45◦ Ω ◦ Ig 6/33
[b] ig lags vg by 45◦ : 2πf = 5000π;
f = 2500 Hz;
T = 1/f = 400 µs
45◦ · . . ig lags vg by (400 µs) = 50 µs 360◦
Problems P 9.20
1 1 = −j20 Ω = −6 jωC (1 × 10 )(50 × 103 ) jωL = j50 × 103 (1.2 × 10−3 ) = j60 Ω Vg = 40/0◦ V
Ze = −j20 + 30j60 = 24 − j8 Ω Ig =
40/0◦ = 1.5 + j0.5 mA 24 − j8
Vo = (30j60)Ig =
30(j60) (1.5 + j0.5) = 30 + j30 = 42.43/45◦ V 30 + j60
vo = 42.43 cos(50,000t + 45◦ ) V P 9.21
[a] Z1 = R1 − j
1 ωC1
R2 R2 /jωC2 R2 − jωR22 C2 = = R2 + (1/jωC2 ) 1 + jωR2 C2 1 + ω 2 R22 C22
Z2 =
Z1 = Z2
when
R1 =
1 ωR22 C2 = ωC1 1 + ω 2 R22 C22 [b] R1 =
R2 1 + ω 2 R22 C22
or
C1 =
and
1 + ω 2 R22 C22 ω 2 R22 C2
1000 = 200 Ω 1 + (40 × 103 )2 (1000)2 (50 × 10−9 )2
C1 =
1 + (40 × 103 )2 (1000)2 (50 × 10−9 )2 = 62.5 nF (40 × 103 )2 (1000)2 (50 × 10−9 )
9–15
9–16 P 9.22
CHAPTER 9. Sinusoidal Steady State Analysis [a] Y2 =
1 + jωC2 R2 jωC1 1 ω 2 R1 C12 + jωC1 = = R1 + (1/jωC1 ) 1 + jωR1 C1 1 + ω 2 R12 C12
Y1 =
Therefore R2 =
1 + ω 2 R12 C12 ω 2 R1 C12
and
when C2 =
C1 1 + ω 2 R12 C12
1 + (50 × 103 )2 (1000)2 (40 × 10−9 )2 = 1250 Ω (50 × 103 )2 (1000)(40 × 10−9 )2
[b] R2 =
C2 = P 9.23
Y1 = Y2
40 × 10−9 = 8 nF 1 + (50 × 103 )2 (1000)2 (40 × 10−9 )2
[a] Z1 = R1 + jωL1 R2 (jωL2 ) ω 2 L22 R2 + jωL2 R22 Z2 = = R2 + jωL2 R2 + ω 2 L22 Z1 = Z2
ω 2 L22 R2 R22 + ω 2 L22
and
(5000)2 (1.25) = 625 mH 50002 + 40002 (1.25)2
L1 = [a] Y2 =
R1 =
(4000)2 (1.25)2 (5000) = 2500 Ω 50002 + 40002 (1.25)2
[b] R1 =
P 9.24
when
1 j − R2 ωL2 1 R1 − jωL1 = 2 R1 + jωL1 R1 + ω 2 L21
Y1 =
Therefore R2 = [b] R2 = L2 =
Y2 = Y1
R12 + ω 2 L21 R1
and
when L2 =
R12 + ω 2 L21 ω 2 L1
80002 + 10002 (4)2 = 10 kΩ 8000 80002 + 10002 (4)2 = 20 H 10002 (4)
L1 =
R22 L2 R22 + ω 2 L22
9–17
Problems P 9.25
Vg = 500/30◦ V; Z=
Ig = 0.1/83.13◦ mA
Vg = 5000/− 53.13◦ Ω = 3000 − j4000 Ω Ig
32 × 103 z = 3000 + j ω − ω ω−
32 × 103 = −4000 ω
ω 2 + 4000ω − 32 × 103 = 0 ω = 7.984 rad/s P 9.26
[a] Zeq =
50,000 −j20 × 106 + (1200 + j0.2ω) 3 ω (1200 + j0.2ω) 50,000 −j20 × 106 + = 6 3 ω 1200 + j[0.2ω − 20×10 ] ω =
50,000 + 3
−j20×106 (1200 ω
+ j0.2ω) 1200 − j 0.2ω −
12002 + 0.2ω −
20×106 ω
2
20×106 ω
20 × 106 20 × 106 20 × 106 (1200)2 − Im(Zeq ) = − 0.2ω 0.2ω − ω ω ω
−20 × 10 (1200) − 20 × 10 6
2
6
20 × 106 0.2ω 0.2ω − ω
20 × 106 −(1200) = 0.2ω 0.2ω − ω
2
0.22 ω 2 − 0.2(20 × 106 ) + 12002 = 0 ω 2 = 64 × 106 .·. [b] Zeq =
.·.
ω = 8000 rad/s
f = 1273.24 Hz 50,000 + −j2500(1200 + j1600) 3 =
50,000 (−j2500)(1200 + j1600) + = 20,000 Ω 3 1200 − j900
30/0◦ = 1.5/0◦ mA Ig = 20,000 ig (t) = 1.5 cos 8000t mA
=0
=0
9–18 P 9.27
CHAPTER 9. Sinusoidal Steady State Analysis [a] Find the equivalent impedance seen by the source, as a function of L, and set the imaginary part of the equivalent impedance to 0, solving for L: ZC =
−j = −j1000 Ω (500)(2 × 10−6 )
Zeq = −j1000 + j500L2000 = −j1000 + = −j1000 +
2000(j500L)(2000 − j500L) 20002 + (500L)2
Im(Zeq ) = −1000 +
20002 (500L) =0 20002 + (500L)2
.·.
20002 (500L) = 1000 20002 + (500L)2
.·.
1 5002 L2 − 20002 L + 20002 = 0 2
Solving the quadratic equation, [b] Ig =
2000(j500L) 2000 + j500L
L = 4H
100/0◦ 100/0◦ = = 0.1/0◦ A −j1000 + j20002000 1000
ig (t) = 0.1 cos 500t A P 9.28
[a] jωL + R(−j/ωC) = jωL +
−jR/ωC R − j/ωC
= jωL +
−jR ωCR − j1
= jωL +
−jR(ωCR + j1) ω 2 C 2 R2 + 1
Im(Zab ) = ωL −
ωCR2 =0 ω 2 C 2 R2 + 1
CR2 ω 2 C 2 R2 + 1
.·.
L=
.·.
ω 2 C 2 R2 + 1 =
.·.
(25×10 )(100) −1 (CR2 /L) − 1 160×10−6 = = 900 × 108 ω = −9 2 2 2 C R (25 × 10 ) (100)2
CR2 L −9
2
ω = 300 krad/s
2
Problems [b] Zab (300 × 103 ) = j48 + P 9.29
(100)(−j133.33) = 64 Ω 100 − j133.33
jωL = j100 × 103 (0.6 × 10−3 ) = j60 Ω −j 1 = = −j25 Ω jωC (100 × 103 )(0.4 × 10−6 )
VT = −j25IT + 5I∆ − 30I∆ I∆ =
−j60 IT 30 + j60
VT = −j25IT + 25
j60 IT 30 + j60
VT = Zab = 20 − j15 = 25/− 36.87◦ Ω IT P 9.30
[a] Z1 = 400 − j
106 = 400 − j800 Ω 500(2.5)
Z2 = 2000j500L =
j106 L 2000 + j500L
ZT = Z1 + Z2 = 400 − j800 + = 400 +
j106 L 2000 + j500L
500 × 106 L2 2 × 109 L − j800 + j 20002 + 5002 L2 20002 + 5002 L2
ZT is resistive when 2 × 109 L = 800 or 5002 L2 − 25 × 105 L + 20002 = 0 20002 + 5002 L2 Solving, L1 = 8 H and L2 = 2 H.
9–19
9–20
CHAPTER 9. Sinusoidal Steady State Analysis [b] When L = 8 H: ZT = 400 +
500 × 106 (8)2 = 2000 Ω 20002 + 5002 (8)2
200/0◦ = 100/0◦ mA Ig = 2000 ig = 100 cos 500t mA When L = 2 H: ZT = 400 +
500 × 106 (2)2 = 800 Ω 20002 + 500(2)2
200/0◦ = 250/0◦ mA 800
Ig =
ig = 250 cos 500t mA P 9.31
[a] Y1 =
11 = 4.4 × 10−6 S 2500 × 103
Y2 = =
1 14,000 + j5ω 14,000 5ω −j 6 2 196 × 10 + 25ω 196 × 106 + 25ω 2
Y3 = jω2 × 10−9 Y T = Y 1 + Y2 + Y3 For ig and vo to be in phase the j component of YT must be zero; thus, ω2 × 10−9 =
5ω 196 × 106 + 25ω 2
or 25ω 2 + 196 × 106 =
5 2 × 10−9
.·. 25ω 2 = 2304 × 106 [b] YT = 4.4 × 10−6 +
.·. ω = 9600 rad/s
14,000 = 10 × 10−6 S 196 × 106 + 25(9600)2
.·. ZT = 100 kΩ Vo = (0.25 × 10−3 /0◦ )(100 × 103 ) = 25/0◦ V vo = 25 cos 9600t V
Problems
P 9.32
[a] Zg = 500 − j
103 (j0.5ω) 106 + 3 ω 10 + j0.5ω
= 500 − j
106 500jω(1000 − j0.5ω) + ω 106 + 0.25ω 2
= 500 − j
250ω 2 106 5 × 105 ω + 6 + j ω 10 + 0.25ω 2 106 + 0.25ω 2
.·. If Zg is purely real,
5 × 105 ω 106 = 6 ω 10 + 0.25ω 2
2(106 + 0.25ω 2 ) = ω 2
.·.
.·.
4 × 106 = ω 2
ω = 2000 rad/s
[b] When ω = 2000 rad/s Zg = 500 − j500 + (j10001000) = 1000 Ω .·. Ig =
20/0◦ = 20/0◦ mA 1000
Vo = Vg − Ig Z1 Z1 = 500 − j500 Ω Vo = 20/0◦ − (0.02/0◦ )(500 − j500) = 10 + j10 = 14.14/45◦ V vo = 14.14 cos(2000t + 45◦ ) V P 9.33
Zab = 1 − j8 + (2 + j4)(10 − j20) + (40j20) = 1 − j8 + 3 + j4 + 8 + j16 = 12 + j12 Ω = 16.971/45◦ Ω
P 9.34
First find the admittance of the parallel branches Yp =
1 1 1 1 + + + = 0.625 − j1.875 S 2 − j6 12 + j4 2 j0.5
Zp =
1 1 = 0.16 + j0.48 Ω = Yp 0.625 − j1.875
Zab = −j4.48 + 0.16 + j0.48 + 2.84 = 3 − j4 Ω Yab =
1 1 = 120 + j160 mS = Zab 3 − j4
= 200/53.13◦ mS
9–21
9–22 P 9.35
CHAPTER 9. Sinusoidal Steady State Analysis Simplify the top triangle using series and parallel combinations: (1 + j1)(1 − j1) = 1 Ω Convert the lower left delta to a wye:
Z1 =
(j1)(1) = j1 Ω 1 + j1 − j1
Z2 =
(−j1)(1) = −j1 Ω 1 + j1 − j1
Z3 =
(j1)(−j1) = 1Ω 1 + j1 − j1
Convert the lower right delta to a wye:
Z4 =
(−j1)(1) = −j1 Ω 1 + j1 − j1
Z5 =
(−j1)(j1) = 1Ω 1 + j1 − j1
Z6 =
(j1)(1) = j1 Ω 1 + j1 − j1
The resulting circuit is shown below:
Simplify the middle portion of the circuit by making series and parallel combinations: (1 + j1 − j1)(1 + 1) = 12 = 2/3 Ω Zab = −j1 + 2/3 + j1 = 2/3 Ω
Problems P 9.36
V o = Vg
Zo 500 − j1000 (100/0◦ ) = 111.8/− 100.3◦ V = ZT 300 + j1600 + 500 − j1000
vo = 111.8 cos(8000t − 100.3◦ ) V P 9.37
1 = −j400 Ω jωC jωL = j1200 Ω Let Z1 = 200 − j400 Ω;
Z2 = 600 + j1200 Ω
Ig = 400/0◦ mA Io =
Z2 600 + j1200 (0.4/0◦ ) Ig = Z1 + Z2 800 + j800
= 450 + j150 mA = 474.34/18.43◦ mA io = 474.34 cos(20,000t + 18.43◦ ) mA P 9.38
V1 = j5(−j2) = 10 V −25 + 10 + (4 − j3)I1 = 0
.·.
I1 =
15 = 2.4 + j1.8 A 4 − j3
I2 = I1 − j5 = (2.4 + j1.8) − j5 = 2.4 − j3.2 A
9–23
9–24
CHAPTER 9. Sinusoidal Steady State Analysis VZ = −j5I2 + (4 − j3)I1 = −j5(2.4 − j3.2) + (4 − j3)(2.4 + j1.8) = −1 − j12 V −25 + (1 + j3)I3 + (−1 − j12) = 0
.·.
I3 = 6.2 − j6.6 A
IZ = I3 − I2 = (6.2 − j6.6) − (2.4 − j3.2) = 3.8 − j3.4 A Z= P 9.39
−1 − j12 VZ = 1.42 − j1.88 Ω = IZ 3.8 − j3.4
Is = 3/0◦ mA 1 = −j0.4 Ω jωC jωL = j0.4 Ω After source transformation we have
Vo =
−j0.4j0.45 (66 × 10−3 ) = 10 mV 28 + −j0.4j0.45
vo = 10 cos 200t mV P 9.40
[a]
Va = (50 + j150)(2/0◦ ) = 100 + j300 V Ib =
100 + j300 = j2.5 A = 2.5/90◦ A 120 − j40
Ic = 2/0◦ + j2.5 + 6 + j3.5 = 8 + j6 A = 10/36.87◦ A Vg = 5Ic + Va = 5(8 + j6) + 100 + j300 = 140 + j330 V = 358.47/67.01◦ V
Problems [b] ib = 2.5 cos(800t + 90◦ ) A ic = 10 cos(800t + 36.87◦ ) A vg = 358.47 cos(800t + 67.01◦ ) V P 9.41
[a] jωL = j(1000)(100) × 10−3 = j100 Ω 106 1 = −j = −j100 Ω jωC (1000)(10)
Using voltage division, Vab =
(100 + j100)(−j100) (247.49/45◦ ) = 350/0◦ j100 + (100 + j100)(−j100)
VTh = Vab = 350/0◦ V [b] Remove the voltage source and combine impedances in parallel to find ZTh = Zab : Yab =
1 1 1 + + = 5 − j5 mS j100 100 + j100 −j100
ZTh = Zab = [c]
1 = 100 + j100 Ω Yab
9–25
9–26 P 9.42
CHAPTER 9. Sinusoidal Steady State Analysis Using voltage division: VTh =
36 (240) = 216 − j72 = 227.68/− 18.43◦ V 36 + j60 − j48
Remove the source and combine impedances in series and in parallel: ZTh = 36(j60 − j48) = 3.6 + j10.8 Ω P 9.43
Open circuit voltage:
V2 − 15 V2 V2 + 88Iφ + =0 10 −j50 Iφ =
5 − (V2 /5) 200
Solving, V2 = −66 + j88 = 110/126.87◦ V = VTh Find the Thévenin equivalent impedance using a test source:
IT =
VT 0.8Vt + 88Iφ + 10 −j50
Iφ =
−VT /5 200
I T = VT
1/5 0.8 1 − 88 + 10 200 −j50
Problems .·. IN =
9–27
VT = 30 − j40 = ZTh IT VTh −66 + j88 = −2.2 + j0 A = 2.2/180◦ A = ZTh 30 − j40
The Norton equivalent circuit:
P 9.44
Short circuit current
Iβ =
−6Iβ 2
2Iβ = −6Iβ ; I1 = 0;
.·. Iβ = 0
.·. Isc = 10/−45◦ A = IN
The Norton impedance is the same as the Thévenin impedance. Find it using a test source
VT = 6Iβ + 2Iβ = 8Iβ ,
Iβ =
j1 IT 2 + j1
9–28
CHAPTER 9. Sinusoidal Steady State Analysis ZTh =
P 9.45
VT 8Iβ j8 = 1.6 + j3.2 Ω = = IT [(2 + j1)/j1]Iβ 2 + j1
Using current division: IN = Isc =
50 (4) = 1.6 − j1.2 = 2/− 36.87◦ A 80 + j60
ZN = −j100(80 + j60) = 100 − j50 Ω The Norton equivalent circuit:
P 9.46
ω = 2π(200/π) = 400 rad/s Zc =
−j = −j2500 Ω 400(10−6 )
VT = (10,000 − j2500)IT + 100(200)IT ZTh =
VT = 30 − j2.5 kΩ IT
P 9.47
IN =
5 − j15 + (1 − j3) mA, ZN
ZN in kΩ
Problems
IN =
−18 − j13.5 + 4.5 − j6 mA, ZN
ZN in kΩ
5 − j15 −18 − j13.5 + 1 − j3 = + (4.5 − j6) ZN ZN 23 − j1.5 = 3.5 − j3 ZN IN =
P 9.48
.·.
ZN = 4 + j3 kΩ
5 − j15 + 1 − j3 = −j6 mA = 6/− 90◦ mA 4 + j3
Open circuit voltage:
V1 − 250 V1 =0 − 0.03Vo + 20 + j10 50 − j100 .·. Vo =
−j100 V1 50 − j100
9–29
9–30
CHAPTER 9. Sinusoidal Steady State Analysis j3V1 V1 250 V1 + + = 20 + j10 50 − j100 50 − j100 20 + j10 V1 = 500 − j250 V;
Vo = 300 − j400 V = VTh = 500/− 53.13◦ V
Short circuit current:
250/0◦ = 3.5 − j0.5 A Isc = 70 + j10 ZTh =
VTh 300 − j400 = 100 − j100 Ω = Isc 3.5 − j0.5
The Thévenin equivalent circuit:
P 9.49
Open circuit voltage:
(9 + j4)Ia − Ib = −60/0◦
Problems −Ia + (9 − j4)Ib = 60/0◦ Solving, Ia = −5 + j2.5 A;
Ib = 5 + j2.5 A
VTh = 4Ia + (4 − j4)Ib = 10/0◦ V Short circuit current:
(9 + j4)Ia − 1Ib − 4Isc = −60 −1Ia + (9 − j4)Ib − (4 − j4)Isc = 60 −4Ia − (4 − j4)Ib + (8 − j4)Isc = 0 Solving, Isc = 2.07/0◦ ZTh =
VTh 10/0◦ = = 4.83 Ω Isc 2.07/0◦
9–31
9–32
CHAPTER 9. Sinusoidal Steady State Analysis Alternate calculation for ZTh :
Z = 4 + 1 + 4 − j4 = 9 − j4
Z1 =
4 9 − j4
Z2 =
4 − j4 9 − j4
Z3 =
16 − j16 9 − j4
Za = 4 + j4 +
56 + j20 4 = 9 − j4 9 − j4
Zb = 4 +
40 − j20 4 − j4 = 9 − j4 9 − j4
Za Zb =
2640 − j320 864 − j384
Z3 + Za Zb =
4176 − j1856 16 − j16 2640 − j320 + = = 4.83 Ω 9 − j4 864 − j384 864 − j384
Problems P 9.50
[a]
IT =
VT − αVT VT + 1000 −j1000
1 IT (1 − α) j−1+α = − = VT 1000 j1000 j1000 .·. ZTh =
VT j1000 = IT α−1+j
ZTh is real when α = 1. [b] ZTh = 1000 Ω [c] ZTh = 500 − j500 = =
j1000 α−1+j
1000(α − 1) 1000 +j 2 (α − 1) + 1 (α − 1)2 + 1
Equate the real parts: 1000 = 500 (α − 1)2 + 1 .·.
.·.
(α − 1)2 = 1 so
(α − 1)2 + 1 = 2 α=0
Check the imaginary parts: (α − 1)1000 = −500 (α − 1)2 + 1 α=1
Thus, α = 0. 1000(α − 1) 1000 +j [d] ZTh = 2 (α − 1) + 1 (α − 1)2 + 1 For Im(ZTh ) > 0, α must be greater than 1. So ZTh is inductive for 1 < α ≤ 10.
9–33
9–34
CHAPTER 9. Sinusoidal Steady State Analysis
P 9.51
V1 V1 − 240 V1 + + =0 j10 50 30 + j10 Solving for V1 yields V1 = 198.63/− 24.44◦ V Vo = P 9.52
30 (V1 ) = 188.43/− 42.88◦ V 30 + j10
jωL = j(2000)(1 × 10−3 ) = j2 Ω 1 106 = −j = −j5 Ω jωC (2000)(100) Vg1 = 20/− 36.87◦ = 16 − j12 V Vg2 = 50/−106.26◦ = −14 − j48 V
Vo − (16 − j12) Vo Vo − (−14 − j48) + + =0 j2 10 −j5 Solving, Vo = 36/0◦ V vo (t) = 36 cos 2000t V
Problems P 9.53
From the solution to Problem 9.52 the phasor-domain circuit is
Making two source transformations yields
Ig1 =
16 − j12 = −6 − j8 A j2
Ig2 =
−14 − j48 = 9.6 − j2.8 A −j5
Y =
1 1 1 + + = (0.1 − j0.3) S j2 10 −j5
Z=
1 = 1 + j3 Ω Y
Ie = Ig1 + Ig2 = 3.6 − j10.8 A Hence the circuit reduces to
Vo = ZIe = (1 + j3)(3.6 − j10.8) = 36/0◦ V .·. vo (t) = 36 cos 2000t V
9–35
9–36 P 9.54
CHAPTER 9. Sinusoidal Steady State Analysis The circuit with the mesh currents identified is shown below:
The mesh current equations are: −20/− 36.87◦ + j2I1 + 10(I1 − I2 ) = 0 50/− 106.26◦ + 10(I2 − I1 ) − j5I2 = 0 In standard form: I1 (10 + j2) + I2 (−10) = 20/− 36.87◦ I1 (−10) + I2 (10 − j5) = −50/− 106.26◦ = 50/73.74◦ Solving on a calculator yields: I1 = −6 + j10A;
I2 = −9.6 + j10A
Thus, Vo = 10(I1 − I2 ) = 36V and vo (t) = 36 cos 2000tV P 9.55
From the solution to Problem 9.52 the phasor-domain circuit with the right-hand source removed is
Vo =
10 − j5 (16 − j12) = 18 − j26 V j2 + 10 − j5
Problems With the left hand source removed
Vo =
10j2 (−14 − j48) = 18 + j26 V −j5 + 10j2
Vo = Vo + Vo = 18 − j26 + 18 + j26 = 36 V vo (t) = 36 cos 2000t V P 9.56
Write a KCL equation at the top node: Vo − 2.4I∆ Vo Vo + + − (10 + j10) = 0 −j8 j4 5 The constraint equation is: I∆ =
Vo −j8
Solving, Vo = j80 = 80/90◦ V P 9.57
Write node voltage equations: Left Node: V1 V1 − Vo /8 + = 0.025/0◦ 40 j20 Right Node: Vo Vo + + 16Io = 0 50 j25
9–37
9–38
CHAPTER 9. Sinusoidal Steady State Analysis The constraint equation is Io =
V1 − Vo /8 j20
Solution: Vo = (4 + j4) = 5.66/45◦ V V1 = (0.8 + j0.6) = 1.0/36.87◦ V Io = (5 − j15) = 15.81/− 71.57◦ mA
P 9.58
(10 + j5)Ia − j5Ib = 100/0◦ −j5Ia − j5Ib = j100 Solving, Ia = −j10 A;
Ib = −20 + j10 A
Io = Ia − Ib = 20 − j20 = 28.28/− 45◦ A io (t) = 28.28 cos(50,000t − 45◦ ) A P 9.59
(12 − j12)Ia − 12Ig − 5(−j8) = 0
Problems −12Ia + (12 + j4)Ig + j20 − 5(j4) = 0 Solving, Ig = 4 − j2 = 4.47/− 26.57◦ A P 9.60
Set up the frequency domain circuit to use the node voltage method:
V1 − V2 V1 − 20/90◦ + =0 −j8 −j4
At V1 :
− 5/0◦ +
At V2 :
V2 − V1 V2 V2 − 20/90◦ + =0 + −j8 j4 12
In standard form:
V1
V1
1 1 1 + + V2 − −j8 −j4 −j8
= 5/0◦ +
1 1 1 1 + + − + V2 −j8 −j8 j4 12
20/90◦ −j4
20/90◦ = 12
Solving on a calculator: 4 8 V1 = − + j V 3 3
V2 = −8 + j4 V
Thus 56 8 = 18.86/− 98.13◦ V V0 = V1 − 20/90◦ = − − j 3 3
9–39
9–40 P 9.61
CHAPTER 9. Sinusoidal Steady State Analysis jωL = j5000(60 × 10−3 ) = j300 Ω −j 1 = = −j100 Ω jωC (5000)(2 × 10−6 )
−400/0◦ + (50 + j300)Ia − 50Ib − 150(Ia − Ib ) = 0 (150 − j100)Ib − 50Ia + 150(Ia − Ib ) = 0 Solving, Ia = −0.8 − j1.6 A;
Ib = −1.6 + j0.8 A
Vo = 100Ib = −160 + j80 = 178.89/153.43◦ V vo = 178.89 cos(5000t + 153.43◦ ) V P 9.62
10/0◦ = (1 − j1)I1 − 1I2 + j1I3 −5/0◦ = −1I1 + (1 + j1)I2 − j1I3
Problems 1 = j1I1 − j1I2 + I3 Solving, I1 = 11 + j10 A;
I2 = 11 + j5 A;
I3 = 6 A
Ia = I3 − 1 = 5 A = 5/0◦ A Ib = I1 − I3 = 5 + j10 A = 11.18/63.43◦ A Ic = I2 − I3 = 5 + j5 A = 7.07/45◦ A Id = I1 − I2 = j5 A = 5/90◦ A P 9.63
Va − (100 − j50) Va Va − (140 + j30) + + =0 20 j5 12 + j16 Solving, Va = 40 + j30 V IZ + (30 + j20) −
140 + j30 (40 + j30) − (140 + j30) + =0 −j10 12 + j16
Solving, IZ = −30 − j10 A Z=
(100 − j50) − (140 + j30) = 2 + j2 Ω −30 − j10
9–41
9–42 P 9.64
CHAPTER 9. Sinusoidal Steady State Analysis [a]
1 = −j50 Ω jωC jωL = j120 Ω Ze = 100 − j50 = 20 − j40 Ω Ig = 2/0◦ Vg = Ig Ze = 2(20 − j40) = 40 − j80 V
Vo =
j120 (40 − j80) = 90 − j30 = 94.87/− 18.43◦ V 80 + j80
vo = 94.87 cos(16 × 105 t − 18.43◦ ) V [b] ω = 2πf = 16 × 105 ;
f=
8 × 105 π
π 1 = = 1.25π µs f 8 × 105 18.43 (1.25π µs) = 201.09 ns .·. 360 .·. vo lags ig by 201.09 ns
T =
P 9.65
jωL = j106 (10 × 10−6 ) = j10 Ω −j 1 = = −j10 Ω 6 jωC (10 )(0.1 × 10−6 ) Va = 50/− 90◦ = −j50 V Vb = 25/90◦ = j25 V
(10 − j10)I1 + j10I2 − 10I3 = −j50
Problems j10I1 + 10I2 − 10I3 = 0 −10I1 − 10I2 + 20I3 = j25 Solving, I1 = 0.5 − j1.5 A;
I3 = −1 + j0.5 A
Ia = −I1 = −0.5 + j1.5 = 1.58/108.43◦ A Ib = −I3 = 1 − j0.5 = 1.12/− 26.57◦ A ia = 1.58 cos(106 t + 108.43◦ ) A ib = 1.12 cos(106 t − 26.57◦ ) A P 9.66
[a] jωL1 = j(5000)(2 × 10−3 ) = j10 Ω jωL2 = j(5000)(8 × 10−3 ) = j40 Ω jωM = j10 Ω
70 = (10 + j10)Ig + j10IL 0 = j10Ig + (30 + j40)IL Solving, Ig = 4 − j3 A;
IL = −1 A
ig = 5 cos(5000t − 36.87◦ ) A iL = 1 cos(5000t − 180◦ ) A [b] k = √
2 M = √ = 0.5 L1 L2 16
I2 = −2.5 A
9–43
9–44
CHAPTER 9. Sinusoidal Steady State Analysis [c] When t = 100π µs, 5000t = (5000)(100π) × 10−6 = 0.5π = π/2 rad = 90◦ ig (100πµs) = 5 cos(53.13◦ ) = 3 A iL (100πµs) = 1 cos(−90◦ ) = 0 A 1 1 1 w = L1 i21 + L2 i22 + M i1 i2 = (2 × 10−3 )(9) + 0 + 0 = 9 mJ 2 2 2 When t = 200π µs, 5000t = π rad = 180◦ ig (200πµs) = 5 cos(180◦ − 36.87◦ ) = −4 A iL (200πµs) = 1 cos(180◦ − 180◦ ) = 1 A 1 1 w = (2 × 10−3 )(16) + (8 × 10−3 )(1) + 2 × 10−3 (−4)(1) = 12 mJ 2 2
P 9.67
Remove the voltage source to find the equivalent impedance:
ZTh
20 = 45 + j125 + |5 + j5|
2
(5 − j5) = 85 + j85 Ω
Using voltage division:
VTh = Vcd
P 9.68
425 = j20I1 = j20 5 + j5
= 850 + j850 V = 1202.1/45◦ V
[a] jωL1 = j(200 × 103 )(10−3 ) = j200 Ω jωL2 = j(200 × 103 )(4 × 10−3 ) = j800 Ω −j 1 = = −j400 Ω 3 jωC (200 × 10 )(12.5 × 10−9 ) .·. Z22 = 100 + 200 + j800 − j400 = 300 + j400 Ω ∗ = 300 − j400 Ω .·. Z22
M = k L1 L2 = 2k × 10−3
Problems ωM = (200 × 103 )(2k × 10−3 ) = 400k
400k Zr = 500
2
(300 − j400) = k 2 (192 − j256) Ω
Zin = 200 + j200 + 192k 2 − j256k 2 1
|Zin | = [(200 + 192k 2 )2 + (200 − 256k 2 )2 ] 2 1 d|Zin | 1 = [(200 + 192k 2 )2 + (200 − 256k 2 )2 ]− 2 × dk 2 [2(200 + 192k 2 )384k + 2(200 − 256k 2 )(−512k)] d|Zin | = 0 when dk 768k(200 + 192k 2 ) − 1024k(200 − 256k 2 ) = 0 √ .·. k = 0.125 = 0.3536 .·. k 2 = 0.125; [b] Zin (min) = 200 + 192(0.125) + j[200 − 0.125(256)] = 224 + j168 = 280/36.87◦ Ω 560/0◦ I1 (max) = = 2/− 36.87◦ A 224 + j168 .·. i1 (peak) = 2 A Note — You can test that the k value obtained from setting d|Zin |/dk = 0 leads to a minimum by noting 0 ≤ k ≤ 1. If k = 1, Zin = 392 − j56 = 395.98/− 8.13◦ Ω Thus, |Zin |k=1 > |Zin |k=√0.125 If k = 0, Zin = 200 + j200 = 282.84/45◦ Ω Thus, |Zin |k=0 > |Zin |k=√0.125 P 9.69
jωL1 = j50 Ω jωL2 = j32 Ω
9–45
9–46
CHAPTER 9. Sinusoidal Steady State Analysis 1 = −j20 Ω jωC
jωM = j(4 × 103 )k (12.5)(8) × 10−3 = j40k Ω Z22 = 5 + j32 − j20 = 5 + j12 Ω ∗ Z22 = 5 − j12 Ω
40k Zr = |5 + j12|
2
(5 − j12) = 47.337k 2 − j113.609k 2
Zab = 20 + j50 + 47.337k 2 − j113.609k 2 = (20 + 47.337k 2 ) + j(50 − 113.609k 2 ) Zab is resistive when 50 − 113.609k 2 = 0
or
k 2 = 0.44 so
k = 0.66
.·. Zab = 20 + (47.337)(0.44) = 40.83 Ω P 9.70
[a] jωLL = j100 Ω jωL2 = j500 Ω Z22 = 300 + 500 + j100 + j500 = 800 + j600 Ω ∗ Z22 = 800 − j600 Ω
ωM = 270 Ω
270 Zr = 1000
2
[800 − j600] = 58.32 − j43.74 Ω
[b] Zab = R1 + jωL1 + Zr = 41.68 + j180 + 58.32 − j43.74 = 100 + j136.26 Ω P 9.71
ZL =
V3 = 80/60◦ Ω I3
Problems V3 V2 = ; 10 1
10I2 = 1I3
V2 V1 =− ; 8 1 Zab =
9–47
8I1 = −1I2
V1 I1
Substituting, Zab =
= P 9.72
82 V2 V1 −8V2 = = I1 −I2 /8 I2 (8)2 (10)2 V3 82 (10V3 ) = = (8)2 (10)2 ZL = 512, 000/60◦ Ω I3 /10 I3
In Eq. 9.69 replace ω 2 M 2 with k 2 ω 2 L1 L2 and then write Xab as Xab
k 2 ω 2 L1 L2 (ωL2 + ωLL ) = ωL1 − 2 R22 + (ωL2 + ωLL )2
= ωL1
k 2 ωL2 (ωL2 + ωLL ) 1− 2 R22 + (ωL2 + ωLL )2
For Xab to be negative requires 2 R22 + (ωL2 + ωLL )2 < k 2 ωL2 (ωL2 + ωLL )
or 2 R22 + (ωL2 + ωLL )2 − k 2 ωL2 (ωL2 + ωLL ) < 0
which reduces to 2 R22 + ω 2 L22 (1 − k 2 ) + ωL2 ωLL (2 − k 2 ) + ω 2 L2L < 0
But k ≤ 1 hence it is impossible to satisfy the inequality. Therefore Xab can never be negative if XL is an inductive reactance.
9–48 P 9.73
CHAPTER 9. Sinusoidal Steady State Analysis [a]
Zab =
Vab V2 V2 = = I1 + I2 I1 + I2 (1 + N1 /N2 )I1
N1 I1 = N2 I2 , V1 N1 = , V2 N2
I2 = V1 =
N1 I1 N2
N1 V2 N2
N1 V1 + V2 = ZL I1 = + 1 V2 N2 Zab =
I1 ZL (N1 /N2 + 1)(1 + N1 /N2 )I1
.·. Zab =
ZL [1 + (N1 /N2 )]2
Q.E.D.
[b] Assume dot on the N2 coil is moved to the lower terminal. Then V1 = −
N1 V2 N2
As before V2 Zab = I1 + I2 .·. Zab = Zab =
N1 I1 N2
and
I2 = −
and
V1 + V2 = ZL I1
V2 Z L I1 = (1 − N1 /N2 )I1 [1 − (N1 /N2 )]2 I1
ZL [1 − (N1 /N2 )]2
Q.E.D.
Problems P 9.74
[a]
Zab =
Vab V1 + V2 = I1 I1
V1 V2 = , N1 N2
V2 =
N1 I1 = N2 I2 ,
N2 V1 N1
I2 =
N1 I1 N2
V2 = (I1 + I2 )ZL = I1
N1 1+ ZL N2
N1 N1 V1 + V2 = + 1 V2 = 1 + N2 N2 .·. Zab =
ZL I1
(1 + N1 /N2 )2 ZL I1 I1
Zab
2
N1 = 1+ N2
2
ZL
Q.E.D.
[b] Assume dot on N2 is moved to the lower terminal, then −V2 V1 = , N1 N2 N1 I1 = −N2 I2 ,
V1 =
−N1 V2 N2
I2 =
−N1 I1 N2
and
Zab =
As in part [a] V2 = (I2 + I1 )ZL Zab =
V1 + V2 I1
(1 − N1 /N2 )V2 (1 − N1 /N2 )(1 − N1 /N2 )ZL I1 = I1 I1
Zab = [1 − (N1 /N2 )]2 ZL
Q.E.D.
9–49
9–50 P 9.75
CHAPTER 9. Sinusoidal Steady State Analysis [a] I =
240 240 + = (10 − j7.5) A 24 j32
Vs = 240/0◦ + (0.1 + j0.8)(10 − j7.5) = 247 + j7.25 = 247.11/1.68◦ V [b] Use the capacitor to eliminate the j component of I, therefore Ic = j7.5 A,
Zc =
240 = −j32 Ω j7.5
Vs = 240 + (0.1 + j0.8)10 = 241 + j8 = 241.13/1.90◦ V [c] Let Ic denote the magnitude of the current in the capacitor branch. Then I = (10 − j7.5 + jIc ) = 10 + j(Ic − 7.5) A Vs = 240/α = 240 + (0.1 + j0.8)[10 + j(Ic − 7.5)] = (247 − 0.8Ic ) + j(7.25 + 0.1Ic ) It follows that 240 cos α = (247 − 0.8Ic ) and
240 sin α = (7.25 + 0.1Ic )
Now square each term and then add to generate the quadratic equation I2c − 605.77Ic + 5325.48 = 0;
Ic = 302.88 ± 293.96
Therefore Ic = 8.92 A (smallest value) and Zc = 240/j8.92 = −j26.90 Ω. P 9.76
The phasor domain equivalent circuit is
Vo =
Vm /0◦ − IRx ; 2
I=
Vm /0◦ Rx − jXC
As Rx varies from 0 to ∞, the amplitude of vo remains constant and its phase angle decreases from 0◦ to −180◦ , as shown in the following phasor diagram:
Problems
P 9.77
9–51
[a]
I =
120 120 + = 16 − j10 A 7.5 j12
V = (0.15 + j6)(16 − j10) = 62.4 + j94.5 = 113.24/56.56◦ V Vs = 120/0◦ + V = 205.43/27.39◦ V [b]
[c] I =
120 120 + = 48 − j30 A 2.5 j4
V = (0.15 + j6)(48 − j30) = 339.73/56.56◦ V Vs = 120 + V = 418.02/42.7◦ V
The amplitude of Vs must be increased from 205.43 V to 418.02 V (more than doubled) to maintain the load voltage at 120 V.
9–52
CHAPTER 9. Sinusoidal Steady State Analysis [d] I =
120 120 120 + + = 48 + j30 A 2.5 j4 −j2
V = (0.15 + j6)(48 + j30) = 339.73/120.57◦ V Vs = 120 + V = 297.23/100.23◦ V
The amplitude of Vs must be increased from 205.43 V to 297.23 V to maintain the load voltage at 120 V. P 9.78
1 = −j20 kΩ jωC Let Va = voltage across the capacitor, positive at upper terminal Then:
Vg = 4/0◦ V;
Va Va Va − 4/0◦ + + = 0; 20,000 −j20,000 20,000 0 − Va 0 − Vo + = 0; 20,000 10,000
Vo = −
.·. Va = (1.6 − j0.8) V
Va 2
.·. Vo = −0.8 + j0.4 = 0.89/153.43◦ V vo = 0.89 cos(200t + 153.43◦ ) V P 9.79
[a]
Va − 4/0◦ Va + jωCo Va + =0 20,000 20,000 Va =
4 2 + j20,000ωCo
Vo = −
Va 2
(see solution to Prob. 9.78)
Problems Vo =
−2 2/180◦ = 2 + j4 × 106 Co 2 + j4 × 106 Co
.·. denominator angle = 45◦ so [b] Vo =
4 × 106 Co = 2
.·.
Co = 0.5 µF
2/180◦ = 0.707/135◦ V 2 + j2
vo = 0.707 cos(200t + 135◦ ) V P 9.80
1 = −j10 kΩ jωC1 1 = −j100 kΩ jωC2
Va Va Va − Vo Va − 2 + + + =0 5000 −j10,000 20,000 100,000 20Va − 40 + j10Va + 5Va + Va − Vo = 0 .·.
(26 + j10)Va − Vo = 40
0 − Vo 0 − Va + =0 20,000 −j100,000 j5Va − Vo = 0 Solving, Vo = 1.43 + j7.42 = 7.55/79.11◦ V vo (t) = 7.55 cos(106 t + 79.11◦ ) V
9–53
9–54 P 9.81
CHAPTER 9. Sinusoidal Steady State Analysis [a] Vg = 25/0◦ V 20 Vg = 5/0◦ ; 100 5 − Vo 5 + =0 80,000 Zp
Vp =
Vn = Vp = 5/0◦ V
Zp = −j80,00040,000 = 32,000 − j16,000 Ω Vo =
5Zp + 5 = 7 − j1 = 7.07/− 8.13◦ V 80,000
vo = 7.07 cos(50,000t − 8.13◦ ) V [b] Vp = 0.2Vm /0◦ ;
Vn = Vp = 0.2Vm /0◦
0.2Vm − Vo 0.2Vm + =0 80,000 32,000 − j16,000 .·. Vo = 0.2Vm +
32,000 − j16,000 Vm (0.2) = 0.2Vm (1.4 − j0.2) 80,000
.·. |0.2Vm (1.4 − j0.2)| ≤ 10 .·. Vm ≤ 35.36 V P 9.82
[a]
1 = −j20 Ω jωC Vn Vn − Vo + =0 20 −j20 Vo Vn Vn = + −j20 20 −j20 Vo = −j1Vn + Vn = (1 − j1)Vn Vp =
Vg Vg (1/jωCo ) = 5 + (1/jωCo ) 1 + j(5)(105 )Co
Vg = 6/0◦ V Vp =
6/0◦ = Vn 1 + j5 × 105 Co
.·. Vo = |Vo | =
(1 − j1)6/0◦ 1 + j5 × 105 Co √ 2(6) 1 + 25 × 1010 Co2
Solving, Co = 2 µF
=6
Problems 6(1 − j1) = −j6 V 1 + j1
[b] Vo =
vo = 6 cos(105 t − 90◦ ) V P 9.83
[a]
Because the op-amps are ideal Iin = Io , thus Zab =
Vab Vab = ; Iin Io
Io =
Vo1 = Vab ;
Vo2
Vab − Vo Z
R2 =− Vo1 = −KVo1 = −KVab R1
Vo = Vo2 = −KVab .·. Io =
Vab − (−KVab ) (1 + K)Vab = Z Z
.·. Zab =
P 9.84
Vab Z Z= (1 + K)Vab (1 + K) 1 ; jωC(1 + K)
[b] Z =
1 ; jωC
[a] I1 =
240 120 + = 23.29 − j13.71 = 27.02/−30.5◦ A 24 8.4 + j6.3
Zab =
.·. Cab = C(1 + K)
I2 =
120 120 − = 5/0◦ A 12 24
I3 =
240 120 + = 28.29 − j13.71 = 31.44/−25.87◦ A 12 8.4 + j6
I4 =
120 = 5/0◦ A; 24
I6 =
240 = 18.29 − j13.71 = 22.86/−36.87◦ A 8.4 + j6.3
I5 =
120 = 10/0◦ A 12
9–55
9–56
CHAPTER 9. Sinusoidal Steady State Analysis
[b]
I1 = 0
I3 = 15 A
I5 = 10 A
I2 = 10 + 5 = 15 A
I4 = −5 A
I6 = 5 A
[c] The clock and television set were fed from the uninterrupted side of the circuit, that is, the 12 Ω load includes the clock and the TV set. [d] No, the motor current drops to 5 A, well below its normal running value of 22.86 A. [e] After fuse A opens, the current in fuse B is only 15 A. P 9.85
[a] The circuit is redrawn, with mesh currents identified:
The mesh current equations are: 120/0◦ = 23Ia − 2Ib − 20Ic 120/0◦ = −2Ia + 43Ib − 40Ic 0 = −20Ia − 40Ib + 70Ic Solving, Ia = 24/0◦ A
Ib = 21.96/0◦ A
The branch currents are: I1 = Ia = 24/0◦ A I2 = Ia − Ib = 2.04/0◦ A I3 = Ib = 21.96/0◦ A I4 = Ic = 19.40/0◦ A I5 = Ia − Ic = 4.6/0◦ A I6 = Ib − Ic = 2.55/0◦ A
Ic = 19.40/0◦ A
Problems
9–57
[b] Let N1 be the number of turns on the primary winding; because the secondary winding is center-tapped, let 2N2 be the total turns on the secondary. From Fig. 9.58, 13,200 240 N2 1 = or = N1 2N2 N1 110 The ampere turn balance requires N1 Ip = N2 I1 + N2 I3 Therefore, N2 1 (24 + 21.96) = 0.42/0◦ A (I1 + I3 ) = Ip = N1 110 Check voltages — V4 = 10I4 = 194/0◦ V V5 = 20I5 = 92/0◦ V V6 = 40I6 = 102/0◦ V All of these voltages are low for a reasonable distribution circuit. P 9.86
[a]
The three mesh current equations are 120/0◦ = 23Ia − 2Ib − 20Ic 120/0◦ = −2Ia + 23Ib − 20Ic 0 = −20Ia − 20Ib + 50Ic Solving, Ia = 24/0◦ A;
Ib = 24/0◦ A;
.·. I2 = Ia − Ib = 0 A [b] Ip = =
N2 N2 (I1 + I3 ) = (Ia + Ib N1 N1 1 (24 + 24) = 0.436/0◦ A 110
Ic = 19.2/0◦ A
9–58
CHAPTER 9. Sinusoidal Steady State Analysis [c] Check voltages — V4 = 10I4 = 10Ic = 192/0◦ V V5 = 20I5 = 20(Ia − Ic ) = 96/0◦ V V6 = 40I6 = 20(Ib − Ic ) = 96/0◦ V Where the two loads are equal, the current in the neutral conductor (I2 ) is zero, and the voltages V5 and V6 are equal. The voltages V4 , V5 , and V6 are too low for a reasonable dirtribution circuit.
P 9.87
[a]
125 = (R + 0.05 + j0.05)I1 − (0.03 + j0.03)I2 − RI3 125 = −(0.03 + j0.03)I1 + (R + 0.05 + j0.05)I2 − RI3 Subtracting the above two equations gives 0 = (R + 0.08 + j0.08)I1 − (R + 0.08 + j0.08)I2 .·. I1 = I2
so
In = I1 − I2 = 0 A
[b] V1 = R(I1 − I3 );
V2 = R(I2 − I3 )
Since I1 = I2 (from part [a]) V1 = V2 [c]
Problems
9–59
250 = (660.04 + j0.04)Ia − 660Ib 0 = −660Ia + 670Ib Solving, Ia = 25.28/− 0.23◦ = 25.28 − j0.10 A Ib = 24.90/− 0.23◦ = 24.90 − j0.10 A I1 = Ia − Ib = 0.377 − j0.00153 A V1 = 60I1 = 22.63 − j0.0195 = 22.64/− 0.23◦ V V2 = 600I1 = 226.3 − j0.915 = 226.4/− 0.23◦ V [d]
125 = (60.05 + j0.05)I1 − (0.03 + j0.03)I2 − 60I3 125 = −(0.03 + j0.03)I1 + (600.05 + j0.05)I2 − 600I3 0 = −60I1 − 600I2 + 670I3 Solving, I1 = 26.97/− 0.24◦ = 26.97 − j0.113 A I2 = 25.10/− 0.24◦ = 25, 10 − j0.104 A I3 = 24.90/− 0.24◦ = 24.90 − j0.104 A V1 = 60(I1 − I3 ) = 124.4/− 0.27◦ V V2 = 600(I2 − I3 ) = 124.6/− 0.20◦ V [e] Because an open neutral can result in severely unbalanced voltages across the 125 V loads.
9–60 P 9.88
CHAPTER 9. Sinusoidal Steady State Analysis [a] Let N1 = primary winding turns and 2N2 = secondary winding turns. Then 250 14,000 = ; N1 2N2
.·.
N2 1 =a = N1 112
In part c), Ip = 2aIa .·. Ip = =
2N2 Ia 1 = Ia N1 56 1 (25.28 − j0.10) 56
Ip = 451.4 − j1.8 mA = 451.4/− 0.23◦ mA In part d), Ip N1 = I1 N2 + I2 N2 .·. Ip =
N2 (I1 + I2 ) N1
=
1 (26.97 − j0.11 + 25.10 − j0.10) 112
=
1 (52.07 − j0.22) 112
Ip = 464.9 − j1.9 mA = 464.9/− 0.24◦ mA [b] Yes, because the neutral conductor carries non-zero current whenever the load is not balanced.
10 Sinusoidal Steady State Power Calculations
Assessment Problems AP 10.1 [a] V = 100/− 45◦ V, I = 20/15◦ A Therefore 1 P = (100)(20) cos[−45 − (15)] = 500 W, 2 Q = 1000 sin −60◦ = −866.03 VAR, [b] V = 100/− 45◦ ,
B→A
I = 20/165◦
P = 1000 cos(−210◦ ) = −866.03 W,
B→A
Q = 1000 sin(−210◦ ) = 500 VAR, [c] V = 100/− 45◦ ,
A→B
I = 20/− 105◦
P = 1000 cos(60◦ ) = 500 W,
A→B
Q = 1000 sin(60◦ ) = 866.03 VAR, [d] V = 100/0◦ ,
A→B
A→B
I = 20/120◦
P = 1000 cos(−120◦ ) = −500 W,
B→A
Q = 1000 sin(−120◦ ) = −866.03 VAR,
B→A
AP 10.2 pf = cos(θv − θi ) = cos[15 − (75)] = cos(−60◦ ) = 0.5 leading rf = sin(θv − θi ) = sin(−60◦ ) = −0.866
10–1
10–2
CHAPTER 10. Sinusoidal Steady State Power Calculations Iρ 0.18 Ieff = √ = √ A 3 3
AP 10.3 From Ex. 9.4
2 P = Ieff R=
0.0324 (5000) = 54 W 3
AP 10.4 [a] Z = (39 + j26)(−j52) = 48 − j20 = 52/− 22.62◦ Ω Therefore I =
250/0◦ = 4.85/18.08◦ A(rms) 48 − j20 + 1 + j4
VL = ZI = (52/− 22.62◦ )(4.85/18.08◦ ) = 252.20/− 4.54◦ V(rms) IL =
VL = 5.38/− 38.23◦ A(rms) 39 + j26
[b] SL = VL I∗L = (252.20/− 4.54◦ )(5.38/+ 38.23◦ ) = 1357/33.69◦ = (1129.09 + j752.73) VA PL = 1129.09 W;
QL = 752.73 VAR
[c] P = |I |2 1 = (4.85)2 · 1 = 23.52 W;
Q = |I |2 4 = 94.09 VAR
[d] Sg (delivering) = 250I∗ = (1152.62 − j376.36) VA Therefore the source is delivering 1152.62 W and absorbing 376.36 magnetizing VAR. (252.20)2 |VL |2 = = −1223.18 VAR −52 −52 Therefore the capacitor is delivering 1223.18 magnetizing VAR.
[e] Qcap =
Check:
94.09 + 752.73 + 376.36 = 1223.18 VAR and 1129.09 + 23.52 = 1152.62 W
AP 10.5 Series circuit derivation: S = 250I∗ = (40,000 − j30,000) Therefore I∗ = 160 − j120 = 200/− 36.87◦ A(rms) I = 200/36.87◦ A(rms) Z=
250 V = = 1.25/− 36.87◦ = (1 − j0.75) Ω I 200/36.87◦
Therefore R = 1 Ω,
XC = −0.75 Ω
Problems
10–3
Parallel circuit derivation: P =
(250)2 ; R
therefore R =
Q=
(250)2 ; XC
therefore XC =
(250)2 = 1.5625 Ω 40,000 (250)2 = −2.083 Ω −30,000
AP 10.6 S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA S2 = 6000(0.8) + j6000(0.6) = 4800 − j3600 VA ST = S1 + S2 = 13,800 + j8400 VA ST = 200I∗ ;
therefore I∗ = 69 + j42
I = 69 − j42 A
Vs = 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.64/15.91◦ V(rms) AP 10.7 [a] The phasor domain equivalent circuit and the Thévenin equivalent are shown below: Phasor domain equivalent circuit:
Thévenin equivalent:
VTh = 3
−j800 = 48 − j24 = 53.67/− 26.57◦ V 20 − j40
ZTh = 4 + j18 +
−j800 = 20 + j10 = 22.36/26.57◦ Ω 20 − j40
For maximum power transfer, ZL = (20 − j10) Ω
10–4
CHAPTER 10. Sinusoidal Steady State Power Calculations [b] I =
53.67/− 26.57◦ = 1.34/− 26.57◦ A 40
Therefore P =
1.34 √ 2
2
20 = 18 W
[c] RL = |ZTh | = 22.36 Ω 53.67/− 26.57◦ = 1.23/− 39.85◦ A [d] I = 42.36 + j10
Therefore P =
1.23 √ 2
2
(22.36) = 17 W
AP 10.8
Mesh current equations: 660 = (34 + j50)I1 + j100(I1 − I2 ) + j40I1 + j40(I1 − I2 ) 0 = j100(I2 − I1 ) − j40I1 + 100I2 Solving, I1 = 3.536/− 45◦ A, I2 = 3.5/0◦ A;
1 .·. P = (3.5)2 (100) = 612.50 W 2
AP 10.9 [a]
248 = j400I1 − j500I2 + 375(I1 − I2 ) 0 = 375(I2 − I1 ) + j1000I2 − j500I1 + 400I2 Solving, I1 = 0.80 − j0.62 A;
I2 = 0.4 − j0.3 = 0.5/− 36.87◦ A
1 .·. P = (0.25)(400) = 50 W 2
Problems [b] I1 − I2 = 0.4 − j0.32 A 1 P375 = |I1 − I2 |2 (375) = 49.20 W 2 1 [c] Pg = (248)(0.8) = 99.20 W 2
Pabs = 50 + 49.2 = 99.20 W
V2 = 14 V1 ; AP 10.10 [a] VTh = 210/0◦ V; Short circuit equations:
(checks) I1 = 14 I2
840 = 80I1 − 20I2 + V1 0 = 20(I2 − I1 ) − V2 .·. I2 = 14 A;
[b] Pmax
210 = 30
2
RTh =
210 = 15 Ω 14
15 = 735 W
AP 10.11 [a] VTh = −4(146/0◦ ) = −584/0◦ V(rms) = 584/180◦ V(rms) V2 = 4V1 ;
I1 = −4I2
Short circuit equations: 146/0◦ = 80I1 − 20I2 + V1 0 = 20(I2 − I1 ) + V2 .·. I2 = −146/365 = −0.40 A;
−584 [b] P = 2920
2
1460 = 58.40 W
RTh =
−584 = 1460 Ω −0.4
10–5
10–6
CHAPTER 10. Sinusoidal Steady State Power Calculations
Problems P 10.1
1 [a] P = (100)(10) cos(50 − 15) = 500 cos 35◦ = 409.58 W 2 Q = 500 sin 35◦ = 286.79 VAR
(abs)
(abs)
1 [b] P = (40)(20) cos(−15 − 60) = 400 cos(−75◦ ) = 103.53 W 2 Q = 400 sin(−75◦ ) = −386.37 VAR
(abs)
(del)
1 [c] P = (400)(10) cos(30 − 150) = 2000 cos(−120◦ ) = −1000 W 2 Q = 2000 sin(−120◦ ) = −1732.05 VAR
(del)
1 [d] P = (200)(5) cos(160 − 40) = 500 cos(120◦ ) = −250 W 2 Q = 500 sin(120◦ ) = 433.01 VAR P 10.2
p = P + P cos 2ωt − Q sin 2ωt; dp = 0 when dt
cos 2ωt = √
P ; + Q2
(del)
(abs)
dp = −2ωP sin 2ωt − 2ωQ cos 2ωt dt
− 2ωP sin 2ωt = 2ωQ cos 2ωt or
P2
(del)
sin 2ωt = − √
tan 2ωt = −
Q P
Q + Q2
P2
Let θ = tan−1 (−Q/P ), then p is maximum when 2ωt = θ and p is minimum when 2ωt = (θ + π). Therefore pmax = P + P · √
and
pmin = P − P · √
P Q(−Q) √ − = P + P 2 + Q2 P 2 + Q2 P 2 + Q2
P Q √ − Q · = P − P 2 + Q2 P 2 + Q2 P 2 + Q2
Problems P 10.3
[a] hair dryer = 600 W sun lamp = 279 W television = 240 W
vacuum = 630 W air conditioner = 860 W
P = 2609 W
2609 = 21.74 A 120 Yes, the breaker will trip.
Therefore Ieff =
[b]
1700 = 14.17 A 120 Yes, the breaker will not trip if the current is reduced to 14.17 A.
P = 2609 − 909 = 1700 W;
P 10.4
[a] Ieff = 40/115 ∼ = 0.35 A;
P 10.5
Wdc =
Vdc2 T; R
.·.
to +T 2 vs Vdc2 T = dt R R to
Vdc2
1 to +T 2 = vs dt T to
Vdc = P 10.6
Ws =
to +T to
Ieff =
[b] Ieff = 130/115 ∼ = 1.13 A vs2 dt R
1 to +T 2 vs dt = Vrms = Veff T to
[a] Area under one cycle of vg2 : A = (52 )(2)(30 × 10−6 ) + 22 (2)(37.5 × 10−6 ) = 1800 × 10−6 Mean value of vg2 : 1800 × 10−6 A = =9 200 × 10−6 200 × 10−6 √ .·. Vrms = 9 = 3 V(rms) M.V. =
[b] P = P 10.7
2 Vrms 32 = 4W = R 2.25
i(t) = 200t
0 ≤ t ≤ 75 ms
i(t) = 60 − 600t
Irms =
1 0.1 =
0.075 0
75 ms ≤ t ≤ 100 ms (200)2 t2 dt +
0.1 0.075
10(5.625) + 10(1.875) =
(60 − 600t)2 dt √
75 = 8.66 A(rms)
10–7
10–8
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.8
2 R P = Irms
P 10.9
Ig = 40/0◦ mA
.·. R =
jωL = j10,000 Ω;
Io =
3 × 103 = 40 Ω 75
1 = −j10,000 Ω jωC
j10,000 (40/0◦ ) = 80/90◦ mA 5000
1 1 P = |Io |2 (5000) = (0.08)2 (5000) = 16 W 2 2 1 Q = |Io |2 (−10,000) = −32 VAR 2 S = P + jQ = 16 − j32 VA |S| = 35.78 VA P 10.10 Ig = 4/0◦ mA;
1 = −j1250 Ω; jωC
jωL = j500 Ω
Zeq = 500 + [−j1250(1000 + j500)] = 1500 − j500 Ω 1 1 Pg = − |I|2 Re{Zeq } = − (0.004)2 (1500) = −12 mW 2 2 The source delivers 12 mW of power to the circuit.
Problems P 10.11 jωL = j105 (0.5 × 10−3 ) = j50 Ω;
−4 +
Vo Vo − 50I∆ =0 + j50 40 − j30
I∆ =
Vo j50
1 1 = −j30 Ω = 5 jωC j10 [(1/3) × 10−6 ]
Place the equations in standard form:
Vo
Vo
1 −50 1 + I∆ + j50 40 − j30 40 − j30
=4
1 + I∆ (−1) = 0 j50
Solving, Vo = 200 − j400 V;
I∆ = −8 − j4 A
Io = 4 − (−8 − j4) = 12 + j4 A 1 1 P40Ω = |Io |2 (40) = (160)(40) = 3200 W 2 2 P 10.12 [a] line loss = 7500 − 2500 = 5 kW line loss = |Ig |2 20
|Ig | =
√
250 A
10–9
.·. |Ig |2 = 250
10–10
CHAPTER 10. Sinusoidal Steady State Power Calculations |Ig |2 RL = 2500 |Ig |2 XL = −5000
.·. RL = 10 Ω .·. XL = −20 Ω
Thus,
|Z| =
(30)2 + (X − 20)2
|Ig | =
500 900 + (X − 20)2
25 × 104 = 1000 250 (X − 20) = ±10.
.·. 900 + (X − 20)2 = Solving, Thus,
X = 10 Ω
or
X = 30 Ω
[b] If X = 30 Ω: 500 = 15 − j5 A Ig = 30 + j10 Sg = −500I∗g = −7500 − j2500 VA Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars. Qj30 = |Ig |2 X = 250(30) = 7500 VAR Therefore the line reactance is absorbing 7500 magnetizing vars. Q−j20 = |Ig |2 XL = 250(−20) = −5000 VAR Therefore the load reactance is generating 5000 magnetizing vars.
Qgen = 7500 VAR =
Qabs
If X = 10 Ω: 500 = 15 + j5 A Ig = 30 − j10 Sg = −500I∗g = −7500 + j2500 VA Thus, the voltage source is delivering 7500 W and absorbing 2500 magnetizing vars. Qj10 = |Ig |2 (10) = 250(10) = 2500 VAR Therefore the line reactance is absorbing 2500 magnetizing vars. The load continues to generate 5000 magnetizing vars.
Qgen = 5000 VAR =
Qabs
Problems P 10.13 Zf = −j10,00020,000 = 4000 − j8000 Ω Zi = 2000 − j2000 Ω .·.
Zf 4000 − j8000 = 3 − j1 = Zi 2000 − j2000
Vo = −
Zf Vg ; Zi
Vg = 1/0◦ V
Vo = (3 − j1)(1) = 3 − j1 = 3.16/− 18.43◦ V P =
1 (10) 1 Vm2 = = 5 × 10−3 = 5 mW 2 R 2 1000
P 10.14 [a] P = −
1 (240)2 = 60 W 2 480
1 −9 × 106 = = −360 Ω ωC (5000)(5)
Q=
1 (240)2 = −80 VAR 2 (−360)
pmax = P + = 60 + (60)2 + (80)2 = 160 W(del) + √ [b] pmin = 60 − 602 + 802 = −40 W(abs) P2
[c] P = 60 W
Q2
from (a)
[d] Q = −80 VAR from (a) [e] generate, because Q < 0 [f] pf = cos(θv − θi ) I=
240 240 + = 0.5 + j0.67 = 0.83/53.13◦ A 480 −j360
.·. pf = cos(0 − 53.13◦ ) = 0.6 leading [g] rf = sin(−53.13◦ ) = −0.8
10–11
10–12
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.15 [a]
The mesh equations are: (10 − j20)I1 + (j20)I2 = 170 (j20)I1 + (12 − j4)I2 = 0 Solving, I1 = 4 + j1 A;
I2 = 3.5 − j5.5 A
S = −Vg I∗1 = −(170)(4 − j1) = −680 + j170 VA [b] Source is delivering 680 W. [c] Source is absorbing 170 magnetizing VAR. √ [d] P10Ω = ( 17)2 (10) = 170 W √ P12Ω = ( 42.5)2 (12) = 510 W (I1 − I2 ) = 0.5 + j6.5 A √ √ Q−j20Ω = ( 42.5)2 (20) = −850 VAR |I1 − I2 | = 42.5 √ Qj16Ω = ( 42.5)2 (16) = 680 VAR [e]
Pdel = 680 W
.·. [f]
Pdiss = 170 + 510 = 680 W
Pdel =
Pdiss = 680 W
Qabs = 170 + 680 = 850 VAR
.·.
Qdev = 850 VAR
mag VAR dev =
mag VAR abs = 850
Problems P 10.16 [a]
1 = −j40 Ω; jωC
jωL = j80 Ω
Zeq = 40 − j40 + j80 + 60 = 80 + j60 Ω Ig =
40/0◦ = 0.32 − j0.24 A 80 + j60
1 1 Sg = − Vg I∗g = − 40(0.32 + j0.24) = −6.4 − j4.8 VA 2 2 P = 6.4 W(del);
Q = 4.8 VAR(del)
|S| = |Sg | = 8 VA [b] I1 =
−j40 Ig = 0.04 − j0.28 A 40 − j40
1 P40Ω = |I1 |2 (40) = 1.6 W 2 1 P60Ω = |Ig |2 (60) = 4.8 W 2
Pdiss = 1.6 + 4.8 = 6.4 W =
Pdev
[c] I−j40Ω = Ig − I1 = 0.28 + j0.04 A 1 Q−j40Ω = |I−j40Ω |2 (−40) = −1.6 VAR(del) 2 1 Qj80Ω = |Ig |2 (80) = 6.4 VAR(abs) 2
Qabs = 6.4 − 1.6 = 4.8 VAR =
P 10.17 [a] Z1 = 240 + j70 = 250/16.26◦ Ω pf = cos(16.26◦ ) = 0.96 lagging rf = sin(16.26◦ ) = 0.28
Qdev
10–13
10–14
CHAPTER 10. Sinusoidal Steady State Power Calculations Z2 = 160 − j120 = 200/− 36.87◦ Ω pf = cos(−36.87◦ ) = 0.80 leading rf = sin(−36.87◦ ) = −0.60 Z3 = 30 − j40 = 50/− 53.13◦ Ω pf = cos(−53.13◦ ) = 0.6 leading rf = sin(−53.13◦ ) = −0.8
[b] Y = Y1 + Y2 + Y3 Y1 =
1 ; 250/16.26◦
Y2 =
1 ; 200/− 36.87◦
Y3 =
1 50/− 53.13◦
Y = 19.84 + j17.88 mS Z=
1 = 37.44/− 42.03◦ Ω Y
pf = cos(−42.03◦ ) = 0.74 leading rf = sin(−42.03◦ ) = −0.67 P 10.18 [a] S1 = 16 + j18 kVA;
S2 = 6 − j8 kVA;
S3 = 8 + j0 kVA
ST = S1 + S2 + S3 = 30 + j10 kVA 250I∗ = (30 + j10) × 103 ; Z=
.·. I = 120 − j40 A
250 = 1.875 + j0.625 Ω = 1.98/18.43◦ Ω 120 − j40
[b] pf = cos(18.43◦ ) = 0.9487 lagging P 10.19 [a] From the solution to Problem 10.18 we have IL = 120 − j40 A(rms) .·. Vs = 250/0◦ + (120 − j40)(0.01 + j0.08) = 254.4 + j9.2 = 254.57/2.07◦ V(rms) [b] |IL | =
16,000
P = (16,000)(0.01) = 160 W [c] Ps = 30,000 + 160 = 30.16 kW 30 (100) = 99.47% [d] η = 30.16
Q = (16,000)(0.08) = 1280 VAR Qs = 10,000 + 1280 = 11.28 kVAR
Problems P 10.20 ST = 4500 − j S1 =
4500 (0.28) = 4500 − j1312.5 VA 0.96
2700 (0.8 + j0.6) = 2700 + j2025 VA 0.8
S2 = ST − S1 = 1800 − j3337.5 = 3791.95/− 61.66◦ VA pf = cos(−61.66◦ ) = 0.4747 leading P 10.21
2400I∗1 = 60,000 + j40,000 I∗1 = 25 + j16.67;
.·. I1 = 25 − j16.67 A(rms)
2400I∗2 = 20,000 − j10,000 I∗2 = 8.33 − j4, 167; I3 =
.·. I2 = 8.33 + j4.167 A(rms)
2400/0◦ = 16.67 + j0 A; 144
I4 =
2400/0◦ = 0 − j25 A j96
Ig = I1 + I2 + I3 + I4 = 50 − j37.5 A Vg = 2400 + (j4)(50 − j37.5) = 2550 + j200 = 2557.83/4.48◦ V(rms) P 10.22 [a] S1 = 60,000 − j70,000 VA S2 =
|VL |2 (2500)2 = 240,000 + j70,000 VA = Z2∗ 24 − j7
S1 + S2 = 300,000 VA 2500I∗L = 300,000;
.·. IL = 120/0◦ A(rms)
Vg = VL + IL (0.1 + j1) = 2500 + (120)(0.1 + j1) = 2512 + j120 = 2514.86/2.735◦ V(rms)
10–15
10–16
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] T =
1 1 = = 16.67 ms f 60
t 2.735◦ = ; ◦ 360 16.67 ms
.·. t = 126.62 µs
[c] VL lags Vg by 2.735◦ or 126.62 µs
P 10.23 [a] From the solution to Problem 9.56 we have:
Vo = j80 = 80/90◦ V 1 1 Sg = − Vo I∗g = − (j80)(10 − j10) = −400 − j400 VA 2 2 Therefore, the independent current source is delivering 400 W and 400 magnetizing vars. Vo = j16 A 5 1 = (16)2 (5) = 640 W 2
I1 = P5Ω
Therefore, the 8 Ω resistor is absorbing 640 W. I∆ =
Vo = −10 A −j8
1 Qcap = (10)2 (−8) = −400 VAR 2 Therefore, the −j8 Ω capacitor is developing 400 magnetizing vars. 2.4I∆ = −24 V I2 =
j80 + 24 Vo − 2.4I∆ = j4 j4
= 20 − j6 A = 20.88/− 16.7◦ A
Problems
10–17
1 Qj4 = |I2 |2 (4) = 872 VAR 2 Therefore, the j4 Ω inductor is absorbing 872 magnetizing vars. Sd.s. = 12 (2.4I∆ )I∗2 = 12 (−24)(20 + j6) = −240 − j72 VA Thus the dependent source is delivering 240 W and 72 magnetizing vars.
[b] [c]
Pgen = 400 + 240 = 640 W =
Pabs
Qgen = 400 + 400 + 72 = 872 VAR =
Qabs
P 10.24 [a] From the solution to Problem 9.58 we have
Ia = −j10 A;
Ib = −20 + j10 A;
Io = 20 − j20 A
1 S100V = − (100)I∗a = −50(j10) = −j500 VA 2 Thus, the 100 V source is developing 500 magnetizing vars. Sj100V = − 12 (j100)I∗b = −j50(−20 − j10) = −500 + j1000 VA Thus, the j100 V source is developing 500 W and absorbing 1000 magnetizing vars. 1 P10Ω = |Ia |2 (10) = 500 W 2 Thus the 10 Ω resistor is absorbing 500 W. 1 Q−j10Ω = |Ib |2 (−10) = −2500 VAR 2 Thus the −j10 Ω capacitor is developing 2500 magnetizing vars. 1 Qj5Ω = |Io |2 (5) = 2000 VAR 2 Thus the j5 Ω inductor is absorbing 2000 magnetizing vars.
[b]
Pdev = 500 W =
Pabs
10–18 [c]
CHAPTER 10. Sinusoidal Steady State Power Calculations
Qdev = 500 + 2500 = 3000 VAR
Qabs = 1000 + 2000 = 3000 VAR =
Qdev
465/0◦ P 10.25 [a] I = = 2.4 − j1.8 = 3/− 36.87◦ A(rms) 124 + j93 P = (3)2 (4) = 36 W [b] YL =
1 = 5.33 − j4 mS 120 + j90
.·. XC =
1 = −250 Ω −4 × 10−3
1 = 187.5 Ω 5.33 × 10−3 465/0◦ [d] I = = 2.43/− 0.9◦ A 191.5 + j3 [c] ZL =
P = (2.43)2 (4) = 23.58 W 23.58 (100) = 65.5% 36 Thus the power loss after the capacitor is added is 65.6% of the power loss before the capacitor is added.
[e] % =
P 10.26 [a]
250I∗1 = 7500 + j2500;
.·. I1 = 30 − j10 A(rms)
250I∗2 = 2800 − j9600;
.·. I2 = 11.2 + j38.4 A(rms)
I3 =
500 500 + = 40 − j10 A(rms) 12.5 j50
Ig1 = I1 + I3 = 70 − j20 A Sg1 = 250(70 + j20) = 17,500 + j5000 VA
Problems
10–19
Thus the Vg1 source is delivering 17.5 kW and 5000 magnetizing vars. Ig2 = I2 + I3 = 51.2 + j28.4 A(rms) Sg2 = 250(51.2 − j28.4) = 12,800 − j7100 VA Thus the Vg2 source is delivering 12.8 kW and absorbing 7100 magnetizing vars.
[b]
Pgen = 17.5 + 12.8 = 30.3 kW
Pabs = 7500 + 2800 +
(500)2 = 30.3kW = Pgen 12.5
Qdel = 9600 + 5000 = 14.6 kVAR Qabs
(500)2 = 14.6 kVAR = = 2500 + 7100 + Qdel 50
P 10.27 S1 = 1200 + 1196 = 2396 + j0 VA .·. I1 =
2396 = 19.97 A 120
S2 = 860 + 600 + 240 = 1700 + j0 VA .·. I2 =
1700 = 14.167 A 120
S3 = 4474 + 12,200 = 16,674 + j0 VA .·. I3 =
16,674 = 69.48 A 240
Ig1 = I1 + I3 = 89.44 A Ig2 = I2 + I3 = 83.64 A Breakers will not trip since both feeder currents are less than 100 A. P 10.28 [a]
I1 =
4000 − j1000 = 32 − j8 A (rms) 125
10–20
CHAPTER 10. Sinusoidal Steady State Power Calculations I2 =
5000 − j2000 = 40 − j16 A (rms) 125
I3 =
10,000 + j0 = 40 + j0 A (rms) 250
.·. Ig1 = I1 + I3 = 72 − j8 A (rms) In = I1 − I2 = −8 + j8 A (rms) Ig2 = I2 + I3 = 80 − j16 A(rms) Vg1 = 0.05Ig1 + 125 + 0.15In = 127.4 + j0.8 V(rms) Vg2 = −0.15In + 125 + 0.05Ig2 = 130.2 − j2 V(rms) Sg1 = [(127.4 + j0.8)(72 + j8)] = [9166.4 + j1076.8] VA Sg2 = [(130.2 − j2)(80 + j16)] = [10,448 + j1923.2] VA Note: Both sources are delivering average power and magnetizing VAR to the circuit. [b] P0.05 = |Ig1 |2 (0.05) = 262.4 W P0.15 = |In |2 (0.15) = 19.2 W P0.05 = |Ig2 |2 (0.05) = 332.8 W
Pdis = 262.4 + 19.2 + 332.8 + 4000 + 5000 + 10,000 = 19,614.4 W Pdev = 9166.4 + 10,448 = 19,614.4 W =
Pdis
Qabs = 1000 + 2000 = 3000 VAR Qdel = 1076.8 + 1923.2 = 3000 VAR =
P 10.29 [a] Let VL = Vm /0◦ :
SL = 600(0.8 + j0.6) = 480 + j360 VA I∗ =
480 360 +j ; Vm Vm
I =
480 360 −j Vm Vm
Qabs
Problems
480 360 −j (1 + j2) 120/θ = Vm + Vm Vm 120Vm /θ = Vm2 + (480 − j360)(1 + j2) = Vm2 + 1200 + j600 120Vm cos θ = Vm2 + 1200;
120Vm sin θ = 600
(120)2 Vm2 = (Vm2 + 1200)2 + 6002 14,400Vm2 = Vm4 + 2400Vm2 + 18 × 105 or Vm4 − 12,000Vm2 + 18 × 105 = 0 Solving, Vm = 108.85 V and Vm = 12.326 V If Vm = 108.85 V: sin θ =
600 = 0.045935; (108.85)(120)
.·. θ = 2.63◦
If Vm = 12.326 V: sin θ =
600 = 0.405647; (12.326)(120)
.·. θ = 23.93◦
[b]
P 10.30 [a] SL = 20,000(0.85 + j0.53) = 17,000 + j10,535.65 VA 125I∗L = (17,000 + j10,535.65);
I∗L = 136 + j84.29 A(rms)
.·. IL = 136 − j84.29 A(rms) Vs = 125 + (136 − j84.29)(0.01 + j0.08) = 133.10 + j10.04 = 133.48/4.31◦ V(rms) |Vs | = 133.48 V(rms) [b] P = |I |2 (0.01) = (160)2 (0.01) = 256 W
10–21
10–22
CHAPTER 10. Sinusoidal Steady State Power Calculations
[c]
(125)2 = −10,535.65; XC −
1 = −1.48; ωC
XC = −1.483 Ω C=
1 = 1788.59 µF (1.48)(120π)
[d] I = 136 + j0 A(rms) Vs = 125 + 136(0.01 + j0.08) = 126.36 + j10.88 = 126.83/4.92◦ V(rms) |Vs | = 126.83 V(rms) [e] P = (136)2 (0.01) = 184.96 W P 10.31
IL =
153,600 − j115,200 = 32 − j24 A(rms) 4800
IC =
4800 4800 =j = jIC −jXC XC
I = 32 − j24 + jIC = 32 + j(IC − 24) Vs = 4800 + (2 + j10)[32 + j(IC − 24)] = (5104 − 10IC ) + j(272 + 2IC ) |Vs |2 = (5104 − 10IC )2 + (272 + 2IC )2 = (4800)2 .·. 104IC2 − 100,992IC + 3,084,800 = 0 Solving,
IC = 31.57 A(rms);
IC = 939.51 A(rms)
*Select the smaller value of IC to minimize the magnitude of I . .·. XC = − .·. C =
4800 = −152.04 31.57
1 = 17.45 µF (152.04)(120π)
Problems
10–23
P 10.32 ZL = |ZL |/θ◦ = |ZL | cos θ◦ + j|ZL | sin θ◦ Thus
|VTh |
|I| =
(RTh + |ZL | cos θ)2 + (XTh + |ZL | sin θ)2
Therefore P =
0.5|VTh |2 |ZL | cos θ (RTh + |ZL | cos θ)2 + (XTh + |ZL | sin θ)2
Let D = demoninator in the expression for P, then (0.5|VTh |2 cos θ)(D · 1 − |ZL |dD/d|ZL |) dP = d|ZL | D2 dD = 2(RTh + |ZL | cos θ) cos θ + 2(XTh + |ZL | sin θ) sin θ d|ZL | dP = 0 when d|ZL |
dD D = |ZL | d|ZL |
Substituting the expressions for D and (dD/d|ZL |) into this equation gives us the 2 2 relationship RTh + XTh = |ZL |2 or |ZTh | = |ZL |. P 10.33 [a] ZTh = j4040 − j40 = 20 − j20 ∗ .·. ZL = ZTh = 20 + j20 Ω
[b] VTh =
40 (120) = 60 − j60 V 40 + j40
60 − j60 = 1.5 − j1.5 A 40 1 Pload = |I|2 (20) = 45 W 2
I=
P 10.34 [a]
115.2 + j33.6 − 240 115.2 + j33.6 =0 + ZTh 80 − j60 .·. ZTh = 40 − j100 Ω .·. ZL = 40 + j100 Ω
10–24
CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] I =
240 = 3 A(rms) 80
P = (3)2 (40) = 360 W P 10.35 [a] ZTh = [(3 + j4) − j8] + 7.32 − j17.24 = 15 − j15 Ω .·. R = |ZTh | = 21.21 Ω [b] VTh =
I=
−j8 (112.5) = 144 − j108 V(rms) 3 − j4
144 − j108 = 4.45 − j1.14 35.21 − j15
P = |I|2 (21.21) = 447.35 W P 10.36 [a] Open circuit voltage:
V1 = 5Iφ = 5
100 − 5Iφ 25 + j10
(25 + j10)Iφ = 100 − 5Iφ Iφ =
100 = 3 − j1 A 30 + j10
VTh =
j3 (5Iφ ) = 15/0◦ V 1 + j3
Problems Short circuit current:
V2 = 5Iφ =
100 − 5Iφ 25 + j10
Iφ = 3 − j1 A 5Iφ = 15 − j5 A 1 15 = 0.9 + j0.3 Ω = 15 − j5
Isc = ZTh
∗ ZL = ZTh = 0.9 − j0.3 Ω
IL =
15 = 8.33 A(rms) 1.8
P = |IL |2 (0.9) = 62.5 W [b] VL = (0.9 − j0.3)(8.33) = 7.5 − j2.5 V(rms)
I1 =
VL = −0.833 − j2.5 A(rms) j3
10–25
10–26
CHAPTER 10. Sinusoidal Steady State Power Calculations I2 = I1 + IL = 7.5 − j2.5 A(rms) 5Iφ = I2 + VL
.·.
Iφ = 3 − j1 A
Id.s. = Iφ − I2 = −4.5 + j1.5 A Sg = −100(3 + j1) = −300 − j100 VA Sd.s. = 5(3 − j1)(−4.5 − j1.5) = −75 + j0 VA Pdev = 300 + 75 = 375 W % developed =
62.5 (100) = 16.67% 375
Checks: P25Ω = (10)(25) = 250 W P1Ω = (62.5)(1) = 62.5 W P0.9Ω = 62.5 W
Pabs = 250 + 62.5 + 62.5 = 375 W =
Qj10 = (10)(10) = 100 VAR Qj3 = (6.94)(3) = 20.83 VAR Q−j0.3 = (69.4)(−0.3) = −20.83 VAR Qsource = −100 VAR
Q = 100 + 20.83 − 20.83 − 100 = 0
P 10.37 [a] Open circuit voltage:
Vφ − 100 Vφ + − 0.1Vφ = 0 5 j5 .·. Vφ = 40 + j80 V(rms)
Pdev
Problems VTh = Vφ + 0.1Vφ (−j5) = Vφ (1 − j0.5) = 80 + j60 V(rms) Short circuit current:
Isc = 0.1Vφ +
Vφ = (0.1 + j0.2)Vφ −j5
Vφ Vφ − 100 Vφ + =0 + j5 −j5 5 .·. Vφ = 100 V(rms) Isc = (0.1 + j0.2)(100) = 10 + j20 A(rms) ZTh =
VTh 80 + j60 = 4 − j2 Ω = Isc 10 + j20
.·. Ro = |ZTh | = 4.47 Ω [b]
80 + j60 √ = 7.36 + j8.82 A(rms) 4 + 20 − j2 √ P = (11.49)2 ( 20) = 590.17 W
I=
10–27
10–28
CHAPTER 10. Sinusoidal Steady State Power Calculations
[c]
I=
80 + j60 = 10 + j7.5 A(rms) 8
P = (102 + 7.52 )(4) = 625 W [d]
Vφ − 100 Vφ Vφ − (25 + j50) + + =0 5 j5 −j5 Vφ = 50 + j25 V(rms) 0.1Vφ = 5 + j2.5 5 + j2.5 + IC = 10 + j7.5 IC = 5 + j5 A(rms) IL =
Vφ = 5 − j10 A(rms) j5
IR = IC + IL = 10 − j5 A(rms) Ig = IR + 0.1Vφ = 15 − j2.5 A(rms) Sg = −100I∗g = −1500 − j250 VA 100 = 5(5 + j2.5) + Vcs + 25 + j50
.·.
Vcs = 50 − j62.5 V(rms)
Scs = (50 − j62.5)(5 − j2.5) = 93.75 − j437.5 VA Thus,
Pdev = 1500
% delivered to Ro =
625 (100) = 41.67% 1500
Problems
10–29
P 10.38 [a] First find the Thévenin equivalent: jωL = j3000 Ω ZTh = 600012,000 + j3000 = 4000 + j3000 Ω VTh =
12,000 (180) = 120/0◦ V 6000 + 12,000
−j = −j1000 Ω ωC
I=
120 = 18 − j6 mA 6000 + j2000
1 P = |I|2 (2000) = 360 mW 2 [b] Set Co = 0.1 µF so −j/ωC = −j2000 Ω Set Ro as close as possible to √ Ro = 40002 + 10002 = 4123.1 Ω .·. Ro = 4000 Ω [c] I =
120 = 14.77 − j1.85 mA 8000 + j1000
1 P = |I|2 (4000) = 443.1 mW 2 Yes; 443.1 mW > 360 mW 120 = 15 mA [d] I = 8000 1 P = (0.015)2 (4000) = 450 mW 2 [e] Ro = 4000 Ω; [f] Yes;
Co = 66.67 nF
450 mW > 443.1 mW
j3000 − j2000 = j1000 Ω
10–30
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.39 [a] Set Co = 0.1 µF, so −j/ωC = −j2000 Ω; also set Ro = 4123.1 Ω I=
120 = 14.55 − j1.79 mA 8123.1 + j1000
1 P = |I|2 (4123.1) = 443.18 mW 2 [b] Yes;
443.18 mW > 360 mW
[c] Yes;
443.18 mW < 450 mW
1 1 = 100 Ω; C= = 26.53 µF ωC (100)(120π) 13,800 13,800 [b] Iwo = + = 46 − j138 A(rms) 300 j100
P 10.40 [a]
Vswo = 13,800 + (46 − j138)(1.5 + j12) = 15,525 + j345 = 15,528.83/1.27◦ V(rms) Iw =
13,800 = 46 A(rms) 300
Vsw = 13,800 + 46(1.5 + j12) = 13,869 + j552 = 13,879.98/2.28◦ V(rms)
% increase =
15,528.82 − 1 (100) = 11.88% 13,879.98
[c] Pwo = |46 − j138|2 1.5 = 31.74 kW Pw = 462 (1.5) = 3174 W
31,740 − 1 (100) = 900% % increase = 3174 P 10.41 [a] So = original load = 1600 + j Sf = final load = 1920 + j
1600 (0.6) = 1600 + j1200 kVA 0.8
1920 (0.28) = 1920 + j560 kVA 0.96
.·. Qadded = 560 − 1200 = −640 kVAR [b] deliver [c] Sa = added load = 320 − j640 = 715.54/− 63.43◦ kVA pf = cos(−63.43) = 0.4472 leading
Problems [d] I∗L =
(1600 + j1200) × 103 = 666.67 + j500 A 2400
IL = 666.67 − j500 = 833.33/− 36.87◦ A(rms) |IL | = 833.33 A(rms) [e] I∗L =
(1920 + j560) × 103 = 800 + j233.33 2400
IL = 800 − j233.33 = 833.33/− 16.26◦ A(rms) |IL | = 833.33 A(rms) P 10.42 [a] Pbefore = Pafter = (833.33)2 (0.05) = 34,722.22 W [b] Vs (before) = 2400 + (666.67 − j500)(0.05 + j0.4) = 2633.33 + j241.67 = 2644.4/5.24◦ V(rms) |Vs (before)| = 2644.4 V(rms) Vs (after) = 2400 + (800 + j233.33)(0.05 + j0.4) = 2346.67 + j331.67 = 2369.99/8.04◦ V(rms) |Vs (after)| = 2369.99 V(rms) P 10.43 [a]
180 = 3I1 + j4I1 + j3(I2 − I1 ) + j9(I1 − I2 ) − j3I1 0 = 9I2 + j9(I2 − I1 ) + j3I1 Solving, I1 = 18 − j18 A(rms);
I2 = 12/0◦ A(rms)
.·. Vo = (12)(9) = 108/0◦ V(rms) [b] P = (12)2 (9) = 1296 W [c] Sg = −(180)(18 + j18) = −3240 − j3240 VA % delivered =
1296 (100) = 40% 3240
.·. Pg = −3240 W
10–31
10–32
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.44 [a] Open circuit voltage:
180 = 3I1 + j4I1 − j3I1 + j9I1 − j3I1 .·. I1 =
180 = 9.31 − j21.72 A(rms) 3 + j7
VTh = j9I1 − j3I1 = j6I1 = 130.34 + j55.86 V = 141.81/23.20◦ V(rms) Short circuit current:
180 = 3I1 + j4I1 + j3(Isc − I1 ) + j9(I1 − Isc ) − j3I1 0 = j9(Isc − I1 ) + j3I1 Solving, Isc = 20 − j20 A ZTh =
IL =
I1 = 30 − j20 A
VTh 130.34 + j55.86 = 1.86 + j4.66 Ω = Isc 20 − j20
130.34 + j55.86 = 35 + j15 = 38.08/23.20◦ A 3.72
Problems PL = (38.12)2 (1.86) = 2700 W [b] I1 =
Zo + j9 1.86 − j4.66 + j9 I2 = (35 + j15) = 30/0◦ A(rms) j6 j6
Pdev = (180)(30) = 5400 W P 10.45 [a]
54 = I1 + j2(I1 − I2 ) + j3I2 0 = 7I2 + j2(I2 − I1 ) − j3I2 + j8I2 + j3(I1 − I2 ) Solving, I1 = 12 − j21 A(rms);
I2 = −3 A(rms)
Vo = 7I2 = −21/180◦ V(rms) [b] P = |I2 |2 (7) = 63 W [c] Pg = (54)(12) = 648 W % delivered =
63 (100) = 9.72% 648
P 10.46 [a]
Open circuit: VTh = −j3I1 + j2 I1 = −jI1 I1 =
54 = 10.8 − j21.6 A 1 + j2
VTh = −21.6 − j10.8 V
10–33
10–34
CHAPTER 10. Sinusoidal Steady State Power Calculations Short circuit: 54 = I1 + j2(I1 − Isc ) + j3Isc 0 = j2(Isc − I1 ) − j3Isc + j8Isc + j3(I1 − Isc ) Solving, Isc = −3.32 + j5.82 ZTh =
VTh −21.6 − j10.8 = 0.2 + 3.6j = 3.6/86.82◦ Ω = Isc −3.32 + j5.82
.·. RL = |ZTh | = 3.606 Ω [b]
I=
−21.6 − j10.8 = 4.610/163.2◦ A 3.806 + j3.6
P = |I|2 (3.6) = 76.6 W,
which is greater than when RL = 7 Ω
P 10.47 [a]
54 = I1 + j2(I1 − I2 ) + j4kI2 0 = 7I2 + j2(I2 − I1 ) − j4kI2 + j8I2 + j4k(I1 − I2 ) Place the equations in standard form: 54 = (1 + j2)I1 + j(4k − 2)I2 0 = j(4k − 2)I1 + [7 + j(10 − 8k)]I2 I1 =
54 − I2 j(4k − 2) (1 + j2)
Substituting, I2 = −
j54(4k − 2) [7 + j(10 − 8k)](1 + j2) + (4k − 2)2
For Vo = 0, I2 = 0, so if 4k − 2 = 0, then k = 0.5.
Problems
10–35
[b] When I2 = 0 I1 =
54 = 10.8 − j21.6 A(rms) 1 + j2
Pg = (54)(10.8) = 583.2 W Check: Ploss = |I1 |2 (1) = 583.2 W P 10.48 [a] From Problem 9.67, ZTh = 85 + j85 Ω and VTh = 850 + j850 V. Thus, for ∗ maximum power transfer, ZL = ZTh = 85 − j85 Ω:
I2 =
850 + j850 = 5 + j5 A 170
425/0◦ = (5 + j5)I1 − j20(5 + j5) .·. I1 =
325 + j100 = 42.5 − j22.5 A 5 + j5
Sg (del) = 425(42.5 + j22.5) = 18,062.5 + j9562.5 VA Pg = 18,062.5 W [b] Ploss = |I1 |2 (5) + |I2 |2 (45) = 11,562.5 + 2250 = 13,812.5 W % loss in transformer =
18,062.5 − 13,812.5 (100) = 23.53% 18,062.5
10–36
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.49 [a] From Problem 9.70, Zab = 100 + j136.26
so
I1 =
50 50 = = 160 − j120 mA 100 + j13.74 + 100 + 136.26 200 + j150
I2 =
jωM j270 (0.16 − j0.12) = 51.84 + j15.12 mA I1 = Z22 800 + j600
VL = (300 + j100)(51.84 + j15.12)103 = 14.04 + j9.72 V |VL | = 17.08 V [b] Pg (ideal) = 50(0.16) = 8 W Pg (practical) = 8 − |I1 |2 (100) = 4 W PL = |I2 |2 (300) = 874.8 mW % delivered =
0.8748 (100) = 21.87% 4
P 10.50 [a]
Open circuit: 120 VTh = (j10) = 36 + j48 V 16 + j12 Short circuit: (16 + j12)I1 − j10Isc = 120 −j10I1 + (11 + j23)Isc = 0 Solving, Isc = 2.4/0◦ A ZTh =
36 + j48 = 15 + j20 Ω 2.4
∗ .·. ZL = ZTh = 15 − j20 Ω
IL =
VTh 36 + j48 = 1.2 + j1.6 A(rms) = 2.0/53.13◦ A(rms) = ZTh + ZL 30
PL = |IL |2 (15) = 60 W
Problems [b] I1 =
26 + j3 Z22 I2 = (1.2 + j1.6) = 5.23/− 30.29◦ A)rms) jωM j10
Ptransformer = (120)(5.23) cos(−30.29◦ ) − (5.23)2 (4) = 432.8 W % delivered =
60 (100) = 13.86% 432.8
P 10.51 [a] jωL1 = j(10,000)(1 × 10−3 ) = j10 Ω jωL2 = j(10,000)(1 × 10−3 ) = j10 Ω jωM = j10 Ω
200 = (5 + j10)Ig + j5IL 0 = j5Ig + (15 + j10)IL Solving, Ig = 10 − j15 A;
IL = −5 A
Thus, ig = 18.03 cos(10,000t − 56.31◦ ) A iL = 5 cos(10,000t − 180◦ ) A M 0.5 = √ = 0.5 L1 L2 1 [c] When t = 50π µs:
[b] k = √
10,000t = (10,000)(50π) × 10−6 = 0.5π rad = 90◦ ig (50π µs) = 18.03 cos(90◦ − 56.31◦ ) = 15 A iL (50π µs) = 5 cos(90◦ − 180◦ ) = 0 A 1 1 1 w = L1 i21 + L2 i22 + M i1 i2 = (10−3 )(15)2 + 0 + 0 = 112.5 mJ 2 2 2 When t = 100π µs: 10,000t = (104 )(100π) × 10−6 = π = 180◦ ig (100π µs) = 18.03 cos(180 − 56.31◦ ) = −10 A iL (100π µs) = 5 cos(180 − 180◦ ) = 5 A 1 1 w = (10−3 )(10)2 + (10−3 )(5)2 + 0.5 × 10−3 (−10)(5) = 37.5 mJ 2 2
10–37
10–38
CHAPTER 10. Sinusoidal Steady State Power Calculations
[d] From (a), Im = 5 A, 1 .·. P = (5)2 (15) = 187.5 W 2 [e] Open circuit: VTh =
200 (−j5) = −80 − j40 V 5 + j10
Short circuit: 200 = (5 + j10)I1 + j5Isc 0 = j10Isc + j5I1 Solving, Isc = −11.094/123.69◦ A; ZTh =
I1 = 22.188/− 56.31◦ A
VTh −80 − j40 = = 1 + j8 Ω Isc 11.094/123.69◦
.·. RL = 8.962 Ω [f]
I=
−80 − j40 = 7.399/165.13◦ A 9.062 + j8
1 P = (7.399)2 (8.062) = 220.70 W 2 ∗ = 1 − j8 Ω [g] ZL = ZTh −80 − j40 = 44.72/− 153.43◦ A [h] I = 2
1 P = (44.72)2 (1) = 1000 W 2
Problems
10–39
P 10.52 [a]
10 = j1(I1 − I2 ) + j1(I3 − I2 ) − j1(I1 − I3 ) 0 = 1I2 + j2(I2 − I3 ) + j1(I2 − I1 ) + j1(I2 − I1 ) + j1(I2 − I3 ) 0 = 1I3 − j1(I3 − I1 ) + j2(I3 − I2 ) + j1(I1 − I2 ) Solving, I1 = 6.25 + j7.5 A(rms);
I2 = 5 + j2.5 A(rms);
I3 = 5 − j2.5 A(rms)
Ia = I1 = 6.25 + j7.5 A
Ib = I1 − I2 = 1.25 + j5 A
Ic = I2 = 5 + j2.5 A
Id = I3 − I2 = −j5 A
Ie = I1 − I3 = 1.25 + j10 A
If = I3 = 5 − j2.5 A
Va = 10 V
Vb = j1Ib + j1Id = j1.25 V
Vc = 1Ic = 5 + j2.5 V
Vd = j2Id + j1Ib = 5 + j1.25 V
Ve = −j1Ie = 10 − j1.25 V
Vf = 1If = 5 − j2.5 V
[b]
Sa = −10I∗a = −62.5 + j75 VA Sb = Vb I∗b = 6.25 + j1.5625 VA
10–40
CHAPTER 10. Sinusoidal Steady State Power Calculations Sc = Vc I∗c = 31.25 + j0 VA Sd = Vd I∗d = −6.25 + j25 VA Se = Ve I∗e = 0 − j101.5625 VA Sf = Vf I∗f = 31.25 VA
[c]
Pdev = 62.5 W
Pabs = 6.25 + 31.25 − 6.25 + 31.25 = 62.5 W
Note that the total power absorbed by the coupled coils is zero: 6.25 − 6.25 = 0 = Pb + Pd
[d]
Qdev = 101.5625 VAR The capacitor is developing magnetizing vars.
Qabs = 75 + 1.5625 + 25 = 101.5625 VAR Q absorbed by the coupled coils is Qb + Qd = 26.5625 VAR
P 10.53 Open circuit voltage:
I1 =
10/0◦ = 2 − j4 A 1 + j2
VTh = j2I1 + j1.2I1 = j3.2I1 = 12.8 + j6.4 = 14.31/26.57◦ V Short circuit current:
10/0◦ = (1 + j2)I1 − j3.2Isc
Problems 0 = −j3.2I1 + j5.4Isc Solving, Isc = 5.89/− 5.92◦ A ZTh =
14.31/26.57◦ = 2.43/32.49◦ = 2.048 + j1.304 Ω 5.89/− 5.92◦
.·. I2 =
14.31/26.57◦ = 3.49/26.57◦ A 4.096
10/0◦ = (1 + j2)I1 − j3.2I2 10 + j3.2(3.49/26.57◦ ) 10 + j3.2I2 = = 5/0◦ A 1 + j2 1 + j2
.·.
I1 =
Zg =
10/0◦ = 2 + j0 = 2/0◦ Ω 5/0◦
P 10.54 [a]
272/0◦ = 2Ig + j10Ig + j14(Ig − I2 ) − j6I2 +j14Ig − j8I2 + j20(Ig − I2 ) 0 = j20(I2 − Ig ) − j14Ig + j8I2 + j4I2 +j8(I2 − Ig ) − j6Ig + 8I2
10–41
10–42
CHAPTER 10. Sinusoidal Steady State Power Calculations Solving, I2 = 24/0◦ A(rms)
Ig = 20 − j4 A(rms); P8Ω = (24)2 (8) = 4608 W
[b] Pg (developed) = (272)(20) = 5440 W Vg 272 − 2 = 11.08 + j2.62 = 11.38/13.28◦ Ω [c] Zab = −2= Ig 20 − j4 [d] P2Ω = |Ig |2 (2) = 832 W
Pdiss = 832 + 4608 = 5440 W =
Pdev
P 10.55 [a]
300 = 60I1 + V1 + 20(I1 − I2 ) 0 = 20(I2 − I1 ) + V2 + 40I2 1 V2 = V1 ; 4
I2 = −4I1
Solving, V1 = 260 V(rms);
V2 = 65 V(rms)
I1 = 0.25 A(rms);
I2 = −1.0 A(rms)
V5A = V1 + 20(I1 − I2 ) = 285 V(rms) .·. P = −(285)(5) = −1425 W Thus 1425 W is delivered by the current source to the circuit. [b] I20Ω = I1 − I2 = 1.25 A(rms) .·. P20Ω = (1.25)2 (20) = 31.25 W
Problems
10–43
P 10.56
30Vo = Va ;
Io = Ia ; 30
−Va Vb = ; 1 20
Ib = −20Ia ;
Vo = 10Io therefore
therefore
Va = 9 kΩ Ia
Vb 9000 = 22.5 Ω = Ib 400
Therefore Ib = [50/(2.5 + 22.5)] = 2 A (rms); since the ideal transformers are lossless, P10Ω = P22.5Ω , and the power delivered to the 22.5 Ω resistor is 22 (22.5) or 90 W. Vb a2 10 = = 2.5 Ω; therefore a2 = 100, Ib 400 50 = 10 A; P = (100)(2.5) = 250 W [b] Ib = 5
P 10.57 [a]
P 10.58 [a] ZTh
200 = 720 + j1500 + 50
2
(40 − j30) = 1360 + j1020 = 1700/36.87◦ Ω
.·. Zab = 1700 Ω Zab =
ZL (1 + N1 /N2 )2
(1 + N1 /N2 )2 = 6800/1700 = 4 .·. N1 /N2 = 1 [b] VTh
or
N2 = N1 = 1000 turns
255/0◦ (j200) = 1020/53.13◦ V = 40 + j30
IL =
1020/53.13◦ = 0.316/34.7◦ A(rms) 3060 + j1020
Since the transformer is ideal, P6800 = P1700 . P = |IL |2 (1700) = 170 W
a = 10
10–44
CHAPTER 10. Sinusoidal Steady State Power Calculations
[c]
255/0◦ = (40 + j30)I1 − j200(0.26 + j0.18) .·. I1 = 4.13 − j1.80 A(rms) Pgen = (255)(4.13) = 1053 W Ptrans = 1053 − 170 = 883 W % transmitted =
883 (100) = 83.85% 1053
P 10.59 [a]
For maximum power transfer, Zab = 90 kΩ
Zab = 1 + .·. 1+
N1 N2
N1 1+ N2
2
2
ZL =
90,000 = 225 400
N1 = ±15; N2
[b] P = |Ii |2 (90,000) =
N1 = 15 − 1 = 14 N2
180 180,000
2
180 [c] V1 = Ri Ii = (90,000) 180,000 [d]
(90,000) = 90 mW
= 90 V
Vg = (2.25 × 10−3 )(100,00080,000) = 100 V
Problems Pg (del) = (2.25 × 10−3 )(100) = 225 mW % delivered =
P 10.60 [a] Zab = 1 + .·. I1 = I2 =
N1 N2
90 (100) = 40% 225
2
(1 − j2) = 25 − j50 Ω
100/0◦ = 2.5/0◦ A 15 + j50 + 25 − j50
N1 I1 = 10/0◦ A N2
.·. IL = I1 + I2 = 12.5/0◦ A(rms) P1Ω = (12.5)2 (1) = 156.25 W P15Ω = (2.5)2 (15) = 93.75 W [b] Pg = −100(2.5/0◦ ) = −250 W
Pabs = 156.25 + 93.75 = 250 W =
P 10.61 [a] 25a21 + 4a22 = 500 I25 = a1 I; I4 = a2 I;
P25 = a21 I2 (25) P4 = a22 I2 (4)
P4 = 4P25 ; .·.
a22 I2 4 = 100a21 I2
100a21 = 4a22
25a21 + 100a21 = 500; 25(4) + 4a22 = 500; [b] I =
a1 = 2 a2 = 10
2000/0◦ = 2/0◦ A(rms) 500 + 500
I25 = a1 I = 4 A P25Ω = (16)(25) = 400 W [c] I4 = a2 I = 10(2) = 20 A(rms) V4 = (20)(4) = 80/0◦ V(rms)
Pdev
10–45
10–46
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.62 [a] Open circuit voltage:
500 = 100I1 + V1 V2 = 400I2 V2 V1 = 1 2
.·.
V2 = 2V1
I1 = 2I2 Substitute and solve: .·.
2V1 = 400I1 /2 = 200I1 500 = 100I1 + 100I1 .·.
.·.
V1 = 100I1 I1 = 500/200 = 2.5 A
1 I2 = I1 = 1.25 A 2
V1 = 100(2.5) = 250 V;
V2 = 2V1 = 500 V
VTh = 20I1 + V1 − V2 + 40I2 = −150 V(rms) Short circuit current:
500 = 80(Isc + I1 ) + 360(Isc + 0.5I1 ) 2V1 = 40
I1 + 360(Isc + 0.5I1 ) 2
500 = 80(I1 + Isc ) + 20I1 + V1
Problems
10–47
Solving, Isc = −1.47 A; RTh =
P =
I1 = 4.41 A;
V1 = 176.47 V
VTh −150 = 102 Ω = Isc −1.47
752 = 55.15 W 102
[b]
500 = 80[I1 − (75/102)] − 75 + 360[I2 − (75/102)] 575 +
6000 27,000 + = 80I1 + 180I1 102 102
.·.
I1 = 3.456 A
Psource = (500)[3.456 − (75/102)] = 1360.29 W % delivered =
[c] P80Ω
55.15 (100) = 4.05% 1360.29
75 = 80 I1 − 102
2
= 592.13 W
P20Ω = 20I21 = 238.86 W P40Ω = 40I22 = 119.43 W P102Ω =
752 = 55.15 W 102
P360Ω
75 = 360 I2 − 102
2
= 354.73 W
Pabs = 592.13 + 238.86 + 119.43 + 55.15 + 354.73 = 1360.3 W =
Pdev
10–48
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.63 [a] Open circuit voltage:
40/0◦ = 4(I1 + I3 ) + 12I3 + VTh I1 = −I3 ; 4
I1 = −4I3
Solving, VTh = 40/0◦ V Short circuit current:
40/0◦ = 4I1 + 4I3 + I1 + V1 4V1 = 16(I1 /4) = 4I1 ;
.·. V1 = I1
.·. 40/0◦ = 6I1 + 4I3 Also, 40/0◦ = 4(I1 + I3 ) + 12I3 Solving, I1 = 6 A; RTh =
I3 = 1 A;
VTh 40 = 16 Ω = Isc 2.5
Isc = I1 /4 + I3 = 2.5 A
Problems
10–49
40/0◦ I= = 1.25/0◦ A(rms) 32 P = (1.25)2 (16) = 25 W [b]
40 = 4(I1 + I3 ) + 12I3 + 20 4V1 = 4I1 + 16(I1 /4 + I3 );
.·. V1 = 2I1 + 4I3
40 = 4I1 + 4I3 + I1 + V1 .·. I1 = 6 A;
I1 + I3 = 5.75/0◦ A;
I3 = −0.25 A;
V1 = 11/0◦ V
P40V (developed) = 40(5.75) = 230 W .·. % delivered = [c] PRL = 25 W;
25 (100) = 10.87% 230
P16Ω = (1.5)2 (16) = 36 W
P4Ω = (5.75)2 (4) = 132.25 W;
P1Ω = (6)2 (1) = 36 W
P12Ω = (−0.25)2 (12) = 0.75 W
Pabs = 25 + 36 + 132.25 + 36 + 0.75 = 230 W =
Pdev
P 10.64 [a] Replace the circuit to the left of the primary winding with a Thévenin equivalent: VTh = (15)(20j10) = 60 + j120 V ZTh = 2 + 20j10 = 6 + j8 Ω
10–50
CHAPTER 10. Sinusoidal Steady State Power Calculations Transfer the secondary impedance to the primary side: Zp =
1 XC (100 + jXC ) = 4 + j Ω 25 25
Now maximize I by setting (XC /25) = −8 Ω: .·. C = [b] I =
1 = 0.25 µF 200(20 × 103 )
60 + j120 = 6 + j12 A 10
P = |I|2 (4) = 720 W Ro = 6 Ω; .·. Ro = 150 Ω 25 60 + j120 = 5 + j10 A [d] I = 12 [c]
P = |I|2 (6) = 750 W
P 10.65 [a] Zab
N1 = 50 − j400 = 1 − N2
2
2800 ZL = 1 − 700
1 .·. ZL = (50 − j400) = 5.556 − j44.444 Ω 9 [b]
I1 =
24 = 240/0◦ mA 100
2
ZL = 9ZL
Problems
10–51
N1 I1 = −N2 I2 I2 = −4I1 = 960/180◦ mA IL = I1 + I2 = 720/180◦ mA(rms) VL = (5.556 − j44.444)IL = −4 + j32 = 32.25/97.13◦ V(rms) P 10.66 [a] Begin with the MEDIUM setting, as shown in Fig. 10.31, as it involves only the resistor R2 . Then, Pmed = 500 W =
V2 1202 = R2 R2
Thus, 1202 = 28.8 Ω 500 [b] Now move to the LOW setting, as shown in Fig. 10.30, which involves the resistors R1 and R2 connected in series: R2 =
Plow
V2 V2 = 250 W = = R1 + R2 R1 + 28.8
Thus, 1202 − 28.8 = 28.8 Ω 250 [c] Note that the HIGH setting has R1 and R2 in parallel: R1 =
Phigh
V2 1202 = = = 1000 W R1 R2 28.828.8
If the HIGH setting has required power other than 1000 W, this problem sould not have been solved. In other words, the HIGH power setting was chosen in such a way that it would be satisfied once the two resistor values were calculated to satisfy the LOW and MEDIUM power settings.
10–52
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.67 [a] PL =
V2 ; R1 + R2
R1 + R2 =
PM =
V2 ; R2
PH =
V 2 (R1 + R2 ) R1 R2
PH = V 2
[b] PH =
V2 ; PL
R1 =
V 2 V 2 /PL
PL
PH =
V2 PM
R2 =
R1 + R2 =
−
2
V PM
2
V2 PL
V PM
=
V2 V2 − PL PM PM PL PM PL (PM − PL )
2 PM PM − PL
(750)2 = 1125 W (750 − 250)
P 10.68 First solve the expression derived in P10.67 for PM as a function of PL and PH . Thus P M − PL =
2 PM PH
2 PM − PM + PL = 0 PH
or
2 − PM P H + PL P H = 0 PM
PH ± .·. PM = 2
PH 2
2
− PL PH
1 PL PH ± PH − = 2 4 PH
For the specified values of PL and PH √ PM = 500 ± 1000 0.25 − 0.24 = 500 ± 100 .·. PM 1 = 600 W;
PM 2 = 400 W
Note in this case we design for two medium power ratings If PM 1 = 600 W R2 =
(120)2 = 24 Ω 600
Problems R1 + R2 =
(120)2 = 60 Ω 240
R1 = 60 − 24 = 36 Ω CHECK: PH =
(120)2 (60) = 1000 W (36)(24)
If PM 2 = 400 W R2 =
(120)2 = 36 Ω 400
R1 + R2 = 60 Ω (as before) R1 = 24 Ω CHECK: PH = 1000 W P 10.69 R1 + R2 + R3 = R2 + R3 =
(120)2 = 24 Ω 600
(120)2 = 16 Ω 900
.·. R1 = 24 − 16 = 8 Ω R3 + R1 R2 =
(120)2 = 12 Ω 1200
.·. 16 − R2 +
8R2 = 12 8 + R2
R2 −
8R2 =4 8 + R2
8R2 + R22 − 8R2 = 32 + 4R2 R22 − 4R2 − 32 = 0 R2 = 2 ±
√
4 + 32 = 2 ± 6
.·. R2 = 8 Ω;
.·. R3 = 8 Ω
10–53
10–54
CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.70 R2 =
(220)2 = 96.8 Ω 500
R1 + R2 =
(220)2 = 193.6 Ω 250
.·. R1 = 96.8 Ω CHECK: R1 R2 = 48.4 Ω PH =
(220)2 = 1000 W 48.4
11 Balanced Three-Phase Circuits Assessment Problems AP 11.1 Make a sketch:
We know VAN and wish to find VBC . To do this, write a KVL equation to find VAB , and use the known phase angle relationship between VAB and VBC to find VBC . VAB = VAN + VNB = VAN − VBN Since VAN , VBN , and VCN form a balanced set, and VAN = 240/− 30◦ V, and the phase sequence is positive, VBN = |VAN |//VAN − 120◦ = 240/− 30◦ − 120◦ = 240/− 150◦ V Then, VAB = VAN − VBN = (240/− 30◦ ) − (240/− 150◦ ) = 415.46/0◦ V Since VAB , VBC , and VCA form a balanced set with a positive phase sequence, we can find VBC from VAB : VBC = |VAB |/(/VAB − 120◦ ) = 415.69/0◦ − 120◦ = 415.69/− 120◦ V Thus, VBC = 415.69/− 120◦ V 11–1
11–2
CHAPTER 11. Balanced Three-Phase Circuits
AP 11.2 Make a sketch:
We know VCN and wish to find VAB . To do this, write a KVL equation to find VBC , and use the known phase angle relationship between VAB and VBC to find VAB . VBC = VBN + VNC = VBN − VCN Since VAN , VBN , and VCN form a balanced set, and VCN = 450/− 25◦ V, and the phase sequence is negative, VBN = |VCN |//VCN − 120◦ = 450/− 23◦ − 120◦ = 450/− 145◦ V Then, VBC = VBN − VCN = (450/− 145◦ ) − (450/− 25◦ ) = 779.42/− 175◦ V Since VAB , VBC , and VCA form a balanced set with a negative phase sequence, we can find VAB from VBC : VAB = |VBC |//VBC − 120◦ = 779.42/− 295◦ V But we normally want phase angle values between +180◦ and −180◦ . We add 360◦ to the phase angle computed above. Thus, VAB = 779.42/65◦ V AP 11.3 Sketch the a-phase circuit:
Problems
11–3
[a] We can find the line current using Ohm’s law, since the a-phase line current is the current in the a-phase load. Then we can use the fact that IaA , IbB , and IcC form a balanced set to find the remaining line currents. Note that since we were not given any phase angles in the problem statement, we can assume that the phase voltage given, VAN , has a phase angle of 0◦ . 2400/0◦ = IaA (16 + j12) so IaA
2400/0◦ = 96 − j72 = 120/− 36.87◦ A = 16 + j12
With an acb phase sequence, /IbB = /IaA + 120◦
and
/IcC = /IaA − 120◦
so IaA = 120/− 36.87◦ A IbB = 120/83.13◦ A IcC = 120/− 156.87◦ A [b] The line voltages at the source are Vab Vbc , and Vca . They form a balanced set. To find Vab , use the a-phase circuit to find VAN , and use the relationship between phase voltages and line voltages for a y-connection (see Fig. 11.9[b]). From the a-phase circuit, use KVL: Van = VaA + VAN = (0.1 + j0.8)IaA + 2400/0◦ = (0.1 + j0.8)(96 − j72) + 2400/0◦ = 2467.2 + j69.6 2468.18/1.62◦ V From Fig. 11.9(b), √ Vab = Van ( 3/− 30◦ ) = 4275.02/− 28.38◦ V With an acb phase sequence, /Vbc = /Vab + 120◦
and
/Vca = /Vab − 120◦
so Vab = 4275.02/− 28.38◦ V Vbc = 4275.02/91.62◦ V Vca = 4275.02/− 148.38◦ V
11–4
CHAPTER 11. Balanced Three-Phase Circuits [c] Using KVL on the a-phase circuit Va n = Va a + Van = (0.2 + j0.16)IaA + Van = (0.02 + j0.16)(96 − j72) + (2467.2 + j69.9) = 2480.64 + j83.52 = 2482.05/1.93◦ V With an acb phase sequence, /Vb n = /Va n + 120◦
and
/Vc n = /Va n − 120◦
so Va n = 2482.05/1.93◦ V Vb n = 2482.05/121.93◦ V
AP 11.4 IcC
Vc n = 2482.05/− 118.07◦ V √ √ = ( 3/− 30◦ )ICA = ( 3/− 30◦ ) · 8/− 15◦ = 13.86/− 45◦ A
AP 11.5 IaA = 12/(65◦ − 120◦ ) = 12/− 55◦ 1 /− 30◦ ◦ √ /− 30 IaA = √ IAB = · 12/− 55◦ 3 3 = 6.93/− 85◦ A
AP 11.6 [a] IAB =
1 √ /30◦ [69.28/− 10◦ ] = 40/20◦ A 3
Therefore Zφ =
[b] IAB =
4160/0◦ = 104/− 20◦ Ω 40/20◦
1 √ /− 30◦ [69.28/− 10◦ ] = 40/− 40◦ A 3
Therefore Zφ = 104/40◦ Ω AP 11.7 Iφ =
110 110 + = 30 − j40 = 50/− 53.13◦ A 3.667 j2.75
Therefore |IaA | =
√
3Iφ =
√
3(50) = 86.60 A
Problems AP 11.8 [a] |S| = Q=
√
3(208)(73.8) = 26,587.67 VA
(26,587.67)2 − (22,659)2 = 13,909.50 VAR
22,659 = 0.8522 lagging 26,587.67
[b] pf =
2450 AP 11.9 [a] VAN = √ /0◦ V; VAN I∗aA = Sφ = 144 + j192 kVA 3 Therefore (144 + j192)1000 √ I∗aA = = (101.8 + j135.7) A 2450/ 3 IaA = 101.8 − j135.7 = 169.67/− 53.13◦ A |IaA | = 169.67 A [b] P =
(2450)2 ; R
Q=
(2450)2 ; X
therefore
R=
therefore
(2450)2 = 41.68 Ω 144,000
X=
(2450)2 = 31.26 Ω 192,000
√ VAN 2450/ 3 [c] Zφ = = = 8.34/53.13◦ = (5 + j6.67) Ω IaA 169.67/− 53.13◦ .·. R = 5 Ω,
X = 6.67 Ω
11–5
11–6
CHAPTER 11. Balanced Three-Phase Circuits
Problems P 11.1
[a] First, convert the cosine waveforms to phasors: Va = 208/27◦ ;
Vb = 208/147◦ ;
Vc = 208/− 93◦
Subtract the phase angle of the a-phase from all phase angles: /Va = 27◦ − 27◦ = 0◦ /Vb = 147◦ − 27◦ = 120◦ /Vc = −93◦ − 27◦ = −120◦ Compare the result to Eqs. 11.1 and 11.2: Therefore
acb
[b] First, convert the cosine waveforms to phasors: Va = 4160/− 18◦ ;
Vb = 4160/− 138◦ ;
Vc = 4160/+ 102◦
Subtract the phase angle of the a-phase from all phase angles: /Va = −18◦ + 18◦ = 0◦ /Vb = −138◦ + 18◦ = −120◦ /Vc = 102◦ + 18◦ = 120◦ Compare the result to Eqs. 11.1 and 11.2: Therefore P 11.2
abc
[a] Va = 180/0◦ V Vb = 180/− 120◦ V Vc = 180/− 240◦ = 180/120◦ V Balanced, positive phase sequence [b] Va = 180/− 90◦ V Vb = 180/30◦ V Vc = 180/− 210◦ V = 180/150◦ V Balanced, negative phase sequence [c] Va = 400/− 270◦ V = 400/90◦ V Vb = 400/120◦ V Vc = 400/− 30◦ V Unbalanced, phase angle in b-phase
Problems
11–7
[d] Va = 200/30◦ V Vb = 201/150◦ V Vc = 200/270◦ V = 200/− 90◦ V Unbalanced, unequal amplitude in the b-phase [e] Va = 208/42◦ V Vb = 208/− 78◦ V Vc = 208/− 201◦ V = 208/159◦ V Unbalanced, phase angle in the c-phase [f] Unbalanced; the frequencies of the waveforms are not the same for the positive sequence of Eq. 11.1 P 11.3
Va = Vm /0◦ = Vm + j0 Vb = Vm /− 120◦ = −Vm (0.5 + j0.866) Vc = Vm /120◦ = Vm (−0.5 + j0.866) Va + Vb + Vc = (Vm )(1 + j0 − 0.5 − j0.866 − 0.5 + j0.866) = Vm (0) = 0 For the negative sequences of Eq. 11.2, Vb and Vc are interchanged, but the sum is still zero. Va + Vb + Vc =0 3(RW + jXW )
P 11.4
I=
P 11.5
[a] IaA =
200 = 8/0◦ A 25
IbB =
200/− 120◦ = 4/− 66.87◦ A 30 − j40
IcC =
200/120◦ = 2/83.13◦ A 80 + j60
The magnitudes are unequal and the phase angles are not 120◦ apart. b] Io = IaA + IbB + IcC = 9.96/− 9.79◦ A
11–8
P 11.6
CHAPTER 11. Balanced Three-Phase Circuits [a] IaA =
277/0◦ = 2.77/− 36.87◦ A 80 + j60
IbB =
277/− 120◦ = 2.77/− 156.87◦ A 80 + j60
IcC =
277/120◦ = 2.77/83.13◦ A 80 + j60
Io = IaA + IbB + IcC = 0 [b] VAN = (78 + j54)IaA = 262.79/− 2.17◦ V [c] VAB = VAN − VBN VBN = (77 + j56)IbB = 263.73/− 120.84◦ V VAB = 262.79/− 2.17◦ − 263.73/− 120.84◦ = 452.89/28.55◦ V [d] Unbalanced — see conditions for a balanced circuit on p. 504 of the text! P 11.7
Zga + Zla + ZLa = 60 + j80 Ω Zgb + Zlb + ZLb = 40 + j30Ω Zgc + Zlc + ZLc = 20 + j15Ω VN − 240 VN − 240/120◦ VN − 240/− 120◦ VN + + + =0 60 + j80 40 + j30 20 + j15 10 Solving for VN yields VN = 42.94/− 156.32◦ V Io =
P 11.8
VN = 4.29/− 156.32◦ A 10
Make a sketch of the load in the frequency domain. Note that we convert the time domain line-to-neutral voltages to phasors:
Problems
11–9
Note that these three voltages form a balanced set with an abc phase sequence. First, use KVL to find VAB : VAB = VAN + VNB = VAN − VBN = (169.71/26◦ ) − (169.71/− 94◦ ) = 293.95/56◦ V With an abc phase sequence, /VBC = /VAB − 120◦
and
/VCA = /VAB + 120◦
so VAB = 293.95/56◦ V VBC = 293.95/− 64◦ V VCA = 293.95/176◦ V To get back to the time domain, perform an inverse phasor transform of the three line voltages, using a frequency of ω: vAB (t) = 293.95 cos(ωt + 56◦ ) V vBC (t) = 293.95 cos(ωt − 64◦ ) V vCA (t) = 293.95 cos(ωt + 176◦ ) V P 11.9
Make a sketch of the three-phase line and load:
Z = 0.25 + j2 Ω/φ ZL = 30.48 + j22.86 Ω/φ
11–10
CHAPTER 11. Balanced Three-Phase Circuits
[a] The line currents are IaA , IbB , and IcC . To find IaA , first find VAN and use Ohm’s law for the a-phase load impedance. Since we are only concerned with finding voltage and current magnitudes, the phase sequence doesn’t matter and we arbitrarily assume a positive phase sequence. Since we are not given any phase angles in the problem statement, we can assume the angle of VAB is 0◦ . Use Fig. 11.9(a) to find VAN from VAB . 660 VAN = √ /(0 − 30◦ ) = 381.05/− 30◦ V 3 Now find IaA using Ohm’s law: IaA =
VAN 381.05/− 30◦ = = 3.993 − j9.20 = 10/− 66.87◦ V ZL 30.48 + j22.86
Thus, the magnitude of the line current is |IaA | = 10 A [b] The line voltage at the source is Vab . From KVL on the top loop of the three-phase circuit, Vab = VaA + VAB + VBb = Z IaA + VAB + Z IBb = Z IaA + VAB − Z IbB = (0.25 + j2)(10/− 66.87◦ ) + 660/0◦ − (0.25 + j2)(10/− 173.13◦ ) = 684.71/2.10◦ V Thus, the magnitude of the line voltage at the source is |Vab | = 684.71 V P 11.10 Make a sketch of the a-phase:
[a] Find the a-phase line current from the a-phase circuit: IaA =
125/0◦ 125/0◦ = 0.1 + j0.8 + 19.9 + j14.2 20 + j15
= 4 − j3 = 4/− 36.87◦ A
Problems
11–11
Find the other line currents using the acb phase sequence: IbB = 5/− 36.87◦ + 120◦ = 5/83.13◦ A IcC = 5/− 36.87◦ − 120◦ = 5/− 156.87◦ A [b] The phase voltage at the source is Van = 125/0◦ V. Use Fig. 11.9(b) to find the line voltage, Van , from the phase voltage: √ Vab = Van ( 3/− 30◦ ) = 216.51/− 30◦ V Find the other line voltages using the acb phase sequence: Vbc = 216.51/− 30◦ + 120◦ = 216.51/90◦ V Vca = 216.51/− 30◦ − 120◦ = 216.51/− 150◦ V [c] The phase voltage at the load in the a-phase is VAN . Calculate its value using IaA and the load impedance: VAN = IaA ZL = (4 − j3)(19.9 + j14.2) = 122.2 − j2.9 = 122.23/− 1.36◦ V Find the phase voltage at the load for the b- and c-phases using the acb sequence: VBN = 122.23/− 1.36◦ + 120◦ = 122.23/118.64◦ V VCN = 122.23/− 1.36◦ − 120◦ = 122.23/− 121.36◦ V [d] The line voltage at the load in the a-phase is VAB . Find this line voltage from the phase voltage at the load in the a-phase, VAN , using Fig, 11.9(b): √ VAB = VAN ( 3/− 30◦ ) = 211.71/− 31.36◦ V Find the line voltage at the load for the b- and c-phases using the acb sequence: VBC = 211.71/− 31.36◦ + 120◦ = 211.71/88.69◦ V VCA = 211.71/− 31.36◦ − 120◦ = 211.71/− 151.36◦ V P 11.11 [a] IAB =
480 = 6.4/− 36.87◦ A 60 + j45
IBC = 6.4/− 156.87◦ A ICA = 6.4/83.13◦ A √ [b] IaA = 3/− 30◦ IAB = 11.09/− 66.87◦ A IbB = 11.09/173.13◦ A IcC = 11.09/53.13◦ A
11–12
CHAPTER 11. Balanced Three-Phase Circuits
[c] Transform the ∆-connected load to a Y-connected load: 60 + j45 Z∆ = = 20 + j15 Ω 3 3 The single-phase equivalent circuit is: ZY =
Van = 277.13/− 30◦ + (0.8 + j0.6)(11.09/− 66.87◦ ) = 288.21/− 30◦ V Vab =
√
3/30◦ Van = 499.20/0◦ V
Vbc = 499.20/− 120◦ V Vca = 499.20/120◦ V P 11.12 [a]
IaA =
7650 7650 + = 252.54/− 6.49◦ A 72 + j21 50
|IaA | = 252.54 A √ 7650 3/30◦ = 88.33/30◦ A [b] IAB = 150 |IAB | = 88.33 A
Problems [c] IAN =
7650/0◦ = 102/− 16.26◦ A 72 + j21
|IAN | = 102 A [d] Van = (252.54/− 6.49◦ )(j1) + 7650/0◦ = 7682.66/1.87◦ V √ |Vab | = 3(7682.66) = 13,306.76 V P 11.13 [a] Since the phase sequence is acb (negative) we have: Van = 2399.47/30◦ V Vbn = 2399.47/150◦ V Vcn = 2399.47/− 90◦ V 1 ZY = Z∆ = 0.9 + j4.5 Ω/φ 3
√ [b] Vab = 2399.47/30◦ − 2399.47/150◦ = 2399.47 3/0◦ = 4156/0◦ V Since the phase sequence is negative, it follows that Vbc = 4156/120◦ V Vca = 4156/− 120◦ V
11–13
11–14
CHAPTER 11. Balanced Three-Phase Circuits
[c]
Iba =
4156 = 301.87/− 78.69◦ A 2.7 + j13.5
Iac = 301.87/− 198.69◦ A IaA = Iba − Iac = 522.86/− 48.69◦ A Since we have a balanced three-phase circuit and a negative phase sequence we have: IbB = 522.86/71.31◦ A IcC = 522.86/− 168.69◦ A
Problems
11–15
[d]
IaA =
2399.47/30◦ = 522.86/− 48.69◦ A 0.9 + j4.5
Since we have a balanced three-phase circuit and a negative phase sequence we have: IbB = 522.86/71.31◦ A IcC = 522.86/− 168.69◦ A P 11.14 [a]
[b] IaA =
2399.47/30◦ = 1.2/46.26◦ A 1920 − j560
VAN = (1910 − j636)(1.2/46.26◦ ) = 2415.19/27.84◦ V √ |VAB | = 3(2415.19) = 4183.23 V
11–16
CHAPTER 11. Balanced Three-Phase Circuits
1.2 [c] |Iab | = √ = 0.69 A 3 [d] Van = (1919.1 − j564.5)(1.2/46.26◦ ) = 2400/29.87◦ V √ |Vab | = 3(2400) = 4156.92 V P 11.15 [a]
[b] IaA = √
13,800 = 2917/− 29.6◦ A 3(2.375 + j1.349)
|IaA | = 2917 A [c] VAN = (2.352 + j1.139)(2917/− 29.6◦ ) = 7622.94/− 3.76◦ V √ |VAB | = 3|VAN | = 13,203.31 V [d] Van = (2.372 + j1.319)(2917/− 29.6◦ ) = 7616.93/− 0.52◦ V √ |Vab | = 3|Van | = 13,712.52 V |IaA | [e] |IAB | = √ = 1684.13 A 3 [f] |Iab | = |IAB | = 1684.13 A P 11.16 [a] IAB =
4160/0◦ = 20.8/− 36.87◦ A 160 + j120
IBC = 20.8/83.13◦ A ICA = 20.8/− 156.87◦ A √ [b] IaA = 3/30◦ IAB = 36.03/− 6.87◦ A IbB = 36.03/113.13◦ A IcC = 36.03/− 126.87◦ A [c] Iba = IAB = 20.8/− 36.87◦ A; Icb = IBC = 20.8/83.13◦ A; Iac = ICA = 20.8/− 156.87◦ A;
Problems P 11.17 [a] IAB =
480/0◦ = 192/16.26◦ A 2.4 − j0.7
IBC =
480/120◦ = 48/83.13◦ A 8 + j6
ICA =
480/− 120◦ = 24/− 120◦ A 20
[b] IaA = IAB − ICA = 210/20.79◦ A IbB = IBC − IAB = 178.68/− 178.04◦ A IcC = ICA − IBC = 70.7/− 104.53◦ A P 11.18 From the solution to Problem 11.17 we have: SAB = (480/0◦ )(192/− 16.26◦ ) = 88,473.6 − j25,804.8 VA SBC = (480/120◦ )(48/− 83.13◦ ) = 18,432.0 + j13,824.0 VA SCA = (480/− 120◦ )(24/120◦ ) = 11,520 + j0 VA P 11.19 [a]
√ 24,000 3/0◦ = 66.5 − j49.9 A I1 = 400 + j300 √ 24,000 3/0◦ I2 = = 33.3 + j24.9 A 800 − j600 I∗3 =
57,600 + j734,400 √ = 1.4 + j17.7 24,000 3
I3 = 1.4 − j17.7 A
11–17
11–18
CHAPTER 11. Balanced Three-Phase Circuits IaA = I1 + I2 + I3 = 101.2 − j42.7 A = 109.8/− 22.8◦ A √ Van = (2 + j16)(101.2 − j42.7) + 24,000 3 = 42,456.2 + j1533.2 V Sφ = Van I∗aA = (42,456.2 + j1533.8)(101.2 + j42.7) = 4,229.2 + j1964.0 kVA
ST = 3Sφ = 12,687.7 + j9892.1 kVA √ [b] S1/φ = 24,000 3(66.5 + j49.9) = 2765.0 + j2073.8 kVA √ S2/φ = 24,000 3(33.3 − j24.9) = 1382.5 − j1036.9 kVA S3/φ = 57.6 + j734.4 kVA Sφ (load) = 4205.1 + j1771.3 kVA
4205.1 % delivered = (100) = 99.4% 4229.2 P 11.20 [a]
IaA =
1365/0◦ = 27.3/− 53.13◦ A 30 + j40
IaA ICA = √ /150◦ = 15.76/96.87◦ A 3 [b] Sg/φ = −1365I∗aA = −22,358.7 − j29,811.6 VA .·. Pdeveloped/phase = 22.359 kW Pabsorbed/phase = |IaA |2 28.5 = 21.241 kW % delivered =
21.241 (100) = 95% 22.359
Problems
11–19
P 11.21 Let pa , pb , and pc represent the instantaneous power of phases a, b, and c, respectively. Then assuming a positive phase sequence, we have pa = van iaA = [Vm cos ωt][Im cos(ωt − θφ )] pb = vbn ibB = [Vm cos(ωt − 120◦ )][Im cos(ωt − θφ − 120◦ )] pc = vcn icC = [Vm cos(ωt + 120◦ )][Im cos(ωt − θφ + 120◦ )] The total instantaneous power is pT = pa + pb + pc , so pT = Vm Im [cos ωt cos(ωt − θφ ) + cos(ωt − 120◦ ) cos(ωt − θφ − 120◦ ) + cos(ωt + 120◦ ) cos(ωt − θφ + 120◦ )] Now simplify using trigonometric identities. In simplifying, collect the coefficients of cos(ωt − θφ ) and sin(ωt − θφ ). We get pT = Vm Im [cos ωt(1 + 2 cos2 120◦ ) cos(ωt − θφ ) +2 sin ωt sin2 120◦ sin(ωt − θφ )] = 1.5Vm Im [cos ωt cos(ωt − θφ ) + sin ωt sin(ωt − θφ )] = 1.5Vm Im cos θφ P 11.22 [a] S1/φ = 40,000(0.96) − j40,000(0.28) = 38,400 − j11,200 VA S2/φ = 60,000(0.8) + j60,000(0.6) = 48,000 + j36,000 VA S3/φ = 33,600 + j5200 VA ST /φ = S1 + S2 + S3 = 120,000 + j30,000 VA
11–20
CHAPTER 11. Balanced Three-Phase Circuits 120,000 + j30,000 = 50 + j12.5 2400 = 50 − j12.5 A
.·. I∗aA = .·. IaA
Van = 2400 + (50 − j12.5)(1 + j8) = 2550 + j387.5 = 2579.27/8.64◦ V √ |Vab | = 3(2579.27) = 4467.43 V [b] Sg/φ = (2550 + j387.5)(50 + j12.5) = 122,656.25 + j51,250 VA % efficiency =
120,000 (100) = 97.83% 122,656.25
P 11.23 [a] S1 = (4.864 + j3.775) kVA S2 = 17.636(0.96) + j17.636(0.28) = (16.931 + j4.938) kVA √ 13,853 3VL IL sin θ3 = 13,853; sin θ3 = √ = 0.521 3(208)(73.8) Therefore
cos θ3 = 0.854
Therefore 13,853 × 0.854 = 22,693.58 W P3 = 0.521 S3 = 22.694 + j13.853 kVA ST = S1 + S2 + S3 = 44.49 + j22.57 kVA 1 ST /φ = ST = 14.83 + j7.52 kVA 3 208 ∗ √ IaA = (14.83 + j7.52)103 ; I∗aA = 123.49 + j62.64 A 3 IaA = 123.49 − j62.64 = 138.46/− 26.90◦ A [b] pf = cos(−26.90◦ ) = 0.892 lagging P 11.24
4000I∗1 = (210 + j280)103
(rms)
Problems I∗1 =
210 280 +j = 52.5 + j70 A 4 4
I1 = 52.5 − j70 A I2 =
4000/0◦ = 240 + j70 A 15.36 − j4.48
.·. IaA = I1 + I2 = 292.5 + j0 A Van = 4000 + j0 + 292.5(0.1 + j0.8) = 4036.04/3.32◦ V |Vab | =
√
3|Van | = 6990.62 V
P 11.25 [a] POUT = 746 × 100 = 74,600 W PIN = 74,600/(0.97) = 76,907.22 W √ 3VL IL cos θ = 76,907.22 76,907.22 = 242.58 A 3(208)(0.88) √ √ [b] Q = 3VL IL sin φ = 3(208)(242.58)(0.475) = 41,510.12 VAR IL = √
P 11.26 [a]
I∗aA
(160 + j46.67)103 = 133.3 + j38.9 = 1200
IaA = 133.3 − j38.9 A Van = 1200 + (133.3 − j38.9)(0.18 + j1.44) = 1280 + j185 V
IC =
1280 + j185 = −3.1 + j21.3 A −j60
Ina = (IaA + IC ) = −130.3 − j17.6 = 131.4/7.7◦ A
11–21
11–22
CHAPTER 11. Balanced Three-Phase Circuits
[b] Sg/φ = (1280 + j185)(−130.3 − j17.6) = −163,472 − j46,567.4 VA SgT = 3Sg/φ = −490.4 − j139.7 kVA Therefore, the source is delivering 490.4 kW and 139.7 kvars. [c] Pdel = 490.4 kW Pabs = 3(160,000) + 3|IaA |2 (0.18) = 490.4 kW = Pdel [d] Qdel = 3|IC |2 (60) + 139.7 × 103 = 223.3 kVAR Qabs = 3(46,666) + 3|IaA |2 (1.44) = 223.3 kVAR = Qdel P 11.27 [a]
1 Ss/φ = (60)(0.96 − j0.28) × 103 = 19.2 − j5.6 kVA 3 S1/φ = 15 kVA S2/φ = Ss/φ − S1/φ = 4.2 − j5.6 kVA 4200 − j5600 √ = 11.547 − j15.396 A .·. I∗2 = 630/ 3 I2 = 11.547 + j15.396 A √ 630/0◦ / 3 = 11.34 − j15.12 Ω Zy = I2 Z∆ = 3Zy = 34.02 − j45.36 Ω
Problems √ (630/ 3)2 = 31.5 Ω; R∆ = 3R = 94.5 Ω [b] R = 4200 √ (630/ 3)2 XL = = −23.625 Ω; X∆ = 3XL = −70.875 Ω −5600
P 11.28 Assume a ∆-connect load (series): 1 Sφ = (96 × 103 )(0.8 + j0.6) = 25,600 + j19,200 VA 3 ∗ Z∆φ
|480|2 = 5.76 − j4.32 Ω = 25,600 + j19,200
Z∆φ = 5.76 + 4.32 Ω
Now assume a Y-connected load (series): 1 ZY φ = Z∆φ = 1.92 + j1.44 Ω 3
11–23
11–24
CHAPTER 11. Balanced Three-Phase Circuits
Now assume a ∆-connected load (parallel): Pφ =
|480|2 R∆
R∆φ =
Qφ =
|480|2 = 9Ω 25,600
|480|2 X∆
X∆ φ =
|480|2 = 12 Ω 19,200
Now assume a Y-connected load (parallel): 1 RY φ = R∆φ = 3 Ω 3 1 XY φ = X∆φ = 4 Ω 3
Problems 1 P 11.29 Sg/φ = (41.6)(0.707 + j0.707) × 103 = 9803.73 + j9806.69 VA 3 I∗aA =
9803.73 + j9803.73 √ = 70.75 + j70.77 A 240/ 3
IaA = 70.75 − j70.77 A
240 VAN = √ − (0.04 + j0.03)(70.75 − j70.77) 3 = 133.61 + j0.71 = 133.61/0.30◦ V |VAB | =
√
3(133.61) = 231.42 V
[b] SL/φ = (133.61 + j0.71)(70.76 + j70.76) = 9403.1 + j9506.3 VA SL = 3SL/φ = 28,209 + j28,519 VA Check: Sg = 41,600(0.707 + j0.707) = 29,411 + j29,420 VA P = 3|IaA |2 (0.04) = 1202 W Pg = PL + P = 28,209 + 1202 = 29,411 W
(checks)
Q = 3|IaA |2 (0.03) = 901 VAR Qg = QL + Q = 28,519 + 901 = 29,420 VAR
(checks)
11–25
11–26
CHAPTER 11. Balanced Three-Phase Circuits
P 11.30 [a]
I∗aA IaA
1 720 (0.6) 103 = 240,000 + j180,000 VA 720 + j 3 0.8 240,000 + j180,000 = = 100 + j75 A 2400 = 100 − j75 A
SL/φ =
Van = 2400 + (0.8 + j6.4)(100 − j75) = 2960 + j580 = 3016.29/11.09◦ V √ |Vab | = 3(3016.29) = 5224.37 V [b]
I1 = 100 − j75 A
(from part [a])
1 S2 = 0 − j (576) × 103 = −j192,000 VAR 3 −j192,000 = −j80 A I∗2 = 2400 .·. I2 = j80 A IaA = 100 − j75 + j80 = 100 + j5 A Van = 2400 + (100 + j5)(0.8 + j6.4) = 2448 + j644 = 2531.29/14.74◦ V √ |Vab | = 3(2531.29) = 4384.33 V
Problems [c] |IaA | = 125 A Ploss/φ = (125)2 (0.8) = 12,500 W Pg/φ = 240,000 + 12,500 = 252.5 kW %η =
240 (100) = 95.05% 252.5
[d] |IaA | = 100.125 A P/φ = (100.125)2 (0.8) = 8020 W %η =
240,000 (100) = 96.77% 248,200
[e] Zcap/Y = −j
24002 = −j30 Ω −192,000
Zcap/∆ = 3Zcap/Y = −j90 Ω .·.
1 = 90; ωC
C=
1 = 29.47 µF (90)(120π)
P 11.31 [a] From Assessment Problem 11.9, IaA = (101.8 − j135.7) A Therefore Icap = j135.7 A √ 2450/ 3 = −j10.42 Ω Therefore ZCY = j135.7 Therefore CY =
1 = 254.5 µF (10.42)(2π)(60)
ZC∆ = (−j10.42)(3) = −j31.26 Ω Therefore C∆ = [b] CY = 254.5 µF [c] |IaA | = 101.8 A
254.5 = 84.84 µF 3
11–27
11–28
CHAPTER 11. Balanced Three-Phase Circuits
P 11.32 Zφ = |Z|/θ =
VAN IaA
θ = /VAN − /IaA θ1 = /VAB − /IaA For a positive phase sequence, /VAB = /VAN + 30◦ Thus, θ1 = /VAN + 30◦ − /IaA = θ + 30◦ Similarly, Zφ = |Z|/θ =
VCN IcC
θ = /VCN − /IcC θ2 = /VCB − /IcC For a positive phase sequence, /VCB = /VBA − 120◦ = /VAB + 60◦ /IcC = /IaA + 120◦ Thus, θ2 = /VAB + 60◦ − /IaA − 120◦ = θ1 − 60◦ = θ + 30◦ − 60◦ = θ − 30◦ P 11.33 Use values from the negative sequence part of Example 11.1 — part (g): VAB = 199.58/− 31.19◦ V IaA = 2.5/− 36.87◦ A Wm1 = |VAB ||IaA | cos(/VAB − /IaA ) = (199.58)(2.4) cos(5.68◦ ) = 476.63 W Wm2 = |VCB ||IcC | cos(/VCB − /IcC ) = (199.58)(2.4) cos(65.68◦ ) = 197.29 W CHECK: W1 + W2 = 673.9 = (2.4)2 (39)(3) = 673.9 W
Problems P 11.34 [a] W2 − W1 = VL IL [cos(θ − 30◦ ) − cos(θ + 30◦ )] = VL IL [cos θ cos 30◦ + sin θ sin 30◦ − cos θ cos 30◦ + sin θ sin 30◦ ] = 2VL IL sin θ sin 30◦ = VL IL sin θ, therefore
√
3(W2 − W1 ) =
√
3VL IL sin θ = QT
[b] Zφ = (8 + j6) Ω √ QT = 3[2476.25 − 979.75] = 2592 VAR, QT = 3(12)2 (6) = 2592 VAR; Zφ = (8 − j6) Ω √ QT = 3[979.75 − 2476.25] = −2592 VAR, QT = 3(12)2 (−6) = −2592 VAR; √ Zφ = 5(1 + j 3) Ω √ QT = 3[2160 − 0] = 3741.23 VAR, √ QT = 3(12)2 (5 3) = 3741.23 VAR; Zφ = 10/− 75◦ Ω √ QT = 3[−645.53 − 1763.63] = −4172.80 VAR, QT = 3(12)2 [−10 sin 75◦ ] = −4172.80 VAR P 11.35 IaA = (VAN /Zφ ) = |IL |/−θφ A, Zφ = |Z|/θφ ,
VBC = |VL |/− 90◦ V,
Wm = |VL | |IL | cos[−90◦ − (−θφ )] = |VL | |IL | cos(θφ − 90◦ ) = |VL | |IL | sin θφ , therefore
√
3Wm =
√
3|VL | |IL | sin θφ = Qtotal
11–29
11–30
CHAPTER 11. Balanced Three-Phase Circuits
P 11.36 [a] Z = 16 − j12 = 20/− 36.87◦ Ω VAN = 680/0◦ V;
.·. IaA = 34/36.87◦ A √ = 680 3/− 90◦ V
VBC = VBN − VCN √ Wm = (680 3)(34) cos(−90 − 36.87◦ ) = −24,027.0 W √ 3Wm = −41,616.0 VAR [b] Qφ = (342 )(−12) = −13,872 VAR √ QT = 3Qφ = −41,616 VAR = 3Wm P 11.37 [a] Zφ = 160 + j120 = 200/36.87◦ Ω Sφ =
41602 = 69,222.4 + j51,916.8 VA 160 − j120
ST = 3Sφ = 207,667.2 + j155,750.4 VA [b] Wm1 = (4160)(36.03) cos(0 + 6.87◦ ) = 148,808.64 W Wm2 = (4160)(36.03) cos(−60◦ + 126.87◦ ) = 58,877.55 W Check: P 11.38 [a] I∗aA =
PT = 207.7 kW = Wm1 + Wm2 .
144(0.96 − j0.28)103 = 20/− 16.26◦ A 7200
VBN = 7200/− 120◦ V; VBC = VBN − VCN
VCN = 7200/120◦ V √ = 7200 3/− 90◦ V
IbB = 20/− 103.74◦ A √ Wm1 = (7200 3)(20) cos(−90◦ + 103.74◦ ) = 242,278.14 W [b] Current coil in line aA, measure IaA . Voltage coil across AC, measure VAC . [c] IaA = 20/16.76◦ A
√ VAC = VAN − VCN = 7200 3/− 30◦ V √ Wm2 = (7200 3)(20) cos(−30◦ − 16.26◦ ) = 172,441.86 W
[d] Wm1 + Wm2 = 414.72kW PT = 432,000(0.96) = 414.72 kW = Wm1 + Wm2
Problems P 11.39 [a] W1 = |VBA ||IbB | cos θ1 Negative phase sequence: √ VBA = 240 3/150◦ V IaA =
240/0◦ = 18/30◦ A 13.33/− 30◦
IbB = 18/150◦ A √ W1 = (18)(240) 3 cos 0◦ = 7482.46 W W2 = |VCA ||IcC | cos θ2 √ VCA = 240 3/− 150◦ V IcC = 18/− 90◦ A √ W2 = (18)(240) 3 cos(−60◦ ) = 3741.23 W [b] Pφ = (18)2 (40/3) cos(−30◦ ) = 3741.23 W PT = 3Pφ = 11,223.69 W W1 + W2 = 7482.46 + 3741.23 = 11,223.69 W .·. W1 + W2 = PT
(checks)
P 11.40 [a] Negative phase sequence: √ VAB = 240 3/− 30◦ V √ VBC = 240 3/90◦ V √ VCA = 240 3/− 150◦ V √ 240 3/− 30◦ IAB = = 20.78/− 60◦ A 20/30◦ √ 240 3/90◦ = 6.93/90◦ A IBC = 60/0◦ √ 240 3/− 150◦ = 10.39/− 120◦ A ICA = 40/− 30◦ IaA = IAB + IAC = 18/− 30◦ A IcC = ICB + ICA = ICA − IBC = 16.75/− 108.06◦ A √ Wm1 = 240 3(18) cos(−30 + 30◦ ) = 7482.46 W √ Wm2 = 240 3(16.75) cos(−90 + 108.07◦ ) = 6621.23 W
11–31
11–32
CHAPTER 11. Balanced Three-Phase Circuits
[b] Wm1 + Wm2 = 14,103.69 W √ PA = (12 3)2 (20 cos 30◦ ) = 7482.46 W √ PB = (4 3)2 (60) = 2880 W √ PC = (6 3)2 [40 cos(−30◦ )] = 3741.23 W PA + PB + PC = 14,103.69 = Wm1 + Wm2 √ 3(W2 − W1 ) P 11.41 tan φ = = 0.7498 W1 + W2 .·. φ = 36.86◦ .·. 2400|IL | cos 66.87◦ = 40,823.09 |IL | = 43.3 A √ 2400/ 3 = 32 Ω |Zφ | = 43.3
.·. Zφ = 32/36.87◦ Ω
1 P 11.42 [a] Z = Z∆ = 4.48 + j15.36 = 16/73.74◦ Ω 3 IaA =
600/0◦ = 37.5/− 73.74◦ A 16/73.74◦
IbB = 37.5/− 193.74◦ A √ VAC = 600 3/− 30◦ V √ VBC = 600 3/− 90◦ V √ W1 = (600 3)(37.5) cos(−30 + 73.74◦ ) = 28,156.15 W √ W2 = (600 3)(37.5) cos(−90 + 193.74◦ ) = −9256.15 W [b] W1 + W2 = 18,900 W PT = 3(37.5)2 (13.44/3) = 18,900 W √ [c] 3(W1 − W2 ) = 64,800 VAR QT = 3(37.5)2 (46.08/3) = 64,800 VAR
Problems P 11.43 From the solution to Prob. 11.17 we have IaA = 210/20.79◦ A
and
IbB = 178.68/− 178.04◦ A
[a] W1 = |Vac | |IaA | cos(θac − θaA ) = 480(210) cos(60◦ − 20.79◦ ) = 78,103.2 W [b] W2 = |Vbc | |IbB | cos(θbc − θbB ) = 480(178.68) cos(120◦ + 178.04◦ ) = 40,317.7 W [c] W1 + W2 = 118,421 W PAB = (192)2 (2.4) = 88,473.6 W PBC = (48)2 (8) = 18,432 W PCA = (24)2 (20) = 11,520 W PAB + PBC + PCA = 118,425.7 therefore W1 + W2 ≈ Ptotal P 11.44 [a] For one phase,
[b]
[c]
(round-off differences)
11–33
11–34
CHAPTER 11. Balanced Three-Phase Circuits
[d]
P 11.45 [a] Q =
|V|2 XC
.·. |XC | =
(13,800)2 = 158.70 Ω 1.2 × 106
1 1 = 158.70; C= = 16.71 µF ωC 2π(60)(158.70) √ (13,800/ 3)2 1 [b] |XC | = = (158.70) 6 1.2 × 10 3 .·.
.·. C = 3(16.71) = 50.14 µF P 11.46 If the capacitors remain connected when the substation drops its load, the expression for the line current becomes 13,800 ∗ √ IaA = −j1.2 × 106 3 or
I∗aA = −j150.61 A IaA = j150.61 A
Hence Now, Van =
13,800 ◦ √ /0 + (0.6 + j4.8)(j150.61) = 7244.49 + j90.37 = 7245.05/0.71◦ V 3
The magnitude of the line-to-line voltage at the generating plant is |Vab | =
√
3(7245.05) = 12,548.80 V.
This is a problem because the voltage is below the acceptable minimum of 13 kV. Thus when the load at the substation drops off, the capacitors must be switched off.
Problems
11–35
P 11.47 Before the capacitors are added the total line loss is PL = 3|150.61 + j150.61|2 (0.6) = 81.66 kW After the capacitors are added the total line loss is PL = 3|150.61|2 (0.6) = 40.83 kW Note that adding the capacitors to control the voltage level also reduces the amount of power loss in the lines, which in this example is cut in half. P 11.48 [a]
13,800 ∗ √ IaA = 80 × 103 + j200 × 103 − j1200 × 103 3 √ √ 3 − j1000 3 80 = 10.04 − j125.51 A I∗aA = 13.8 .·. IaA = 10.04 + j125.51 A 13,800 ◦ √ /0 + (0.6 + j4.8)(10.04 + j125.51) 3 = 7371.01 + j123.50 = 7372.04/0.96◦ V
Van =
.·. |Vab | =
√
3(7372.04) = 12,768.75 V
[b] Yes, the magnitude of the line-to-line voltage at the power plant is less than the allowable minimum of 13 kV. P 11.49 [a]
13,800 ∗ √ IaA = (80 + j200) × 103 3 √ √ 80 3 + j200 3 ∗ = 10.04 + j25.1 A IaA = 13.8 .·. IaA = 10.04 − j25.1 A 13,800 ◦ √ /0 + (0.6 + j4.8)(10.04 − j25.1) 3 = 8093.95 + j33.13 = 8094.02/0.23◦ V
Van =
.·. |Vab | = [b] Yes:
√
3(8094.02) = 14,019.25 V
13 kV < 14,019.25 < 14.6 kV
[c] Ploss = 3|10.04 + j125.51|2 (0.6) = 28.54 kW [d] Ploss = 3|10.04 + j25.1|2 (0.6) = 1.32 kW [e] Yes, the voltage at the generating plant is at an acceptable level and the line loss is greatly reduced.
12 Introduction to the Laplace Transform Assessment Problems eβt + e−βt 2 Therefore, 1 ∞ −(s−β)t [e + e−(s+β)t ]dt L{cosh βt} = 2 0−
AP 12.1 [a] cosh βt =
1 e−(s−β)t ∞ e−(s+β)t = + 2 −(s − β) 0− −(s + β)
=
1 2
1 1 + s−β s+β
=
s2
∞ − 0
s − β2
eβt − e−βt 2 Therefore, 1 ∞ −(s−β)t e − e−(s+β)t dt L{sinh βt} = 2 0−
[b] sinh βt =
1 e−(s−β)t = 2 −(s − β) =
1 2
∞ 0−
1 e−(s+β)t − 2 −(s + β)
1 1 − s−β s+β
AP 12.2 [a] Let f (t) = te−at : F (s) = L{te−at } = Now,
1 (s + a)2
L{tf (t)} = −
dF (s) ds
12–1
=
∞
β (s2 − β 2 )
0−
12–2
CHAPTER 12. Introduction to the Laplace Transform
So,
−at
L{t · te
f (t) = e−at sinh βt,
[b] Let
L{f (t)} = F (s) =
df (t) L dt [c] Let
1 d 2 }=− = 2 ds (s + a) (s + a)2 then
β (s + a)2 − β 2
= sF (s) − f (0− ) =
s(β) βs − 0 = (s + a)2 − β 2 (s + a)2 − β 2
f (t) = cos ωt. Then
F (s) =
s 2 (s + ω 2 )
and
dF (s) −(s2 − ω 2 ) = 2 ds (s + ω 2 )2
Therefore L{t cos ωt} = −
s2 − ω 2 dF (s) = 2 ds (s + ω 2 )2
K1 K2 K3 6s2 + 26s + 26 = + + AP 12.3 F (s) = (s + 1)(s + 2)(s + 3) s+1 s+2 s+3 K1 =
6 − 26 + 26 = 3; (1)(2)
K3 =
54 − 78 + 26 =1 (−2)(−1)
K2 =
24 − 52 + 26 =2 (−1)(1)
Therefore f (t) = [3e−t + 2e−2t + e−3t ] u(t) AP 12.4 F (s) =
K1 K2 K3 7s2 + 63s + 134 = + + (s + 3)(s + 4)(s + 5) s+3 s+4 s+5
K1 =
63 − 189 + 134 = 4; 1(2)
K3 =
175 − 315 + 134 = −3 (−2)(−1)
K2 =
f (t) = [4e−3t + 6e−4t − 3e−5t ]u(t) AP 12.5 F (s) =
10(s2 + 119) (s + 5)(s2 + 10s + 169)
s1,2 = −5 +
√
25 − 169 = −5 + j12
112 − 252 + 134 =6 (−1)(1)
Problems F (s) =
K1 =
K2 K2∗ K1 + + s + 5 s + 5 − j12 s + 5 + j12
10(25 + 119) = 10 25 − 50 + 169
10[(−5 + j12)2 + 119] K2 = = j4.167 = 4.167/90◦ (j12)(j24) Therefore f (t) = [10e−5t + 8.33e−5t cos(12t + 90◦ )] u(t) = [10e−5t − 8.33e−5t sin 12t] u(t) AP 12.6 F (s) = K0 =
K1 K0 K2 4s2 + 7s + 1 + = + 2 2 s(s + 1) s (s + 1) s+1
1 = 1; (1)2
K1 =
d 4s2 + 7s + 1 K2 = ds s =
4−7+1 =2 −1
s=−1
s(8s + 7) − (4s2 + 7s + 1) = s2 s=−1
1+2 =3 1
Therefore f (t) = [1 + 2te−t + 3e−t ] u(t) AP 12.7 F (s) = =
(s2
K1∗ K1 K2 + + (s + 2 − j1)2 (s + 2 − j1) (s + 2 + j1)2 +
K1 =
40 40 = 2 + 4s + 5) (s + 2 − j1)2 (s + 2 + j1)2
K2∗ (s + 2 + j1)
40 = −10 = 10/180◦ 2 (j2)
40 d K2 = ds (s + 2 + j1)2 K2∗ = j10
and
= s=−2+j1
K1∗ = −10 −80 = −j10 = 10/− 90◦ 3 (j2)
12–3
12–4
CHAPTER 12. Introduction to the Laplace Transform Therefore f (t) = [20te−2t cos(t + 180◦ ) + 20e−2t cos(t − 90◦ )] u(t) = 20e−2t [sin t − t cos t] u(t)
AP 12.8 F (s) =
5s2 + 29s + 32 s+8 5s2 + 29s + 32 = =5− 2 (s + 2)(s + 4) s + 6s + 8 (s + 2)(s + 4)
K1 K2 s+8 = + (s + 2)(s + 4) s+2 s+4 K1 =
−2 + 8 = 3; 2
K2 =
−4 + 8 = −2 −2
Therefore, F (s) = 5 −
2 3 + s+2 s+4
f (t) = 5δ(t) + [−3e−2t + 2e−4t ]u(t) AP 12.9 F (s) =
4(s + 1) 2s3 + 8s2 + 2s − 4 4 = 2s − 2 + = 2s − 2 + 2 s + 5s + 4 (s + 1)(s + 4) s+4
f (t) = 2 AP 12.10
dδ(t) − 2δ(t) + 4e−4t u(t) dt
7s3 [1 + (9/s) + (134/7s2 )] lim sF (s) = lim 3 =7 s→∞ s→∞ s [1 + (3/s)][1 + (4/s)][1 + (5/s)] .·. f (0+ ) = 7
7s3 + 63s2 + 134s lim sF (s) = lim =0 s→0 s→0 (s + 3)(s + 4)(s + 5) .·. f (∞) = 0
s3 [4 + (7/s) + (1/s2 )] lim sF (s) = lim =4 s→∞ s→∞ s3 [1 + (1/s)]2 .·. f (0+ ) = 4
Problems
4s2 + 7s + 1 lim sF (s) = lim =1 s→0 s→0 (s + 1)2 .·. f (∞) = 1
40s lim sF (s) = lim 4 =0 s→∞ s→∞ s [1 + (4/s) + (5/s2 )]2 .·. f (0+ ) = 0
40s lim sF (s) = lim =0 s→0 s→0 (s2 + 4s + 5)2 .·. f (∞) = 0
12–5
12–6
CHAPTER 12. Introduction to the Laplace Transform
Problems P 12.1
[a] f (t) = 5t[u(t) − u(t − 2)] + 10[u(t − 2) − u(t − 6)]+ (−5t + 40)[u(t − 6) − u(t − 8)] [b] f (t) = (10 sin πt)[u(t) − u(t − 2)] [c] f (t) = 4t[u(t) − u(t − 5)]
P 12.2
[a] (10 + t)[u(t + 10) − u(t)] + (10 − t)[u(t) − u(t − 10)] = (t + 10)u(t + 10) − 2tu(t) + (t − 10)u(t − 10) [b] (−24 − 8t)[u(t + 3) − u(t + 2)] − 8[u(t + 2) − u(t + 1)] + 8t[u(t + 1) − u(t − 1)] +8[u(t − 1) − u(t − 2)] + (24 − 8t)[u(t − 2) − u(t − 3)] = −8(t + 3)u(t + 3) + 8(t + 2)u(t + 2) + 8(t + 1)u(t + 1) − 8(t − 1)u(t − 1) −8(t − 2)u(t − 2) + 8(t − 3)u(t − 3)
P 12.3
P 12.4
[a]
Problems
12–7
[b] f (t) = −20t[u(t) − u(t − 1)] − 20[u(t − 1) − u(t − 2)] +20 cos( π2 t)[u(t − 2) − u(t − 4)] +(100 − 20t)[u(t − 4) − u(t − 5)]
P 12.5
P 12.6
1 1 1 bh = (2ε) =1 [a] A = 2 2 ε [b] 0; [c] ∞ [a] I =
3 −1
(t + 2)δ(t) dt + 3
3 −1
8(t3 + 2)δ(t − 1) dt
= (03 + 2) + 8(13 + 2) = 2 + 8(3) = 26 [b] I =
2 −2
t2 δ(t) dt +
2 −2
t2 δ(t + 1.5) dt +
2 −2
t2 δ(t − 3) dt
= 02 + (−1.5)2 + 0 = 2.25 P 12.7 P 12.8
1 4 + j0 jt0 2 1 ∞ (4 + jω) jtω · πδ(ω) · e dω = πe = f (t) = 2π −∞ (9 + jω) 2π 9 + j0 9
As ε → 0 the amplitude → ∞; the duration → 0; and the area is independent of ε, i.e., A=
P 12.9
∞
ε 1 dt = 1 π ε2 + t2
−∞
F (s) =
ε −ε
1 −st esε − e−sε e dt = 2ε 2εs
1 sesε + se−sε 1 2s · =1 F (s) = = lim ε→0 2s 1 2s 1 dv = δ (t − a) dt,
P 12.10 [a] Let
v = δ(t − a)
du = f (t) dt
u = f (t), Therefore ∞
−∞
f (t)δ (t − a) dt = f (t)δ(t − a)
∞
−∞
−
∞ −∞
δ(t − a)f (t) dt
= 0 − f (a)
[b] L{δ (t)} =
∞ 0−
−st
δ (t)e
d(e−st ) dt = − dt
= − −se−st t=0
t=0
=s
12–8
CHAPTER 12. Introduction to the Laplace Transform
P 12.11
F (s) =
−ε/2 −ε
ε/2 ε 4 −st −4 −st 4 −st e dt + e dt + e dt 3 3 ε ε −ε/2 ε/2 ε3
Therefore F (s) =
4 sε [e − 2esε/2 + 2e−sε/2 − e−sε ] 3 sε
L{δ (t)} = lim F (s) ε→0
After applying L’Hopital’s rule three times, we have
2s s s 2s 3s sesε − esε/2 − e−sε/2 + se−sε = lim ε→0 3 4 4 3 2
Therefore L{δ (t)} = s2
dn f (t) P 12.12 L dtn
= sn F (s) − sn−1 f (0− ) − sn−2 f (0− ) − · · · ,
Therefore L{δ (n) (t)} = sn (1) − sn−1 δ(0− ) − sn−2 δ (0− ) − · · · = sn P 12.13 [a] L{t} =
1 ; s2
therefore L{te−at } =
ejωt − e−jωt j2 Therefore
1 (s + a)2
[b] sin ωt =
L{sin ωt} = =
1 j2
s2
ω + ω2
1 1 − s − jω s + jω
=
1 j2
2jω 2 s + ω2
Problems [c] sin(ωt + θ) = (sin ωt cos θ + cos ωt sin θ) Therefore L{sin(ωt + θ)} = cos θL{sin ωt} + sin θL{cos ωt} ω cos θ + s sin θ = s2 + ω 2 ∞ ∞ e−st 1 1 [d] L{t} = te−st dt = 2 (−st − 1) = 0 − 2 (0 − 1) = 2 s s s 0 0 [e] f (t) = cosh t cosh θ + sinh t sinh θ From Assessment Problem 12.1(a) s L{cosh t} = 2 s −1 From Assessment Problem 12.1(b) 1 L{sinh t} = 2 s −1
s 1 · + sinh θ 2 . . L{cosh(t + θ)} = cosh θ 2 s −1 s −1 = −at
P 12.14 [a] L{te
}=
∞ 0−
te−(s+a)t dt
e−(s+a)t = (s + a)2
[b]
sinh θ + s[cosh θ] s2 − 1
− (s + a)t − 1
= 0+
1 (s + a)2
.·. L{te−at } =
1 (s + a)2
∞ 0−
d −at s (te ) = L −0 dt (s + a)2 s = (s + a)2 [c]
d −at (te ) = −ate−at + e−at dt −a −a 1 s+a = L{−ate−at + e−at } = + + 2 2 (s + a) (s + a) (s + a) (s + a)2
d −at s (te ) = .·. L dt (s + a)2
CHECKS
12–9
12–10
CHAPTER 12. Introduction to the Laplace Transform
P 12.15 [a] L{f (t)} =
ε
e−st dt + −ε 2ε
∞ ε
− ae−a(t−ε) e−st dt
a 1 sε (e − e−sε ) − e−sε = F (s) 2sε s+a s a = lim F (s) = 1 − ε→0 s+a s+a =
1 s+a
[b] L{e−at } =
Therefore L{f (t)} = sF (s) − f (0− ) = −at
P 12.16 L{e
f (t)} =
t
∞ 0−
−at
[e
−st
f (t)]e
dt =
∞ 0−
s s −0= s+a s+a
f (t)e−(s+a)t dt = F (s + a)
1 F (s) = s s(s + a) t 1 1 1 y dy = = 3 [b] L 2 s s s 0− t −at 1 e [c] e−ax dx = − a a 0−
P 12.17 [a] L
0−
−ax
e
dx =
1 e−at L − a a
t
t2 y dy = ; 2 0−
d sin ωt P 12.18 [a] L dt
d3 (t2 ) [c] L dt3 [d]
t2 L 2
d cos ωt [b] L dt
=
s2
=
=s
1 1 1 1 = = − a s s+a s(s + a) =
1 2 1 · 3 = 3 2 s s
sω −0 + ω2
s2 −0 s2 + ω 2
3
2 − s2 (0) − s(0) − 2(0) = 2 s3
d sin ωt = (cos ωt) · ω, dt
L{ω cos ωt} =
ωs s2 + ω 2
d cos ωt = −ω sin ωt + δ(t) dt L{−ω sin ωt + δ(t)} = − d2 (t2 ) = 2u(t); dt2
ω2 s2 + 1 = s2 + ω 2 s2 + ω 2
d3 (t2 ) = 2δ(t); dt3
L{2δ(t)} = 2
Problems P 12.19 [a] f (t) = 5t[u(t) − u(t − 2)] +(20 − 5t)[u(t − 2) − u(t − 6)] +(5t − 40)[u(t − 6) − u(t − 8)] = 5tu(t) − 10(t − 2)u(t − 2) +10(t − 6)u(t − 6) − 5(t − 8)u(t − 8) 5[1 − 2e−2s + 2e−6s − e−8s ] .·. F (s) = s2 [b]
f (t) = 5[u(t) − u(t − 2)] − 5[u(t − 2) − u(t − 6)] +5[u(t − 6) − u(t − 8)] = 5u(t) − 10u(t − 2) + 10u(t − 6) − 5u(t − 8) L{f (t)} =
5[1 − 2e−2s + 2e−6s − e−8s ] s
[c]
f (t) = 5δ(t) − 10δ(t − 2) + 10δ(t − 6) − 5δ(t − 8) L{f (t)} = 5[1 − 2e−2s + 2e−6s − e−8s ]
12–11
12–12
CHAPTER 12. Introduction to the Laplace Transform t
P 12.20 [a]
0−
t2 2
x dx =
t2 L 2
1 2
=
∞ 0−
t2 e−st dt ∞
1 e−st 2 2 (s t + 2st + 2) = − 2 −s3 0
1 1 (2) = 2s3 s3
= .·. L [b] L
t
t
0−
x dx =
0−
.·. L
x dx = t 0−
1 s3
L{t} 1/s2 1 = = 3 s s s
x dx =
1 s3
CHECKS 40e−3s (s + 8)
P 12.21 [a] L{40e−8(t−3) u(t − 3)} = [b] First rewrite f (t) as
f (t) = (5t − 10)u(t − 2) + (40 − 10t)u(t − 4) +(10t − 80)u(t − 8) + (50 − 5t)u(t − 10) = 5(t − 2)u(t − 2) − 10(t − 4)u(t − 4) +10(t − 8)u(t − 8) − 5(t − 10)u(t − 10) .·. F (s) = P 12.22 L{f (at)} =
∞
Let
u = at,
and
u=∞
0−
5[e−2s − 2e−4s + 2e−8s − e−10s ] s2 f (at)e−st dt u = 0−
du = a dt, when
t = 0−
t=∞
Therefore L{f (at)} = P 12.23 [a] f1 (t) = e−at sin ωt;
when
∞ 0−
f (u)e−(u/a)s
F1 (s) =
F (s) = sF1 (s) − f1 (0− ) =
1 du = F (s/a) a a
ω (s + a)2 + ω 2
sω −0 (s + a)2 + ω 2
Problems [b] f1 (t) = e−at cos ωt; F (s) = [c]
F1 (s) =
s+a (s + a)2 + ω 2
s+a F1 (s) = s s[(s + a)2 + ω 2 ]
d −at [e sin ωt] = ωe−at cos ωt − ae−at sin ωt dt Therefore F (s) = t 0−
ωs ω(s + a) − ωa = 2 2 (s + a) + ω (s + a)2 + ω 2
e−ax cos ωx dx =
Therefore
−ae−at cos ωt + ωe−at sin ωt + a a2 + ω 2
−a(s + a) ω2 a 1 + + F (s) = 2 2 2 2 2 2 a + ω (s + a) + ω (s + a) + ω s = P 12.24 [a]
s+a s[(s + a)2 + ω 2 ]
d dF (s) = ds ds
∞ 0−
f (t)e−st dt = −
Therefore L{tf (t)} = −
0−
tf (t)e−st dt
dF (s) ds
d2 F (s) ∞ 2 = t f (t)e−st dt; [b] ds2 0− Therefore
∞
d3 F (s) ∞ 3 = −t f (t)e−st dt ds3 0−
∞ dn F (s) n = (−1) tn f (t)e−st dt = (−1)n L{tn f (t)} dsn 0−
[c] L{t5 } = L{t4 t} = (−1)4 d L{t sin βt} = (−1)1 ds
d4 ds4
1 s2
β 2 s + β2
=
120 s6
=
(s2
L{te−t cosh t}: From Assessment Problem 12.1(a), s F (s) = L{cosh t} = 2 s −1 (s2 − 1)1 − s(2s) s2 + 1 dF = = − ds (s2 − 1)2 (s2 − 1)2 Therefore
−
s2 + 1 dF = 2 ds (s − 1)2
2βs + β 2 )2
12–13
12–14
CHAPTER 12. Introduction to the Laplace Transform Thus L{t cosh t} =
s2 + 1 (s2 − 1)2
L{e−t t cosh t} = P 12.25 [a]
∞ s
F (u)du = =
(s + 1)2 + 1 s2 + 2s + 2 = [(s + 1)2 − 1]2 s2 (s + 2)2
∞ ∞ 0−
s
∞
−ut
f (t)e
∞
0−
f (t)
s
dt du =
e−ut du dt =
∞
∞ ∞ 0−
t sin βt therefore L t
=
∞ s
f (t)e
du dt
e−tu ∞ f (t) dt −t s 0−
∞
−e−st f (t) = f (t) dt = L −t t 0− 2βs [b] L{t sin βt} = 2 (s + β 2 )2
s
−ut
2βu du 2 (u + β 2 )2
Let ω = u2 + β 2 , then ω = s2 + β 2 when u = s, and ω = ∞ when u = ∞; also dω = 2u du, thus
t sin βt L t P 12.26 Ig (s) =
1.2s ; s2 + 1
=β
∞
s2 +β 2
dω −1 ∞ β = β 2 2= 2 2 ω ω s + β2 s +β
1 = 1.6; RC
1 = 1; LC
1 = 1.6 C
V (s) 1 V (s) + + C[sV (s) − v(0− )] = Ig (s) R L s V (s)
1 1 + + sC = Ig (s) R Ls
V (s) =
=
1 R
Ig (s) = 1 + Ls + sC
1 sIg (s) LsIg (s) C = 1 L s2 + RC s+ s + 1 + s2 LC R
1.92s2 (1.6)(1.2)s2 = (s2 + 1.6s + 1)(s2 + 1) (s2 + 1.6s + 1)(s2 + 1)
1t dvo vo − Vdc + =0 vo dx + C P 12.27 [a] R L 0 dt .·.
1 LC
R t dvo = Vdc vo dx + RC vo + L 0 dt
Problems [b] Vo +
R Vo Vdc + RCsVo = L s s
.·.
sLVo + RVo + RCLs2 Vo = LVdc
.·.
Vo (s) =
[c] io =
(1/RC)Vdc + (1/RC)s + (1/LC)
1t vo dx L 0 (1/RCL)Vdc Vo = 2 sL s[s + (1/RC)s + (1/LC)]
Io (s) = P 12.28 [a]
s2
1 1 = = 50 × 106 −3 LC (200 × 10 )(100 × 10−9 ) 1 1 = = 2000 RC (5000)(100 × 10−9 ) Vo (s) =
s2
70,000 + 2000s + 50 × 106
s1,2 = −1000 ± j7000 rad/s Vo (s) =
70,000 (s + 1000 − j7000)(s + 1000 + j7000)
=
K1∗ K1 + s + 1000 − j7000 s + 1000 + j7000
K1 =
70,000 = 5/− 90◦ j14,000
vo (t) = 10e−1000t cos(7000t − 90◦ )u(t) V = 10e−1000t sin(7000t)u(t) V [b] Io (s) = =
35(10,000) s(s + 1000 − j7000)(s + 1000 + j7000) K2 K2∗ K1 + + s s + 1000 − j7000 s + 1000 + j7000
K1 =
35(10,000) = 7 mA 50 × 106
K2 =
35(10,000) = 3.54/171.87◦ mA (−1000 + j7000)(j14,000)
io (t) = [7 + 7.07e−1000t cos(7000t + 171.87◦ )]u(t) mA
12–15
12–16
CHAPTER 12. Introduction to the Laplace Transform
dvo 1t vo +C vo dx + P 12.29 [a] Idc = L 0 R dt Vo (s) Vo (s) Idc = + + sCVo (s) [b] s sL R .·. Vo (s) = [c] io = C
s2
Idc /C + (1/RC)s + (1/LC)
dvo dt
.·. Io (s) = sCVo (s) = P 12.30 [a]
s2
sIdc + (1/RC)s + (1/LC)
1 1 = 500 = 3 RC (1 × 10 )(2 × 10−6 ) 1 1 = = 40,000 LC (12.5)(2 × 10−6 ) Vo (s) =
500,000Idc s + 500s + 40,000
=
500,000Idc (s + 100)(s + 400)
=
15,000 (s + 100)(s + 400)
=
K1 K2 + s + 100 s + 400
K1 =
15,000 = 50; 300
Vo (s) =
K2 =
15,000 = −50 −300
50 50 − s + 100 s + 400
vo (t) = [50e−100t − 50e−400t ]u(t) V [b] Io (s) =
0.03s (s + 100)(s + 400)
=
K2 K1 + s + 100 s + 400
K1 =
0.03(−100) = −0.01 300
K2 =
0.03(−400) = 0.04 −300
Problems Io (s) =
12–17
−0.01 0.04 + s + 100 s + 400
io (t) = (40e−400t − 10e−100t )u(t) mA [c] io (0) = 40 − 10 = 30 mA Yes. The initial inductor current is zero by hypothesis, the initial resistor current is zero because the initial capacitor voltage is zero by hypothesis. Thus at t = 0 the source current appears in the capacitor. P 12.31 [a] C
dv1 v1 − v2 + = ig dt R
1t v2 − v1 =0 v2 dτ + L 0 R or dv1 v1 v2 + − = ig C dt R R −
v1 v2 1t + v2 dτ = 0 + R R L 0
[b] CsV1 (s) + −
V1 (s) V2 (s) − = Ig (s) R R
V1 (s) V2 (s) V2 (s) + + =0 R R sL
or (RCs + 1)V1 (s) − V2 (s) = RIg (s) −sLV1 (s) + (R + sL)V2 (s) = 0 Solving, V2 (s) = P 12.32
1 = 5 × 106 ; C V2 (s) =
C[s2
sIg (s) + (R/L)s + (1/LC)] 1 = 25 × 106 ; LC
R = 8000 L
(6 × 10−3 )(5 × 106 ) s2 + 8000s + 25 × 106
s1,2 = −4000 ± j3000 V2 (s) =
30,000 (s + 4000 − j3000)(s + 4000 + j3000)
K1∗ K1 + = s + 4000 − j3000 s + 4000 + j3000
12–18
CHAPTER 12. Introduction to the Laplace Transform
K1 =
30,000 = −j5 = 5/− 90◦ j6000
v2 (t) = 10e−4000t cos(3000t − 90◦ ) = [10e−4000t sin 3000t]u(t) V P 12.33 [a] For t ≥ 0+ : dvo vo +C + io = 0 R dt vo = L
dio ; dt
dvo d2 io =L 2 dt dt
d2 io L dio + LC 2 R dt dt
.·.
1 d2 io 1 dio + io = 0 + 2 dt RC dt LC 1 1 Io (s) = 0 [b] s2 Io (s) − sIdc − 0 + [sIo (s) − Idc ] + RC LC
1 1 s+ = Idc (s + 1/RC) Io (s) s2 + RC LC or
Io (s) = P 12.34
1 = 8000; RC Io (s) =
s2
Idc [s + (1/RC)] [s2 + (1/RC)s + (1/LC)] 1 = 16 × 106 LC
0.005(s + 8000) + 8000s + 16 × 106
s1,2 = −4000 Io (s) =
0.005(s + 8000) K1 K2 = + 2 2 (s + 4000) (s + 4000) s + 4000
K1 = 0.005(s + 8000) K2 =
s=−4000
= 20
d [0.005(s + 8000)]s=−4000 = 0.005 ds
Io (s) =
20 0.005 + 2 (s + 4000) s + 4000
io (t) = [20te−4000t + 0.005e−4000t ]u(t) V
Problems P 12.35 [a] 300 = 60i1 + 25 0=5
d di1 d di1 + 10 (i2 − i1 ) + 5 (i1 − i2 ) − 10 dt dt dt dt
di1 d (i2 − i1 ) + 10 + 40i2 dt dt
Simplifying the above equations gives: 300 = 60i1 + 10 0 = 40i2 + 5 [b]
di2 di1 +5 dt dt
di2 di1 +5 dt dt
300 = (10s + 60)I1 (s) + 5sI2 (s) s 0 = 5sI1 (s) + (5s + 40)I2 (s)
[c] Solving the equations in (b), I1 (s) =
60(s + 8) s(s + 4)(s + 24)
I2 (s) =
−60 (s + 4)(s + 24)
[d] I1 (s) =
K2 K3 K1 + + s s + 4 s + 24
K1 =
(60)(8) = 5; (4)(24)
K3 =
(60)(−16) = −2 (−24)(−20)
I1 (s) =
(60)(4) = −3 (−4)(20)
K2 =
3 2 5 − − s s + 4 s + 24
i1 (t) = (5 − 3e−4t − 2e−24t )u(t) A I2 (s) = K1 =
K1 K2 + s + 4 s + 24
−60 = −3; 20
K2 =
3 −3 + I2 (s) = s + 4 s + 24
−60 =3 −20
i2 (t) = (3e−24t − 3e−4t )u(t) A [e] i1 (∞) = 5 A;
i2 (∞) = 0 A
12–19
12–20
CHAPTER 12. Introduction to the Laplace Transform
[f] Yes, at t = ∞ 300 = 5A 60 Since i1 is a dc current at t = ∞ there is no voltage induced in the 10 H inductor; hence, i2 = 0. Also note that i1 (0) = 0 and i2 (0) = 0. Thus our solutions satisfy the condition of no initial energy stored in the circuit. i1 =
P 12.36 From Problem 12.26: V (s) =
1.92s2 (s2 + 1.6s + 1)(s2 + 1)
s2 + 1.6s + 1 = (s + 0.8 + j0.6)(s + 0.8 − j0.6);
s2 + 1 = (s − j1)(s + j1)
Therefore V (s) = =
1.92s2 (s + 0.8 + j0.6)(s + 0.8 − j0.6)(s − j1)(s + j1) K1∗ K2 K2∗ K1 + + + s + 0.8 − j0.6 s + 0.8 + j0.6 s − j1 s + j1
1.92s2 = 1/− 126.87◦ K1 = 2 (s + 0.8 + j0.6)(s + 1) s=−0.8+j0.6 1.92s2 K2 = = 0.6/0◦ 2 (s + j1)(s + 1.6s + 1) s=j1
Therefore v(t) = [2e−0.8t cos(0.6t − 126.87◦ ) + 1.2 cos(t)]u(t) V P 12.37 [a] F (s) =
K2 K3 K1 + + s+1 s+2 s+4
8s2 + 37s + 32 K1 = =1 (s + 2)(s + 4) s=−1 8s2 + 37s + 32 K2 = =5 (s + 1)(s + 4) s=−2 8s2 + 37s + 32 K3 = =2 (s + 1)(s + 2) s=−4
f (t) = [e−t + 5e−2t + 2e−4t ]u(t)
Problems [b] F (s) =
K2 K3 K4 K1 + + + s s+2 s+3 s+5
8s3 + 89s2 + 311s + 300 = 10 K1 = (s + 2)(s + 3)(s + 5) s=0 8s3 + 89s2 + 311s + 300 K2 = =5 s(s + 3)(s + 5) s=−2
K3 =
8s3 + 89s2 + 311s + 300 = −8 s(s + 2)(s + 5) s=−3
8s3 + 89s2 + 311s + 300 K4 = =1 s(s + 2)(s + 3) s=−5
f (t) = [10 + 5e−2t − 8e−3t + e−5t ]u(t) [c] F (s) =
K2 K2∗ K1 + + s+1 s+2−j s+2+j
22s2 + 60s + 58 K1 = = 10 s2 + 4s + 5 s=−1 22s2 + 60s + 58 = 6 + j8 = 10/53.13◦ K2 = (s + 1)(s + 2 + j) s=−2+j
f (t) = [10e−t + 20e−2t cos(t + 53.13◦ )]u(t) [d] F (s) =
K1 K2 K2∗ + + s s+7−j s+7+j
250(s + 7)(s + 14) = 490 K1 = s2 + 14s + 50 s=0
K2 =
250(s + 7)(s + 14) = 125/− 163.74◦ s(s + 7 + j) s=−7+j
f (t) = [490 + 250e−7t cos(t − 163.74◦ )]u(t) P 12.38 [a] F (s) =
K3 K1 K2 + + 2 s s s+5
100 = 20 K1 = s + 5 s=0
d 100 −100 = = −4 K2 = ds s + 5 (s + 5)2 s=0
K3 =
100 =4 s2 s=−5
f (t) = [20t − 4 + 4e−5t ]u(t)
12–21
12–22
CHAPTER 12. Introduction to the Laplace Transform
[b] F (s) =
K2 K3 K1 + + 2 s (s + 1) s+1
50(s + 5) = 250 K1 = (s + 1)2 s=0
K2 =
50(s + 5) = −200 s s=−1
50 50(s + 5) d 50(s + 5) − K3 = = ds s s s2
= −250 s=−1
f (t) = [250 − 200te−t − 250e−t ]u(t) [c] F (s) =
K3 K3∗ K1 K2 + + + s2 s s+3−j s+3+j
100(s + 3) K1 = 2 = 30 s + 6s + 10 s=0
d 100(s + 3) K2 = ds s2 + 6s + 10
100(s + 3)(2s + 6) 100 − = 2 s + 6s + 10 (s2 + 6s + 10)2
= 10 − 18 = −8 s=0
100(s + 3) = 4 + j3 = 5/36.87◦ K3 = 2 s (s + 3 + j) s=−3+j
f (t) = [30t − 8 + 10e−3t cos(t + 36.87◦ )]u(t) [d] F (s) =
K2 K1 K3 K4 + + + s (s + 1)3 (s + 1)2 s + 1
K1 =
5(s + 2)2 = 20 (s + 1)3 s=0
K2 =
5(s + 2)2 = −5 s s=−1
10(s + 2) 5(s + 2)2 d 5(s + 2)2 − K3 = = ds s s s2 = −10 − 5 = −15
1 d 10(s + 2) 5(s + 2)2 − K4 = 2 ds s s2
s=−1
1 10 10(s + 2) 10(s + 2) 10(s + 2)2 − = − + 2 s s2 s2 s3
s=−1
Problems
12–23
1 = (−10 − 10 − 10 − 10) = −20 2 f (t) = [20 − 2.5t2 e−t − 15te−t − 20e−t ]u(t) [e] F (s) = K1 = K2 =
K2 K3∗ K2∗ K3 K1 + + + + s (s + 2 − j)2 (s + 2 + j)2 s + 2 − j s + 2 − j
400 = 16 (s2 + 4s + 5)2 s=0
400 = 44.72/26.57◦ s(s + 2 + j)2 s=−2+j
400 −400 d −800 K3 = = 2 + 2 2 ds s(s + 2 + j) s (s + 2 + j) s(s + 2 + j)3
s=−2+j
= 12 + j16 − 20 + j40 = −8 + j56 = 56.57/98.13◦ f (t) = [16 + 89.44te−2t cos(t + 26.57◦ ) + 113.14e−2t cos(t + 98.13◦ )]u(t) P 12.39 [a]
5 F (s) = s2 + 6s + 8
5s2 + 38s + 80 5s2 + 30s + 40 8s + 40
F (s) = 5 +
8s + 40 K1 K2 =5+ + + 6s + 8 s+2 s+4
s2
8s + 40 K1 = = 12 s + 4 s=−2
8s + 40 = −4 K2 = s + 2 s=−4
f (t) = 5δ(t) + [12e−2t − 4e−4t ]u(t) [b]
10 F (s) = s2 + 48s + 625
10s2 + 512s + 7186 10s2 + 480s + 6250 32s + 936
F (s) = 10 + K1 =
K2∗ K1 32s + 936 + = 10 + s2 + 48s + 625 s + 24 − j7 s + 24 + j7
32s + 936 = 16 − j12 = 20/− 36.87◦ s + 24 + j7 s=−24+j7
f (t) = 10δ(t) + [40e−24t cos(7t − 36.87◦ )]u(t)
12–24
CHAPTER 12. Introduction to the Laplace Transform s − 10
[c] F (s) = s2 + 15s + 50
s3 + 5s2 − 50s − 100 s3 + 15s2 + 50s −10s2 − 100s − 100 −10s2 − 150s − 500 50s + 400
F (s) = s − 10 +
K2 K1 + s + 5 s + 10
50s + 400 K1 = = 30 s + 10 s=−5
K2 =
50s + 400 = 20 s + 5 s=−10
f (t) = δ (t) − 10δ(t) + [30e−5t + 20e−10t ]u(t) P 12.40 [a] F (s) =
K3 K3∗ K1 K2 + + + s2 s s + 1 − j2 s + 1 + j2
100(s + 1) = 20 K1 = 2 s + 2s + 5 s=0
100(s + 1)(2s + 2) 100 d 100(s + 1) − = 2 K2 = 2 ds s + 2s + 5 s + 2s + 5 (s2 + 2s + 5)2 = 20 − 8 = 12
100(s + 1) K3 = 2 = −6 + j8 = 10/126.87◦ s (s + 1 + j2) s=−1+j2
f (t) = [20t + 12 + 20e−t cos(2t + 126.87◦ )]u(t) [b] F (s) = K1 =
K2 K1 K3 K4 + + + s (s + 5)3 (s + 5)2 s + 5
500 =4 (s + 5)3 s=0
K2 =
500 = −100 s s=−5
K3 =
d 500 −500 = = −20 ds s s2 s=−5
K4 =
1 d −500 1 1000 = = −4 2 ds s2 2 (s3 ) s=−5
f (t) = [4 − 50t2 e−5t − 20te−5t − 4e−5t ]u(t)
s=0
Problems [c] F (s) =
12–25
K2 K3 K4 K1 + + + 3 2 s (s + 1) (s + 1) s+1
40(s + 2) = 80 K1 = (s + 1)3 s=0
40(s + 2) K2 = = −40 s s=−1
40 40(s + 2) d 40(s + 2) − = K3 = ds s s s2
1 d 40 40(s + 2) − K4 = 2 ds s s2
1 −40 40 80(s + 2) = − 2 + 2 s2 s s3
s=−1
= −40 − 40 = −80 s=−1
1 = (−40 − 40 − 80) = −80 2
f (t) = [80 − 20t2 e−t − 80te−t − 80e−t ]u(t) [d] F (s) =
K2 K1 K3 K4 K5 + + + + 4 3 2 s (s + 1) (s + 1) (s + 1) s+1
(s + 5)2 = 25 K1 = (s + 1)4 s=0
K2 =
(s + 5)2 = −16 s s=−1
2(s + 5) (s + 5)2 d (s + 5)2 − = K3 = ds s s s2 = −8 − 16 = −24
1 d 2(s + 5) (s + 5)2 − K4 = 2 ds s s2
s=−1
1 2 2(s + 5) 2(s + 5) 2(s + 5)2 − = − + s2 s2 s3 2 s
s=−1
1 = (−2 − 8 − 8 − 32) = −25 2
1 d 2 2(s + 5) 2(s + 5) 2(s + 5)2 K5 = − − + 6 ds s s2 s2 s3
1 −2 2 4(s + 5) 2 4(s + 5) 4(s + 5) 6(s + 5)2 = − + − + + − 6 s2 s2 s3 s2 s3 s3 s4 1 = (−2 − 2 − 16 − 2 − 16 − 16 − 96) = −25 6 f (t) = [25 − (8/3)t3 e−t − 12t2 e−t − 25te−t − 25e−t ]u(t)
s=−1
12–26
CHAPTER 12. Introduction to the Laplace Transform −1
P 12.41 f (t) = L
K∗ K + s + α − jβ s + α + jβ
= Ke−αt ejβt + K ∗ e−αt e−jβt = |K|e−αt [ejθ ejβt + e−jθ e−jβt ] = |K|e−αt [ej(βt+θ) + e−j(βt+θ) ] = 2|K|e−αt cos(βt + θ) n
n
P 12.42 [a] L{t f (t)} = (−1) Let
f (t) = 1,
dn F (s) dsn
1 F (s) = , s
then
Therefore L{tn } = (−1)n It follows that and
L{t(r−1) } =
L{t(r−1) e−at } =
thus
dn F (s) (−1)n n! = dsn s(n+1)
(−1)n n! n! = (n+1) (n+1) s s
(r − 1)! sr
(r − 1)! (s + a)r
Therefore
Ktr−1 e−at K K L{tr−1 e−at } = = L (r − 1)! (s + a)r (r − 1)!
−1
[b] f (t) = L
K K∗ + (s + α − jβ)r (s + α + jβ)r
Therefore f (t) = =
Ktr−1 −(α−jβ)t K ∗ tr−1 −(α+jβ)t e + e (r − 1)! (r − 1)! |K|tr−1 e−αt jθ jβt e e + e−jθ e−jβt (r − 1)!
2|K|tr−1 e−αt = cos(βt + θ) (r − 1)!
1.92s3 lim 4 =0 P 12.43 [a] s→∞ lim sV (s) = s→∞ s [1 + (1.6/s) + (1/s2 )][1 + (1/s2 )] Therefore v(0+ ) = 0 [b] No, V has a pair of poles on the imaginary axis.
Problems P 12.44 [a] sF (s) =
8s3 + 37s2 + 32s (s + 1)(s + 2)(s + 4)
lim sF (s) = 0,
s→0
lim sF (s) = 8,
s→∞
[b] sF (s) =
.·. f (∞) = 0 .·. f (0+ ) = 8
8s3 + 89s2 + 311s + 300 (s + 2)(s2 + 8s + 15)
lim sF (s) = 10;
.·. f (∞) = 10
lim sF (s) = 8,
.·. f (0+ ) = 8
s→0
s→∞
[c] sF (s) =
22s3 + 60s2 + 58s (s + 1)(s2 + 4s + 5)
lim sF (s) = 0,
s→0
lim sF (s) = 22,
s→∞
[d] sF (s) =
12–27
.·. f (∞) = 0 .·. f (0+ ) = 22
250(s + 7)(s + 14) (s2 + 14s + 50)
250(7)(14) = 490, .·. f (∞) = 490 s→0 50 .·. f (0+ ) = 250 lim sF (s) = 250, s→∞ lim sF (s) =
100 s(s + 5) F (s) has a second-order pole at the origin so we cannot use the final value theorem.
P 12.45 [a] sF (s) =
lim sF (s) = 0,
s→∞
[b] sF (s) =
.·. f (0+ ) = 0
50(s + 5) (s + 1)2
lim sF (s) = 250,
s→0
lim sF (s) = 0,
s→∞
.·. f (∞) = 250 .·. f (0+ ) = 0
100(s + 3) s(s2 + 6s + 10) F (s) has a second-order pole at the origin so we cannot use the final value theorem.
[c] sF (s) =
lim sF (s) = 0,
s→∞
.·. f (0+ ) = 0
12–28
CHAPTER 12. Introduction to the Laplace Transform
[d] sF (s) =
5(s + 2)2 (s + 1)3
lim sF (s) = 20,
.·. f (∞) = 20
lim sF (s) = 0,
.·. f (0+ ) = 0
s→0
s→∞
[e] sF (s) =
(s2
400 + 4s + 5)2
lim sF (s) = 16,
.·. f (∞) = 16
lim sF (s) = 0,
.·. f (0+ ) = 0
s→0
s→∞
P 12.46 All of the F (s) functions referenced in this problem are improper rational functions, and thus the corresponding f (t) functions contain impulses (δ(t)). Thus, neither the initial value theorem nor the final value theorem may be applied to these F (s) functions! P 12.47 sVo (s) =
sVdc /RC s2 + (1/RC)s + (1/LC)
lim sVo (s) = 0,
s→0
lim sVo (s) = 0,
s→∞
sIo (s) =
s2
Vdc Vdc /RLC = , 1/LC R
lim sIo (s) = 0,
s→∞
P 12.48 sVo (s) =
s2
lim sVo (s) = 0, lim sVo (s) = 0,
s→∞
.·. io (0+ ) = 0
.·. vo (∞) = 0 .·. vo (0+ ) = 0
s2 Idc s2 + (1/RC)s + (1/LC)
lim sIo (s) = 0,
s→0
lim sIo (s) = Idc ,
s→∞
.·. io (∞) =
(Idc /C)s + (1/RC)s + (1/LC)
s→0
sIo (s) =
.·. vo (0+ ) = 0
Vdc /RCL) + (1/RC)s + (1/LC)
lim sIo (s) =
s→0
.·. vo (∞) = 0
.·. io (∞) = 0 .·. vo (0+ ) = Idc
Vdc R
Problems
12–29
100(s + 1) s(s2 + 2s + 5) F (s) has a second-order pole at the origin, so we cannot use the final value theorem here.
P 12.49 [a] sF (s) =
.·. f (0+ ) = 0
lim sF (s) = 0,
s→∞
[b] sF (s) =
500 (s + 5)3 .·. f (∞) = 4
lim sF (s) = 4,
s→0
.·. f (0+ ) = 0
lim sF (s) = 0,
s→∞
[c] sF (s) =
40(s + 2) (s + 1)3
lim sF (s) = 80,
.·. f (∞) = 80
lim sF (s) = 0,
.·. f (0+ ) = 0
s→0
s→∞
[d] sF (s) =
(s + 5)2 (s + 1)4
lim sF (s) = 25,
.·. f (∞) = 25
lim sF (s) = 0,
.·. f (0+ ) = 0
s→0
s→∞
P 12.50 sIo (s) =
s2
Idc s[s + (1/RC)] + (1/RC)s + (1/LC)
lim sIo (s) = 0,
s→0
lim sIo (s) = Idc ,
s→∞
.·. io (∞) = 0 .·. io (0+ ) = Idc
13 The Laplace Transform in Circuit Analysis
Assessment Problems AP 13.1 [a] Y =
1 1 C[s2 + (1/RC)s + (1/LC) + + sC = R sL s
106 1 = = 80,000; RC (500)(0.025)
1 = 25 × 108 LC
25 × 10−9 (s2 + 80,000s + 25 × 108 ) s √ = −40,000 ± 16 × 108 − 25 × 108 = −40,000 ± j30,000 rad/s
Therefore Y = [b] −z1,2
−z1 = −40,000 − j30,000 rad/s −z2 = −40,000 + j30,000 rad/s −p1 = 0 rad/s AP 13.2 [a] Z = 2000 +
4 × 107 s 1 = 2000 + 2 Y s + 80,000s + 25 × 108
2000(s2 + 105 s + 25 × 108 ) 2000(s + 50,000)2 = s2 + 80,000s + 25 × 108 s2 + 80,000s + 25 × 108 [b] −z1 = −z2 = −50,000 rad/s =
−p1 = −40,000 − j30,000 rad/s −p2 = −40,000 + j30,000 rad/s
13–1
CHAPTER 13. The Laplace Transform in Circuit Analysis
13–2
AP 13.3 [a] At t = 0− ,
0.2v1 = 0.8v2 ;
Therefore v1 (0− ) = 80 V = v1 (0+ );
I=
v1 + v2 = 100 V
v1 = 4v2 ;
v2 (0− ) = 20 V = v2 (0+ )
20 × 10−3 (80/s) + (20/s) = 5000 + [(5 × 106 )/s] + (1.25 × 106 /s) s + 1250
80 5 × 106 − V1 = s s
20 × 10−3 s + 1250
20 1.25 × 106 − V2 = s s
[b] i = 20e−1250t u(t) mA;
=
20 × 10−3 s + 1250
80 s + 1250
=
20 s + 1250
v1 = 80e−1250t u(t) V
v2 = 20e−1250t u(t) V AP 13.4 [a]
I=
Vdc /s Vdc /L = 2 R + sL + (1/sC) s + (R/L)s + (1/LC)
Vdc = 40; L I=
s2
R = 1.2; L
40 + 1.2s + 1
1 = 1.0 LC
Problems [b] I =
K1 K1∗ 40 = + (s + 0.6 − j0.8)(s + 0.6 + j0.8) s + 0.6 − j0.8 s + 0.6 + j0.8
K1 =
40 = −j25 = 25/− 90◦ ; j1.6
K1∗ = 25/90◦
i = 50e−0.6t cos(0.8t − 90◦ ) = [50e−0.6t sin 0.8t]u(t) A [c] V = sLI = = K1 =
160s 160s = s2 + 1.2s + 1 (s + 0.6 − j0.8)(s + 0.6 + j0.8) K1∗ K1 + s + 0.6 − j0.8 s + 0.6 + j0.8
160(−0.6 + j0.8) = 100/36.87◦ j1.6
[d] v(t) = [200e−0.6t cos(0.8t + 36.87◦ )]u(t) V AP 13.5 [a]
The two node voltage equations are 5 V1 − V2 V2 V2 − V1 V2 − (15/s) + + =0 + V1 s = and s s 3 s 15 Solving for V1 and V2 yields V1 =
5(s + 3) , 2 s(s + 2.5s + 1)
V2 =
2.5(s2 + 6) s(s2 + 2.5s + 1)
[b] The partial fraction expansions of V1 and V2 are 15 15 50/3 5/3 125/6 25/3 − + and V2 = − + s s + 0.5 s + 2 s s + 0.5 s + 2 It follows that 50 5 v1 (t) = 15 − e−0.5t + e−2t u(t) V and 3 3 V1 =
125 −0.5t 25 −2t e v2 (t) = 15 − + e u(t) V 6 3 [c] v1 (0+ ) = 15 −
50 5 + =0 3 3
v2 (0+ ) = 15 −
125 25 + = 2.5 V 6 3
13–3
CHAPTER 13. The Laplace Transform in Circuit Analysis
13–4
[d] v1 (∞) = 15 V;
v2 (∞) = 15 V
AP 13.6 [a]
With no load across terminals a–b, Vx = 20/s: 20 1 20 − VTh = 0 − VTh s + 1.2 2 s s 20(s + 2.4) therefore VTh = s(s + 2)
Vx = 5IT
and
ZTh =
VT IT
Solving for IT gives (VT − 5IT )s + VT − 6IT 2 Therefore
IT =
14IT = VT s − 5sIT + 2VT ;
therefore
[b]
I=
20(s + 2.4) VTh = ZTh + 2 + s s(s + 3)(s + 6)
ZTh =
5(s + 2.8) s+2
Problems AP 13.7 [a] i2 = 1.25e−t − 1.25e−3t ; Therefore
di2 = −1.25e−t + 3.75e−3t dt
therefore
di2 = 0 when dt
1.25e−t = 3.75e−3t
or
e2t = 3,
t = 0.5(ln 3) = 549.31 ms
i2 (max) = 1.25[e−0.549 − e−3(0.549) ] = 481.13 mA [b] From Eqs. 13.68 and 13.69, we have ∆ = 12(s2 + 4s + 3) = 12(s + 1)(s + 3) and Therefore I1 =
N1 = 60(s + 2)
5(s + 2) N1 = ∆ (s + 1)(s + 3)
A partial fraction expansion leads to the expression 2.5 2.5 + s+1 s+3 Therefore we get I1 =
i1 = 2.5[e−t + e−3t ]u(t) A di1 = −2.5[e−t + 3e−3t ]; dt [d] When i2 is at its peak value,
[c]
di2 =0 dt
Therefore L2 [e] i2 (max) =
di2 dt
di1 (0.54931) = −2.89 A/s dt
= 0 and
M i2 = − 12
−2(−2.89) = 481.13 mA 12
di1 dt
(Checks)
AP 13.8 [a] The s-domain circuit with the voltage source acting alone is
V V s V − (20/s) + + =0 2 1.25s 20
13–5
CHAPTER 13. The Laplace Transform in Circuit Analysis
13–6
V =
100/3 100/3 200 = − (s + 2)(s + 8) s+2 s+8
v =
100 −2t [e − e−8t ]u(t) V 3
[b] With the current source acting alone,
V V s 5 V + + = 2 1.25s 20 s V =
50/3 50/3 100 = − (s + 2)(s + 8) s+2 s+8
v =
50 −2t [e − e−8t ]u(t) V 3
[c] v = v + v = [50e−2t − 50e−8t ]u(t) V Vo s Vo + = Ig ; s+2 10 [b] −z1 = −2 rad/s;
AP 13.9 [a]
AP 13.10 [a] Vo =
Vo 10(s + 2) = H(s) = 2 Ig s + 2s + 10 −p1 = −1 + j3 rad/s; −p2 = −1 − j3 rad/s therefore
10(s + 2) 1 K0 K1 K1∗ · = + + s2 + 2s + 10 s s s + 1 − j3 s + 1 + j3
K0 = 2;
K1 = (5/3)/− 126.87◦ ;
K1∗ = (5/3)/126.87◦
vo = [2 + (10/3)e−t cos(3t − 126.87◦ )]u(t) V [b] Vo =
K2 K2∗ 10(s + 2) · 1 = + s2 + 2s + 10 s + 1 − j3 s + 1 + j3
K2 = 5.27/− 18.43◦ ;
K2∗ = 5.27/18.43◦
vo = [10.54e−t cos(3t − 18.43◦ )]u(t) V AP 13.11 [a] H(s) = L{h(t)} = L{vo (t)} vo (t) = 10,000 cos θe−70t cos 240t − 10,000 sin θe−70t sin 240t = 9600e−70t cos 240t − 2800e−70t sin 240t
Problems Therefore H(s) = = [b] Vo (s) = H(s) ·
s2
9600s + 140s + 62,500
9600 1 = 2 s s + 140s + 62,500 =
K1 =
9600(s + 70) 2800(240) − 2 2 (s + 70) + (240) (s + 70)2 + (240)2
K1∗ K1 + s + 70 − j240 s + 70 + j240
9600 = −j20 = 20/− 90◦ j480
Therefore vo (t) = [40e−70t cos(240t − 90◦ )]u(t) V = [40e−70t sin 240t]u(t) V AP 13.12 From Assessment Problem 13.9: H(s) =
10(s + 2) + 2s + 10
s2
Therefore H(j4) =
10(2 + j4) = 4.47/− 63.43◦ 10 − 16 + j8
Thus, vo = (10)(4.47) cos(4t − 63.43◦ ) = 44.7 cos(4t − 63.43◦ ) V AP 13.13 [a] Let R1 = 10 kΩ,
R2 = 50 kΩ,
C = 400 pF,
Vg R2 R2 + (1/sC)
then
V1 = V2 =
Also
V1 − Vg V1 − Vo + =0 R1 R1
therefore Vo = 2V1 − Vg Now solving for Vo /Vg , we get It follows that
H(j50,000) =
H(s) =
R2 Cs − 1 R2 Cs + 1
j−1 = j1 = 1/90◦ j+1
Therefore vo = 10 cos(50,000t + 90◦ ) V
R2 C = 2 × 10−5
13–7
13–8
CHAPTER 13. The Laplace Transform in Circuit Analysis [b] Replacing R2 by Rx gives us
H(s) =
Rx Cs − 1 Rx Cs + 1
Therefore H(j50,000) =
Rx + j50,000 j20 × 10−6 Rx − 1 = −6 j20 × 10 Rx + 1 Rx − j50,000
Thus, 50,000 = tan 60◦ = 1.7321, Rx
Rx = 28,867.51 Ω
Problems
Problems P 13.1
−I0 −LI0 = ; ZN = sL sL s Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4.
P 13.2
1t vdτ + I0 ; i= L 0−
P 13.3
VTh = Vab = CVo
P 13.4
L[s2 + (R/L)s + (1/LC)] 1 = [a] Z = R + sL + sC s
Iscab = IN =
1 sC
therefore
1 I= L
Vo ; s
ZTh =
=
V s
+
V I0 I0 = + s sL s
1 sC
0.0025[s2 + 16 × 107 s + 1010 ] s [b] Zeros at −62.5 rad/s and −1.6 × 108 rad/s Pole at 0. =
P 13.5
[a] Y =
C[s2 + (1/RC)s + (1/LC)] 1 1 + sC = + R sL s
Z=
s/C 4 × 106 s 1 = 2 = 2 Y s + (1/RC)s + (1/LC) s + 2000s + 64 × 104
[b] zero at −z1 = 0 poles at −p1 = −400 rad/s and −p2 = −1600 rad/s P 13.6
[a]
Z=
(1/C)(s + R/L) (R + sL)(1/sC) = 2 R + sL + (1/sC) s + (R/L)s + (1/LC)
250 R = = 3125; L 0.08 Z=
1 1 = = 25 × 106 LC (0.08)(0.5 × 10−6 )
2 × 106 (s + 3125) s2 + 3125s + 25 × 106
13–9
13–10
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Z =
2 × 106 (s + 3125) (s + 1562.5 − j4749.6)(s + 1562.5 + j4749.6)
−z1 = −3125 rad/s;
−p1 = −1562.5 + j4749.6 rad/s
−p2 = −1562.5 − j4749.6 rad/s P 13.7
Transform the Y-connection of the two resistors and the capacitor into the equivalent delta-connection:
where Za =
(1/s)(1) + (1)(1/s) + (1)(1) =s+2 1/s
Zb = Zc =
s+2 (1/s)(1) + (1)(1/s) + (1)(1) = 1 s
Then Zab = Za [(sZc ) + (sZb )] = Za 2(sZb ) sZb =
s(s + 2) s+2 = 2 s + (s + 2)/s s +s+2
Zab = (s + 2) =
2s(s + 2)2 2s(s + 2) = s2 + s + 2 (s + 2)(s2 + s + 2) + 2s(s + 2)
2s 2s(s + 2) = s2 + 3s + 2 s+1
One zero at the origin (0 rad/s); one pole at −1 rad/s. P 13.8
Z1 =
16 4s 4(s2 + 4s + 16) 16 + s4 = + = s s s+4 s(s + 4)
Zab = 4
16(s2 + 4s + 16) 4(s2 + 4s + 16) = s(s + 4) 8s2 + 32s + 64
Problems
=
13–11
2(s + 2 + j3.46)(s + 2 − j3.46) 2(s2 + 4s + 16) = 2 s + 4s + 8 (s + 2 + j2)(s + 2 − j2)
Zeros at −2 + j3.46 rad/s and −2 − j3.46 rad/s; poles at −2 + j2 rad/s and −2 − j2 rad/s. P 13.9
[a] For t > 0:
−150 2.5s [b] Vo = 5 (16 × 10 )/s + 5000 + 2.5s s =
s2
−150s + 2000s + 64 × 104
−150s (s + 400)(s + 1600) K2 K1 + [c] Vo = s + 400 s + 1600 =
K1 =
−150s = 50 s + 1600 s=−400
−150s = −200 K2 = s + 400 s=−1600
Vo =
50 200 − s + 400 s + 1600
vo (t) = (50e−400t − 200e−1600t )u(t) V
13–12
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.10 [a] For t < 0:
1 1 1 1 = + + = 0.1875; Re 8 80 20
Re = 5.33 Ω
v1 = (9)(5.33) = 48 V iL (0− ) =
48 = 2.4 A 20
vC (0− ) = −v1 = −48 V For t = 0+ :
s-domain circuit:
where R = 20 Ω;
C = 6.25 µF;
L = 6.4 mH; [b]
and
γ = −48 V;
ρ = −2.4 A
Vo ρ Vo + Vo sC − γC + − =0 R sL s .·. Vo =
s2
γ[s + (ρ/γC)] + (1/RC)s + (1/LC)
ρ −2.4 = = 8000 γC (−48)(6.25 × 10−6 )
Problems 1 1 = = 8000 RC (20)(6.25 × 10−6 ) 1 1 = = 25 × 106 −3 LC (6.4 × 10 )(6.25 × 10−6 ) Vo = [c] IL =
s2
−48(s + 8000) + 8000s + 25 × 106
ρ Vo 2.4 Vo − = + sL s 0.0064s s =
[d] Vo = =
s2
s(s2
−7500(s + 8000) 2.4 2.4(s + 4875) + = 2 6 + 8000s + 25 × 10 ) s (s + 8000s + 25 × 106 )
−48(s + 8000) + 8000s + 25 × 106
K1∗ K1 + s + 4000 − j3000 s + 4000 + j3000
−48(s + 8000) = 40/126.87◦ K1 = s + 4000 + j3000 s=−4000+j3000
vo (t) = [80e−4000t cos(3000t + 126.87◦ )]u(t) V [e] IL = =
s2
2.4(s + 4875) + 8000s + 25 × 106
K1∗ K1 + s + 4000 − j3000 s + 4000 + j3000
2.4(s + 4875) = 1.25/− 16.26◦ K1 = s + 4000 + j3000 s=−4000+j3000
iL (t) = [2.5e−4000t cos(3000t − 16.26◦ )]u(t) A P 13.11 For t < 0:
vo (0− ) − 500 vo (0− ) vo (0− ) + + =0 5 25 100
13–13
CHAPTER 13. The Laplace Transform in Circuit Analysis
13–14
25vo (0− ) = 10,000 iL (0− ) =
.·.
vo (0− ) = 400 V
400 vo (0− ) = = 16 A 25 25
For t > 0 :
Vo Vo − (400/s) Vo + 400 + + =0 25 + 25s 100 100/s
Vo
1 s 1 400 + + =4− 25 + 25s 100 100 25 + 25s
.·.
Io =
Vo =
400(s − 3) s2 + 2s + 5
−20s − 20 Vo − (400/s) = 2 100/s s + 2s + 5 =
K1 K1∗ + s + 1 − j2 s + 1 + j2
−20(s + 1) = −10 K1 = s + 1 + j2 s=−1+j2
io (t) = [−20e−t cos 2t]u(t) A
Problems P 13.12 [a] For t < 0:
V2 =
10 (450) = 90 V 10 + 40
For t > 0:
[b] V1 =
25(450/s) (125,000/s) + 25 + 1.25 × 10−3 s =
9 × 106 9 × 106 = s2 + 20, 000s + 108 (s + 10,000)2
v1 (t) = (9 × 106 te−10,000t )u(t) V [c] V2 =
(25,000/s)(450/s) 90 − s (125,000/s) + 1.25 × 10−3 s + 25 = =
s2
90(s + 20,000) + 20,000s + 108
900,000 90 + 2 (s + 10,000) s + 10,000
v2 (t) = [9 × 105 te−10,000t + 90e−10,000t ]u(t) V
13–15
13–16
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.13 [a] For t < 0:
iL (0− ) = i1 =
−100 −100 = = −5 A 4 + 1040 + 8 20
10 (5) = 1 A 50
vC (0− ) = 10(1) + 4(5) − 100 = −70 V For t > 0:
[b] (20 + 2s + 100/s)I = 10 + 5(s + 7) + 10s + 50
.·.
I=
Vo =
70 100 I− s s
s2
70 s
=
−70(s + 20/7) −70s2 − 200s = 2 2 s + 10s + 50 s(s + 10s + 50)
=
K1 K1∗ + s + 5 − j5 s + 5 + j5
−70(s + 20/7) K1 = = 38.1/− 156.8◦ s + 5 + j5 s=−5+j5
[c] vo (t) = 76.2e−5t cos(5t − 156.8◦ )u(t) V
Problems P 13.14 [a] iL (0− ) = iL (0+ ) =
24 = 8A 3
directed upward
20(10/s) 200 25IT (10/s) + VT = 25Iφ + IT = IT 20 + (10/s) 20 + (10/s) 10 + 20s 45 VT 250 + 200 = =Z= IT 20s + 10 2s + 1 Vo Vo (2s + 1) Vo 8 + + = 5 45 5.625s s [9s + (2s + 1)s + 8]Vo 8 = 45s s Vo [2s2 + 10s + 8] = 360 Vo = [b] Vo =
2s2
180 360 = 2 + 10s + 8 s + 5s + 4
K1 K2 180 = + (s + 1)(s + 4) s+1 s+4
K1 =
180 = 60; 3
Vo =
60 60 − s+1 s+4
K2 =
180 = −60 −3
vo (t) = [60e−t − 60e−4t ]u(t) V
13–17
13–18
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.15 [a]
Vo − 35/s Vo − 8Iφ + 0.4V∆ + =0 2 s + (250/s)
Vo − 8Iφ V∆ = s; s + (250/s)
Iφ =
(35/s) − Vo 2
Solving for Vo yields: Vo =
29.4s2 + 56s + 1750 29.4s2 + 56s + 1750 = s(s2 + 2s + 50) s(s + 1 − j7)(s + 1 + j7)
Vo =
K2 K2∗ K1 + + s s + 1 − j7 s + 1 + j7
29.4s2 + 56s + 1750 K1 = = 35 s2 + 2s + 50 s=0
K2 =
29.4s2 + 56s + 1750 s(s + 1 + j7) s=−1+j7
= −2.8 + j0.6 = 2.86/167.91◦ .·. vo (t) = [35 + 5.73e−t cos(7t + 167.91◦ )]u(t) V [b] At t = 0+
vo = 35 + 5.73 cos(167.91◦ ) = 29.4 V
vo − 35 + 0.4v∆ = 0; 2
vo − 35 + 0.8v∆ = 0
vo = v∆ + 8iφ = v∆ + 8(0.4v∆ ) = 4.2v∆ vo + (0.8)
vo = 35; 4.2
.·. vo (0+ ) = 29.4 V(Checks)
Problems At t = ∞, the circuit is
v∆ = 0,
iφ = 0
.·. vo = 35 V(Checks)
P 13.16 [a] For t < 0:
Vc Vc − 137.5 Vc − 50 + + =0 400 1200 500
Vc
1 1 137.5 1 50 + + + = 400 1200 500 400 500
Vc = 75 V iL (0− ) =
75 − 137.5 = −0.125 A 500
For t > 0:
75 5 × 105 I+ [b] Vo = s s 0=−
5 × 105 75 137.5 + 100I + I+ − 1.25 × 10−3 + 0.01sI s s s
13–19
CHAPTER 13. The Laplace Transform in Circuit Analysis
13–20
5 × 105 62.5 + 0.01s = + 1.25 × 10−3 I 100 + s s .·. I =
s2
6250 + 0.125s + 104 s + 5 × 107
5 × 105 Vo = s = [c] Vo =
6250 + 0.125s 75 + 2 4 7 s + 10 s + 5 × 10 s
75s2 + 812,500s + 6875 × 106 s(s2 + 104 s + 5 × 107 )
K1 K2 K2∗ + + s s + 5000 − j5000 s + 5000 + j5000
K1 =
75s2 + 812,500s + 6875 × 106 = 137.5 s2 + 104 s + 5 × 107 s=0
75s2 + 812,500s + 6875 × 106 K2 = = 40.02/141.34◦ s(s + 5000 + j5000) s=−5000+j5000
vo (t) = [137.5 + 80.04e−5000t cos(5000t + 141.34◦ )]u(t) V P 13.17
5 × 10−3 Vo Vo −3 V + = + 3.75 × 10 φ s 200 + 4 × 106 /s 0.04s Vφ = .·.
4 × 106 /s 4 × 106 Vo V = o 200 + 4 × 106 /s 200s + 4 × 106 Vo s 15,000Vo 25Vo 5 × 10−3 = + + 6 6 s 200s + 4 × 10 200s + 4 × 10 s
.·. Vo =
s2
s + 20,000 K1 K2 = + 8 2 + 20,000s + 10 (s + 10,000) s + 10,000
K1 = 10,000; Vo =
K2 = 1
10,000 1 + 2 (s + 10,000) s + 10,000
vo (t) = [10,000te−10,000t + e−10,000t ]u(t) V
Problems P 13.18 vo (0− ) = vo (0+ ) = 0
−
0.05 Vo Vo Vo Vo + + − 21 + 7 =0 s 1000 25 1000 10 /s
Vo .·.
s 20 + 7 1000 10 Vo =
=
0.05 s
2.5 2.5 500,000 = − s(s + 200,000) s s + 200,000
vo (t) = [2.5 − 2.5e−200,000t ]u(t) V P 13.19 [a] io (0− ) =
Io =
20 = 5 mA 4000
20/s + Lρ R + sL + 1/sC =
s2
40 + s(0.005) 20/L + sρ = 2 + sR/L + 1/LC s + 8000s + 16 × 106
Vo = −Lρ + sLIo = −0.0025 + =
−40,000 (s + 4000)2
vo (t) = −40,000te−4000t u(t) V
s2
0.0025s(s + 8000) + 8000s + 16 × 106
13–21
13–22
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Io =
s2 =
0.005(s + 8000) + 8000s + 16 × 106 K1 K2 + 2 (s + 4000) s + 4000
K1 = 20
K2 = 0.005
io (t) = [20te−4000t + 0.005e−4000t ]u(t) A P 13.20
240 5 × 106 /s Vo = 6 1120 + 0.8s + 5 × 10 /s s
=
12 × 108 s(0.8s2 + 1120s + 5 × 106 )
=
15 × 108 s(s2 + 1400s + 625 × 104 )
=
K2 K2∗ K1 + + s s + 700 − j2400 s + 700 + j2400
K1 = 240;
K2 = 125/163.74◦
vo (t) = [240 + 250e−700t cos(2400t + 163.74◦ )]u(t) V P 13.21
Vo − 240/s Vo s + 14.4 × 10−6 = 0 + 1120 + 0.8s 5 × 106
Problems
Vo
s 1 + 1120 + 0.8s 5 × 106
Vo =
=
240/s − 14.4 × 10−6 0.8s + 1120
−72s2 − 100,800s + 15 × 108 s(s2 + 1400s + 625 × 104 ) =
.·.
240 162.5/163.74◦ 162.5/− 163.74◦ + + s s + 700 − j2400 s + 700 + j2400
vo (t) = [240 + 325e−700t cos(2400t + 163.74◦ )]u(t) V
P 13.22 [a]
Vo =
Ig /C (1/sC)(LIg ) = 2 R + sL + (1/sC) s + (R/L)s + (1/LC)
15 Ig = = 150 C 0.1 R = 7; L Vo = [b] sVo =
1 = 10 LC
150 s2 + 7s + 10 s2
150s + 7s + 10
lim sVo = 0;
s→0
lim sVo = 0;
s→∞
[c] Vo =
.·. vo (∞) = 0 .·. vo (0+ ) = 0
50 −50 150 = + (s + 2)(s + 5) s+2 s+5
vo = [50e−2t − 50e−5t ]u(t) V
13–23
13–24
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.23 IL =
Vo Ig Ig − = − sCVo s 1/sC s
IL =
−10 15 15s 15 25 − = − + s (s + 2)(s + 5) s s+2 s+5
iL (t) = [15 + 10e−2t − 25e−5t ]u(t) A Check: iL (0+ ) = 0 (ok);
iL (∞) = 15 (ok)
P 13.24 [a]
[b] I1 =
25/s 0.01 = 2500 + (125,000/s) s + 50
V1 =
1000 (100,000/s)(25/s) = 2500 + (125,000/s) s(s + 50)
V2 =
(25,000/s)(25/s) 250 = 2500 + (125,000/s) s(s + 50)
[c] i1 (t) = 10e−50t u(t) mA V1 =
20 20 − s s + 50
V2 =
5 5 − s s + 50
.·. .·.
v1 (t) = (20 − 20e−50t )u(t) V v2 (t) = (5 − 5e−50t )u(t) V
[d] i1 (0+ ) = 10 mA i1 (0+ ) =
25 = 10 mA(Checks) 2.5 × 10−3
v1 (0+ ) = 0; v1 (∞) = 20 V;
v2 (0+ ) = 0(Checks) v2 (∞) = 5 V(Checks)
Problems v1 (∞) + v2 (∞) = 25 V(Checks) (10 × 10−6 )v1 (∞) = 200 µC (40 × 10−6 )v2 (∞) = 200 µC(Checks) P 13.25 [a]
1005s =
100s 500s = 5s + 100 s + 20
50 100s 5000s Vo = = 2 s + 20 (s + 25) (s + 20)(s + 25)2 Io =
Vo 50s = 100 (s + 20)(s + 25)2
IL =
1000 Vo = 5s (s + 20)(s + 25)2
[b] Vo =
K2 K1 K3 + + 2 s + 20 (s + 25) s + 25
K1 =
5000s = −4000 (s + 25)2 s=−20
5000s K2 = = 25,000 (s + 20) s=−25
d 5000s K3 = ds s + 20
s=−25
5000s 5000 − = s + 20 (s + 20)2
= 4000 s=−25
vo (t) = [−4000e−20t + 25,000te−25t + 4000e−25t ]u(t) V Io =
K2 K1 K3 + + s + 20 (s + 25)2 s + 25
50s = −40 K1 = (s + 25)2 s=−20 50s K2 = = 250 (s + 20) s=−25
50s d K3 = ds s + 20
s=−25
50s 50 − = s + 20 (s + 20)2
= 40 s=−25
13–25
CHAPTER 13. The Laplace Transform in Circuit Analysis
13–26
io (t) = [−40e−20t + 250te−25t + 40e−25t ]u(t) V IL =
K2 K1 K3 + + 2 s + 20 (s + 25) s + 25
1000 = 40 K1 = (s + 25)2 s=−20 1000 K2 = = −200 (s + 20) s=−25
d 1000 K3 = ds s + 20
s=−25
1000 = − (s + 20)2
= −40 s=−25
iL (t) = [40e−20t − 200te−25t − 40e−25t ]u(t) V P 13.26
10 = (s + 1)I1 − sI2 s
1 0 = −sI1 + s + 1 + I2 s In standard form, s(s + 1)I1 − s2 I2 = 10 −s2 I1 + (s2 + s + 1)I2 = 0
∆=
s(s + 1) −s2
N1 =
N2 =
10 0
−s 2 (s + s + 1) 2
−s 2 (s + s + 1)
s(s + 1) −s2
= 2s(s2 + s + 0.5)
2
= 10(s2 + s + 1)
10 0
= 10s2
Problems I1 =
N1 ; ∆
.·.
I2 =
Io =
=
N2 ; ∆
I0 = I1 − I2
5(s + 1) N 1 − N2 = 2 ∆ s(s + s + 0.5)
K2 K2∗ K1 + + s s + 0.5 − j0.5 s + 0.5 + j0.5
K1 =
5 = 10 0.5
K2 =
5(−0.5 + j0.5 + 1) = 5/− 180◦ (−0.5 + j0.5)(j1)
io (t) = [10 − 10e−t/2 cos 0.5t]u(t) A P 13.27 [a]
5 0 = 2.5s(I1 − 6/s) + (I1 − I2 ) + 10I1 s −75 5 = (I2 − I1 ) + 5(I2 − 6/s) s s or (s2 + 4s + 2)I1 − 2I2 = 6s −I1 + (s + 1)I2 = −9 ∆=
2 (s
+ 4s + 2) −1
−2
(s + 1)
= s(s + 2)(s + 3)
13–27
13–28
CHAPTER 13. The Laplace Transform in Circuit Analysis 6s −9
N1 =
I1 =
−2 (s + 1)
= 6(s2 + s − 3)
6(s2 + s − 3) N1 = ∆ s(s + 2)(s + 3) 2 (s
N2 =
+ 4s + 2) 6s −1
−9
= −9s2 − 30s − 18
N2 −9s2 − 30s − 18 = I2 = ∆ s(s + 2)(s + 3) [b] sI1 =
6(s2 + s − 3) (s + 2)(s + 3) lim sI1 = i1 (∞) = −3 A
lim sI1 = i1 (0+ ) = 6 A;
s→∞
sI2 =
s→0
−9s2 − 30s − 18 (s + 2)(s + 3)
lim sI2 = i2 (0+ ) = −9 A;
lim sI2 = i2 (∞) = −3 A
s→∞
[c] I1 =
s→0
K1 K2 K3 6(s2 + s − 3) = + + s(s + 2)(s + 3) s s+2 s+3
K1 =
6(−3) = −3; 6
K3 =
6(9 − 3 − 3) =6 (−3)(−1)
K2 =
6(4 − 2 − 3) =3 (−2)(1)
i1 (t) = [−3 + 3e−2t + 6e−3t ]u(t) A I2 =
K1 K2 K3 −9s2 − 30s − 18 = + + s(s + 2)(s + 3) s s+2 s+3
K1 =
−18 = −3; 6
K3 =
−81 + 90 − 18 = −3 (−3)(−1)
K2 =
−36 + 60 − 18 = −3 (−2)(1)
i2 (t) = [−3 − 3e−2t − 3e−3t ]u(t) A
Problems P 13.28 [a]
54 = 2I1 − I2 − I3 s
45 0 = −I1 + 2 + I2 − I3 s 0 = −I1 − I2 + (2 + 0.36s)I3
∆=
−1 (2s + 45)/s −1 2 −1 −1
N2 =
2 −1 −1
N3 =
−1
2
−1 (54/s) 0 0
−1 −1 (0.36s + 2)
−1 −1 (0.36s + 2)
−1 (2s + 45)/s −1
(54/s) 0 0
1.08(s + 5)(s + 25) s
=
162 (0.12s + 1) s
=
162 (s + 15) s2
I2 =
150(0.12s + 1) N2 = ∆ (s + 5)(s + 25)
Vo =
45 6750(0.12s + 1) I2 = s s(s + 5)(s + 25)
I3 =
150(s + 15) N3 = = Io ∆ s(s + 5)(s + 25)
[b] Vo =
=
K2 K3 K1 + + s s + 5 s + 25
K1 =
6750 = 54; 125
K2 =
6750(−0.6 + 1) = −27 (−5)(20)
13–29
13–30
CHAPTER 13. The Laplace Transform in Circuit Analysis K3 =
6750(−3 + 1) = −27 (−25)(−20)
.·. vo (t) = [54 − 27e−5t − 27e−25t ]u(t) V Io =
K2 K3 K1 + + s s + 5 s + 25
K1 =
150(15) = 18; (5)(25)
K3 =
150(−10) = −3 (−25)(−20)
K2 =
150(10) = −15 (−5)(20)
.·. io (t) = [18 − 15e−5t − 3e−25t ]u(t) A [c] At t = 0+ the circuit is
Both vo and io are zero, which agrees with our solutions in part (a). At t = ∞ the circuit is
Our solutions predict vo (∞) = 54 V and io (∞) = 18 A. Also observe from the circuit at t = 0+ that the voltage across the inductor is 54 V. Our solution predicts dio (0+ ) = 0.36(75 + 75) = 54 V dt At t = 0+ the current in the capacitive branch is (1/2)(54/1.5) = 18 A. From our solution we have 150(0.12 + 1/s) sI2 = and s→∞ lim sI2 = i2 (0+ ) = 150(0.12) = 18 A (1 + 5/s)(1 + 25/s)
vL (0+ ) = 0.36
Problems P 13.29 [a]
120 250 = 50(I1 − 0.05Vφ ) + (I1 − I2 ) s s
120 250 250 250 I1 − I2 ; (I2 − I1 ) + = 50I1 − 2.5 s s s s 0=
250 (I2 − I1 ) + 20s(I2 − 0.05Vφ ) + 700I2 s
250 250 (I2 − I1 ) + 20s I2 − 0.05 0= (I2 − I1 ) Vφ ) + 700I2 s s Simplifying, (50s + 875)I1 − 875I2 = 120 250(s − 1)I1 + (20s2 + 450s + 250)I2 = 0 ∆=
(50s + 875) 250(s − 1)
N1 =
N2 =
120 0
(20s2 + 450s + 250)
−875
−875 2 (20s + 450s + 250)
(50s + 875) 250(s − 1)
= 1200(2s2 + 45s + 25)
120
0
= −30,000(s − 1)
I1 =
1.2(2s2 + 45s + 25) N1 = ∆ s(s2 + 40s + 625)
I2 =
−30(s − 1) N2 = 2 ∆ s(s + 40s + 625)
Io = I2 − 0.05Vφ = I2 − 0.05 I2 − I1 =
= 1000s(s2 + 40s + 625)
−2.4s(s + 35) s(s2 + 40s + 625)
250 (I2 − I1 ) s
13–31
13–32
CHAPTER 13. The Laplace Transform in Circuit Analysis 250 −600(s + 35) (I2 − I1 ) = s s(s2 + 40s + 625) .·.
[b] sIo =
Io =
(s2
−30(s − 1) 30(s + 35) 1080 + = 2 2 + 40s + 625) s(s + 40s + 625) s(s + 40s + 625)
s(s2
1080 + 40s + 625)
io (0+ ) = lim sIo = 0 s→∞ io (∞) = lim sIo = s→0
1080 = 1728 mA 625
[c] At t = 0+ the circuit is
i0 (0+ ) = 0 (Checks) At t = ∞ the circuit is
120 = 50(ia − i1 ) + 700ia = 50(ia − 0.05vφ ) + 700ia = 750ia − 2.5vφ vφ = −700ia
.·.
120 = 750ia + 1750ia = 2500ia
120 = 48 mA 2500 vφ = −700ia = −33.60 V ia =
io (∞) = 48 × 10−3 − 0.05(−33.60) = 48 × 10−3 + 1.68 = 1728 mA (Checks)
Problems [d] Io =
13–33
K1 K2 K2∗ 1080 = + + s(s2 + 40s + 625) s s + 20 − j15 s + 20 + j15
K1 =
1080 = 1.728 625
K2 =
1080 = 1.44/126.87◦ (−20 + j15)(j30)
io (t) = [1728 + 2880e−20t cos(15t + 126.87◦ )]u(t) mA Check:
io (0+ ) = 0 mA;
io (∞) = 1728 mA
P 13.30 [a]
V1 V1 − 50/s V1 − Vo + + =0 10 25/s 4s −5 Vo − V1 Vo − 50/s + + =0 s 4s 30 Simplifying, (4s2 + 10s + 25)V1 − 25Vo = 200s −15V1 + (2s + 15)Vo = 400 ∆=
(4s2
No =
+ 10s + 25)
(4s2
−15
−25 (2s + 15)
+ 10s + 25) −15
200s 400
= 8s(s + 5)2
= 200(8s2 + 35s + 50)
Vo =
200(8s2 + 35s + 50) K2 No K1 K3 = + = + 2 2 ∆ 8s(s + 5) s (s + 5) s+5
K1 =
(25)(50) = 50; 25
K2 =
d 8s2 + 35s + 50 K3 = 25 ds s
25(200 − 175 + 50) = −375 −5
s=−5
s(16s + 35) − (8s2 + 35s + 50) = 25 s2
s=−5
13–34
CHAPTER 13. The Laplace Transform in Circuit Analysis = −5(−45) − 75 = 150 50 150 375 + − 2 s (s + 5) s+5
.·. Vo =
[b] vo (t) = [50 − 375te−5t + 150e−5t ]u(t) V [c] At t = 0+ :
vo (0+ ) = 50 + 150 = 200 V(Checks) At t = ∞:
vo (∞) − 50 vo (∞) −5+ =0 10 30 .·. 3vo (∞) − 150 + vo (∞) − 50 = 0; .·.
vo (∞) = 50 V(Checks)
.·. 4vo (∞) = 200
Problems P 13.31 [a]
10 10 I1 + (I1 − I2 ) + 10(I1 − 9/s) = 0 s s 10 10 (I2 − 9/s) + (I2 − I1 ) + 10I2 = 0 s s Simplifying, (s + 2)I1 − I2 = 9 −I1 + (s + 2)I2 = ∆=
(s + 2) −1
N1 =
9 9/s
9 s
−1 (s + 2)
−1 (s + 2)
=
= s2 + 4s + 3 = (s + 1)(s + 3) 9 9s2 + 18s + 9 = (s + 1)2 s s
9 (s + 1)2 N1 9(s + 1) = I1 = = ∆ s (s + 1)(s + 3) s(s + 3) N2 =
(s + 2) −1
9
9/s
=
18 (s + 1) s
I2 =
18(s + 1) 18 N2 = = ∆ s(s + 1)(s + 3) s(s + 3)
Ia =
6 6 9 9 9(s + 1) − I1 = − = − s s s(s + 3) s s+3
Ib = I1 =
3 6 9(s + 1) = + s(s + 3) s s+3
13–35
13–36
CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] ia (t) = 6(1 − e−3t )u(t) A ib (t) = 3(1 + 2e−3t )u(t) A
[c] Va = =
10 10 3 6 Ib = + s s s s+3
30 60 30 20 20 + = 2 + − 2 s s(s + 3) s s s+3
Vb =
10 10 (I2 − I1 ) = s s
=
6 6 6 3 − + − s s+3 s s+3
12 40 10 3 30 40 − + = 2 − s s s+3 s s s+3
6 10 10 9 6 − + Vc = (9/s − I2 ) = s s s s s+3 =
20 30 20 − + 2 s s s+3
[d] va (t) = [30t + 20 − 20e−3t ]u(t) V vb (t) = [30t − 40 + 40e−3t ]u(t) V vc (t) = [30t + 20 − 20e−3t ]u(t) V [e] Calculating the time when the capacitor voltage drop first reaches 1000 V: 30t + 20 − 20e−3t = 1000 or
30t − 40 + 40e−3t = 1000
Note that in either of these expressions the exponential term over time becomes is negligible when compared to the other terms. Thus, 30t + 20 = 1000 or
30t − 40 = 1000
Thus, 980 1040 t= = 32.67 s or t = = 34.67 s 30 30 Therefore, the breakdown will occur at t = 32.67 s.
Problems P 13.32 [a]
20Iφ + 25s(Io − Iφ ) + 25(Io − I1 ) = 0 50 Iφ + 5I1 + 25(I1 − Io ) + 25s(Iφ − Io ) = 0 s Iφ − I1 =
100 s
.·.
I1 = Iφ −
100 s
Simplifying, (−25s − 5)Iφ + (25s + 25)Io = −2500/s (50/s + 25s + 30)Iφ + (−25s − 25)Io = 3000/s
−5(5s + 1)
25(s + 1)
= −625(s + 1)(1 + 2/s) ∆= 5 (5s2 + 6s + 10) −25(s + 1) s
−5(5s + 1) N2 = 5 (5s2 + 6s + 10) s Io =
−2500/s 3000/s
= −12,500
20(s2 − 4.8s − 10) N2 = ∆ s(s + 1)(s + 2)
lim sIo = 20 A [b] io (0+ ) = s→∞ io (∞) = lim sIo = s→0
−200 = −100 A 2
s2 − 4.8s − 10 s2
13–37
13–38
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] At t = 0+ the circuit is
20Iφ + 5I1 = 0;
Iφ − I1 = 100
.·. 20Iφ + 5(Iφ − 100) = 0;
25Iφ = 500
.·. Iφ = Io (0+ ) = 20 A(Checks) At t = ∞ the circuit is
Io (∞) = −100 A(Checks) [d] Io =
K1 K2 K3 20(s2 − 4.8s − 10) = + + s(s + 1)(s + 2) s s+1 s+2
K1 =
−200 = −100; (1)(2)
K3 =
20(4 + 9.6 − 10) = 36 (−2)(−1)
Io =
K2 =
84 36 −100 + + s s+1 s+2
20(1 + 4.8 − 10) = 84 (−1)(1)
Problems io (t) = (−100 + 84e−t + 36e−2t )u(t) A io (∞) = −100 A(Checks) io (0+ ) = −100 + 84 + 36 = 20 A(Checks) P 13.33 vC = 12 × 105 te−5000t V,
dvC iC = C dt
iC > 0 when and
therefore
C = 5 µF;
= 6e−5000t (1 − 5000t) A 1 > 5000t or
iC < 0 when
0 < t < 200 µs
iC < 0 when t > 200 µs
iC = 0 when
1 − 5000t = 0,
or
t = 200 µs
dvC = 12 × 105 e−5000t [1 − 5000t] dt .·. iC = 0 when
dvC =0 dt
P 13.34 [a] The s-domain equivalent circuit is
I=
Vg /L Vg = , R + sL s + (R/L)
I=
K1 K1∗ K0 + + s + R/L s − jω s + jω
K0 =
Vm (ωL cos φ − R sin φ) , R 2 + ω 2 L2
Vg =
Vm (ω cos φ + s sin φ) s2 + ω 2
K1 =
Vm /φ − 90◦ − θ(ω) √ 2 R 2 + ω 2 L2
where tan θ(ω) = ωL/R. Therefore, we have i(t) =
Vm (ωL cos φ − R sin φ) −(R/L)t Vm sin[ωt + φ − θ(ω)] √ e + R 2 + ω 2 L2 R 2 + ω 2 L2
[b] iss (t) = √
Vm sin[ωt + φ − θ(ω)] R 2 + ω 2 L2
13–39
13–40
CHAPTER 13. The Laplace Transform in Circuit Analysis
Vm (ωL cos φ − R sin φ) −(R/L)t e R 2 + ω 2 L2 Vg [d] I = , Vg = Vm /φ − 90◦ R + jωL [c] itr =
Therefore I = √
Vm /φ − 90◦ Vm =√ 2 /φ − 90◦ − θ(ω) 2 2 2 R + ω L /θ(ω) R + ω 2 L2
Therefore iss = √
R2
Vm sin[ωt + φ − θ(ω)] + ω 2 L2
[e] The transient component vanishes when ωL cos φ = R sin φ or
tan φ =
ωL R
or
φ = θ(ω)
P 13.35
VTh =
40 400 40 10s · = = 10s + 1000 s 10s + 1000 s + 100
ZTh = 1000 + 100010s = 1000 +
I=
(5 × =
105 )/s
10,000s 2000(s + 50) = 10s + 1000 s + 100
40/(s + 100) 40s = 2 + 2000(s + 50)/(s + 100) 2000s + 600,000s + 5 × 107
K1 K1∗ 0.02s = + s2 + 300s + 25,000 s + 150 − j50 s + 150 + j50
0.02s K1 = = 31.62 × 10−3 /71.57◦ s + 150 + j50 s=−150+j50
i(t) = 63.25e−150t cos(50t + 71.57◦ )u(t) mA
Problems P 13.36 [a]
180 = (100 + 15s)I1 + 10sI2 s 0 = 10sI1 + (20s + 200)I2 ∆=
15s + 100 10s
N2 =
I2 =
15s + 100 10s
180/s 0
= 200(s + 5)(s + 20)
= −1800
−9 N2 = ∆ (s + 5)(s + 20)
Vo = 160I2 = [b] sVo =
20s + 200
10s
−1440 (s + 5)(s + 20)
−1440s (s + 5)(s + 20)
lim sVo = vo (∞) = 0 V
s→0
lim sVo = vo (0+ ) = 0 V
s→∞
[c] Vo =
96 −96 + s + 5 s + 20
vo (t) = [−96e−5t + 96e−20t ]u(t) V P 13.37
180 = (100 + 15s)I1 − 10sI2 s 0 = −10sI1 + (20s + 200)I2
13–41
13–42
CHAPTER 13. The Laplace Transform in Circuit Analysis
∆=
15s + 100 −10s
N2 =
I2 =
−10s 20s + 200
15s + 100 −10s
180/s 0
= 200(s + 5)(s + 20)
= 1800
9 N2 = ∆ (s + 5)(s + 20)
Vo = 160I2 =
96 96 1440 = − (s + 5)(s + 20) s + 5 s + 20
vo (t) = [96e−5t − 96e−20t ]u(t) V 1 1 P 13.38 [a] W = L1 i21 + L2 i22 + M i1 i2 2 2 W = 4(15)2 + 9(100) + 150(6) = 2700 J [b] 120i1 + 8
di2 di1 −6 =0 dt dt
270i2 + 18
di2 di1 −6 =0 dt dt
Laplace transform the equations to get 120I1 + 8(sI1 − 15) − 6(sI2 + 10) = 0 270I2 + 18(sI2 + 10) − 6(sI1 − 15) = 0 In standard form, (8s + 120)I1 − 6sI2 = 180 −6sI1 + (18s + 270)I2 = −270 ∆=
N1 =
N2 =
8s + 120 −6s = 108(s + 10)(s + 30) −6s 18s + 270 180 −6s = 1620(s + 30) −270 18s + 270 8s + 120 180 = −1080(s + 30) −6s −270
Problems I1 =
1620(s + 30) 15 N1 = = ∆ 108(s + 10)(s + 30) s + 10
I2 =
−1080(s + 30) −10 N2 = = ∆ 108(s + 10)(s + 30) s + 10
[c] i1 (t) = 15e−10t u(t) A; [d] W120Ω =
∞
W270Ω =
0
−20t
(225e
∞ 0
i2 (t) = −10e−10t u(t) A
e−20t ∞ )(120) dt = 27,000 = 1350 J −20 0
(100e−20t )(270) dt = 27,000
W120Ω + W270Ω = 2700 J
e−20t ∞ = 1350 J −20 0
(Checks)
1 1 [e] W = L1 i21 + L2 i22 + M i1 i2 = 900 + 900 − 900 = 900 J 2 2 With the dot reversed the s-domain equations are (8s + 120)I1 + 6sI2 = 60 6sI1 + (18s + 270)I2 = −90 As before, N1 =
N2 =
I1 =
∆ = 108(s + 10)(s + 30). Now,
60 6s = 1620(s + 10) −90 18s + 270 8s + 120 60 = −1080(s + 10) 6s −90
N1 15 = ; ∆ s + 30
i1 (t) = 15e−30t u(t) A; W270Ω = W120Ω =
∞ 0
∞ 0
I2 =
N2 −10 = ∆ s + 30
i2 (t) = −10e−30t u(t) A
(100e−60t )(270) dt = 450 J (225e−60t )(120) dt = 450 J
W120Ω + W270Ω = 900 J
(Checks)
13–43
13–44
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.39 [a] s-domain equivalent circuit is
i2 (0+ ) = −
Note: [b]
20 = −2 A 10
24 = (120 + 3s)I1 + 3sI2 + 6 s 0 = −6 + 3sI1 + (360 + 15s)I2 + 36 In standard form, (s + 40)I1 + sI2 = (8/s) − 2 sI1 + (5s + 120)I2 = −10 ∆=
s + 40 s
(8/s) − 2 −10
N1 =
I1 = [c] sI1 =
s 5s + 120
= 4(s + 20)(s + 60)
5s + 120
s
=
−200(s − 4.8) s
N1 −50(s − 4.8) = ∆ s(s + 20)(s + 60) −50(s − 4.8) (s + 20)(s + 60)
lim sI1 = i1 (0+ ) = 0 A
s→∞
lim sI1 = i1 (∞) =
s→0
[d] I1 =
(−50)(−4.8) = 0.2 A (20)(60)
K2 K3 K1 + + s s + 20 s + 60
K1 =
240 = 0.2; 1200
K2 =
−50(−20) + 240 = −1.55 (−20)(40)
Problems K3 =
−50(−60) + 240 = 1.35 (−60)(−40)
i1 (t) = [0.2 − 1.55e−20t + 1.35e−60t ]u(t) A P 13.40 For t < 0:
For t > 0+ :
18 × 4 = 72;
18 × 3 = 54
20I1 − 72 + 4sI1 + s(I2 − I1 ) + 10(I1 − I2 ) = 0 −54 + 3sI2 + 10(I2 − I1 ) + s(I1 − I2 ) = 0 In standard form, (3s + 30)I1 + (s − 10)I2 = 72
13–45
13–46
CHAPTER 13. The Laplace Transform in Circuit Analysis
(s − 10)I1 + (2s + 10)I2 = 54 .·. ∆ =
N1 =
N2 =
(3s + 30) (s − 10)
72 54
= 5(s + 2)(s + 20)
(s − 10)
(2s + 10)
(3s + 30) (s − 10)
Io = I1 − I2 =
=
(s − 10) (2s + 10)
72 54
= 90s + 1260
= 90s + 2340
−1080 N1 N2 − = ∆ ∆ 5(s + 2)(s + 20)
12 12 −216 − − (s + 2)(s + 20) s + 2 s + 20
io (t) = [12e−2t + 12e−20t ]u(t) A P 13.41 The s-domain equivalent circuit is
V1 V1 − 48/s V1 + 9.6 + =0 + 4 + (100/s) 0.8s 0.8s + 20 V1 =
−1200 s2 + 10s + 125
Vo =
20 −30,000 V1 = 0.8s + 20 (s + 25)(s + 5 − j10)(s + 5 + j10) K2 K2∗ K1 + + = s + 25 s + 5 − j10 s + 5 + j10
Problems −30,000 = −60 K1 = 2 s + 10s + 125 s=−25 −30,000 = 67.08/63.43◦ K2 = (s + 25)(s + 5 + j10) s=−5+j10
vo (t) = [−60e−25t + 134.16e−5t cos(10t + 63.43◦ )]u(t) V P 13.42 [a] Voltage source acting alone:
Vo1 Vo1 − 60/s Vo1 s + + =0 10 80 20 + 10s .·. Vo1 =
480(s + 2) s(s + 4)(s + 6)
Current source acting alone:
Vo2 Vo2 s Vo2 − 30/s + + =0 10 80 10(s + 2) .·. Vo2 =
240 s(s + 4)(s + 6)
Vo = Vo1 + Vo2 =
480(s + 2.5) 480(s + 2) + 240 = s(s + 4)(s + 6) s(s + 4)(s + 6)
13–47
CHAPTER 13. The Laplace Transform in Circuit Analysis
13–48
[b] Vo =
K2 K3 K1 + + s s+4 s+6 (480)(2.5) = 50; (4)(6)
K1 =
K2 =
480(−1.5) = 90; (−4)(2)
vo (t) = [50 + 90e−4t − 140e−6t ]u(t) V P 13.43 ∆ =
Y11 Y12
N2 =
V2 =
Y12 Y22
Y11 Y12
= Y11 Y22 − Y122
[(Vg /R1 ) + γC − (ρ/s)] (Ig − γC)
N2 ∆
Substitution and simplification lead directly to Eq. 13.90. P 13.44
Va Va − Vo Va − 4.8/s + + =0 0.8 1/s 1/s 0 − Va 0 − Vo + =0 1/s 1.25 Va =
−Vo 1.25s
Va (2s + 1.25) − sVo = 6/s
K3 =
480(−3.5) = −140 (−6)(−2)
Problems
−Vo
−Vo
Vo =
(2s + 1.25) + s = 6/s 1.25s
125s2 + 2s + 1.25 = 6/s 1.25s
1.25s2 =
K1 =
−7.5 −6 = 2 + 2s + 1.25 s + 1.6s + 1
K1∗ K1 + s + 0.8 − j0.6 s + 0.8 + j0.6
−6 = 5/90◦ s + 0.8 + j0.6 s=−0.8+j0.6
vo (t) = 10e−0.8t cos(0.6t + 90◦ )u(t) V = −10e−0.8t sin(0.6t)u(t) V P 13.45 [a] Vo = −
Zf Vg Zi
Zf =
1010 /s 1010 107 107 1000 = 7 = = s 10 /s + 1000 1000s + 107 s + 104
Zi =
2 × 106 400s + 2 × 106 400 + 400 = = (s + 5000) s s s
Vg =
20,000 s2
.·. Vo = [b] Vo =
−107 /(s + 104 ) 20,000 −5 × 108 = · (400/s)(s + 5000) s2 s(s + 5000)(s + 10,000)
K1 K2 K3 + + s s + 5000 s + 10,000
−5 × 108 = −10 K1 = (s + 5000)(s + 10,000) s=0 −5 × 108 K2 = = 20 s(s + 10,000) s=−5000 −5 × 108 K3 = = −10 s(s + 5000) s=−10,000
.·. vo (t) = [−10 + 20e−5000t − 10e−10,000t ]u(t) V
13–49
13–50
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] −10 + 20e−5000ts − 10e−10,000ts = −5 Let x = e−5000ts . Then 10x2 − 20x + 5 = 0 Solving, x = 0.292893 e−5000ts = 0.292893 [d] vg = m tu(t);
.·.
Vg =
ts = 245.6 µs
m s2
m −107 s · 2 Vo = 400(s + 5000)(s + 10,000) s = K1 =
−25,000m s(s + 5000)(s + 10,000) −25,000m = −5 × 10−4 m (5000)(10,000) .·. m = 10,000 V/s
.·. −5 = −5 × 10−4 m P 13.46 [a]
Vp =
50/s 50 Vg2 = Vg2 5 + 50/s 5s + 50
Vp − 40/s Vp − Vo Vp − Vo + + =0 20 5 100/s
Vp
1 s s 1 1 2 + + + − Vo = 20 5 100 5 100 s
s 50 16 2 1 s + 20 s + 25 − = Vo + = Vo 100 5s + 50 s s 5 100 100
Problems
2 16(s + 25) 100 −40s + 2000 − Vo = = s + 20 10(s + 10)(s) s s(s + 10)(s + 20) =
K2 K3 K1 + + s s + 10 s + 20 K2 = −24;
K1 = 10; .·.
K3 = 14
vo (t) = [10 − 24e−10t + 14e−20t ]u(t) V
[b] 10 − 24e−10t + 14e−20t = 5 Let x = e−10ts . Then 10 − 24x + 14x2 = 5 14x2 − 24x + 5 = 0 x = 0.242691 e−10ts = 0.242691
.·.
ts = 141.60 ms
P 13.47 Let vo1 equal the output voltage of the first op amp. Then Vo1 =
−Zf 1 Vg ZA1
ZA1 = 25,000 +
= .·.
where Zf 1 = 25 × 103 Ω 25,000(20 × 104 /s) 25,000 + (20 × 104 /s)
25,000(s + 16) Ω (s + 8)
Vo1 =
−(s + 8) Vg (s + 16)
Also, Vo =
−Zf 2 Vo1 ZA2
.·.
−8000 −8000 −(s + 8) Vo1 = Vg Vo = s s (s + 16)
where Zf 2 =
2 × 108 Ω and ZA2 = 25,000 Ω s
=
8000(s + 8) Vg s(s + 16)
vg (t) = 16u(t) mV;
.·.
Vg =
16 × 10−3 s
13–51
13–52
CHAPTER 13. The Laplace Transform in Circuit Analysis
Vo =
K1 K2 K3 128(s + 8) = 2 + + 2 s (s + 16) s s s + 16
K1 =
128(8) = 64 16
d s+8 K2 = 128 ds s + 16 K3 =
s=0
=4
128(−8) = −4 256
vo (t) = [64t + 4 − 4e−16t ]u(t) V The op amp will saturate when vo = ±6 V. Hence, saturation will occur when 64t + 4 − 4e−16t = 6 or
16t − 0.5 = e−16t
This equation can be solved by trial and error. First note that t > 0.5/16 or t > 31.25 ms. Try 40 ms: 0.64 − 0.5 = 0.14;
e−0.64 = 0.53
Try 50 ms: 0.80 − 0.5 = 0.30;
e−0.80 = 0.45
Try 60 ms: 0.96 − 0.5 = 0.46;
e−0.96 = 0.38
Further trial and error gives tsat ∼ = 56.5 ms P 13.48 [a] Let va be the voltage across the 0.5 µF capacitor, positive at the upper terminal. Let vb be the voltage across the 100 kΩ resistor, positive at the upper terminal. Also note 2 × 106 106 = 0.5s s
and
106 4 × 106 = ; 0.25s s
Va sVa Va − (0.5/s) + =0 + 6 2 × 10 200,000 200,000
Vg =
0.5 s
Problems sVa + 10Va − Va =
5 + 10Va = 0 s
5 s(s + 20)
(0 − Vb )s 0 − Va + =0 200,000 4 × 106 .·.
Vb = −
20 −100 Va = 2 s s (s + 20)
(Vb − 0)s (Vb − Vo )s Vb + + =0 100,000 4 × 106 4 × 106 40Vb + sVb + sVb = sVo 2(s + 20)Vb ; Vo = s
.·.
−100 −200 Vo = 2 = 3 s s3
[b] vo (t) = −100t2 u(t) V [c] −100t2 = −4; P 13.49 [a]
1/sC Vo = Vi R + 1/sC H(s) =
[b]
t = 0.2 s = 200 ms
200 (1/RC) = ; s + (1/RC) s + 200
−p1 = −200 rad/s
RCs s R Vo = = = Vi R + 1/sC RCs + 1 s + (1/RC)
s ; z1 = 0, −p1 = −200 rad/s s + 200 s s sL Vo = = = [c] Vi R + sL s + R/L s + 8000 =
z1 = 0; [d]
−p1 = −8000 rad/s
Vo R/L 8000 R = = = Vi R + sL s + (R/L) s + 8000 −p1 = −8000 rad/s
[e]
Vo − Vi Vo s Vo + =0 + 6 4 × 10 10,000 40,000
13–53
13–54
CHAPTER 13. The Laplace Transform in Circuit Analysis sVo + 400Vo + 100Vo = 100Vi H(s) =
Vo 100 = Vi s + 500
−p1 = −500 rad/s P 13.50 [a] Let R1 = 250 kΩ;
R2 = 125 kΩ;
C2 = 1.6 nF;
and
(R2 + 1/sC2 )1/sCf (s + 1/R2 C2 ) = Zf = C +C R2 + sC1 2 + sC1 f Cf s s + C22Cf Rf2 1 = 2.5 × 109 Cf 1 62.5 × 107 = = 5000 rad/s R2 C2 125 × 103 C2 + Cf 2 × 10−9 = 25,000 rad/s = C2 Cf R2 (0.64 × 10−18 )(125 × 103 ) .·. Zf =
2.5 × 109 (s + 5000) Ω s(s + 25,000)
Zi = R1 = 250 × 103 Ω H(s) =
−Zf −104 (s + 5000) Vo = = Vg Zi s(s + 25,000)
[b] −z1 = −5000 rad/s −p1 = 0;
−p2 = −25,000 rad/s
P 13.51 [a]
sVa Va − Vg (Va − Vo )s + + =0 6 1000 5 × 10 5 × 106 5000Va − 5000Vg + 2sVa − sVo = 0
Cf = 0.4 nF. Then
Problems (5000 + 2s)Va − sVo = 5000Vg (0 − Va )s 0 − Vo + =0 5 × 106 5000 .·.
−sVa − 1000Vo = 0;
Va =
−1000 Vo s
−1000 (2s + 5000) Vo − sVo = 5000Vg s 1000Vo (2s + 5000) + s2 Vo = −5000sVg Vo (s2 + 2000s + 5 × 106 ) = −5000sVg Vo −5000s = 2 Vg s + 2000s + 5 × 106 √ s1,2 = −1000 ± 106 − 5 × 106 = −1000 ± j2000 −5000s Vo = Vg (s + 1000 − j2000)(s + 1000 + j2000) −p1 = −1000 + j2000;
[b] z1 = 0;
P 13.52 [a] Zi = 1000 + Zf =
−p2 = −1000 − j2000
1000(s + 5000) 5 × 106 = s s
40 × 106 40 × 106 40,000 = s s + 1000
H(s) = −
−40,000s Zf −40 × 106 /(s + 1000) = = Zi 1000(s + 5000)/s (s + 1000)(s + 5000)
[b] Zero at s = 0;
Poles at −p1 = −1000 rad/s and −p2 = −5000 rad/s
P 13.53 [a]
Va =
s Vi (500,000) = Vi 6 500,000 + [(20 × 10 )/s] s + 40
0.2Vi = Vo + Va
13–55
13–56
CHAPTER 13. The Laplace Transform in Circuit Analysis .·.
Vo = 0.2Vi −
s Vi s + 40
Vo 0.2(s + 40) − s −0.8s + 8 −0.8(s − 10) = = = Vi s + 40 s + 40 s + 40 [b] −z1 = 10 rad/s −p1 = −40 rad/s P 13.54
Vg = 25sI1 − 35sI2
16 × 106 0 = −35sI1 + 50s + 10,000 + I2 s
∆=
N2 =
I2 =
25s −35s 25s −35s
50s + 10,000 + 16 × 106 /s
−35s
Vg 0
= 25(s + 2000)(s + 8000)
= 35sVg
35sVg N2 = ∆ 25(s + 2000)(s + 8000)
16 × 106 22.4 × 106 Vg I2 = Vo = s (s + 2000)(s + 8000) H(s) = .·.
22.4 × 106 Vo = Vg (s + 2000)(s + 8000)
−p1 = −2000 rad/s;
−p2 = −8000 rad/s
Problems
13–57
P 13.55 [a]
Vo Vo + + Vo (10−7 )s = Ig 5000 0.2s .·. Vo = Ig =
s2
10 × 106 s · Ig s2 + 2000s + 50 × 106
0.1s ; + 108
.·. H(s) =
Io =
Vo s 10 × 106
s2 s2 + 2000s + 50 × 106
(s2 )(0.1s) (s + 1000 − j7000)(s + 1000 + j7000)(s2 + 108 )
[b] Io =
Io =
0.1s3 (s + 1000 − j7000)(s + 1000 + j7000)(s + j104 )(s − j104 )
[c] Damped sinusoid of the form M e−1000t cos(7000t + θ1 ) [d] Steady-state sinusoid of the form N cos(104 t + θ2 ) [e] Io =
K1∗ K2 K1 K2∗ + + + s + 1000 − j7000 s + 1000 + j7000 s − j104 s + j104
K1 =
0.1(−1000 + j7000)3 = 46.90 × 10−3 /− 140.54◦ (j14,000)(−1000 − j3000)(−1000 + j17,000)
K2 =
0.1(j104 )3 = 92.85 × 10−3 /21.80◦ (j20,000)(1000 + j3000)(1000 + j17,000)
io (t) = [93.8e−1000t cos(7000t − 140.54◦ ) + 185.7 cos(104 t + 21.80◦ )] mA Test: io (0) = 93.8 cos(−140.54◦ ) + 185.7 cos(21.80◦ ) mA = 100 mA Z=
1 ; Y
Y =
1 1 1 2 + j5 + + = 5000 j2000 −j1000 10,000
10,000 = 1856.95/− 68.2◦ Ω .·. Z = 2 + j5
13–58
CHAPTER 13. The Laplace Transform in Circuit Analysis Vo = Ig Z = (0.1/0◦ )(1856.95/− 68.2◦ ) = 185.695/− 68.2◦ V Io =
Vo = 185.7/21.80◦ mA −j1000
ioss = 185.7 cos(104 t + 21.80◦ ) mA(Checks) P 13.56 [a]
2000(Io − Ig ) + 8000Io + µ(Ig − Io )(2000) + 2sIo = 0 .·. Io =
1000(1 − µ) Ig s + 1000(5 − µ)
.·. H(s) =
1000(1 − µ) s + 1000(5 − µ)
[b] µ < 5 [c] µ
H(s)
−3 4000/(s + 8000)
Io 20,000/s(s + 8000)
0 1000/(s + 5000)
5000/s(s + 5000)
4 −3000/(s + 1000)
−15,000/s(s + 1000)
5 −4000/s
−20,000/s2
6 −5000/(s − 1000) −25,000/s(s − 1000) µ = −3: 2.5 2.5 − ; io = [2.5 − 2.5e−8000t ]u(t) A Io = s (s + 8000) µ = 0: 1 1 ; io = [1 − e−5000t ]u(t) A Io = − s s + 5000 µ = 4: 15 −15 + ; io = [−15 + 15e−1000t ]u(t) A Io = s s + 1000 µ = 5: −20,000 Io = ; io = −20,000t u(t) A s2
Problems µ = 6: Io =
25 25 − ; s s − 1000
1 Vo = ; Vi s+1 0 ≤ t ≤ 1:
h(t) = e−t
P 13.57 H(s) = For
t
vo = For
0
e−λ dλ = (1 − e−t ) V
1 ≤ t ≤ ∞: t
vo =
io = 25[1 − e1000t ]u(t) A
t−1
P 13.58 H(s) =
e−λ dλ = (e − 1)e−t V
1 Vo s =1− ; = Vi s+1 s+1
h(t) = δ(t) − e−t
h(λ) = δ(λ) − e−λ For
0 ≤ t ≤ 1: t
vo =
0
[δ(λ) − e−λ ] dλ = 1 + [e−λ ] |t0 = e−t V
For 1 ≤ t ≤ ∞: t
vo =
t−1
−λ
(−e
−λ
) dλ = e
t
t−1
= (1 − e)e−t V
13–59
13–60
CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.59 [a]
t0
t>0
13–106
CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] vo1 = 169.71 cos(120πt − 85.35◦ )V vo2 = 172.61 cos(120πt − 78.50◦ )V
t0
Problems
13–107
13–108
CHAPTER 13. The Laplace Transform in Circuit Analysis
14 Introduction to Frequency-Selective Circuits
Assessment Problems AP 14.1 fc = 8 kHz, ωc =
1 ; RC
.·. C =
ωc = 2πfc = 16π krad/s R = 10 kΩ;
1 1 = = 1.99 nF ωc R (16π × 103 )(104 )
AP 14.2 [a] ωc = 2πfc = 2π(2000) = 4π krad/s L=
5000 R = = 0.40 H ωc 4000π
[b] H(jω) =
4000π ωc = ωc + jω 4000π + jω
When ω = 2πf = 2π(50,000) = 100,000π rad/s H(j100,000π) =
1 4000π = = 0.04/− 87.71◦ 4000π + j100,000π 1 + j25
.·. |H(j100,000π)| = 0.04 [c] .·. θ(100,000π) = −87.71◦ AP 14.3 ωc =
5000 R = = 1.43 Mrad/s L 3.5 × 10−3
14–1
14–2
CHAPTER 14. Introduction to Frequency-Selective Circuits 106 106 1 = = = 10 krad/s RC R 100 106 = 200 rad/s [b] ωc = 5000 106 [c] ωc = = 33.33 rad/s 3 × 104
AP 14.4 [a] ωc =
AP 14.5 Let Z represent the parallel combination of (1/sC) and RL . Then Z=
RL (RL Cs + 1)
Thus
H(s) =
RL Z = R+Z R(RL Cs + 1) + RL (1/RC)
=
where K = AP 14.6 ωo2 =
1 LC
s+
1 RC
=
(1/RC) s+
1 K
1 RC
RL R + RL L=
so
ωo ωo = β R/L
Q=
R+RL RL
so
1 ωo2 C
=
R=
1 (24π ×
103 )2 (0.1
× 10−6 )
= 1.76 mH
(24π × 103 )(1.76 × 10−3 ) ωo L = = 22.10 Ω Q 6
AP 14.7 ωo = 2π(2000) = 4000π rad/s; β = 2π(500) = 1000π rad/s; β=
1 RC
so
C=
ωo2 =
1 LC
so
L=
AP 14.8 ωo2 =
1 LC
so
L=
β=
1 RC
so
R=
R = 250 Ω
1 1 = = 1.27 µF βR (1000π)(250) 1 106 = = 4.97 mH ωo2 C (4000π)2 (1.27) 1 ωo2 C
=
1 (104 π)2 (0.2
× 10−6 )
= 5.07 mH
1 1 = = 3.98 kΩ βC 400π(0.2 × 10−6 )
Problems AP 14.9 ωo2 =
1 LC
so
L=
1 ωo2 C
=
1 (4000π)2 (0.2
× 10−6 )
= 31.66 mH
5 × 103 fo = = 25 = ωo RC Q= β 200 .·. R =
25 Q = = 9.95 kΩ ωo C (4000π)(0.2 × 10−6 )
AP 14.10 ωo = 8000π rad/s C = 500 nF 1
ωo2 =
1 LC
Q=
ωo ωo L 1 = = β R ωo CR
.·. R =
so
L=
ωo2 C
= 3.17 mH
1 1 = = 15.92 Ω ωo CQ (8000π)(500 × 10−9 )(5)
AP 14.11 ωo = 2πfo = 2π(20,000) = 40π krad/s; Q=
ωo L ωo = β RC
ωo2 =
1 LC
so
so
C=
L=
R = 100 Ω;
Q=5
RQ 100 = = 3.98 mH ωo 40π × 103
1 1 = = 15.92 nF 2 3 2 ωo L (40π × 10 ) (3.98 × 10−3 )
14–3
14–4
CHAPTER 14. Introduction to Frequency-Selective Circuits
Problems P 14.1
[a] ωc =
127 R = = 12.7 krad/s L 10 × 10−3
.·. fc = [b] H(s) =
12,700 ωc = = 2021.27 Hz 2π 2π
ωc 12,700 = s + ωc s + 12,700
H(jω) = H(jωc ) =
12,700 12,700 + jω 12,700 = 0.7071/− 45◦ 12,700 + j12,700
H(j0.2ωc ) = H(j5ωc ) =
12,700 = 0.981/− 11.31◦ 12,700 + j2540
12,700 = 0.196/− 78.69◦ 12,700 + j63,500
[c] vo (t)|ωc = 7.07 cos(12,700t − 45◦ ) V vo (t)|0.2ωc = 9.81 cos(2540t − 11.31◦ ) V vo (t)|5ωc = 1.96 cos(63,500t − 78.69◦ ) V P 14.2
[a] ωo =
R = 2000π rad/s L
R = Lωo = (0.005)(2000π) = 31.42 Ω [b] Re = 31.42270 = 28.14 Ω Re = 5628 rad/s L ωloaded = 895.77 Hz .·. floaded = 2π ωloaded =
P 14.3
Note: add the resistor to the cirucit in Fig. 14.4(a). [a] H(s) =
Vo R (R/L) = = Vi sL + R + Rl s + (R + Rl )/L
Problems (R/L) [b] H(jω) = R+R l + jω L |H(jω)| =
(R/L)
R+Rl 2 L
+ ω2
|H(jω)|max occurs when ω = 0 R R + Rl R R/L [d] |H(jωc )| = √ = 2(R + Rl ) R+Rl 2 [c] |H(jω)|max =
L
R + Rl .·. ωc2 = L
2
;
+ ωc2
.·. ωc = (R + Rl )/L
[e] Note – add 75 Ω resistor in series with the 10 mH inductor. ωc =
127 + 75 = 20,200 rad/s 0.01
H(jω) =
12,700 20,200 + jω
H(j0) = 0.6287 H(j20,200) = H(j6060) =
12,700 = 0.6022/− 16.70◦ 20,200 + j6060
H(j60,600) = P 14.4
[a] ωc = fc =
0.6287 √ /− 45◦ = 0.4446/− 45◦ 2
12,700 = 0.1988/− 71.57◦ 20,200 + j60,600
1 1 = = 10 krad/s 3 RC (10 )(100 × 10−9 ) ωc = 1591.55 Hz 2π
[b] H(jω) =
ωc 10,000 = s + ωc s + 10,000
H(jω) = H(jωc ) =
10,000 10,000 + jω 10,000 = 0.7071/− 45◦ 10,000 + j10,000
14–5
14–6
CHAPTER 14. Introduction to Frequency-Selective Circuits H(j0.1ωc ) =
10,000 = 0.9950/− 5.71◦ 10,000 + j1000
H(j10ωc ) =
10,000 = 0.0995/− 84.29◦ 10,000 + j100,000
[c] vo (t)|ωc = 200(0.7071) cos(10,000t − 45◦ ) = 141.42 cos(10,000t − 45◦ ) mV vo (t)|0.1ωc = 200(0.9950) cos(1000t − 5.71◦ ) = 199.01 cos(1000t − 5.71◦ ) mV vo (t)|10ωc = 200(0.0995) cos(100,000t − 84.29◦ ) = 19.90 cos(100,000t − 84.29◦ ) mV P 14.5
[a] Let Z =
RL RL (1/sC) = RL + 1/sC RL Cs + 1
Then H(s) = = =
Z Z +R RL RRL Cs + R + RL (1/RC) R + RL s+ RRL C (1/RC)
[b] |H(jω)| =
ω 2 + [(R + RL )/RRL C]2
|H(jω)| is maximum at ω = 0 RL R + RL RL (1/RC) [d] |H(jωc )| = √ = 2(R + RL ) ωc2 + [(R + RL )/RRL C]2 [c] |H(jω)|max =
.·. ωc = [e] ωc =
1 R + RL = (1 + (R/RL )) RRL C RC 1
(103 )(10−7 )
H(j0) =
[1 + (103 /104 )] = 10,000(1 + 0.1) = 11,000 rad/s
10,000 = 0.9091/0◦ 11,000
Problems H(jωc ) =
P 14.6
H(j0.1ωc ) =
10,000 = 0.9046/− 5.71◦ 11,000 + j1100
H(j10ωc ) =
10,000 = 0.0905/− 84.29◦ 11,000 + j110,000
[a] fc = [b]
50,000 50 ωc = = × 103 = 7957.75 Hz 2π 2π 2π
1 = 50 × 103 RC
1
R= [c] ωc = .·.
(50 ×
[a]
103 )(0.5
1 R 1+ RC RL R = 0.05 RL
[d] H(j0) = P 14.7
10,000 = 0.6428/− 45◦ 11,000 + j11,000
× 10−6 )
= 40 Ω
.·. RL = 20R = 800 Ω
800 RL = 0.9524 = R + RL 840
1 1 = = 4000 rad/s 3 RC (50 × 10 )(5 × 10−9 )
4000 = 636.62 Hz 2π s jω [b] H(s) = .·. H(jω) = s + ωc 4000 + jω fc =
H(jωc ) = H(j4000) =
j4000 = 0.7071/45◦ 4000 + j4000
H(j0.2ωc ) = H(j800) =
j800 = 0.1961/78.69◦ 4000 + j800
H(j5ωc ) = H(j20, 000) =
j20,000 = 0.9806/11.31◦ 4000 + j20,000
[c] vo (t)|ωc = (0.7071)(500) cos(4000t + 45◦ ) = 353.55 cos(4000t + 45◦ ) mV vo (t)|0.2ωc = (0.1961)(500) cos(800t + 78.69◦ ) = 98.06 cos(800t + 78.69◦ ) mV vo (t)|5ωc = (0.9806)(500) cos(20,000t + 11.31◦ ) = 490.29 cos(20,000t + 11.31◦ ) mV
14–7
14–8
P 14.8
CHAPTER 14. Introduction to Frequency-Selective Circuits [a] H(s) =
R Vo = Vi R + Rc + (1/sC)
R s · R + Rc [s + (1/(R + Rc )C)] jω R · [b] H(jω) = R + Rc jω + (1/(R + Rc )C) =
R ω · 2 R + Rc ω + (R+R1c )2 C 2
|H(jω)| =
The magnitude will be maximum when ω = ∞. R [c] |H(jω)|max = R + Rc Rωc [d] |H(jωc )| = (R + Rc ) ωc2 + [1/(R + Rc )C]2 .·. |H(jω)| = √
[e] ωc =
when
1 (R + Rc )2 C 2
.·. ωc2 = or ωc =
R 2(R + Rc )
1 (R + Rc )C 1
(62.5 ×
103 )(5
× 10−9 )
= 3200 rad/s
R 50 = 0.8 = R + Rc 62.5 .·.
H(jω) =
H(jωc ) =
(0.8)j3200 = 0.5657/45◦ 3200 + j3200
H(j0.2ωc ) = H(j5ωc ) = P 14.9
[a] ωc =
0.8jω 3200 + jω
(0.8)j640 = 0.1569/78.69◦ 3200 + j640
(0.8)j16,000 = 0.7845/11.31◦ 3200 + j16,000
1 = 2π(300) = 600π rad/s RC
.·. R =
1 1 = = 5305.16 Ω ωc C (600π)(100 × 10−9 )
Problems
14–9
[b] Re = 5305.1647,000 = 4767.08 Ω ωc =
1 1 = = 2097.7 rad/s Re C (4767.08)(100 × 10−9 )
fc =
2097.7 ωc = = 333.86 Hz 2π 2π
P 14.10 [a] ωc =
R L
R = ωc L = (25 × 103 )(5 × 10−3 ) = 125 Ω
so
[b] ωc (loaded) =
RL R · = 24,000 rad/s L R + RL
24,000 ωc (loaded) RL = = 0.96 = R + RL ωc (unloaded) 25,000 RL = 0.96(R + RL ) .·. 0.04RL = 0.96R = (0.96)(125) .·.
.·. RL =
(0.96)(125) = 3 kΩ 0.04
P 14.11 By definition Q = ωo /β therefore β = ωo /Q. Substituting this expression into Eqs. 14.34 and 14.35 yields ωc1 = −
ωc2 =
ωo + 2Q
ωo + 2Q
ωo 2
2Q
ωo 2
2Q
+ ωo2
+ ωo2
Now factor ωo out to get
ωc1 = ωo −
1 + 2Q
ωc2 = ωo P 14.12 ωo = fo =
√
1 + 2Q
ωc1 ωc2 =
1+
1+
1 2Q
2
(121)(100) = 110 krad/s
ωo = 17.51 kHz 2π
β = 121 − 100 = 21 krad/s Q=
1 2Q
2
110 ωo = = 5.24 β 21
or
3.34 kHz
14–10 P 14.13 β =
CHAPTER 14. Introduction to Frequency-Selective Circuits ωo 50,000 = = 12.5 krad/s; Q 4
1 ωc2 = 50,000 + 8 fc2 =
1 1+ 8
= 56.64 krad/s
56.64 k = 9.01 kHz 2π
1 ωc1 = 50,000 − + 8 fc1 =
2
12,500 = 1.99 kHz 2π
1+
2
1 8
= 44.14 krad/s
44.14 k = 7.02 kHz 2π
P 14.14 [a] ωo2 = R=
1 LC
so
1 = 79.16 mH [8000(2π)]2 (5 × 10−9 )
L=
8000(2π)(79.16 × 10−3 ) ωo L = = 1.99 kΩ Q 2
1 [b] fc1 = 8000 − + 4
1 [c] fc2 = 8000 + 4
1 1+ = 6.25 kHz 16
1 1+ = 10.25 kHz 16
[d] β = fc2 − fc1 = 4 kHz or 8000 fo = = 4 kHz β= Q 2 P 14.15 [a] ωo2 =
1 1 = = 1010 −3 LC (10 × 10 )(10 × 10−9 )
ωo = 105 rad/s = 100 krad/s 105 ωo = = 15.92 kHz 2π 2π [c] Q = ωo RC = (100 × 103 )(8000)(10 × 10−9 ) = 8
[b] fo =
[d] ωc1 =
ωo −
1 + 2Q
1 1+ 2Q
ωc1 = 14.96 kHz [e] .·. fc1 = 2π
2
1 = 105 − + 16
1 1+ = 93.95 krad/s 256
Problems ωo
[f] ωc2 =
1 + 2Q
1 1+ 2Q
2
1 = 10 + 16 5
1 1+ = 106.45 krad/s 256
ωc2 = 16.94 kHz [g] .·. fc2 = 2π ωo 105 [h] β = = = 12.5 krad/s or 1.99 kHz Q 8 1
P 14.16 [a] L =
ωo2 C
R=
=
1 (50 ×
10−9 )(20
= 50 mH
5 Q = = 5 kΩ 3 ωo C (20 × 10 )(50 × 10−9 )
1 1 [b] ωc2 = ωo + 1 + 2Q 2Q
.·.
= 22.10 krad/s
1 ωc1 = ωo + − 2Q
2
1 + = 20,000 10
fc2 =
1 1+ 2Q .·.
= 18.10 krad/s [c] β =
× 103 )2
2
ωo 20,000 = = 4000 rad/s Q 5
1 1+ 100
ωc2 = 3.52 kHz 2π
1 + = 20,000 − 10
fc1 = or
1 1+ 100
ωc1 = 2.88 kHz 2π 636.62 Hz
1 1 = = 625 × 106 LC (40 × 10−3 )(40 × 10−9 )
P 14.17 [a] ωo2 =
ωo = 25 × 103 rad/s = 25 krad/s fo = [b] Q =
25,000 = 3978.87 Hz 2π
ωo L (25 × 103 )(40 × 10−3 ) =5 = R + Ri 200
[c] ωc1 =
ωo −
1 + 2Q
1 1+ 2Q
= 22.62 krad/s
[d] wc2 = ωo
1 + 2Q
or
1+
= 27.62 krad/s
or
2
1 = 25,000 − + 10
1 1+ 100
3.60 kHz 1 2Q
2
= 25,000
4.40 kHz
14–11
1 + 10
1+
1 100
14–12
CHAPTER 14. Introduction to Frequency-Selective Circuits
[e] β = ωc2 − ωc1 = 27.62 − 22.62 = 5 krad/s or ωo 25,000 β= = = 5 krad/s or 795.77 Hz Q 5 P 14.18 [a] H(s) =
(R/L)s
1 s2 + + LC For the numerical values in Problem 14.17 we have 4500s H(s) = 2 s + 5000s + 625 × 106 (R+Ri ) s L
.·. H(jω) =
(625 ×
4500jω − ω 2 ) + j5000ω
106
j4500(25 × 103 ) = 0.9/0◦ H(jωo ) = 3 j5000(25 × 10 ) .·. vo (t) = 500(0.9) cos 25,000t = 450 cos 25,000t mV [b] From the solution to Problem 14.17, ωc1 = 22.62 krad/s H(j22.62 k) =
j4500(22.62 × 103 ) = 0.6364/45◦ (113.12 + j113.12) × 106
.·. vo (t) = 500(0.6364) cos(22,620t + 45◦ ) = 318.2 cos(22,620t + 45◦ ) mV [c] From the solution to Problem 14.17, ωc2 = 27.62 krad/s H(j27.62 k) =
j4500(27.62 × 103 ) = 0.6364/− 45◦ (−138.12 + j138.12) × 106
.·. vo (t) = 500(0.6364) cos(27,620t − 45◦ ) = 318.2 cos(27,620t − 45◦ ) mV P 14.19 [a]
[b] L =
1 1 = = 20 mH ωo2 C (50 × 103 )2 (20 × 10−9 )
R=
ωo L (50 × 103 )(20 × 10−3 ) = = 160 Ω Q 6.25
Problems [c] Re = 160480 = 120 Ω Re + Ri = 120 + 80 = 200 Ω Qsystem = [d] βsystem =
ωo L (50 × 103 )(20 × 10−3 ) =5 = Re + Ri 200 ωo
Qsystem
50 × 103 = 10 krad/s 5
10,000 = 1591.55 Hz 2π
βsystem (Hz) = P 14.20 [a]
=
1 Vo Z where Z = = Vi Z +R Y and Y = sC + H(s) = = = =
RL s2
RLCs2
+
RL Ls + (R + RL )Ls + RRL
(1/RC)s R+RL RL
1 RC
s+
1 LC
RL R+RL 1 s R+RL R RC L 1 1 L s2 + R+R s + LC RL RC
s2
Kβs , + βs + ωo2
R + RL RL 1 [c] βU = RC [b] βL =
1 LCRL s2 + sL + RL 1 = + sL RL RL Ls
K=
RL R + RL
1 RC
R + RL R .·. βL = βU = 1 + βU RL RL [d] QL =
ωo RC ωo = R+R L β RL
[e] QU = ωo RC .·. QL = [f] H(jω) =
ωo2
RL 1 QU QU = R + RL [1 + (R/RL )]
Kjωβ − ω 2 + jωβ
H(jωo ) = K
14–13
14–14
CHAPTER 14. Introduction to Frequency-Selective Circuits Let ωc represent a cutoff frequency. Then K Kωc β |H(jωc )| = √ = 2 (ωo2 − ωc2 )2 + ωc2 β 2 1 ωc β .·. √ = 2 (ωo2 − ωc2 )2 + ωc2 β 2 Squaring both sides leads to (ωo2 − ωc2 )2 = ωc2 β 2 or (ωo2 − ωc2 ) = ±ωc β .·. ωc2 ± ωc β − ωo2 = 0 or β ωc = ∓ ± 2
β2 + ωo2 4
The two positive roots are β ωc1 = − + 2
where
β = 1+ P 14.21 [a] ωo2 =
R RL
β2 + ωo2 4
and
β ωc2 = + 2
β2 + ωo2 4
1 1 and ωo2 = RC LC
1 1 = = 1012 −3 −12 LC (5 × 10 )(200 × 10 )
ωo = 1 Mrad/s R + RL 1 = [b] β = · RL RC
500 × 103 400 × 103
1 3 (100 × 10 )(200 × 10−12 )
106 ωo = = 16 β 62.5 × 103 RL [d] H(jωo ) = = 0.8/0◦ R + RL [c] Q =
.·. vo (t) = 250(0.8) cos(106 t) = 200 cos 106 t mV
R [e] β = 1 + RL
1 100 = 1+ (50 × 103 ) rad/s RC RL
ωo = 106 rad/s Q=
20 ωo = β 1 + (100/RL )
where RL is in kilohms
= 62.5 krad/s
Problems
14–15
[f]
1 1 = = 1016 −6 LC (2 × 10 )(50 × 10−12 )
P 14.22 ωo2 =
ωo = 100 Mrad/s Qu = ωo RC = (100 × 106 )(2.4 × 103 )(50 × 10−12 ) = 12 .·.
RL 12 = 7.5; R + RL
.·. RL =
7.5 R = 4 kΩ 4.5
P 14.23 [a] In analyzing the circuit qualitatively we visualize vi as a sinusoidal voltage and we seek the steady-state nature of the output voltage vo . At zero frequency the inductor provides a direct connection between the input and the output, hence vo = vi when ω = 0. At infinite frequency the capacitor provides the direct connection, hence vo = vi when ω = ∞. At the resonant frequency of the parallel combination of L and C the impedance of the combination is infinite and hence the output voltage will be zero when ω = ωo . At frequencies on either side of ωo the amplitude of the output voltage will be nonzero but less than the amplitude of the input voltage. Thus the circuit behaves like a band-reject filter. [b] Let Z represent the impedance of the parallel branches L and C, thus Z=
sL(1/sC) sL = 2 sL + 1/sC s LC + 1
Then Vo R(s2 LC + 1) R H(s) = = = Vi Z +R sL + R(s2 LC + 1) =
[s2 + (1/LC)] s2 +
1 RC
s+
1 LC
14–16
CHAPTER 14. Introduction to Frequency-Selective Circuits H(s) =
s2 + ωo2 s2 + βs + ωo2
[c] From part (b) we have H(jω) =
ωo2 − ω 2 ωo2 − ω 2 + jωβ
It follows that H(jω) = 0 when ω = ωo .·. ωo = √
1 LC ωo2 − ω 2
[d] |H(jω)| =
(ωo2 − ω 2 )2 + ω 2 β 2
1 |H(jω)| = √ when ω 2 β 2 = (ωo2 − ω 2 )2 2 or ± ωβ = ωo2 − ω 2 , thus ω 2 ± βω − ωo2 = 0 The two positive roots of this quadratic are 2
β −β ωc1 = + 2 2
+ ωo2
2
β β ωc2 = + 2 2
+ ωo2
Also note that since β = ωo /Q
2
2
1 −1 + 1 + ωc1 = ωo 2Q 2Q
ωc2 = ωo
1 + 2Q
1+
1 2Q
[e] It follows from the equations derived in part (b) that β = 1/RC [f] By definition Q = ωo /β = ωo RC
Problems P 14.24 [a] ωo2 =
1 1 = 1012 = −6 −9 LC (50 × 10 )(20 × 10 )
.·. ωo = 1 Mrad/s ωo = 159.15 kHz 2π [c] Q = ωo RC = (106 )(750)(20 × 10−9 ) = 15
[b] fo =
[d] ωc1 = ωo −
1 + 2Q
1+
1 2Q
= 967.22 krad/s ωc1 = 153.94 kHz [e] fc1 = 2π
[f] ωc2 =
ωo
1 + 2Q
1 1+ 2Q
2
2
= 106 −
1 + 30
1 = 106 + 30
1+
1 900
1 1+ 900
= 1.03 Mrad/s ωc2 = 164.55 kHz [g] fc2 = 2π [h] β = fc2 − fc1 = 10.61 kHz P 14.25 [a] ωo = 2πfo = 8π krad/s L=
1 1 = = 3.17 mH ωo2 C (8000π)2 (0.5 × 10−6 )
R=
Q 5 = = 397.89 Ω ωo C (8000π)(0.5 × 10−6 )
[b] fc2 =
fo
1 + 2Q
1 1+ 2Q
= 4.42 kHz
1 + fc1 = fo − 2Q
2
1 1+ 2Q
= 3.62 kHz [c] β = fc2 − fc1 = 800 Hz or 4000 fo = = 800 Hz β= Q 5
1 = 4000 + 10
2
1 1+ 100
1 + = 4000 − 10
1 1+ 100
14–17
14–18
CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.26 [a] Re = 397.891000 = 284.63 Ω Q = ωo Re C = (8000π)(284.63)(0.5 × 10−6 ) = 3.58 [b] β =
fo 4000 = = 1.12 kHz Q 3.58
1 + [c] fc2 = 4000 7.15
1 + [d] fc1 = 4000 − 7.15 P 14.27 [a] Let Z = Z=
1 1+ = 4.60 kHz 7.152
1 1+ = 3.48 kHz 7.152
RL (sL + (1/sC)) RL + sL + (1/sC)
RL (s2 LC + 1) s2 LC + RL Cs + 1
Then H(s) =
s2 RL CL + RL Vo = Vi (R + RL )LCs2 + RRL Cs + R + RL
Therefore H(s) = =
RL [s2 + (1/LC)] · RRL s 1 R + RL s2 + R+R + L LC L
K(s2 + ωo2 ) s2 + βs + ωo2
RL where K = ; R + RL
ωo2
1 LC RRL 1 [c] β = R + RL L ωo L ωo = [d] Q = β [RRL /(R + RL )]
[b] ωo = √
[e] H(jω) =
K(ωo2 − ω 2 ) (ωo2 − ω 2 ) + jβω
H(jωo ) = 0 [f] H(j0) =
Kωo2 =K ωo2
1 ; = LC
RRL β= R + RL
1 L
Problems
K (ωo /ω)2 − 1
[g] H(jω) = (ωo /ω)2 − 1 + jβ/ω
H(j∞) = [h] H(jω) =
−K =K −1
K(ωo2 − ω 2 ) (ωo2 − ω 2 ) + jβω
H(j0) = H(j∞) = K Let ωc represent a corner frequency. Then K |H(jωc )| = √ 2 K K(ωo2 − ωc2 ) .·. √ = 2 (ωo2 − ωc2 )2 + ωc2 β 2 Squaring both sides leads to (ωo2 − ωc2 )2 = ωc2 β 2 or (ωo2 − ωc2 ) = ±ωc β .·. ωc2 ± ωc β − ωo2 = 0 or β ωc = ∓ ± 2
β2 + ωo2 4
The two positive roots are β ωc1 = − + 2
β2 + ωo2 4
and
β ωc2 = + 2
β2 + ωo2 4
where β= P 14.28 [a] ωo2 =
1 1 RRL · and ωo2 = R + RL L LC 1 1 = 0.25 × 1018 = 25 × 1016 = −12 −6 LC (10 )(4 × 10 )
ωo = 5 × 108 = 500 Mrad/s β=
(30)(150) 1 RRL 1 · −6 = 25 Mrad/s = 3.98 MHz · = 180 10 R + RL L
Q=
500 M ωo = = 20 β 25 M
14–19
14–20
CHAPTER 14. Introduction to Frequency-Selective Circuits
[b] H(j0) =
150 RL = 0.8333 = R + RL 180
H(j∞) =
RL = 0.8333 R + RL
250 1 + [c] fc2 = π 40
1 1+ = 81.59 MHz 1600
250 1 − + fc2 = π 40
1 1+ = 77.61 MHz 1600
Check: β = fc2 − fc1 = 3.98 MHz. 500 × 106 ωo = RRL 1 [d] Q = β ·L R+R L
500(R + RL ) 50 30 = 1+ RRL 3 RL where RL is in ohms. =
[e]
P 14.29 [a] ωo2 =
1 = 1012 LC
.·. L =
1 (1012 )(400
RL = 0.96; R + RL .·.
RL = 24R
× 10−12 )
= 2.5 mH
.·. 0.04RL = 0.96R .·. R =
36,000 = 1.5 kΩ 24
RL 1 [b] β = R · = 576 × 103 R + RL L Q=
ωo 106 = 1.74 = β 576 × 103
Problems
14–21
P 14.30 Refer to Sections E.5 and E.7. [a] ωn = 105 2ζωn = 50,000, ζ = 0.25 √ √ ωo = 2ωp = 2ωn 1 − 2ζ 2 = 132,287.57 rad/s .·. ω = 0 ω = 132,287.57 rad/s
[b] ωp = ωn 1 − 2ζ 2 = 93,541.43 rad/s P 14.31 [a] Use the cutoff frequencies to calculate the bandwidth: ωc1 = 2π(697) = 4379.38 rad/s
ωc2 = 2π(941) = 5912.48 rad/s
β = ωc2 − ωc1 = 1533.10 rad/s
Thus
Calculate inductance using Eq. (14.32) and capacitance using Eq. (14.31): L= C=
600 R = = 0.39 H β 1533.10 1 Lωc1 ωc2
=
1 = 0.10 µF (0.39)(4379.38)(5912.48)
[b] At the outermost two frequencies in the low-frequency group (687 Hz and 941 Hz) the amplitudes are |V697 Hz | = |V941 Hz | =
|Vpeak | √ = 0.707|Vpeak | 2
because these are cutoff frequencies. We calculate the amplitudes at the other two low frequencies using Eq. (14.32): ωβ
|V | = (|Vpeak |)(|H(jω)|) = |Vpeak |
(ωo2 − ω 2 )2 + (ωβ)2
Therefore (4838.05)(1533.10) |V770 Hz | = |Vpeak | (5088.522 − 4838.052 )2 + [(4838.05)(1533.10)]2 = 0.948|Vpeak | and (5353.27)(1533.10) |V852 Hz | = |Vpeak | (5088.522 − 5353.272 )2 + [(5353.27)(1533.10)]2 = 0.948|Vpeak |
14–22
CHAPTER 14. Introduction to Frequency-Selective Circuits It is not a coincidence that these two magnitudes are the same. The frequencies in both bands of the DTMF system were carefully chosen to produce this type of predictable behavior with linear filters. In other words, the frequencies were chosen to be equally far apart with respect to the response produced by a linear filter. Most musical scales consist of tones designed with this same property – note intervals are selected to place the notes equally far apart. That is why the DTMF tones remind us of musical notes! Unlike musical scales, DTMF frequencies were selected to be harmonically unrelated, to lower the risk of misidentifying a tone’s frequency if the circuit elements are not perfectly linear.
[c] The high-band frequency closest to the low-frequency band is 1209 Hz. The amplitude of a tone with this frequency is (7596.37)(1533.10)
|V1209 Hz | = |Vpeak | =
(5088.522
− 7596.372 )2 + [(7596.37)(1533.10)]2
= 0.344|Vpeak | This is less than one half the amplitude of the signals with the low-band cutoff frequencies, ensuring adequate separation of the bands. P 14.32 The cutoff frequencies and bandwidth are ωc1 = 2π(1209) = 7596 rad/s ωc2 = 2π(1633) = 10.26 krad/s β = ωc2 − ωc1 = 2664 rad/s Telephone circuits always have R = 600 Ω. Therefore, the filter’s inductance and capacitance values are L=
R 600 = = 0.225 H β 2664
C=
1 = 0.057 µF ωc1 ωc2 L
At the highest of the low-band frequencies, 941 Hz, the amplitude is ωβ
|Vω | = |Vpeak |
(ωo2 − ω 2 )2 + ω 2 β 2
where |Vω | =
ωo =
√
ωc1 ωc2 . Thus,
|Vpeak |(5912)(2664) [(8828)2 − (5912)2 ]2 + [(5912)(2664)]2
= 0.344 |Vpeak |
Problems
14–23
Again it is not coincidental that this result is the same as the response of the low-band filter to the lowest of the high-band frequencies. P 14.33 From Problem 14.31 the response to the largest of the DTMF low-band tones is 0.948|Vpeak |. The response to the 20 Hz tone is |V20 Hz | =
[(50892
|Vpeak |(125.6)(1533) − 125.62 )2 + [(125.6)(1533)]2 ]1/2
= 0.00744|Vpeak | .·.
0.00744|Vring-peak | = 0.5 0.948|VDTMF-peak |
.·. |Vring-peak | = 63.7|VDTMF-peak | Thus, the 20 Hz signal can be 63.7 times as large as the DTMF tones.
Active Filter Circuits
Assessment Problems AP 15.1 H(s) =
−(R2 /R1 )s s + (1/R1 C)
1 = 1 rad/s; R1 C R2 = 1, R1 .·.
R1 = 1 Ω,
.·. R2 = R1 = 1 Ω
Hprototype (s) =
AP 15.2 H(s) =
.·. C = 1 F
−s s+1
−20,000 −(1/R1 C) = s + (1/R2 C) s + 5000
1 = 20,000; R1 C .·. R1 =
C = 5 µF
1 = 10 Ω (20,000)(5 × 10−6 )
1 = 5000 R2 C .·. R2 =
1 = 40 Ω (5000)(5 × 10−6 )
15–1
15
15–2
CHAPTER 15. Active Filter Circuits
AP 15.3 ωc = 2πfc = 2π × 104 = 20,000π rad/s .·. kf = 20,000π = 62,831.85 C =
C kf km
.·. km =
.·.
0.5 × 10−6 =
(0.5 ×
10−6 )(62,831.85)
1
1 kf km = 31.83
AP 15.4 For a 2nd order prototype Butterworth high pass filter H(s) =
s2 √ s2 + 2s + 1
For the circuit in Fig. 15.25 H(s) =
s2 +
2 R2 C
s2 s+
1 R1 R2 C 2
Equate the transfer functions. For C = 1F, √ 2 = 2, R2 C
.·. R2 =
1 = 1, R1 R2 C 2
√
2 = 1.414 Ω
1 .·. R1 = √ = 0.707 Ω 2
AP 15.5 Q = 8, K = 5, ωo = 1000 rad/s, C = 1 µF For the circuit in Fig 15.26
1 s R1 C H(s) = R1 + R2 2 2 s + s+ R3 C R1 R2 R3 C 2 Kβs = 2 s + βs + ωo2 −
β=
2 , R3 C
β=
ωo 1000 = = 125 rad/s Q 8
.·.
R3 =
2 βC
Problems .·. R3 = Kβ =
2 × 106 = 16 kΩ (125)(1)
1 R1 C
.·. R1 =
1 1 = = 1.6 kΩ KβC 5(125)(1 × 10−6 )
R1 + R2 R1 R2 R3 C 2
ωo2 = 106 =
(1600 + R2 ) (1600)(R2 )(16,000)(10−6 )2
Solving for R2 , R2 =
(1600 + R2 )106 , 256 × 105
AP 15.6 ωo = 1000 rad/s;
246R2 = 16,000,
R2 = 65.04 Ω
Q = 4;
C = 2 µF s2 + (1/R2 C 2 ) 4(1 − σ) 1 2 s + s+ RC R2 C 2 2 2 s + ωo 1 = 2 ; ωo = ; 2 s + βs + ωo RC
H(s) =
R=
1 1 = = 500 Ω ωo C (1000)(2 × 10−6 )
β=
1000 ωo = = 250 Q 4
.·.
β=
4(1 − σ) RC
4(1 − σ) = 250 RC
4(1 − σ) = 250RC = 250(500)(2 × 10−6 ) = 0.25 1−σ =
0.25 = 0.0625; 4
.·.
σ = 0.9375
15–3
CHAPTER 15. Active Filter Circuits
15–4
Problems P 15.1
Summing the currents at the inverting input node yields 0 − Vi 0 − Vo + =0 Zi Zf .·.
Vi Vo =− Zf Zi
.·. H(s) =
P 15.2
[a] Zf =
Zf Vo =− Vi Zi
R2 R2 (1/sC2 ) = [R2 + (1/sC2 )] R2 C2 s + 1
(1/C2 ) s + (1/R2 C2 ) Likewise (1/C1 ) Zi = s + (1/R1 C1 ) =
.·. H(s) =
−(1/C2 )[s + (1/R1 C1 )] [s + (1/R2 C2 )](1/C1 )
=−
C1 [s + (1/R1 C1 )] C2 [s + (1/R2 C2 )]
−C1 jω + (1/R1 C1 ) [b] H(jω) = C2 jω + (1/R2 C2 ) −C1 H(j0) = C2
R2 C2 R1 C1
=
−R2 R1
C1 j −C1 [c] H(j∞) = − = C2 j C2 [d] As ω → 0 the two capacitor branches become open and the circuit reduces to a resistive inverting amplifier having a gain of −R2 /R1 . As ω → ∞ the two capacitor branches approach a short circuit and in this case we encounter an indeterminate situation; namely vn → vi but vn = 0 because of the ideal op amp. At the same time the gain of the ideal op amp is infinite so we have the indeterminate form 0 · ∞. Although ω = ∞ is indeterminate we can reason that for finite large values of ω H(jω) will approach −C1 /C2 in value. In other words, the circuit approaches a purely capacitive inverting amplifier with a gain of (−1/jωC2 )/(1/jωC1 ) or −C1 /C2 .
Problems P 15.3
[a] Zf =
15–5
(1/C2 ) s + (1/R2 C2 )
Zi = R 1 +
1 R1 [s + (1/R1 C1 )] = sC1 s
H(s) = −
s (1/C2 ) · [s + (1/R2 C2 )] R1 [s + (1/R1 C1 )]
=− [b] H(jω) = −
1 s R1 C2 [s + (1/R1 C1 )][s + (1/R2 C2 )] 1 R1 C2 jω +
jω
1 R1 C1
jω +
1 R2 C2
H(j0) = 0 [c] H(j∞) = 0 [d] As ω → 0 the capacitor C1 disconnects vi from the circuit. Therefore vo = vn = 0. As ω → ∞ the capacitor short circuits the feedback network, thus Zf = 0 and therefore vo = 0. P 15.4
R2 R1 1 1 = = 212.21 Ω R2 = ωc C (2π)(103 )(750 × 10−9 )
[a] K = 10(10/20) = 3.16 =
R1 =
R2 212.21 = 67.11 Ω = K 3.16
[b]
P 15.5
[a] R1 =
1 1 = = 5.10 kΩ ωc C (2π)(8 × 103 )(3.9 × 10−9 )
K = 10(14/20) = 5.01 =
R2 R1
.·. R2 = 5.01R1 = 25.57 kΩ
CHAPTER 15. Active Filter Circuits
15–6 [b]
P 15.6
For the RC circuit H(s) =
(1/RC) Vo = Vi s + (1/RC)
R = km R;
C =
.·. R C = km R
C km kf
C 1 1 = RC = km kf kf kf
1 = kf R C H (s) =
kf (1/R C ) = s + (1/R C ) s + kf
H (s) =
1 (s/kf ) + 1
For the RL circuit H(s) =
V0 R/L = Vi s + (R/L)
R = km R;
L =
km L kf
R km R R = km = k f = kf L L L kf H (s) =
kf (R /L ) = s + (R /L ) s + kf
H (s) =
1 (s/kf ) + 1
Problems P 15.7
For the RC circuit H(s) =
s Vo = Vi s + (1/RC)
R = km R; .·. R C = H (s) =
C =
C km kf
RC 1 = ; kf kf
1 = kf R C
s s (s/kf ) = = s + (1/R C ) s + kf (s/kf ) + 1
For the RL circuit H(s) =
s s + (R/L)
R = km R;
L =
km L kf
R R = k = kf f L L H (s) =
P 15.8
s s (s/kf ) = = s + (R /L ) s + kf (s/kf ) + 1
βs (R/L)s = 2 + (R/L)s + (1/LC) s βs + ωo2 For the prototype circuit ωo = 1 and β = ωo /Q = 1/Q. For the scaled circuit
H(s) =
H (s) =
s2
(R /L )s s2 + (R /L )s + (1/L C )
where R = km R; L =
km C L; and C = kf kf km
R km R R · = km = k f = kf β .. L L L kf kf2 1 kf km = kf2 = = km L C LC LC kf
15–7
CHAPTER 15. Active Filter Circuits
15–8
Q =
ωo kf ωo = =Q β kf β
therefore the Q of the scaled circuit is the same as the Q of the unscaled circuit. Also note β = kf β.
.·. H (s) =
s2 +
kf s Q kf s+ Q
H (s) = 2 s kf
P 15.9
1 Q
+
[a] L = 1 H; R=
1 Q
s kf
kf2
s kf
+1
C = 1F
1 1 = = 0.05 Ω Q 20
ωo = 40,000; ωo Thus,
[b] kf =
km =
5000 R = = 100,000 R 0.05
R = km R = (0.05)(100,000) = 5 kΩ L =
km 100,000 (1) = 2.5 H L= kf 40,000
C =
C 1 = 250 pF = km kf (40,000)(100,000)
[c]
P 15.10 [a] Since ωo2 = 1/LC and ωo = 1 rad/s, C=
1 1 = L Q
[b] H(s) =
s2
H(s) =
(R/L)s + (R/L)s + (1/LC)
s2
(1/Q)s + (1/Q)s + 1
Problems [c] In the prototype circuit R = 1 Ω;
L = 16 H;
C=
R · . . km = = 10,000; R
1 = 0.0625 F L
ωo = 25,000 kf = ωo
Thus R = km R = 10 kΩ L =
km 10,000 L= (16) = 6.4 H kf 25,000
C =
C 0.0625 = 250 pF = km kf (10,000)(25,000)
[d]
[e] H (s) = H (s) =
s 25,000
s2
1 16 2
+
s 25,000 1 16
s 25,000
+1
1562.5s + 1562.5s + 625 × 106
P 15.11 [a] Using the first prototype ωo = 1 rad/s; km =
C = 1 F;
L = 1 H;
40,000 R = = 1600; R 25
kf =
R = 25 Ω
ωo = 50,000 ωo
Thus, R = km R = 40 kΩ; C =
L =
km 1600 L= (1) = 32 mH; kf 50,000
C 1 = 12.5 nF = km kf (1600)(50,000)
Using the second prototype ωo = 1 rad/s; L=
1 = 40 mH; 25
C = 25 F R = 1Ω
15–9
CHAPTER 15. Active Filter Circuits
15–10
R = 40,000; R
kf =
ωo = 50,000 ωo
R = km R = 40 kΩ;
L =
km 40,000 (0.04) = 32 mH; L= kf 50,000
km = Thus,
C =
C 25 = 12.5 nF = km kf (40,000)(50,000)
[b]
P 15.12 For the scaled circuit H (s) =
L = .·.
s2 +
s2 +
R L
km L; kf
s+
C =
kf2 1 ; = LC L C
R R = kf .·. L L
1
L C
1
L C
C km kf R = km R
It follows then that 2
H (s) =
s + s2 +
R L
= 2 s kf
kf s +
s kf
+
kf2 LC
2
+
R L
kf2 LC
1 LC
s kf
= H(s)|s=s/kf P 15.13 For the circuit in Fig. 15.31 H(s) =
s2 + s2 +
s RC
1 LC
+
1 LC
+
1 LC
Problems It follows that H (s) =
s2 + L1C s2 + RsC + L1C
where R = km R; C =
.·.
km L; kf
L =
C km kf
kf2 1 = L C LC
kf 1 = RC RC 2
H (s) =
s + s2 +
= 2 s kf
kf RC
+
kf2 LC
s kf
s+
kf2 LC
+
1 LC
2
1 RC
s kf
+
1 LC
= H(s)|s=s/kf P 15.14 [a] For the circuit in Fig. P15.14(a) H(s) =
s+
1 s
s2 + 1 Vo = = 1 1 2+ 1 s+1 Vi s +s+ Q Q s
For the circuit in Fig. P15.14(b) Qs + Qs Vo = H(s) = Vi 1 + Qs + Qs = H(s) =
Q(s2 + 1) Qs2 + s + Q s2 + 1 s2 +
1 Q
s+1
15–11
15–12
CHAPTER 15. Active Filter Circuits ωo = 104 ; Q = 8; ωo Replace s with s/kf .
[b] kf=
H (s) = =
s 104
2
s 104
2
+
1 8
+1
s 104 8
+1
s2 + 10 s2 + 1250s + 108
P 15.15 For prototype circuit (a): H(s) =
Vo Q Q = 1 = Vi Q + s2s+1 Q + s+ 1 s
H(s) =
s2 + 1 Q(s2 + 1) = Q(s2 + 1) + s s2 + Q1 s + 1
For prototype circuit (b): H(s) = =
Vo 1 = Vi 1 + (s(s/Q) 2 +1) s2 + 1 s2 +
1 Q
s+1
P 15.16 From the solution to Problem 14.15, ωo = 100 krad/s and β = 12.5 krad/s. Compute the two scale factors: kf =
ωo 2π(200 × 103 ) = = 4π ωo 100 × 103
km =
1 C 1 10 × 10−9 1 = = −9 kf C 4π 2.5 × 10 π
Thus, R = km R =
8000 = 2546.48 Ω π
L =
km 1/π (10 × 10−3 ) = 253.303 µH L= kf 4π
Calculate the cutoff frequencies: = kf ωc1 = 4π(93.95 × 103 ) = 1180.6 krad/s ωc1 = kf ωc2 = 4π(106.45 × 103 ) = 1337.7 krad/s ωc2
To check, calculate the bandwidth: − ωc1 = 157.1 krad/s = 4πβ (Checks!) β = ωc2
Problems
15–13
P 15.17 From the solution to Problem 14.24, ωo = 106 rad/s and β = 2π(10.61) krad/s. Calculate the scale factors: kf =
ωo 50 × 103 = = 0.05 ωo 106
km =
k f L 0.05(200 × 10−6 ) = 0.2 = L 50 × 10−6
Thus, R = km R = (0.2)(750) = 150 Ω
C =
C 20 × 10−9 = 2 µF = km kf (0.2)(0.05)
Calculate the bandwidth: β = kf β = (0.05)[2π(10.61 × 103 )] = 3333 rad/s To check, calculate the quality factor: Q=
106 ωo = = 15 β 2π(10.61 × 103 )
Q =
ωo 50 × 103 = 15 (Checks) = β 3333
P 15.18 [a] km = L =
[b]
1000 R = = 1000; R 1
kf =
C 1 = 5000 = km C (1000)(200 × 10−9 )
km 1000 (1) = 200 mH (L) = kf 5000
V V V − 10/s + + =0 1000 0.2s 1000 + (5 × 106 /s)
V
5 s 1 + + 1000 s 1000s + 5 × 106
V =
2s2
=
1 100s
5(s + 5000) 10(s + 5000) = 2 6 + 10,000s + 25 × 10 s + 5000s + 12.5 × 106
15–14
CHAPTER 15. Active Filter Circuits Io =
25(s + 5000) V = 2 0.2s s(s + 5000s + 12.5 × 106 ) =
K1 K2 K2∗ + + s s + 2500 − j2500 s + 2500 + j2500
K1 = 0.01;
K2 = −0.005
io (t) = 10 − 10e−2500t cos 2500t mA Since km = 1000 and the source voltage didn’t change, the amplitude of the current is reduced by a factor of 1000. Since kf = 5000 the coefficients of t are multiplied by 5000. 5000 R P 15.19 km = = = 100; R 50 C =
C 4 × 10−3 = 8 nF = km kf (100)(5000)
50 Ω → 5 kΩ; L =
ωo kf = = 5000 ωo
700 Ω → 70 kΩ
km 100 (20) = 0.4 H L= kf 5000
0.05vφ →
0.05 vφ = 5 × 10−4 vφ 100
The original expression for the current: io (t) = 1728 + 2880e−20t cos(15t + 126.87◦ ) mA The frequency components will be multiplied by kf = 5000: 20 → 20(5000) = 105 ;
15 → 15(5000) = 75,000
The magnitudes will be reduced by km = 100: 1728 → 1728/100 = 17.28;
2880 → 2880/100 = 28.80
The expression for the current in the scaled circuit is thus, 5
io (t) = 17.28 + 28.80e−10 t cos(75,000t + 126.87◦ ) mA
Problems P 15.20 [a] From Eq 15.1 we have −Kωc H(s) = s + ωc where K =
R2 , R1
ωc =
.·. H (s) =
−K ωc s + ωc
where K =
R2 R1
1 R2 C
ωc =
1 R2 C
By hypothesis R1 = km R1 ; and C =
R2 = km R2 ,
C . It follows that kf km
K = K and ωc = kf ωc , therefore H (s) =
−Kkf ωc −Kωc =s s + kf ωc + ωc kf
[b] H(s) =
−K (s + 1) −K
[c] H (s) = s kf
+1
=
−Kkf s + kf
P 15.21 [a] From Eq. 15.4 R2 −Ks where K = and H(s) = s + ωc R1 ωc =
1 R1 C
.·. H (s) = and ωc =
−K s R2 where K = s + ωc R1
1 R1 C
By hypothesis R1 = km R1 ;
R2 = km R2 ;
C =
It follows that K = K and ωc = kf ωc .·. H (s) =
−Ks −K(s/kf ) = s s + kf ωc + ωc kf
C km kf
15–15
15–16
CHAPTER 15. Active Filter Circuits
[b] H(s) =
−Ks (s + 1) −K(s/kf )
[c] H (s) = P 15.22 [a] Hhp =
s kf
+1
−s ; s+1
= .·. Hhp
.·. RH =
1 = 1.59 kΩ (2000π)(0.1 × 10−6 )
5000(2π) ωo kf = = = 10,000π ω 1
−10,000π s + 10,000π
1 = 10,000π; RL CL
=
1000(2π) ωo = = 2000π ω 1
−s s + 2000π
−1 Hlp = ; s+1
[b] H (s) =
−Ks (s + kf )
kf =
1 = 2000π; RH CH
= .·. Hlp
=
.·. RL =
1 = 318.3 Ω (10,000π)(0.1 × 10−6 )
−10,000π −s · s + 2000π s + 10,000π 10,000πs (s + 2000π)(s + 10,000π)
√ (2000π)(10,000π) = 1000π 20 rad/s √ 20) (10,000π)(j1000π √ √ H (jωo ) = (2000π + j1000π 20)(10,000π + j1000π 20) √ j10 20 √ √ = 0.8333/0◦ = (2 + j 20)(10 + j 20)
[c] ωo =
√
ωc1 ωc1 =
Problems
15–17
[d] G = 20 log10 (0.8333) = −1.58 dB [e]
P 15.23 [a] For the high-pass section: kf =
4000(2π) ωo = = 8000π ω 1
H (s) = .·.
−s s + 8000π
1 = 8000π; R1 (10 × 10−9 )
R1 = 3.98 kΩ
.·.
R2 = 3.98 kΩ
For the low-pass section: kf =
400(2π) ωo = = 800π ω 1
H (s) = .·.
−800π s + 800π
1 = 800π; R2 (10 × 10−9 )
R2 = 39.8 kΩ
.·.
R1 = 39.8 kΩ
0 dB gain corresponds to K = 1. In the summing amplifier we are free to choose Rf and Ri so long as Rf /Ri = 1. To keep from having many different resistance values in the circuit we opt for Rf = Ri = 39.8 kΩ.
15–18
CHAPTER 15. Active Filter Circuits
[b]
[c] H (s) =
800π s + s + 8000π s + 800π
s2 + 1600πs + 64 × 105 π 2 (s + 800π)(s + 8000π) √ [d] ωo = (8000π)(800π) = 800π 10 √ √ √ −(800π 10)2 + 1600π(j800π 10) + 64 × 105 π 2 √ √ H (j800π 10) = (800π + j800π 10)(8000π + j800π 10) √ j128 × 104 10π 2 √ √ = (800π)2 (1 + j 10)(10 + j 10) √ j2 10 √ √ = (1 + j 10)(10 + j 10) =
= 0.1818/0◦ [e] G = 20 log10 0.1818 = −14.81 dB
Problems
15–19
[f]
P 15.24 [a] H(s) =
(1/RC) (1/sC) = R + (1/sC) s + (1/RC)
H(jω) =
(1/RC) jω + (1/RC)
|H(jω)| = |H(jω)|2 =
(1/RC) ω 2 + (1/RC)2
(1/RC)2 ω 2 + (1/RC)2
[b] Let Va be the voltage across the capacitor, positive at the upper terminal. Then Va Va − Vi =0 + sCVa + R1 R2 + sL Solving for Va yields Va =
R1
LCs2
(R2 + sL)Vi + (R1 R2 C + L)s + (R1 + R2 )
But Vo =
sLVa R2 + sL
Therefore Vo =
R1
H(s) =
LCs2 R1
H(jω) =
sLVi + (L + R1 R2 C)s + (R1 + R2 )
LCs2
sL + (L + R1 R2 C)s + (R1 + R2 )
jωL [(R1 + R2 ) − R1 LCω 2 ] + jω(L + R1 R2 C)
CHAPTER 15. Active Filter Circuits
15–20
|H(jω)| = |H(jω)|2 = =
ωL [R1 + R2 − R1 LCω 2 ]2 + ω 2 (L + R1 R2 C)2
ω 2 L2 (R1 + R2 − R1 LCω 2 )2 + ω 2 (L + R1 R2 C)2 ω 2 L2 R12 L2 C 2 ω 4 + (L2 + R12 R22 C 2 − 2R12 LC + 2R1 R2 LC)ω 2 + (R1 + R2 )2
[c] Let Va be the voltage across R2 positive at the upper terminal. Then Va Va − Vi + + Va sC + Va sC = 0 R1 R2 (0 − Va )sC + (0 − Va )sC + .·. Va =
0 − Vo =0 R3
R2 Vi 2R1 R2 Cs + R1 + R2
and Va = −
Vo 2R3 Cs
It follows directly that H(s) =
−2R2 R3 Cs Vo = Vi 2R1 R2 Cs + (R1 + R2 )
.·. H(jω) =
−2R2 R3 C(jω) (R1 + R2 ) + jω(2R1 R2 C)
|H(jω)| = |H(jω)|2 =
2R2 R3 Cω (R1 + R2 )2 + ω 2 4R12 R22 C 2
4R22 R32 C 2 ω 2 (R1 + R2 )2 + 4R12 R22 C 2 ω 2
P 15.25 ωo = 2πfo = 400π rad/s β = 2π(1000) = 2000π rad/s .·. ωc2 − ωc1 = 2000π √
ωc1 ωc2 = ωo = 400π
Solve for the cutoff frequencies: ωc1 ωc2 = 16 × 104 π 2
Problems ωc2 = .·.
16 × 104 π 2 ωc1
16 × 104 π 2 − ωc1 = 2000π ωc1
or ωc21 + 2000πωc1 − 16 × 104 π 2 = 0 ωc1 = −1000π ±
√
ωc1 = 1000π(−1 ±
106 π 2 + 0.16 × 106 π 2 √
1.16) = 242.01 rad/s
.·. ωc2 = 2000π + 242.01 = 6525.19 rad/s Thus, fc1 = 38.52 Hz
and
fc2 = 1038.52 Hz
β = fc2 − fc1 = 1000Hz
Check: ωc2 =
1 = 6525.19 RL CL
RL =
1 = 30.65 Ω (6525.19)(5 × 10−6 )
ωc1 =
1 = 242.01 RH CH
RH =
1 = 826.43 Ω (242.01)(5 × 10−6 )
P 15.26 ωo = 1000 rad/s; β = 4000 rad/s;
GAIN = 6 C = 0.2 µF
β = ωc2 − ωc1 = 4000 ωo =
√
ωc1 ωc2 = 1000
Solve for the cutoff frequencies: .·. ωc21 + 4000ωc1 − 106 = 0 √ ωc1 = −2000 ± 1000 5 = 236.07 rad/s
15–21
15–22
CHAPTER 15. Active Filter Circuits
ωc2 = 4000 + ωc1 = 4236.07 rad/s β = ωc2 − ωc1 = 4000 rad/s
Check: ωc1 =
1 RL CL
.·. RL =
1 (0.2 ×
10−6 )(236.07)
= 21.81 kΩ
1 = 4236.07 RH CH RH =
1 (0.2 ×
10−6 )(4236.07)
= 1.18 kΩ
Rf =6 Ri If Ri = 1 kΩ
Rf = 6Ri = 6 kΩ
1 = −10 log10 (1 + ω 2n ) 2n 1+ω From the laws of logarithms we have
P 15.27 [a] y = 20 log10 √
−10 y= ln(1 + ω 2n ) ln 10 Thus
−10 2nω 2n−1 dy = dω ln 10 (1 + ω 2n ) x = log10 ω =
ln ω ln 10
.·. ln ω = x ln 10 1 dω = ln 10, ω dx dy = dx
dy dω
dω = ω ln 10 dx dω dx
=
−20nω 2n dB/decade 1 + ω 2n
at ω = ωc = 1 rad/s dy = −10n dB/decade. dx
Problems
15–23
1 = −10n log10 (1 + ω 2 ) [b] y = 20 log10 √ [ 1 + ω 2 ]n =
−10n ln(1 + ω 2 ) ln 10
−10n dy 1 −20nω = 2ω = 2 dω ln 10 1 + ω (ln 10)(1 + ω 2 ) As before dω = ω(ln 10); dx
.·.
At the corner ωc =
√
dy −20nω 2 = dx (1 + ω 2 ) .·. ωc2 = 21/n − 1
21/n − 1
−20n[21/n − 1] dy = dB/decade. dx 21/n [c] For the Butterworth Filter
For the cascade of identical sections
n
dy/dx (dB/decade)
n
dy/dx (dB/decade)
1
−10
1
−10
2
−20
2
−11.72
3
−30
3
−12.38
4
−40
4
−12.73
∞
−∞
∞
−13.86
[d] It is apparent from the calculations in part (c) that as n increases the amplitude characteristic at the cutoff frequency decreases at a much faster rate for the Butterworth filter. Hence the transition region of the Butterworth filter will be much narrower than that of the cascaded sections. P 15.28 [a] n ∼ =
(−0.05)(−30) ∼ = 2.76 log10 (7000/2000)
.·. n = 3 1
[b] Gain = 20 log10
1 + (7000/2000)6
= −32.65 dB
P 15.29 [a] For the scaled circuit H (s) =
1/(R )2 C1 C2 s2 + R2C s + (R )21C C 1
1
2
where R = km R;
C1 = C1 /kf km ;
C2 = C2 /kf km
15–24
CHAPTER 15. Active Filter Circuits It follows that kf2 1 = (R )2 C1 C2 R2 C1 C2 2 R C1
=
2kf RC1
.·. H (s) =
kf2 /R2 C1 C2 s2 +
= 2 s kf
k2
2kf s RC1
+ R2 Cf1 C2 1/R2 C1 C2
+
2 RC1
s kf
+
1 R 2 C1 C2
1 (s + 1)(s2 + s + 1) [b] fc = 2000 Hz; ωc = 4000π rad/s;
P 15.30 [a] H(s) =
H (s) =
kf = 4000π
1 ( ksf
+
1)[( ksf )2
+
s kf
+ 1]
kf3 = (s + kf )(s2 + kf s + kf2 ) =
(4000π)3 (s + 4000π)[s2 + 4000πs + (4000π)2 ]
[c] H (j14,000π) =
64 (4 + j14)(−180 + j56)
= 0.02332/− 236.77◦ Gain = 20 log10 (0.02332) = −32.65 dB P 15.31 [a] In the first-order circuit R = 1 Ω and C = 1 F. km =
1000 R = = 1000; R 1
R = km R = 1000 Ω;
kf = C =
ωo 2π(2000) = 4000π = ωo 1
C 1 = 79.58 nF = km kf (1000)(4000π)
In the second-order circuit R = 1 Ω, 2/C1 = 1 so C1 = 2 F, and C2 = 1/C1 = 0.5 F. Therefore in the scaled second-order circuit R = km R = 1000 Ω; C2 =
C1 =
C1 2 = 159.15 nF = km kf (1000)(4000π)
C2 0.5 = 39.79 nF = km kf (1000)(4000π)
Problems [b]
(−0.05)(−48) = 3.99 .·. n = 4 log10 (2000/500) From Table 15.1 the transfer function of the first section is
P 15.32 [a] n =
s2 H1 (s) = 2 s + 0.765s + 1 For the prototype circuit 2 = 0.765; R2
R2 = 2.61 Ω;
R1 =
1 = 0.383 Ω R2
The transfer function of the second section is s2 s2 + 1.848s + 1 For the prototype circuit H2 (s) =
2 = 1.848; R2
R2 = 1.082 Ω;
R1 =
1 = 0.9240 Ω R2
The scaling factors are: kf =
ωo 2π(2000) = 4000π = ωo 1
C =
C km kf
.·.
km =
.·.
10 × 10−9 =
1 4000πkm
1 = 7957.75 4000π(10 × 10−9 )
Therefore in the first section R1 = km R1 = 3.04 kΩ;
R2 = km R2 = 20.80 kΩ
In the second section R1 = km R1 = 7.35 kΩ;
R2 = km R2 = 8.61 kΩ
15–25
CHAPTER 15. Active Filter Circuits
15–26 [b]
P 15.33 n = 5: 1 + (−1)5 s10 = 0;
s10 = 1
s10 = 1/0 + 36◦ k k
sk+1
0
1/0◦
1 1/36◦ 2 1/72◦ 3 1/108◦ 4 1/144◦ 5 1/180◦ 6 1/216◦ 7 1/252◦ 8 1/288◦ 9 1/324◦
Group by conjugate pairs to form denominator polynomial. (s + 1)[s − (cos 108◦ + j sin 108◦ )][s − (cos 252◦ + j sin 252◦ )] · [s − (cos 144◦ + j sin 144◦ )][s − (cos 216◦ + j sin 216◦ )] = (s + 1)(s + 0.309 − j0.951)(s + 0.309 + j0.951)· (s + 0.809 − j0.588)(s + 0.809 + j0.588) which reduces to (s + 1)(s2 + 0.618s + 1)(s2 + 1.618s + 1)
Problems n = 6: 1 + (−1)6 s12 = 0
s12 = −1
s12 = 1/15◦ + 36◦ k k
sk+1
0 1/15◦ 1 1/45◦ 2 1/75◦ 3 1/105◦ 4 1/135◦ 5 1/165◦ 6 1/195◦ 7 1/225◦ 8 1/255◦ 9 1/285◦ 10 1/315◦ 11 1/345◦
Grouping by conjugate pairs yields (s + 0.2588 − j0.9659)(s + 0.2588 + j0.9659)× (s + 0.7071 − j0.7071)(s + 0.7071 + j0.7071)× (s + 0.9659 − j0.2588)(s + 0.9659 + j0.2588) or (s2 + 0.518s + 1)(s2 + 1.414s + 1)(s2 + 1.932s + 1) P 15.34 H (s) =
H (s) = =
s2 2 s2 + km R2 (C/k s+ m kf )
1 2 k2 ) km R1 km R2 (C 2 /km f
s2 s2 +
2kf s R2 C
(s/kf )2 +
k2
+ R1 Rf2 C 2 (s/kf )2 2 R2 C
s kf
+
1 R1 R2 C 2
15–27
15–28
CHAPTER 15. Active Filter Circuits (−0.05)(−48) = 3.99 .·. n=4 log10 (32/8) From Table 15.1 the transfer function is 1 H(s) = 2 (s + 0.765s + 1)(s2 + 1.848s + 1)
P 15.35 [a] n =
The capacitor values for the first stage prototype circuit are 2 = 0.765 C1 C2 =
.·.
C1 = 2.61 F
1 = 0.38 F C1
The values for the second stage prototype circuit are 2 = 1.848 C1 C2 =
.·.
C1 = 1.08 F
1 = 0.92 F C1
The scaling factors are km =
R = 1000; R
kf =
ωo = 16,000π ωo
Therefore the scaled values for the components in the first stage are R1 = R2 = R = 1000 Ω C1 =
2.61 = 52.01 nF (16,000π)(1000)
C2 =
0.38 = 7.61 nF (16,000π)(1000)
The scaled values for the second stage are R1 = R2 = R = 1000 Ω C1 =
1.08 = 21.53 nF (16,000π)(1000)
C2 =
0.92 = 18.38 nF (16,000π)(1000)
Problems
15–29
[b]
P 15.36 [a] The cascade connection is a bandpass filter. [b] The cutoff frequencies are 2 kHz and 8 kHz. The center frequency is (2)(8) = 4 kHz. The Q is 4/(8 − 2) = 2/3 = 0.67 [c] For the high pass section kf = 4000π. The prototype transfer function is Hhp (s) = .·.
s4 (s2 + 0.765s + 1)(s2 + 1.848s + 1)
(s) Hhp
(s/4000π)4 = [(s/4000π)2 + 0.765(s/4000π) + 1] ·
=
[(s/4000π)2
1 + 1.848(s/4000π) + 1]
s4 (s2 + 3060πs + 16 × 106 π 2 )(s2 + 7392πs + 16 × 106 π 2 )
For the low pass section kf = 16,000π Hlp (s) = .·.
(s2
1 + 0.765s + 1)(s2 + 1.848s + 1)
(s) = Hlp
· =
1 [(s/16,000π)2 + 0.765(s/16,000π) + 1] [(s/16,000π)2
1 + 1.848(s/16,000π) + 1]
(16,000π)4 ([s2 + 12,240πs + (16,000π)2 )][s2 + 29,568πs + (16,000π)2 ]
The cascaded transfer function is H (s) = Hhp (s)Hlp (s)
15–30
CHAPTER 15. Active Filter Circuits For convenience let D1 = s2 + 3060πs + 16 × 106 π 2 D2 = s2 + 7392πs + 16 × 106 π 2 D3 = s2 + 12,240πs + 256 × 106 π 2 D4 = s2 + 29,568πs + 256 × 106 π 2 Then H (s) =
65,536 × 1012 π 4 s4 D1 D2 D3 D4
[d] ωo = 2π(4000) = 8000π rad/s s = j8000π s4 = 4096 × 1012 π 4 D1 = (16 × 106 π 2 − 64 × 106 π 2 ) + j(8000π)(3060π) = 106 π 2 (−48 + j24.48) = 106 π 2 (53.88/152.98◦ ) D2 = (16 × 106 π 2 − 64 × 106 π 2 ) + j(8000π)(7392π) = 106 π 2 (−48 + j59.136) = 106 π 2 (76.16/129.07◦ ) D1 = (256 × 106 π 2 − 64 × 106 π 2 ) + j(8000π)(12,240π) = 106 π 2 (192 + j97.92) = 106 π 2 (215.53/27.02◦ ) D1 = (256 × 106 π 2 − 64 × 106 π 2 ) + j(8000π)(29,568π) = 106 π 2 (192 + j236.544) = 106 π 2 (304.66/50.93◦ ) H (jωo ) =
(65,536)(4096)π 8 × 1024 (π 8 × 1024 )[(53.88)(76.16)(215.53)(304.66)/360◦ ]
= 0.996/− 360◦ = 0.996/0◦ P 15.37 [a] From the statement of the problem, K = 10 ( = 20 dB). Therefore for the prototype bandpass circuit R1 = R2 =
16 Q = = 1.6 Ω K 10 16 Q = Ω −K 502
2Q2
R3 = 2Q = 32 Ω
Problems The scaling factors are kf =
ωo = 2π(6400) = 12,800π ωo
km =
C 1 = 1243.40 = −9 C kf (20 × 10 )(12,800π)
Therefore, R1 = km R1 = (1.6)(1243.40) = 1.99 kΩ R2 = km R2 = (16/502)(1243.40) = 39.63 Ω R3 = km R3 = 32(1243.40) = 39.79 kΩ [b]
P 15.38 From Eq 15.58 we can write H(s) =
−
2 R3 C
s2 +
R3 C 2
2 s R3 C
+
1 s R1 C R1 +R2 R1 R2 R3 C 2
or H(s) =
− s2 +
R3 2R1 2 s R3 C
+
2 s R3 C R1 +R2 R1 R2 R3 C 2
Therefore ωo 2 =β= ; R3 C Q and K =
R1 + R2 = ωo2 ; R1 R2 R3 C 2
R3 2R1
By hypothesis C = 1 F and ωo = 1 rad/s .·.
2 1 or R3 = 2Q = R3 Q
15–31
15–32
CHAPTER 15. Active Filter Circuits
R1 =
Q R3 = 2K K
R1 + R2 =1 R1 R2 R3
Q Q + R2 = (2Q)R2 K K .·. R2 =
Q −K
2Q2
P 15.39 [a] First we will design a unity gain filter and then provide the passband gain with an inverting amplifier. For the high pass section the cut-off frequency is 500 Hz. The order of the Butterworth is (−0.05)(−20) n= = 2.51 log10 (500/200) .·. n = 3 Hhp (s) =
s3 (s + 1)(s2 + s + 1)
For the prototype first-order section R1 = R2 = 1 Ω,
C = 1F
For the prototype second-order section R1 = 0.5 Ω,
R2 = 2 Ω,
C = 1F
The scaling factors are kf =
ωo = 2π(500) = 1000π ωo
km =
106 C 1 = = C kf (15 × 10−9 )(1000π) 15π
In the scaled first-order section R1 = R2 = km R1 =
106 (1) = 21.22 kΩ 15π
C = 15 nF In the scaled second-order section R1 = 0.5km = 10.61 kΩ R2 = 2km = 42.44 kΩ
Problems
15–33
C = 15 nF For the low-pass section the cut-off frequency is 4500 Hz. The order of the Butterworth filter is (−0.05)(−20) n= = 2.51; .·. n = 3 log10 (11,250/4500) Hlp (s) =
1 (s + 1)(s2 + s + 1)
For the prototype first-order section R1 = R2 = 1 Ω,
C = 1F
For the prototype second-order section R1 = R2 = 1 Ω;
C1 = 2 F;
C2 = 0.5 F
The low-pass scaling factors are km =
R = 104 ; R
kf =
ωo = (4500)(2π) = 9000π ωo
For the scaled first-order section C 1 = 3.54 nF C = = R1 = R2 = 10 kΩ; kf km (9000π)(104 ) For the scaled second-order section R1 = R2 = 10 kΩ C1 =
C1 2 = 7.07 nF = kf km (9000π)(104 )
C2 =
C2 0.5 = 1.77 nF = kf km (9000π)(104 )
GAIN AMPLIFIER 20 log10 K = 20 dB,
.·. K = 10
Since we are using 10 kΩ resistors in the low-pass stage, we will use Rf = 100 kΩ and Ri = 10 kΩ in the inverting amplifier stage.
15–34
CHAPTER 15. Active Filter Circuits
[b]
P 15.40 [a] Unscaled high-pass stage s3 Hhp (s) = (s + 1)(s2 + s + 1) The frequency scaling factor is kf = (ωo /ωo ) = 1000π. Therefore the scaled transfer function is (s) = Hhp
=
(s/1000π)3
s 1000π
(s +
+1
s 1000π 3
1000π)[s2
2
+
s 1000π
+1
s + 1000πs + 106 π 2 ]
Unscaled low-pass stage Hlp (s) =
1 (s + 1)(s2 + s + 1)
The frequency scaling factor is kf = (ωo /ωo ) = 9000π. Therefore the scaled transfer function is 1
Hlp (s) = 2 s s s +1 + 9000π + 1 9000π 9000π =
(9000π)3 (s + 9000π)(s2 + 9000πs + 81 × 106 π 2 )
Thus the transfer function for the filter is (s)Hlp (s) = H (s) = 10Hhp
729 × 1010 π 3 s3 D1 D2 D3 D4
Problems where D1 = s + 1000π D2 = s + 9000π D3 = s2 + 1000πs + 106 π 2 D4 = s2 + 9000πs + 81 × 106 π 2 [b] At f = 200 Hz
ω = 400π rad/s
D1 (j400π) = 400π(2.5 + j1) D2 (j400π) = 400π(22.5 + j1) D3 (j400π) = 4 × 105 π 2 (2.1 + j1.0) D4 (j400π) = 4 × 105 π 2 (202.1 + j9) Therefore D1 D2 D3 D4 (j400π) = 256π 6 1014 (28,534.82/52.36◦ ) H (j400π) =
(729π 3 × 1010 )(64 × 106 π 3 ) 256π 6 × 1014 (28,534.82/52.36◦ )
= 0.639/− 52.36◦ .·. 20 log10 |H (j400π)| = 20 log10 (0.639) = −3.89 dB At f = 1500 Hz,
ω = 3000π rad/s
Then D1 (j3000π) = 1000π(1 + j3) D2 (j3000π) = 3000π(3 + j1) D3 (j3000π) = 106 π 2 (−8 + j3) D4 (j3000π) = 9 × 106 π 2 (8 + j3) H (j3000π) =
(729 × π 3 × 1010 )(27 × 109 π 3 ) 27 × 1018 π 6 (730/270◦ )
= 9.99/90◦ .·. 20 log10 |H (j3000π)| = 19.99 dB
15–35
15–36
CHAPTER 15. Active Filter Circuits
[c] From the transfer function the gain is down 19.99 + 3.89 or 23.88 dB at 200 Hz. Because the upper cut-off frequency is nine times the lower cut-off frequency we would expect the high-pass stage of the filter to predict the loss in gain at 200 Hz. For a 3nd order Butterworth 1 GAIN = 20 log10 = −23.89 dB. 1 + (500/200)6 1500 Hz is in the passband for this bandpass filter, and is in fact the center frequency. Hence we expect the gain at 1500 Hz to equal, or nearly equal, 20 dB as specified in Problem 15.39. Thus our scaled transfer function confirms that the filter meets the specifications. P 15.41 [a] From Table 15.1 Hlp (s) =
1 √ (s2 + 0.518s + 1)(s2 + 2s + 1)(s2 + 1.932s + 1)
Hhp (s) = 1
s2
Hhp (s) =
+ 0.518
1 s
+1
1 s2
1 √ 1 + 2 s + 1 s12 + 1.932 1s + 1
s6 √ (s2 + 0.518s + 1)(s2 + 2s + 1)(s2 + 1.932s + 1)
P 15.42 [a] kf = 25,000 (s/25,000)6 [(s/25,000)2 + 0.518(s/25,000) + 1]
(s) = Hhp
· =
[(s/25,000)2
s6 (s2 + 12,950s + 625 × 106 )(s2 + 35,350s + 625 × 106 ) ·
(s2
[b] H (j25,000) = =
1 + 1.414s/25,000 + 1][(s/25,000)2 + 1.932s/25,000 + 1]
1 + 48,300s + 625 × 106 ) −(25,000)6 [12,950(j25,000)][35,350(j25,000)][48,300(j25,000)]
−(25,000)3 (12,950)(25,350)(48,300)j 3
= 0.7067/− 90◦ 20 log10 |H (j25,000)| = −3.02 dB
Problems
15–37
P 15.43 [a] At very low frequencies the two capacitor branches are open and because the op amp is ideal the current in R3 is zero. Therefore at low frequencies the circuit behaves as an inverting amplifier with a gain of R2 /R1 . At very high frequencies the capacitor branches are short circuits and hence the output voltage is zero. [b] Let the node where R1 , R2 , R3 , and C2 join be denoted as a, then (Va − Vi )G1 + Va sC2 + (Va − Vo )G2 + Va G3 = 0 −Va G3 − Vo sC1 = 0 or (G1 + G2 + G3 + sC2 )Va − G2 Vo = G1 Vi Va =
−sC1 Vo G3
Solving for Vo /Vi yields H(s) = = = = =
−G1 G3 (G1 + G2 + G3 + sC2 )sC1 + G2 G3 s2 C1 C2 s2 + s2 + s2
where K = and b1 =
−G1 G3 + (G1 + G2 + G3 )C1 s + G2 G3
−G1 G3 /C1 C2
(G1 +G2 +G3 ) s C2 1 G2 G3 −G G2 C1 C2
(G1 +G2 +G3 ) s C2
+
G2 G3 C1 C2
+
G2 G3 C1 C2
−Kbo + b1 s + bo G1 ; G2
bo =
G2 G3 C1 C2
G1 + G2 + G3 C2
[c] Equating coefficients we see that G1 = KG2 G3 =
bo C1 C2 bo C1 = G2 G2
since by hypothesis C2 = 1 F b1 =
G1 + G2 + G3 = G1 + G2 + G3 C2
15–38
CHAPTER 15. Active Filter Circuits .·. b1 = KG2 + G2 +
bo C1 G2
b1 = G2 (1 + K) +
bo C1 G2
Solving this quadratic equation for G2 we get b1 G2 = ± 2(1 + K) =
b1 ±
b21 − bo C1 4(1 + K) 4(1 + K)2
b21 − 4bo (1 + K)C1 2(1 + K)
For G2 to be realizable b21 4bo (1 + K)
C1 <
[d] 1. Select C2 = 1 F 2. Select C1 such that C1 <
b21 4bo (1 + K)
3. Calculate G2 (R2 ) 4. Calculate G1 (R1 ); G1 = KG2 5. Calculate G3 (R3 ); G3 = bo C1 /G2 P 15.44 [a] In the second order section of a third order Butterworth filter bo = b1 = 1 Therefore, C1 ≤
b21 1 = = 0.05 F 4bo (1 + K) (4)(1)(5)
.·.
C1 = 0.05 F
[b] G2 =
(limiting value)
1 = 0.1 S 2(1 + 4)
G3 =
1 (0.05) = 0.5 S 0.1
G1 = 4(0.1) = 0.4 S Therefore, 1 R1 = = 2.5 Ω; G1
R2 =
1 = 10 Ω; G2
R3 =
1 = 2Ω G3
Problems [c] kf =
15–39
ωo = 2π(2500) = 5000π ωo
km =
C2 1 = = 6366.2 C2 kf (10 × 10−9 )kf
C1 =
0.05 = 0.5 × 10−9 = 500 pF kf km
R1 = (2.5)(6366.2) = 15.92 kΩ R2 = (10)(6366.2) = 63.66 kΩ R3 = (2)(6366.2) = 12.73 kΩ [d] R1 = R2 = (6366.2)(1) = 6.37 kΩ C =
C 1 = 8 = 10 nF kf km 10
[e]
P 15.45 [a] By hypothesis the circuit becomes:
For very small frequencies the capacitors behave as open circuits and therefore vo is zero. As the frequency increases, the capacitive branch impedances become small compared to the resistive branches. When this happens the circuit becomes an inverting amplifier with the capacitor C2 dominating the feedback path. Hence the gain of the amplifier approaches (1/jωC2 )/(1/jωC1 ) or C1 /C2 . Therefore the circuit behaves like a high-pass filter with a passband gain of C1 /C2 .
15–40
CHAPTER 15. Active Filter Circuits
[b] Summing the currents away from the upper terminal of R2 yields Va G2 + (Va − Vi )sC1 + (Va − Vo )sC2 + Va sC3 = 0 or Va [G2 + s(C1 + C2 + C3 )] − Vo sC2 = sC1 Vi Summing the currents away from the inverting input terminal gives (0 − Va )sC3 + (0 − Vo )G1 = 0 or sC3 Va = −G1 Vo ;
Va =
−G1 Vo sC3
Therefore we can write −G1 Vo [G2 + s(C1 + C2 + C3 )] − sC2 Vo = sC1 Vi sC3 Solving for Vo /Vi gives H(s) =
−C1 C3 s2 Vo = Vi [C2 C3 s2 + G1 (C1 + C2 + C3 )s + G1 G2 ]
= =
s2 +
s2
G1 (C1 C2 C3 2
−C1 2 s C2
+ C2 + C3 )s +
G1 G2 C2 C3
−Ks + b1 s + bo
Therefore the circuit implements a second-order high-pass filter with a passband gain of C1 /C2 . [c] C1 = K: b1 =
G1 (K + 2) = G1 (K + 2) (1)(1)
.·. G1 = bo =
b1 ; K +2
K +2 R1 = b1
G1 G2 = G1 G2 (1)(1)
.·. G2 =
bo bo = (K + 2) G1 b1
.·. R2 =
b1 bo (K + 2)
Problems
15–41
[d] From Table 15.1 the transfer function of the second-order section of a third-order high-pass Butterworth filter is H(s) =
Ks2 s2 + s + 1
Therefore b1 = bo = 1 Thus C1 = K = 8 F R1 =
8+2 = 10 Ω 1
R2 =
1 = 0.10 Ω 1(8 + 2)
P 15.46 [a] Low-pass filter: n=
(−0.05)(−30) = 3.77; log10 (1000/400)
.·. n = 4
In the first prototype second-order section: b1 = 0.765, bo = 1, C2 = 1 F C1 ≤
(0.765)2 b21 ≤ 0.0732 ≤ 4bo (1 + K) (4)(2)
choose C1 = 0.03 F G2 =
0.765 ±
(0.765)2 − 4(2)(0.03) 4
=
0.765 ± 0.588 4
Arbitrarily select the larger value for G2 , then G2 = 0.338 S;
.·.
G1 = KG2 = 0.338 S; G3 =
R2 = .·.
1 = 2.96 Ω G2 R1 =
bo C1 (1)(0.03) = 0.089 = G2 0.338
.·.
1 = 2.96 Ω G1 R3 = 1/G3 = 11.3 Ω
Therefore in the first second-order prototype circuit R1 = R2 = 2.96 Ω; C1 = 0.03 F;
R3 = 11.3 Ω
C2 = 1 F
In the second second-order prototype circuit: b1 = 1.848, b0 = 1, C2 = 1 F .·. C1 ≤
(1.848)2 ≤ 0.427 8
15–42
CHAPTER 15. Active Filter Circuits choose C1 = 0.30 F G2 =
1.848 ±
(1.848)2 − 8(0.3) 4
=
1.848 ± 1.008 4
Arbitrarily select the larger value, then G2 = 0.7139 S;
.·.
G1 = KG2 = 0.7139 S; G3 =
R2 = .·.
1 = 1.4008 Ω G2 R1 =
bo C1 (1)(0.30) = 0.4202 S = G2 0.7139
1 = 1.4008 Ω G1 .·.
R3 = 1/G3 = 2.3796 Ω
In the low-pass section of the filter kf =
ωo = 2π(400) = 800π ωo
km =
C 1 125,000 = = −9 C kf (10 × 10 )kf π
Therefore in the first scaled second-order section R1 = R2 = 2.96km = 118 kΩ R3 = 11.3km = 450 kΩ C1 =
0.03 = 300 pF kf km
C2 = 10 nF In the second scaled second-order section R1 = R2 = 1.4008km = 55.74 kΩ R3 = 2.3796km = 94.68 kΩ C1 =
0.3 = 3 nF kf km
C2 = 10 nF High-pass filter section n=
(−0.05)(−30) = 3.77; log10 (6400/2560)
n = 4.
In the first prototype second-order section: b1 = 0.765; bo = 1; C2 = C3 = 1 F C1 = K = 1 F
Problems R1 =
K +2 3 = 3.92 Ω = b1 0.765
R2 =
b1 0.765 = = 0.255 Ω bo (K + 2) 3
In the second prototype second-order section: b1 = 1.848; bo = 1; C2 = C3 = 1 F C1 = K = 1 F R1 =
K +2 3 = 1.623 Ω = b1 1.848
R2 =
1.848 b1 = = 0.616 Ω bo (K + 2) 3
In the high-pass section of the filter ωo kf = = 2π(6400) = 12,800π ωo km =
7812.5 C 1 = = −9 C kf (10 × 10 )(12,800π) π
In the first scaled second-order section R1 = 3.92km = 9.75 kΩ R2 = 0.255km = 634 Ω C1 = C2 = C3 = 10 nF In the second scaled second-order section R1 = 1.623km = 4.04 kΩ R2 = 0.616km = 1.53 kΩ C1 = C2 = C3 = 10 nF In the gain section, let Ri = 10 kΩ and Rf = 10 kΩ.
15–43
15–44
CHAPTER 15. Active Filter Circuits
[b]
P 15.47 [a] The prototype low-pass transfer function is Hlp (s) =
1 + 0.765s + 1)(s2 + 1.848s + 1)
(s2
The low-pass frequency scaling factor is kflp = 2π(400) = 800π The scaled transfer function for the low-pass filter is Hlp (s) = =
s 800π
[s2
2
1
+
0.765s 800π
+1
2 s + 1.848s 800π 800π 8 4
+1
4096 × 10 π + 612πs + (800π)2 ] [s2 + 1478.4πs + (800π)2 ]
The prototype high-pass transfer function is Hhp (s) =
s4 (s2 + 0.765s + 1)(s2 + 1.848s + 1)
The high-pass frequency scaling factor is kfhp = 2π(6400) = 12,800π
Problems
15–45
The scaled transfer function for the high-pass filter is (s) = Hhp
=
s 12,800π
2
(s/12,800π)4 +
0.765s 12,800π
+1
s 12,800π 4
2
+
1.848s 12,800π
+1
s
[s2
+ 9792πs +
(12,800π)2 ][s2
+ 23,654.4πs + (12,800π)2 ]
The transfer function for the filter is
H (s) = Hlp (s) + Hhp (s)
[b] fo =
fc1 fc2 =
(400)(6400) = 1600 Hz
ωo = 2πfo = 3200π rad/s (jωo )2 = −1024 × 104 π 2 (jωo )4 = 1,048,576 × 108 π 4 Hlp (jωo ) =
4096 × 108 π 4 × [−960 × 104 π 2 + j612(3200π 2 )] [−960 ×
=
104 π 2
1 + j1478.4(3200π 2 )]
40,000 (−3000 + j612)(−3000 + j1478.4)
= 3906.2 × 10−6 /37.76◦ 1,048,576 × 108 π 4 Hhp (jωo ) = [15,360 × 104 π 2 + j9792(3200π 2 )] 1 [15,360 × 104 π 2 + j23,654.4(3200π 2 )] =
10.24 × 106 (48,000 + j9792)(48,000 + j23,654.4)
= 3906.2 × 10−6 /− 37.76◦ .·. H (jωo ) = −3906.2 × 10−6 (1/37.76◦ + 1/− 37.76◦ ) = −3906.2 × 10−6 (1.58/0◦ ) = −6176.35 × 10−6 /0◦ G = 20 log10 |H (jωo )| = 20 log10 (6176.35 × 10−6 ) = −44.19 dB P 15.48 [a] At low frequencies the capacitor branches are open; vo = vi . At high frequencies the capacitor branches are short circuits and the output voltage is zero. Hence the circuit behaves like a unity-gain low-pass filter.
15–46
CHAPTER 15. Active Filter Circuits
[b] Let va represent the voltage-to-ground at the right-hand terminal of R1 . Observe this will also be the voltage at the left-hand terminal of R2 . The s-domain equations are (Va − Vi )G1 + (Va − Vo )sC1 = 0 (Vo − Va )G2 + sC2 Vo = 0 or (G1 + sC1 )Va − sC1 Vo = G1 Vi −G2 Va + (G2 + sC2 )Vo = 0 .·. Va = .·. .·.
G2 + sC2 Vo G2
(G2 + sC2 ) (G1 + sC1 ) − sC1 Vo = G1 Vi G2
G1 G2 Vo = Vi (G1 + sC1 )(G2 + sC2 ) − C1 G2 s
which reduces to Vo G1 G2 /C1 C2 bo = 2 G1 G1 G2 = 2 Vi s + b1 s + bo s + C1 s + C1 C2 [c] There are four circuit components and two restraints imposed by H(s); therefore there are two free choices. G1 · [d] b1 = . . G1 = b1 C1 C1 bo =
G1 G2 · bo . . G2 = C2 C1 C2 b1
[e] No, all physically realizeable capacitors will yield physically realizeable resistors. [f] From Table 15.1 we know the transfer function of the prototype 4th order Butterworth filter is 1 H(s) = 2 (s + 0.765s + 1)(s2 + 1.848s + 1) In the first section bo = 1,
b1 = 0.765
.·. G1 = (0.765)(1) = 0.765 S R1 = 1/G1 = 1.307 Ω G2 =
1 (1) = 1.307 S 0.765
Problems R2 = 1/G2 = 0.765 Ω In the second section bo = 1,
b1 = 1.848
.·. G1 = 1.848 S R1 = 1/G1 = 0.541 Ω G2 =
1 (1) = 0.541 S 1.848
R2 = 1/G2 = 1.848 Ω
P 15.49 [a] kf =
ωo = 2π(3000) = 6000π ωo
km =
106 C 1 = = C kf (4.7 × 10−9 )(6000π) 28.2π
In the first section R1 = 1.307km = 14.75 kΩ R2 = 0.765km = 8.64 kΩ In the second section R1 = 0.541km = 6.11 kΩ R2 = 1.848km = 20.86 kΩ
15–47
15–48
CHAPTER 15. Active Filter Circuits
[b]
P 15.50 [a] Interchanging the Rs and Cs yields the following circuit.
At low frequencies the capacitors appear as open circuits and hence the output voltage is zero. As the frequency increases the capacitor branches approach short circuits and va = vi = vo . Thus the circuit is a unity-gain, high-pass filter. [b] The s-domain equations are (Va − Vi )sC1 + (Va − Vo )G1 = 0 (Vo − Va )sC2 + Vo G2 = 0 It follows that Va (G1 + sC1 ) − G1 Vo = sC1 Vi and Va = Thus
(G2 + sC2 )Vo sC2
(G2 + sC2 ) (G1 + sC1 ) − G1 Vo = sC1 Vi sC2
Vo {s2 C1 C2 + sC1 G2 + G1 G2 } = s2 C1 C2 Vi
Problems H(s) =
Vo = Vi
s2
G2 G1 G2 + s+ C2 C1 C2 Vo s2 = = 2 Vi s + b1 s + bo
s2
[c] There are 4 circuit components: R1 , R2 , C1 and C2 . There are two transfer function constraints: b1 and bo . Therefore there are two free choices. G1 G2 G2 [d] bo = ; b1 = C1 C2 C2 .·. G2 = b1 C2 ; G1 =
R2 =
1 b1 C2
bo b1 C1 .·. R1 = b1 bo C1
[e] No, all realizeable capacitors will produce realizeable resistors. [f] The second-order section in a 3rd-order Butterworth high-pass filter is s2 /(s2 + s + 1). Therefore bo = b1 = 1 and R1 =
1 = 1 Ω. (1)(1)
R2 =
1 = 1 Ω. (1)(1)
P 15.51 [a] kf =
ωo = 104 π ωo
km =
C 1 105 = = C kf (75 × 10−9 )(104 π) 75π
C1 = C2 = 75 nF; [b] R = 424.4 Ω;
R1 = R2 = km R = 424.4 Ω
C = 75 nF
15–49
15–50
CHAPTER 15. Active Filter Circuits
[c]
s3 (s + 1)(s2 + s + 1)
[d] Hhp (s) =
Hhp (s) =
(s/104 π)3 [(s/104 π) + 1][(s/104 π)2 + (s/104 π) + 1]
s3 = (s + 104 π)(s2 + 104 πs + 108 π 2 ) [e] Hhp (j104 π) =
.·.
(j104 π)3 = 0.7071/135◦ (j104 π + 104 π)[(j104 π)2 + 104 π(j104 π) + 108 π 2 ]
| = 0.7071 = −3 dB |Hhp
P 15.52 [a] It follows directly from Eq 15.64 that s2 + 1 H(s) = 2 s + 4(1 − σ)s + 1 Now note from Eq 15.69 that (1 − σ) equals 1/4Q, hence H(s) =
s2 + 1 s2 + Q1 s + 1
[b] For Example 15.13, ωo = 5000 rad/s and Q = 5. Therefore kf = 5000 and (s/5000)2 + 1 s 1 2 (s/5000) + +1 5 5000 2 6 s + 25 × 10 = 2 s + 1000s + 25 × 106
H (s) =
P 15.53 [a] ωo = 2000π rad/s .·.
kf =
ωo = 2000π ωo
Problems km =
15–51
105 C 1 = = C kf (15 × 10−9 )(2000π) 3π
R = km R =
105 (1) = 10,610 Ω 3π
R = 5,305 Ω 2 σ =1−
1 1 =1− = 0.9875 4Q 4(20)
σR = 10,478 Ω;
(1 − σ)R = 133 Ω
C = 15 nF 2C = 30 nF [b]
[c] kf = 2000π H(s) = =
(s/2000π)2 + 1 1 (s/2000π)2 + 20 (s/2000π) + 1 s2 + 4 × 106 π 2 s2 + 100πs + 4 × 106 π 2
P 15.54 To satisfy the gain specification of 20 dB at ω = 0 and α = 1 requires R1 + R2 = 10 R1
or
R2 = 9R1
Choose a standard resistor of 11.1 kΩ for R1 and a 100 kΩ potentiometer for R2 . Since (R1 + R2 )/R1 1 the value of C1 is C1 =
1 = 39.79 nF 2π(40)(105 )
CHAPTER 15. Active Filter Circuits
15–52
Choose a standard capacitor value of 39 nF. Using the selected values of R1 and R2 the maximum gain for α = 1 is 20 log10
111.1 11.1
α=1
= 20.01 dB
When C1 = 39 nF the frequency 1/R2 C1 is 1 109 = 5 = 256.41 rad/s = 40.81 Hz R2 C1 10 (39) The magnitude of the transfer function at 256.41 rad/s is |H(j256.41)|α=1 =
|111.1 × 103 + j256.41(11.1)(100)(39)10−3 | = 7.11 |11.1 × 103 + j256.41(11.1)(100)(39)10−3 |
Therefore the gain at 40.81 Hz is 20 log10 (7.11)α=1 = 17.04 dB P 15.55 20 log10 .·.
R1 + R2 R1
R1 + R2 = 5; R1
Choose
= 13.98 .·. R2 = 4R1
R1 = 100 kΩ. Then
1 = 100π rad/s; R2 C1 P 15.56 [a] |H(j0)| =
R2 = 400 kΩ
.·. C1 =
1 = 7.96 nF (100π)(400 × 103 )
R1 + αR2 11.1 + α(100) = R1 + (1 − α)R2 11.1 + (1 − α)100
Problems
15–53
P 15.57 [a] Combine the impedances of the capacitors in series in Fig. P15.53(b) to get Ceq =
1−α α 1 + = sC1 sC1 sC1
which is identical to the impedance of the capacitor in Fig. P15.53(a). [b]
Vx =
α/sC1 V =α (1 − α)/sC1 + α/sC1
Vy =
αR2 = α = Vx (1 − α)R2 + αR2
[c] Since x and y are both at the same potential, they can be shorted together, and the circuit in Fig. 15.34 can thus be drawn as shown in Fig. 15.53(c). [d] The feedback path between Vo and Vs containing the resistance R4 + 2R3 has no effect on the ratio Vo /Vs , as this feedback path is not involved in the nodal equation that defines the voltage ratio. Thus, the circuit in Fig. 15.53(c) can be simplified into the form of Fig. 15.2, where the input impedance is the equivalent impedance of R1 in series with the parallel combination of (1 − α)/sC1 and (1 − α)R2 , and the feedback impedance is the equivalent impedance of R1 in series with the parallel combination of α/sC1 and αR2 : Zi = R1 + =
· (1 − α)R2
(1 − α)R2 +
α sC1
· αR2 αR2 + sCα1
R1 + αR2 + R1 R2 C1 s 1 + R2 C1 s
P 15.58 As ω → 0 |H(iω)| →
(1−α) sC1
R1 + (1 − α)R2 + R1 R2 C1 s 1 + R2 C1 s
Zf = R 1 + =
(1−α) sC1
2R3 + R4 =1 2R3 + R4
15–54
CHAPTER 15. Active Filter Circuits
Therefore the circuit would have no effect on low frequency signals. As ω → ∞ |H(jω)| →
[(1 − β)R4 + Ro ](βR4 + R3 ) [(1 − β)R4 + R3 ](βR4 + Ro )
When β = 1 |H(j∞)|β=1 =
Ro (R4 + R3 ) R3 (R4 + Ro )
If R4 Ro Ro |H(j∞)|β=1 ∼ >1 = R3 Thus, when β = 1 we have amplification or “boost”. When β = 0 |H(j∞)|β=0 =
R3 (R4 + R3 ) Ro (R4 + Ro )
If R4 Ro R3 |H(j∞)|β=0 ∼ 0.5 we have amplification, and if β < 0.5 we have attenuation. Also note the amplification an attenuation are symmetric about β = 0.5. i.e. |H(jω)|β=0.6 =
1 |H(jω)|β=0.4
Yes, the circuit can be used as a treble volume control because • The circuit has no effect on low frequency signals • Depending on β the circuit can either amplify (β > 0.5) or attenuate (β < 0.5) signals in the treble range • The amplification (boost) and attenuation (cut) are symmetric around β = 0.5. When β = 0.5 the circuit has no effect on signals in the treble frequency range.
Problems P 15.59 [a] |H(j∞)|β=1 = .·.
(65.9)(505.9) Ro (R4 + R3 ) = = 9.99 R3 (R4 + Ro ) (5.9)(565.9)
maximum boost = 20 log10 9.99 = 19.99 dB
[b] |H(j∞)|β=0 = .·.
15–55
R3 (R4 + R3 ) Ro (R4 + Ro )
maximum cut = −21.93 dB
[c] R4 = 500 kΩ;
Ro = R1 + R3 + 2R5 = 65.9 kΩ
.·. R4 = 7.59Ro Yes, R4 is significantly greater than Ro . [d] |H(j/R3 C2 )|β=1 =
=
R (2R3 + R4 ) + j Ro3 (R4 + R3 ) (2R3 + R4 ) + j(R4 + Ro ) 511.8 + j 65.9 (505.9) 5.9 511.8 + j565.9
= 7.44 20 log10 |H(j/R3 C2 )|β=1 = 20 log10 7.44 = 17.43 dB [e] When β = 0 |H(j/R3 C2 )|β=0 =
(2R3 + R4 ) + j(R4 + Ro ) Ro (2R3 + R4 ) + j (R4 + R3 ) R3
Note this is the reciprocal of |H(j/R3 C2 )|β=1 . .·. 20 log10 |H(j/R3 C2 )|β+0 = −17.43 dB [f] The frequency 1/R3 C2 is very nearly where the gain is 3 dB off from its maximum boost or cut. Therefore for frequencies higher than 1/R3 C2 the circuit designer knows that gain or cut will be within 3 dB of the maximum.
15–56
CHAPTER 15. Active Filter Circuits
P 15.60 |H(j∞)| = =
[(1 − β)R4 + Ro ][βR4 + R3 ] [(1 − βR4 + R3 ][βR4 + Ro ] [(1 − β)500 + 65.9][β500 + 5.9] [(1 − β)500 + 5.9][β500 + 65.9]
16
Fourier Series
Assessment Problems Vm 1 2T /3 1T 7 AP 16.1 av = Vm dt + dt = Vm = 7π V T 0 T 2T /3 3 9 2 ak = T
=
=
0
4Vm 3kω0 T
2 bk = T
2T /3
2T /3 0
4Vm 3kω0 T
Vm cos kω0 t dt +
4kπ sin 3
2T /3
T
4kπ 1 − cos 3
2T /3
Vm cos kω0 t dt 3
6 4kπ = sin k 3
Vm sin kω0 t dt +
T
Vm sin kω0 t dt 3
6 = k
4kπ 1 − cos 3
AP 16.2 [a] av = 7π = 21.99 V [b] a1 = −5.196 a2 = 2.598 a3 = 0 a4 = −1.299 a5 = 1.039 b1 = 9
b2 = 4.5
b3 = 0
b4 = 2.25
b5 = 1.8
2π = 50 rad/s T [d] f3 = 3f0 = 23.87 Hz [c] ω0 =
[e] v(t) = 21.99 − 5.2 cos 50t + 9 sin 50t + 2.6 cos 100t + 4.5 sin 100t −1.3 cos 200t + 2.25 sin 200t + 1.04 cos 250t + 1.8 sin 250t + · · · V AP 16.3 Odd function with both half- and quarter-wave symmetry.
6Vm t, vg (t) = T
0 ≤ t ≤ T /6; 16–1
av = 0,
ak = 0 for all k
16–2
CHAPTER 16. Fourier Series bk = 0 for k even bk =
8 T
8 = T =
vg (t) =
T /4 0
f (t) sin kω0 t dt,
T /6
0
k odd
6Vm 8 t sin kω0 t dt + T T
12Vm kπ sin 2 2 k π 3
T /4 T /6
Vm sin kω0 t dt
∞ 1 12Vm nπ sin nω0 t V sin π 2 n=1,3,5 n2 3
AP 16.4 [a] Using the results from AP 16.2, and Equation (16.39), A1 = −5.2 − j9 = 10.4/− 120◦ ;
A2 = 2.6 − j4.5 = 5.2/− 60◦
A4 = −1.3 − j2.25 = 2.6/− 120◦
A3 = 0;
A5 = 1.04 − j1.8 = 2.1/− 60◦ θ1 = −120◦ ;
θ2 = −60◦ ;
θ4 = −120◦ ;
θ5 = −60◦
θ3 not defined;
[b] v(t) = 21.99 + 10.4 cos(50t − 120◦ ) + 5.2 cos(100t − 60◦ ) + 2.6 cos(200t − 120◦ ) + 2.1 cos(250t − 60◦ ) + · · · V AP 16.5 The Fourier series for the input voltage is
∞ 1 8A nπ sin sin nω0 (t + T /4) 2 2 π n=1,3,5 n 2
vi =
∞ 1 nπ 8A sin2 cos nω0 t = 2 2 π n=1,3,5 n 2
=
∞ 1 8A cos nω0 t π 2 n=1,3,5 n2
8A 8(281.25π 2 ) = = 2250 mV π2 π2 ω0 =
2π 2π = × 103 = 10 T 200π
Problems .·.
vi = 2250
∞
1 cos 10nt mV 2 n=1,3,5 n
From the circuit we have Vo =
Vi 1 Vi · = R + (1/jωC) jωC 1 + jωRC
Vo =
1/RC 100 Vi = Vi 1/RC + jω 100 + jω
Vi1 = 2250/0◦ mV;
ω0 = 10 rad/s
Vi3 =
2250 ◦ /0 = 250/0◦ mV; 9
Vi5 =
2250 ◦ /0 = 90/0◦ mV; 25
Vo1 =
100 (2250/0◦ ) = 2238.83/− 5.71◦ mV 100 + j10
Vo3 =
100 (250/0◦ ) = 239.46/− 16.70◦ mV 100 + j30
Vo5 =
100 (90/0◦ ) = 80.50/− 26.57◦ mV 100 + j50
.·.
vo = 2238.33 cos(10t − 5.71◦ ) + 239.46 cos(30t − 16.70◦ )
3ω0 = 30 rad/s 5ω0 = 50 rad/s
+ 80.50 cos(50t − 26.57◦ ) + . . . mV AP 16.6 [a] The Fourier series of the input voltage is vg =
∞ 1 4A sin nω0 (t + T /4) π n=1,3,5 n
= 42
∞ 1 n=1,3,5
nπ sin n 2
cos 2000nt V
From the circuit we have Vo − Vg Vo Vo sC + + =0 sL R .·.
Vo s/RC = H(s) = 2 Vg s + (s/RC) + (1/LC)
16–3
16–4
CHAPTER 16. Fourier Series Substituting in the numerical values yields 500s H(s) = 2 s + 500s + 108 Vg1 = 42/0◦
ω0 = 2000 rad/s
Vg3 = 14/180◦
3ω0 = 6000 rad/s
Vg5 = 8.4/0◦
5ω0 = 10,000 rad/s
Vg7 = 6/180◦
7ω0 = 14,000 rad/s
H(j2000) =
j1 500(j2000) = = 0.01042/89.40◦ 108 − 4 × 106 + 500(j2000) 96 + j1
H(j6000) = 0.04682/87.32◦ H(j10,000) = 1/0◦ H(j14,000) = 0.07272/− 85.83◦ Thus, Vo1 = (42/0◦ )(0.01042/89.40◦ ) = 0.4375/89.40◦ V Vo3 = 0.6555/− 92.68◦ V Vo5 = 8.4/0◦ V Vo7 = 0.4363/94.17◦ V Therefore, vo = 0.4375 cos(2000t + 89.40◦ ) + 0.6555 cos(6000t − 92.68◦ ) + 8.4 cos(10,000t) + 0.4363 cos(14,000t + 94.17◦ ) + . . . V [b] The 5th harmonic, that is, the term at 10,000 rad/s, dominates the output voltage. The circuit is a bandpass filter with a center frequency of 10,000 rad/s and a bandwidth of 500 rad/s. Thus, Q is 20 and the filter is quite selective. This causes the attenuation of the fundamental, third, and seventh harmonic terms in the output signal. AP 16.7 ω0 =
2π × 103 = 3 rad/s 2094.4
jω0 k = j3k
Problems VR =
16–5
2 2sVg (Vg ) = 2 2 + s + 1/s s + 2s + 1
VR Vg
H(s) =
=
s2
2s + 2s + 1
H(jω0 k) = H(j3k) =
j6k (1 − 9k 2 ) + j6k Vg1 = 25.98/0◦ V
vg1 = 25.98 sin ω0 t V; H(j3) =
j6 = 0.6/− 53.13◦ ; −8 + j6
VR1 = 15.588/− 53.13◦ V
√ (15.588/ 2)2 = 60.75 W P1 = 2 vg3 = 0,
therefore P3 = 0 W Vg5 = 1.04/180◦
vg5 = −1.04 sin 5ω0 t V;
j30 = 0.1327/− 82.37◦ −224 + j30
H(j15) =
VR5 = (1.04/180◦ )(0.1327/− 82.37◦ ) = 138/97.63◦ mV √ (0.138/ 2)2 P5 = = 4.76 mW; 2
therefore
P ∼ = 60.75 W = P1 ∼
AP 16.8 Odd function with half- and quarter-wave symmetry, therefore av = 0, ak = 0 for all k, bk = 0 for k even; for k odd we have 8 bk = T =
T /8 0
8 πk
8 2 sin kω0 t dt + T
kπ 1 + 3 cos 4
−j4 Therefore Cn = nπ
T /4 T /8
8 sin kω0 t dt
,
k odd
nπ 1 + 3 cos 4
,
n odd
16–6
CHAPTER 16. Fourier Series
2 T 3T T AP 16.9 [a] Irms = (2)2 (2) + (8)2 − T 8 8 8 −j12.5 j1.5 j0.9 [b] C1 = ; C3 = ; C5 = ; π π π C7 =
−j1.8 ; π
Irms =
2 I
dc
C9 =
+2
∞
−j1.4 ; π
C11 =
|Cn |2 ∼ =
n=1,3,5
=
√
34 = 5.831 A
j0.4 π
2 (12.52 + 1.52 + 0.92 + 1.82 + 1.42 + 0.42 ) π2
∼ = 5.777 A 5.777 − 5.831 × 100 = −0.93% 5.831 [d] Using just the terms C1 – C9 ,
[c] % Error =
Irms =
2 I
dc
+2
∞
|Cn |2 ∼ =
n=1,3,5
2 (12.52 + 1.52 + 0.92 + 1.82 + 1.42 ) π2
∼ = 5.774 A % Error =
5.774 − 5.831 × 100 = −0.98% 5.831
Thus, the % error is still less than 1%. AP 16.10 T = 32 ms, therefore 8 ms requires shifting the function T /4 to the right. i=
∞ n=−∞
=
n(odd) ∞
4 π
4 nπ jnω0 (t−T /4) −j 1 + 3 cos e nπ 4
n=−∞
n(odd)
1 nπ −j(n+1)(π/2) jnω0 t 1 + 3 cos e e n 4
Problems
16–7
Problems P 16.1
2π = 31, 415.93 rad/s 200 × 10−6 2π = 157.080 krad/s ωob = 40 × 10−6 1 1 1 = 5000 Hz; fob = = 25,000 Hz [b] foa = = −6 T 200 × 10 40 × 10−6 100(10 × 10−6 ) [c] ava = 0; avb = = 25 V 40 × 10−6 [d] The periodic function in Fig. P16.1(a) has half-wave symmetry. Therefore, [a] ωoa =
ava = 0;
aka = 0 for k even;
bka = 0 for k even
For k odd, 4 T /2 4 T /4 2πkt 2πkt dt + dt aka = 40 cos 80 cos T 0 T T T /4 T =
2πkt T /4 320 T 2πkt 160 T sin sin + T 2πk T 0 T 2πk T
=
πk 160 80 πk sin + sin πk − sin πk 2 πk 2
=− bka =
80 πk sin , πk 2
T /2 T /4
k odd
4 T /2 4 T /4 2πkt 2πkt dt + dt 40 sin 80 sin T 0 T T T /4 T
2πkt −160 T 2πkt T /4 320 T cos − = cos T 2πk T 0 T 2πk T
T /2 T /4
160 −80 (0 − 1) − (−1 − 0) πk πk 240 = πk The periodic function in Fig. P16.1(b) is even; therefore, bk = 0 for all k. Also, =
avb = 25 V akb =
4 T /8 2πkt dt 100 cos T 0 T
2πk T /8 400 T sin t = T 2πk T 0 πk 200 sin = πk 4
16–8
CHAPTER 16. Fourier Series [e] For the periodic function in Fig. P16.1(a), v(t) =
∞ 80 nπ 1 3 cos nωo t + sin nωo t V − sin π n=1,3,5 n 2 n
For the periodic function in Fig. P16.1(b),
∞ nπ 200 1 sin cos nωo t V v(t) = 25 + π n=1 n 4
P 16.2
In studying the periodic function in Fig. P16.2 note that it can be visualized as the combination of two half-wave rectified sine waves, as shown in the figure below. Hence we can use the Fourier series for a half-wave rectified sine wave which is given as the answer to Problem 16.3(c).
v1 (t) =
∞ cos nωo t 100 200 + 50 sin ωo t − V π π n=2,4,6 (n2 − 1)
v2 (t) =
∞ cos nωo (t − T /2) 60 120 + 30 sin ωo (t − T /2) − V π π n=2,4,6 (n2 − 1)
Observe the following, noting that n is even:
2π T sin ωo (t − T /2) = sin ωo t − T 2
= sin(ωo t − π) = − sin ωo t
Problems
2πn T cos nωo (t − T /2) = cos nωo t − T 2
16–9
= cos(nωo t − nπ) = cos nωo t
Using the observations above, v2 (t) =
∞ cos(nωo t) 60 120 − 30 sin ωo t − V π π n=2,4,6 (n2 − 1)
Thus, v(t) = v1 (t) + v2 (t) = P 16.3
∞ cos(nωo t) 160 320 + 20 sin ωo t − V π π n=2,4,6 (n2 − 1)
[a] Odd function with half- and quarter-wave symmetry, av = 0, ak = 0 for all k, bk = 0 for even k; for k odd we have 8 T /4 4Vm , bk = Vm sin kω0 t dt = T 0 kπ
and
k odd
∞ 4Vm 1 sin nω0 t V v(t) = π n=1,3,5 n
[b] Even function: bk = 0 for k 2Vm 2 T /2 π Vm sin t dt = av = T 0 T π
1 1 4 T /2 π 2Vm + Vm sin t cos kω0 t dt = ak = T 0 T π 1 − 2k 1 + 2k
= and
4Vm /π 1 − 4k 2
∞ 2Vm 1 v(t) = 1+2 cos nω0 t V 2 π n=1 1 − 4n
2π 1 T /2 Vm [c] av = Vm sin t dt = T 0 T π
2 T /2 2π Vm 1 + cos kπ ak = Vm sin t cos kω0 t dt = T 0 T π 1 − k2
Note:
ak = 0 for k-odd,
ak =
2Vm π(1 − k 2 )
for k even,
2 T /2 2π Vm sin t sin kω0 t dt = 0 for k = 2, 3, 4, . . . T 0 T Vm ; therefore For k = 1, we have b1 = 2
bk =
v(t) =
∞ Vm Vm 1 2Vm + sin ω0 t + cos nω0 t V π 2 π n=2,4,6 1 − n2
CHAPTER 16. Fourier Series
16–10 P 16.4
Starting with Eq. (16.2), f (t) sin kω0 t = av sin kω0 t +
∞
an cos nω0 t sin kω0 t +
n=1
∞
bn sin nω0 t sin kω0 t
n=1
Now integrate both sides from to to to + T. All the integrals on the right-hand side reduce to zero except in the last summation when n = k, therefore we have to +T to
P 16.5
f (t) sin kω0 t dt = 0 + 0 + bk to +T
[a] I6 =
to
T 2
or
2 to +T bk = f (t) sin kω0 t dt T to
to +T 1 sin mω0 t dt = − cos mω0 t mω0 to
=
−1 [cos mω0 (to + T ) − cos mω0 to ] mω0
=
−1 [cos mω0 to cos mω0 T − sin mω0 to sin mω0 T − cos mω0 to ] mω0
=
−1 [cos mω0 to − 0 − cos mω0 to ] = 0 for all m, mω0 to +T
I7 =
to
to +T 1 cos mω0 to dt = [sin mω0 t] mω0 to
=
1 [sin mω0 (to + T ) − sin mω0 to ] mω0
=
1 [sin mω0 to − sin mω0 to ] = 0 for all m mω0
to +T
1 to +T cos mω0 t sin nω0 t dt = [sin(m + n)ω0 t − sin(m − n)ω0 t] dt [b] I8 = 2 to to But (m + n) and (m − n) are integers, therefore from I6 above, I8 = 0 for all m, n. to +T 1 to +T sin mω0 t sin nω0 t dt = [cos(m − n)ω0 t − cos(m + n)ω0 t] dt [c] I9 = 2 to to If m = n, both integrals are zero (I7 above). If m = n, we get T 1 to +T 1 to +T T dt − cos 2mω0 t dt = − 0 = 2 to 2 to 2 2
I9 = [d] I10 = =
to +T to
cos mω0 t cos nω0 t dt
1 to +T [cos(m − n)ω0 t + cos(m + n)ω0 t] dt 2 to
If m = n, both integrals are zero (I7 above). If m = n, we have I10 =
1 to +T 1 to +T T T dt + cos 2mω0 t dt = + 0 = 2 to 2 to 2 2
Problems
P 16.6
16–11
T /2 1 to +T 1 0 av = f (t) dt = f (t) dt + f (t) dt T to T −T /2 0
Let
t = −x,
dt = −dx,
x=
T 2
when
t=
−T 2
and x = 0 when t = 0 Therefore
10 1 T /2 10 f (t) dt = f (−x)(−dx) = − f (x) dx T −T /2 T T /2 T 0
Therefore av = −
1 T /2 1 T /2 f (t) dt + f (t) dt = 0 T 0 T 0
20 2 T /2 f (t) cos kω0 t dt + f (t) cos kω0 t dt ak = T 0 T −T /2 Again, let t = −x in the first integral and we get 2 T /2 20 f (t) cos kω0 t dt = − f (x) cos kω0 x dx T −T /2 T 0 Therefore ak = 0
for all k.
20 2 T /2 f (t) sin kω0 t + f (t) sin kω0 t dt bk = T −T /2 T 0 Using the substitution t = −x, the first integral becomes 2 T /2 f (x) sin kω0 x dx T 0 4 T /2 f (t) sin kω0 t dt Therefore we have bk = T 0 P 16.7
20 2 T /2 f (t) sin kω0 t dt + f (t) sin kω0 t dt T −T /2 T 0 Now let t = x − T /2 in the first integral, then dt = dx, x = 0 when t = −T /2 and x = T /2 when t = 0, also sin kω0 (x − T /2) = sin(kω0 x − kπ) = sin kω0 x cos kπ. Therefore bk =
20 2 T /2 f (t) sin kω0 t dt = − f (x) sin kω0 x cos kπ dx T −T /2 T 0 bk =
and
T /2 2 (1 − cos kπ) f (x) sin kω0 t dt T 0
Now note that 1 − cos kπ = 0 when k is even, and 1 − cos kπ = 2 when k is odd. Therefore bk = 0 when k is even, and 4 T /2 f (t) sin kω0 t dt when k is odd bk = T 0
16–12 P 16.8
CHAPTER 16. Fourier Series
Because the function is even and has half-wave symmetry, we have av = 0, ak = 0 for k even, bk = 0 for all k and ak =
4 T /2 f (t) cos kω0 t dt, T 0
k odd
The function also has quarter-wave symmetry; therefore f (t) = −f (T /2 − t) in the interval T /4 ≤ t ≤ T /2; thus we write ak =
4 T /4 4 T /2 f (t) cos kω0 t dt + f (t) cos kω0 t dt T 0 T T /4
Now let t = (T /2 − x) in the second integral, then dt = −dx, x = T /4 when t = T /4 and x = 0 when t = T /2. Therefore we get 4 T /2 4 T /4 f (t) cos kω0 t dt = − f (x) cos kπ cos kω0 x dx T T /4 T 0 Therefore we have T /4 4 ak = (1 − cos kπ) f (t) cos kω0 t dt T 0
But k is odd, hence ak = P 16.9
8 T /4 f (t) cos kω0 t dt, T 0
k odd
Because the function is odd and has half-wave symmetry, av = 0, ak = 0 for all k, and bk = 0 for k even. For k odd we have 4 T /2 bk = f (t) sin kω0 t dt T 0 The function also has quarter-wave symmetry, therefore f (t) = f (T /2 − t) in the interval T /4 ≤ t ≤ T /2. Thus we have bk =
4 T /4 4 T /2 f (t) sin kω0 t dt + f (t) sin kω0 t dt T 0 T T /4
Now let t = (T /2 − x) in the second integral and note that dt = −dx, x = T /4 when t = T /4 and x = 0 when t = T /2, thus T /4 4 T /2 4 f (t) sin kω0 t dt = − cos kπ f (x)(sin kω0 x) dx T T /4 T 0
But k is odd, therefore the expression becomes 8 T /4 f (t) sin kω0 t dt bk = T 0
Problems 1 1 = 62.5 Hz = T 16 × 10−3 [b] no, because f (3 ms) = 10 mA but f (−3 ms) = −10 mA.
P 16.10 [a] f =
[c] yes, because f (−t) = −f (t) for all t. [d] yes [e] yes [f] av = 0,
function is odd
ak = 0,
for all k; the function is odd
bk = 0,
for k even, the function has half-wave symmetry
bk =
8 T
8 = T =
T /4 0
f (t) sin kωo t,
T /8 0
k odd
5t sin kωo t dt +
T /4 T /8
0.01 sin kωo t dt
8 {Int1 + Int2} T T /8
Int1 = 5
0
t sin kωo t dt
T /8 1 t = 5 2 2 sin kωo t − cos kωo t k ωo kωo 0
=
5 kπ 0.625T kπ sin cos − k 2 ωo2 4 kωo 4 T /4
Int2 = 0.01
T /8
Int1 + Int2 =
sin kωo t dt =
T /4 0.01 −0.01 kπ cos kωo t = cos kωo kωo 4 T /8
0.01 0.625T kπ 5 + sin − 2 2 k ωo 4 kωo kωo
cos
kπ 4
0.625T = 0.625(16 × 10−3 ) = 0.01 .·.
Int1 + Int2 =
5 k 2 ωo2
sin
kπ 4
5 0.16 8 kπ kπ · 2 2 · T 2 sin = 2 2 sin , bk = T 4π k 4 π k 4 i(t) =
∞ sin(nπ/4) 160 sin nωo t mA π 2 n=1,3,5 n2
k odd
16–13
16–14
CHAPTER 16. Fourier Series 2π = 2π rad/s T
ωo =
P 16.11 [a] T = 1; [b] yes [c] no [d] no
P 16.12 [a] v(t) is even and has both half- and quarter-wave symmetry, therefore av = 0, bk = 0 for all k, ak = 0 for k-even; for odd k we have kπ 8 T /4 4Vm sin Vm cos kω0 t dt = ak = T 0 πk 2
∞ nπ 1 4Vm sin v(t) = cos nω0 t V π n=1,3,5 n 2
[b] v(t) is even and has both half- and quarter-wave symmetry, therefore av = 0, ak = 0 for k-even, bk = 0 for all k; for k-odd we have ak =
8 T /4 4Vp 8Vp t − Vp cos kω0 t dt = − 2 2 T 0 T π k
∞ 1 8Vp Therefore v(t) = − 2 cos nω0 t V π n=1,3,5 n2
P 16.13 [a] i(t) is even, therefore bk = 0 for all k. av =
Im 1 T 1 · · Im · 2 · = A 2 4 T 4
4 ak = T
T /4
0
4Im = T
4Im t cos kωo t dt Im − T
T /4 0
16Im cos kωo t dt − T2
T /4 0
t cos kωo t dt
= Int1 − Int2 Int1 =
kπ 4Im T /4 2Im sin cos kωo t dt = T 0 πk 2
Int2 =
16Im T2
16Im = T2
T /4 0
t cos kωo t dt
T /4 1 t cos kωo t + sin kωo t 2 2 k ωo kωo 0
4Im kπ 2Im kπ = 2 2 cos −1 + sin π k 2 kπ 2
Problems
4Im kπ .·. ak = 2 2 1 − cos π k 2 .·. i(t) =
16–15
A
∞ 1 − cos(nπ/2) Im 4Im cos nωo t A + 2 4 π n=1 n2
[b] Shifting the reference axis to the left is equivalent to shifting the periodic function to the right: cos nωo (t − T /2) = cos nπ cos nωo t Thus ∞ Im 4Im (1 − cos(nπ/2)) cos nπ i(t) = cos nωo t A + 2 4 π n=1 n2
P 16.14 [a]
[b] Even, since f (t) = f (−t) [c] Yes, since f (t) = −f (T /2 − t) in the interval 0 < t < 4. [d] av = 0, bk = 0,
ak = 0, for all k
for k even
(half-wave symmetry)
(function is even)
Because of the quarter-wave symmetry, the expression for ak is 8 T /4 f (t) cos kω0 t dt, ak = T 0
k odd
2
2t k 2 ω02 t2 − 2 8 2 2 4t cos kω0 t dt = 4 2 2 cos kω0 t + sin kω0 t = 8 0 k ω0 k 3 ω03
0
16–16
CHAPTER 16. Fourier Series
2π kπ (2) = kω0 (2) = k 8 2 cos(kπ/2) = 0,
since k is odd
4k 2 ω02 − 2 16k 2 ω02 − 8 ak = 4 0 + sin(kπ/2) = sin(kπ/2) k 3 ω03 k 3 ω03
.·.
π 2π = ; ω0 = 8 4
ak =
ω02
π2 = ; 16
ω03
π3 = 64
k2π2 − 8 (64) sin(kπ/2) k3π3
f (t) = 64
∞
n=1,3,5
n2 π 2 − 8 sin(nπ/2) cos(nω0 t) π 3 n3
[e] cos nω0 (t − 2) = cos(nω0 t − π/2) = sin nω0 t sin(nπ/2) f (t) = 64
∞ n=1,3,5
n2 π 2 − 8 sin2 (nπ/2) sin(nω0 t) π 3 n3
P 16.15 [a]
[b] Odd, since f (−t) = −f (t) [c] f (t) has quarter-wave symmetry, since f (T /2 − t) = f (t) in the interval 0 < t < 4. [d] av = 0, bk = 0, bk =
(half-wave symmetry); for k even
(half-wave symmetry)
8 T /4 f (t) sin kω0 t dt, T 0 =
ak = 0,
8 2 3 t sin kω0 t dt 8 0
k odd
for all k
(function is odd)
Problems
16–17 2
3t2 6 t3 6t cos kω0 t + 3 3 cos kω0 t = 2 2 sin kω0 t − 4 4 sin kω0 t − k ω0 k ω0 kω0 k ω0 kω0 (2) = k
0
2π kπ (2) = 8 2
cos(kπ/2) = 0,
since k is odd
12 6 bk = 2 2 sin(kπ/2) − 4 4 sin(kπ/2) k ω0 k ω0
.·.
2π kω0 = k 8
=
kπ ; 4
k 2 ω02 =
k2π2 ; 16
192 8 bk = 2 2 1 − 2 2 sin(kπ/2), π k π k
.·.
k 4 ω04 =
k4π4 256
k odd
∞ 1 8 192 1 − 2 2 sin(nπ/2) sin nω0 t f (t) = 2 2 π n=1,3,5 n π n
[e] sin nω0 (t − 2) = sin(nω0 t − π/2) = − cos nω0 t sin(nπ/2)
∞ 1 8 −192 1 − 2 2 sin2 (nπ/2) cos nω0 t f (t) = 2 2 π n=1,3,5 n π n
P 16.16 [a]
[b] av = 0; ak =
ak = 0,
for k even;
8 T /4 f (t) cos kω0 t dt, T 0 =
bk = 0,
for all k
for k odd
8 T /4 40 8 T /8 120t cos kω0 t dt + 10 + t cos kω0 t dt T T T /8 T T 0
80 T /4 320 T /4 960 T /8 t cos kω0 t dt + cos kω0 t dt + 2 t cos kω0 t dt = 2 T 0 T T /8 T T /8
16–18
CHAPTER 16. Fourier Series
960 cos kω0 t t sin kω0 t + = 2 T k 2 ω02 kω0
T /8 0
320 cos kω0 t t sin kω0 t + 2 + T k 2 ω02 kω0 kω0
T kπ = ; 4 2
kω0
80 sin kω0 t T /4 + T kω0 T /8
T /4 T /8
T kπ = 8 4
960 cos(kπ/4) T 1 80 [sin(kπ/2) − sin(kπ/4)] bk = 2 + sin(kπ/4) − 2 2 + 2 2 T k ω0 8kω0 k ω0 kω0 T
320 cos(kπ/2) T sin(kπ/2) cos(kπ/4) T sin(kπ/4) + − − + 2 T k 2 ω02 4 kω0 k 2 ω02 8kω0 =
640 160 960 cos(kπ/4) + sin(kπ/2) − 2 2 (kω0 T ) kω0 T (kω0 T )2
kω0 T = 2kπ; ak = [c] ak =
(kω0 T )2 = 4k 2 π 2
240 160 80 sin(kπ/2) − 2 2 cos(kπ/4) + 2 2 π k πk π k
80 [2 cos(kπ/4) + πk sin(kπ/2) − 3] π2k2
a1 =
80 [2 cos(π/4) + πk sin(π/2) − 3] ∼ = 12.61 π2
a3 =
80 [2 cos(3π/4) + πk sin(3π/2) − 3] ∼ = −12.46 9π 2
a5 =
80 [2 cos(5π/4) + πk sin(5π/2) − 3] ∼ = 3.66 25π 2
f (t) = 12.61 cos(ω0 t) − 12.46 cos(3ω0 t) + 3.66 cos(5ω0 t) + . . . [d] t =
T ; 4
ω0 t =
π 2π T · = T 4 2
f (T /4) ∼ = 12.61 cos(π/2) − 12.46 cos(3π/2) + 3.66 cos(5π/2) = 0 The result would have been non-trivial for t = T /8 or if the function had been specified as odd.
Problems
16–19
P 16.17 Let f (t) = v2 (t − T /6). av = −(2Vm /3)(T /3)(1/T ) = −(2Vm /9) and
bk = 0 since f (t) is even
T /6 4 T /6 2Vm 4 2Vm 1 − cos kωo tdt = − sin kωo t ak = T 0 3 T 3 kωo 0
=−
8Vm π sin k 3k2π 3
Therefore,
and
=−
4Vm π sin k 3kπ 3
∞ 1 nπ 2Vm 4Vm − sin v2 (t − T /6) = − cos nωo t 9 3π n=1 n 3
∞ 1 nπ 2Vm 4Vm − sin cos nωo (t + T /6) v2 (t) = − 9 3π n=1 n 3
Then, v(t) = v1 (t) + v2 (t). Simplifying,
∞ 7Vm 4Vm 1 nπ nπ − v(t) = sin cos cos nωo t 9 3π n=1 n 3 3
∞ 4Vm 1 2 nπ + sin sin nωo t V 3π n=1 n 3
If
Vm = 9π
av = 7π = 21.99 (Checks)
then
12 nπ nπ sin cos ak = − n 3 3
12 nπ bk = sin2 n 3
12 = n
a1 = 6 sin(4π/3) = −5.2;
12 =− n
1 2
1 2nπ sin 2 3
2nπ 1 − cos 3
6 = n
b2 = 3[1 − cos(8π/3)] = 4.5
a3 = 2 sin(12π/3) = 0;
b3 = 2[1 − cos(12π/3)] = 0
a4 = 1.5 sin(16π/3) = −1.3;
b4 = 1.5[1 − cos(16π/3)] = 2.25
a5 = 1.2 sin(20π/3) = 1.04;
b5 = 1.2[1 − cos(20π/3)] = 1.8
All coefficients check!
4nπ 1 − cos 3
b1 = 6[1 − cos(4π/3)] = 9
a2 = 3 sin(8π/3) = 2.6;
6 4nπ = sin n 3
16–20
CHAPTER 16. Fourier Series
P 16.18 [a] The voltage has half-wave symmetry. Therefore, av = 0;
ak = bk = 0,
k even
For k odd, ak =
4 T /2 2Im Im − t cos kωo t dt T 0 T
8Im T /2 4 T /2 Im cos kω0 t dt − 2 t cos kω0 t dt = T 0 T 0
t 4Im sin kω0 t T /2 8Im cos kωo t − 2 + sin kω0 T = 2 2 T kω0 T k ω0 kω0 0
=0−
8Im cos kπ 1 − 2 T 2 k 2 ω0 k 2 ω02
8Im = T2 =
T /2 0
1 (1 − cos kπ) 2 k ω02
20 4Im = 2, 2 2 π k k
for k odd
4 T /2 2Im t sin kωo t dt bk = Im − T 0 T
=
8Im T /2 4Im T /2 sin kω0 t dt − 2 t sin kω0 t dt T 0 T 0
T /2
4Im − cos kω0 t = T kω0 =
0
T /2
4Im 1 − cos kπ 8Im −T cos kπ − 2 T kω0 T 2kω0
8Im 1 = 1 + cos kπ kω0 T 2 =
8Im sin kωo t t − 2 − cos kω0 t 2 T k 2 ω0 kω0
10π 2Im = , πk k
for k odd
10 2 20 10π 10 √ ak − jbk = 2 − j = − jπ = 2 π 2 k 2 + 4/− θk k k k k k where i(t) = 10
tan θk = ∞ n=1,3,5
πk 2
(nπ)2 + 4 n2
cos(nω0 t − θn )
0
Problems √ [b] A1 = 10 4 + π 2 ∼ = 37.24 A
tan θ1 =
π 2
16–21
θ1 ∼ = 57.52◦
10 √ 3π tan θ3 = θ3 ∼ 4 + 9π 2 ∼ = 10.71 A = 78.02◦ 9 2 10 √ 5π tan θ5 = θ5 ∼ A5 = 4 + 25π 2 ∼ = 6.33 A = 82.74◦ 25 2 10 √ 7π tan θ7 = θ7 ∼ A7 = 4 + 49π 2 ∼ = 4.51 A = 84.80◦ 49 2 10 √ 9π tan θ9 = θ9 ∼ A9 = 4 + 81π 2 ∼ = 3.50 A = 85.95◦ 81 2 ◦ i(t) ∼ = 37.24 cos(ωo t − 57.52 ) + 10.71 cos(3ωo t − 78.02◦ ) A3 =
+ 6.33 cos(5ωo t − 82.74◦ ) + 4.51 cos(7ωo t − 84.80◦ ) + 3.50 cos(9ωo t − 85.95◦ ) + . . . i(T /4) ∼ = 37.24 cos(90 − 57.52◦ ) + 10.71 cos(270 − 78.02◦ ) + 6.33 cos(450 − 82.74◦ ) + 4.51 cos(630 − 84.80◦ ) + 3.50 cos(810 − 85.95◦ ) ∼ = 26.22 A Actual value: T 1 i = (5π 2 ) ∼ = 24.67 A 4 2 P 16.19 The function has half-wave symmetry, thus ak = bk = 0 for k-even, av = 0; for k-odd ak =
4 T /2 8Vm T /2 −t/RC Vm cos kω0 t dt − e cos kω0 t dt T 0 ρT 0
where
ρ = 1 + e−T /2RC .
Upon integrating we get 4Vm sin kω0 t T /2 ak = T kω0 0
T /2 − cos kω0 t 8Vm e−t/RC + kω − · sin kω t · 0 0 ρT (1/RC)2 + (kω0 )2 RC 0
=
−8Vm RC T [1 + (kω0 RC)2 ]
CHAPTER 16. Fourier Series
16–22
4 bk = T
T /2 0
8Vm Vm sin kω0 t dt − ρT
T /2 0
e−t/RC sin kω0 t dt
4Vm cos kω0 t T /2 =− T kω0 0
T /2 −e−t/RC sin kω0 t 8Vm · + kω − · cos kω t 0 0 2 2 ρT (1/RC) + (kω0 ) RC 0
=
P 16.20 [a]
8kω0 Vm R2 C 2 4Vm − πk T [1 + (kω0 RC)2 ]
a2k
+
b2k
=
a2k
4Vm + kω0 RCak + πk
= a2k [1 + (kω0 RC)2 ] + But
ak =
+
b2k
8Vm πk
2Vm πk
+ kω0 RCak
−8Vm RC T [1 + (kω0 RC)2 ] 64Vm2 R2 C 2 , T 2 [1 + (kω0 RC)2 ]2
Therefore a2k = a2k
2
thus we have
64Vm2 R2 C 2 64Vm2 kω0 R2 C 2 16Vm2 = 2 + 2 2 − T [1 + (kω0 RC)2 ] π k πkT [1 + (kω0 RC)2 ]
Now let α = kω0 RC and note that T = 2π/ω0 , thus the expression for a2k + b2k reduces to a2k + b2k = 16Vm2 /π 2 k 2 (1 + α2 ). It follows that
a2k + b2k =
πk 1 + (kω0 RC)2
[b] bk = kω0 RCak + Thus
4Vm
4Vm πk
4Vm 1 + α2 bk 1 = kω0 RC + =α− =− ak πkak α α
Therefore
ak = −α = −kω0 RC bk
Problems P 16.21
16–23
Since av = 0 (half-wave symmetry), Eq. 16.38 gives us vo (t) =
∞
1 4Vm cos(nω0 t − θn ) where 1 + (nω0 RC)2 1,3,5 nπ
tan θn =
bn an
But from Eq. 16.57, we have tan βk = kω0 RC. It follows from Eq. 16.72 that tan βk = −ak /bk or tan θn = − cot βn . Therefore θn = 90◦ + βn and cos(nω0 t − θn ) = cos(nω0 t − βn − 90◦ ) = sin(nω0 t − βn ), thus our expression for vo becomes vo =
∞ sin(nω0 t − βn ) 4Vm π n=1,3,5 n 1 + (nω0 RC)2
P 16.22 [a] e−x ∼ =1−x −t/RC
e
for small x;
therefore
t ∼ = 1− RC
−T /2RC
and
e
Vm 2Vm [1 − (t/RC)] = vo ∼ = Vm − 2 − (T /2RC) RC ∼ =
Vm RC
T t− 4
T ∼ = 1− 2RC
2t − (T /2) 2 − (T /2RC)
Vm Vm T = t− RC 4RC
−8 −8 [b] ak = Vp = 2 2 π k π2k2
Vm T 4RC
=
for
0≤t≤
T 2
−4Vm πω0 RCk 2
P 16.23 [a] Express vg as a constant plus a symmetrical square wave. The constant is Vm /2 and the square wave has an amplitude of Vm /2, is odd, and has half- and quarter-wave symmetry. Therefore the Fourier series for vg is vg =
∞ 1 Vm 2Vm + sin nω0 t 2 π n=1,3,5 n
The dc component of the current is Vm /2R, and with sin nω0 t = cos(nω0 t − 90◦ ) the kth harmonic phase current is Ik =
2Vm /kπ 2Vm = /− 90◦ − θk ◦ 2 2 R + jkω0 L/− 90 kπ R + (kω0 L) −1
where θk = tan
kω0 L R
Thus the Fourier series for the steady-state current is i=
∞ Vm 2Vm sin(nω0 t − θn ) + A 2R π n=1,3,5 n R2 + (nω0 L)2
16–24
CHAPTER 16. Fourier Series
[b]
The steady-state current will alternate between I1 and I2 in exponential traces as shown. Assuming t = 0 at the instant i increases toward (Vm /R), we have
Vm −t/τ Vm + I1 − e i= R R
0≤t≤
for
T 2
and i = I2 e−[t−(T /2)]/τ for T /2 ≤ t ≤ T, where τ = L/R. Now we solve for I1 and I2 by noting that I1 = I2 e−T /2τ
I2 =
and
Vm Vm −T /2τ + I1 − e R R
These two equations are now solved for I1 . Letting x = T /2τ, we get (Vm /R)e−x 1 + e−x Therefore the equations for i become
I1 =
Vm Vm − i= e−t/τ R R(1 + e−x )
0≤t≤
for
T ≤t≤T 2
Vm i= e−[t−(T /2)]/τ R(1 + e−x )
T 2
for
and
A check on the validity of these expressions shows they yield an average value of (Vm /2R): Iavg
1 = T
T /2 V m 0
=
1 T
=
Vm 2R
Vm −t/τ + I1 − e dt + R R
Vm T Vm + τ (1 − e−x ) I1 − + I2 2R R since
I1 + I2 =
Vm R
T T /2
−[t−(T /2)]/τ
I2 e
dt
Problems
16–25
∞ 1 4A sin nω0 (t + T /4) π n=1,3,5 n
P 16.24 vi =
ω0 =
∞ nπ 1 4A sin cos nω0 t π n=1,3,5 n 2
=
2π × 103 = 500 rad/s; 4π
vi = 60
∞ 1
n
n=1,3,5
sin
4A = 60 π
nπ cos 500nt V 2
From the circuit jω Vi jω · jωL = Vi = Vi R + jωL R/L + jω 1000 + jω
Vo =
Vi1 = 60/0◦ V;
ω = 500 rad/s
Vi3 = −20/0◦ = 20/180◦ V; Vi5 = 12/0◦ V;
3ω = 1500 rad/s
5ω = 2500 rad/s
Vo1 =
j500 (60/0◦ ) = 26.83/63.43◦ V 1000 + j500
Vo3 =
j1500 (20/180◦ ) = 16.64/− 146.31◦ V 1000 + j1500
Vo5 =
j2500 (12/0◦ ) = 11.14/21.80◦ V 1000 + j2500
.·.
vo = 26.83 cos(500t + 63.43◦ ) + 16.64 cos(1500t − 146.31◦ ) + 11.14 cos(2500t + 21.80◦ ) + . . . V
P 16.25 [a] From the solution to Assessment Problem 16.6 the Fourier series for the input voltage is vg = 42
∞ 1 n=1,3,5
nπ sin n 2
cos 2000nt V
Also from the solution to Assessment Problem 16.6 we have Vg1 = 42/0◦
ω0 = 2000 rad/s
16–26
CHAPTER 16. Fourier Series Vg3 = 14/180◦
3ω0 = 6000 rad/s
Vg5 = 8.4/0◦
5ω0 = 10,000 rad/s
Vg7 = 6/180◦
7ω0 = 14,000 rad/s
From the circuit in Fig. P16.26 we have Vo Vo − Vg + + (Vo − Vg )sC = 0 R sL s2 + 1/LC Vo = H(s) = 2 Vg s + (s/RC) + (1/LC)
.·.
Substituting in the numerical values gives H(s) =
s2 + 108 s2 + 500s + 108
H(j2000) =
96 = 0.9999/− 0.60◦ 96 + j1
H(j6000) =
64 = 0.9989/− 2.68◦ 64 + j3
H(j10,000) = 0 H(j14,000) =
96 = 0.9974/4.17◦ 96 − j7
Vo1 = (42/0◦ )(0.9999/− 0.60◦ ) = 41.998/− 0.60◦ V Vo3 = (14/180◦ )(0.9989/− 2.68◦ ) = 13.985/177.32◦ V Vo5 = 0 V Vo7 = (6/180◦ )(0.9974/4.17◦ ) = 5.984/184.17◦ V vo = 41.998 cos(2000t − 0.60◦ ) + 13.985 cos(6000t + 177.32◦ ) + 5.984 cos(14,000t + 184.17◦ ) + . . . V
[b] The 5th harmonic at the frequency 1/LC = 10,000 rad/s has been eliminated from the output voltage by the circuit, which is a bandreject filter with a center frequency of 10,000 rad/s. P 16.26 [a] Note – find io (t) V0 − Vg V0 + V0 (12.5 × 10−6 s) + =0 16s 1000
V0
1 Vg 1 = + 12.5 × 10−6 s + 16s 16s 1000
Problems
16–27
V0 (1000 + 0.2s2 + 16s) = 1000Vg V0 =
5000Vg s2 + 80s + 5000
I0 =
V0 5Vg = 2 1000 s + 80s + 5000
H(s) =
5 I0 = 2 Vg s + 80s + 5000
H(njω0 ) = ω0 =
5 (5000 −
2π = 240π; T
H(jnω0 ) =
n2 ω02 )
+ j80nω0
ω02 = 57,600π 2 ;
80ω0 = 19,200π
5 (5000 − 57,600π 2 n2 ) + j19,200πn
H(0) = 10−3 H(jω0 ) = 8.82 × 10−6 /− 173.89◦ H(j2ω0 ) = 2.20 × 10−6 /− 176.96◦ H(j3ω0 ) = 9.78 × 10−7 /− 177.97◦ H(j4ω0 ) = 5.5 × 10−7 /− 178.48◦
680 1360 1 1 1 1 − cos ω0 t + cos 2ω0 t + cos 3ω0 t + cos 4ω0 t + . . . vg = π π 3 15 35 63 i0 =
680 1360 × 10−3 − (8.82 × 10−6 ) cos(ω0 t − 173.89◦ ) π 3π −
1360 (2.20 × 10−6 ) cos(2ω0 t − 176.96◦ ) 15π
−
1360 (9.78 × 10−7 ) cos(3ω0 t − 177.97◦ ) 35π
−
1360 (5.5 × 10−7 ) cos(4ω0 t − 178.48◦ ) − . . . 63π
= 216.45 × 10−3 − 1.27 × 10−3 cos(ω0 t − 173.89◦ ) − 6.35 × 10−5 cos(2ω0 t − 176.96◦ ) − 1.21 × 10−5 cos(3ω0 t − 177.97◦ ) − 3.8 × 10−6 cos(4ω0 t − 178.48◦ ) − . . .
CHAPTER 16. Fourier Series
16–28
i0 ∼ = 216.45 − 1.27 cos(ω0 t − 173.89◦ ) mA Note that the sinusoidal component is very small compared to the dc component, so i0 ∼ = 216.45 mA
(a dc current)
[b] Yes, the solution makes sense. The circuit is a low-pass filter which nearly eliminates all but the dc component. P 16.27 The function is odd with half-wave and quarter-wave symmetry. Therefore, ak = 0,
for all k; the function is odd
bk = 0,
for k even, the function has half-wave symmetry
8 bk = T 8 = T =
T /4 0
f (t) sin kωo t,
T /10 0
k odd
500t sin kωo t dt +
T /4 T /10
sin kωo t dt
8 {Int1 + Int2} T T /10
Int1 = 500
0
t sin kωo t dt
T /10 1 t = 500 2 2 sin kωo t − cos kωo t k ωo kωo 0
=
Int2 =
500 kπ 50T kπ − sin cos 2 2 k ωo 5 kωo 5 T /4 T /10
sin kωo t dt =
T /4 −1 1 kπ cos kωo t = cos kωo kωo 5 T /10
1 kπ 50T kπ 500 + − cos Int1 + Int2 = 2 2 sin k ωo 5 kωo kωo 5 50T = 50(20 × 10−3 ) = 1 .·.
Int1 + Int2 =
kπ 500 sin k 2 ωo2 5
500 20 8 kπ kπ · 2 2 · T 2 sin = 2 2 sin , bk = T 4π k 5 π k 5
k odd
Problems i(t) =
16–29
∞ 20 sin(nπ/5) sin nωo t A 2 π n=1,3,5 n2
From the circuit, H(s) =
Vo = Zeq Ig
1 1 + sC + R1 R2 + sL
Yeq =
Zeq =
s2
1/C(s + R2 /L) + s(R1 R2 C + L)/R1 LC + (R1 + R2 )/R1 LC
Therefore, H(s) =
320 × 104 (s + 32 × 104 ) s2 + 32.8 × 104 s + 28.8 × 108
We want the output for the third harmonic: ω0 =
2π 2π = = 100π; T 20 × 10−3
Ig3 =
20 3π sin = 0.214/− 90◦ 2 9π 5 sin 3ω0 t
3ω0 = 300π
320 × 104 (j300π + 32 × 104 ) H(j300π) = = 353.6/− 5.96◦ 2 4 8 (j300π) + 32.8 × 10 (j300π) + 28.8 × 10 Therefore, Vo3 = H(j300π)Ig3 = (353.6/− 5.96◦ )(0.214/− 90◦ ) = 75.7/− 90◦ − 5.96◦ V vo3 = 75.7 sin(300πt − 5.96◦ ) V P 16.28 ωo =
2π 2π = × 106 = 200 krad/s T 10π 5 × 106 = 25 0.2 × 106
.·. n =
3 × 106 = 15; 0.2 × 106
H(s) =
(1/RC)s Vo = 2 Vg s + (1/RC)s + (1/LC)
n=
16–30
CHAPTER 16. Fourier Series 1012 1 = = 106 ; RC (250 × 103 )(4)
H(s) =
(103 )(1012 ) 1 = = 25 × 1012 LC (10)(4)
106 s s2 + 106 s + 25 × 1012
H(jω) =
jω × 106 (25 × 1012 − ω 2 ) + j106 ω
15th harmonic input: vg15 = (150)(1/15) sin(15π/2) cos 15ωo t = −10 cos 3 × 106 t V .·. Vg15 = 10/− 180◦ V j3 = 0.1843/79.38◦ 16 + j3
H(j3 × 106 ) =
Vo15 = (10)(0.1843)/− 100.62◦ V vo15 = 1.84 cos(3 × 106 t − 100.62◦ ) V 25th harmonic input: vg25 = (150)(1/25) sin(25π/2) cos 5 × 106 t = 6 cos 5 × 106 t V .·. Vg25 = 6/0◦ V j5 = 1/0◦ 0 + j5
H(j5 × 106 ) = Vo25 = 6/0◦ V
vo25 = 6 cos 5 × 106 t V
T 1 T T 3Im Im + Im = P 16.29 [a] av = 2 2 2 2 4 i(t) =
2Im t, T
i(t) = Im ,
0 ≤ t ≤ T /2 T /2 ≤ t ≤ T
2 T /2 2Im 2T t cos kωo t dt + Im cos kωo t dt ak = T 0 T T T /2
Problems =
Im (cos kπ − 1) π2k2
2 T /2 2Im 2T t sin kωo t dt + Im sin kωo t dt bk = T 0 T T T /2 =
−Im πk
av =
3Im , 4
a3 =
−2Im 9π 2
b1 =
−Im , π
.·.
a1 =
−2Im , π2
b2 =
Irms = Im
a2 = 0
−Im 2π
2 9 1 1 + 4 + 2 + 2 = 0.8040Im 16 π 2π 8π
Irms = 192.95 mA P = (0.19295)2 (1000) = 37.23 W [b] Area under i2 : T /2
A=
0
2 4Im 2 T t dt + I m T2 2
2 3 T /2 t 4Im 2 T +Im = 2 T 3 0 2
=
2 T Im
Irms =
1 3 2 2 + = T Im 6 6 3
1 2 2 · TI = T 3 m
2 Im = 195.96 mA 3
P = (0.19596)2 1000 = 38.4 W [c] Error =
37.23 − 1 (100) = −3.05% 38.40
∞ 1 80 cos nωo t V P 16.30 vg = 10 + 2 π n=1,3,5 n2
ωo =
2π 2π = × 103 = 500 rad/s T 4π
vg = 10 +
80 80 cos 500t + 2 cos 1500t + . . . 2 π 9π
(Eq. 16.81)
16–31
16–32
CHAPTER 16. Fourier Series
Vo − Vg Vo =0 + sCVo + sL R Vo (RLCs2 + Ls + R) = RVg H(s) =
Vo 1/LC = 2 Vg s + s/RC + 1/LC
106 1 = = 106 LC (0.1)(10) √ 106 1 √ = = 1000 2 RC (50 2)(10) H(s) =
106 √ s2 + 1000 2s + 106
H(jω) =
106 √ 106 − ω 2 + j1000ω 2
H(j0) = 1 H(j500) = 0.9701/− 43.31◦ H(j1500) = 0.4061/− 120.51◦ vo = 10(1) + +
80 (0.9701) cos(500t − 43.31◦ ) π2
80 (0.4061) cos(1500t − 120.51◦ ) + . . . 9π 2
vo = 10 + 7.86 cos(500t − 43.31◦ ) + 0.3658 cos(1500t − 120.51◦ ) + . . .
Problems Vrms ∼ =
7.86 102 + √ 2
2
0.3658 √ + 2
16–33
2
= 11.44 V
V2 √ = 1.85 W P ∼ = rms 50 2 Note – the higher harmonics are severely attenuated and can be ignored. For example, the 5th harmonic component of vo is
vo5
80 = (0.1580) cos(2500t − 146.04◦ ) = 0.0512 cos(2500t − 146.04◦ ) V 25π 2
P 16.31 [a] av =
2
1 T V 2 4 m
=
T
4 T /4
ak =
T
Vm 4
4Vm t cos kωo t dt Vm − T
0
4Vm kπ = 2 2 1 − cos π k 2 bk = 0,
all k
av =
60 = 15 V 4
a1 =
240 π2
a2 =
240 120 (1 − cos π) = 4π 2 π2
Vrms = P =
(15)2
240 π2
2
120 + π2
2
= 24.38 V
(24.38)2 = 59.46 W 10
[b] Area under v 2 ; v 2 = 3600 − A=2
T /4
Vrms = √
0 ≤ t ≤ T /4 57,600 2 28,800 t+ t T T2
57,600 2 28,800 t+ 3600 − t dt = 600T T T2
0
P =
1 + 2
√ 1 600T = 600 = 24.49 V T 2
600 /10 = 60 W
16–34
CHAPTER 16. Fourier Series
59.46 [c] Error = − 1 100 = −0.9041% 60.00 P 16.32 [a] v = 15 + 400 cos 500t + 100 cos(1500t − 90◦ ) V i = 2 + 5 cos(500t − 30◦ ) + 3 cos(1500t − 15◦ ) A 1 1 P = (15)(2) + (400)(5) cos(30◦ ) + (100)(3) cos(−75◦ ) = 934.85 W 2 2 [b] Vrms = [c] Irms =
(15)2
400 + √ 2
(2)2
5 + √ 2
2
2
3 + √ 2
P 16.33 [a] Area under v = A = 4
Therefore Vrms =
100 + √ 2
2
=
T /6 0
2
= 291.93 V
2
= 4.58 A
T 36Vm2 2 T − t dt + 2Vm2 2 T 3 6
V 2T 2Vm2 T + m 9 3
1 2Vm2 T
9
T
V 2T + m 3
= Vm
2 1 + = 74.5356 V 9 3
[b] vg = 105.30 sin ω0 t − 4.21 sin 5ω0 t + 2.15 sin 7ω0 t + · · · V
Therefore Vrms ∼ = P 16.34 [a] v(t) =
(105.30)2 + (4.21)2 + (2.15)2 = 74.5306 V 2
1 1 1 1 480 {sin ωo t + sin 3ωo t + sin 5ωo t + sin 7ωo t + sin 9ωo t + · · ·} π 3 5 7 9
Vrms ∼ = = ∼ =
2
480 1 √
2
1 1 + √ + √ π 2 3 2 5 2 1 1 1 480 1 √ 1+ + + + 9 25 49 81 π 2 117.55 V
2
117.55 [b] % error = − 1 (100) = −2.04% 120 1 1 960 sin 5ωo t [c] v(t) = 2 sin ωo t + sin 3ωo t + π 9 25 +
1 1 sin 7ωo t + sin 9ωo t − · · · 49 81
Vrms
960 1 1 1 1 ∼ + + + = 2√ 1 + 81 625 2401 6561 π 2 ∼ = 69.2765 V
1 + √ 7 2
2
1 + √ 9 2
2
Problems 120 Vrms = √ = 69.2820 V 3
69.2765 − 1 (100) = −0.0081% % error = 69.2820
1 340 680 1 P 16.35 [a] v(t) ≈ − cos ωo t + cos 2ωo t + · · · π π 3 15 Vrms ≈
340 2
π
680 + π
2
1 √ 3 2
2
1 √ + 15 2
2
1 1 340 + = 1+4 = 120.0819 V π 18 450 170 [b] Vrms = √ = 120.2082 2
120.0819 − 1 (100) = −0.11% % error = 120.2082 [c] v(t) ≈
170 340 + 85 sin ωo t − cos 2ωo t π 3π
Vrms ≈ Vrms =
170 2
π
85 + √ 2
2
340 + √ 3 2π
2
≈ 84.8021 V
170 = 85 V 2
% error = −0.23% P 16.36 [a] Half-wave symmetry av = 0, ak = bk = 0, even k. For k odd, ak =
4 T
T /4 4I m 0
16Im = T2
T
t cos kω0 t dt =
16Im T2
4 T
T /4 4I m 0
16Im = T2
t cos kω0 t dt
kπ 2Im ak = sin πk 2 bk =
0
T /4 cos kω0 t t + sin kω0 t 2 2 k ω0 kω0 0
1 16Im T kπ − 2 2 = 0+ sin 2 T 4kω0 2 k ω0
T /4
T
2 − , πk
t sin kω0 t dt =
16Im T2
T /4 0
t sin kω0 t dt
T /4 sin kω0 t kπ t 4Im = − cos kω t sin 0 2 2 2 2 k ω0 kω0 π k 2 0
16–35
16–36
CHAPTER 16. Fourier Series
2Im [b] ak − jbk = πk
2Im a1 − jb1 = π a3 − jb3 =
2Im 3π
a5 − jb5 =
2Im 5π
2Im a7 − jb7 = 7π
kπ sin 2
2 2 1− −j π π −1 − 1−
kπ 2 2 sin − − j πk πk 2
= 0.47Im /− 60.28◦
2 2 +j 3π 3π
2 2 −j 5π 5π
2 2 −1 − +j 7π 7π
= 0.26Im /170.07◦
= 0.11Im /− 8.30◦
= 0.10Im /175.23◦
ig = 0.47Im cos(ω0 t − 60.28◦ ) + 0.26Im cos(3ω0 t + 170.07◦ ) + 0.11Im cos(5ω0 t − 8.30◦ ) + 0.10Im cos(7ω0 t + 175.23◦ ) + · · · [c] Ig =
∞ A2n
2
n=1,3,5
(0.47)2 + (0.26)2 + (0.11)2 + (0.10)2 ∼ = 0.39Im = Im 2 2 T /4 2 2 4I 32I t3 T /4 Im T m m t dt = [d] Area under i2g = 2 = 2 T T 3 0 6 0 Ig =
2T
1 Im
6
T
[e] % error =
Im = √ = 0.41Im 6
estimated − 1 100 = exact
0.3927Im √ − 1 100 = −3.8% (Im / 6)
P 16.37 [a] v has half-wave symmetry, quarter-wave symmetry, and is odd .·. av = 0, ak = 0 all k, bk = 0 k-even 8 bk = T 8 = T
T /4 0
f (t) sin kωo t dt,
T /8 V m
4
0
k-odd
sin kωo t dt +
T /4 T /8
Vm sin kωo t dt
T /4
8Vm cos kωo t T /8 8Vm cos kωo t = − + − 4T kωo T kωo 0
T /8
8Vm kπ 8Vm kπ = 1 − cos + cos −0 4kωo T 4 T kωo 4
Problems 8Vm = kωo T 4Vm = πk
kπ kπ 1 1 − cos + cos 4 4 4 4
1 kπ + 0.75 cos 4 4
=
1 [10 + 30 cos(kπ/4)] k
b1 = 10 + 30 cos(π/4) = 31.21 1 b3 = [10 + 30 cos(3π/4)] = −3.74 3 1 b5 = [10 + 30 cos(5π/4)] = −2.24 5 1 b7 = [10 + 30 cos(7π/4)] = 4.46 7
V (rms) ≈ Vm
31.212 + 3.742 + 2.242 + 4.462 = 22.51 2
[b] Area under v 2 = 2 2(2.5π)2
V (rms) =
T 8
+ 100π 2
T 4
= 53.125π 2 T
√ 1 (53.125π 2 )T = 53.125π = 22.90 T
22.51 − 1 (100) = −1.7% [c] % Error = 22.90 P 16.38 [a] From Problem 16.16,
The area under v 2 : A=4
0
T /8
T /4 14,400 2 40t t dt + 10 + 2 T T T /8
2
dt
16–37
16–38
CHAPTER 16. Fourier Series T /4 3200 t2 57,600 t3 T /8 + 400t + = T 2 3 0 T 2 T /8
=
T /4
6400 t3 T /4 + 2 T 3 T /8 T /8
T 3T 7T 575 57,600 T + 400 + 1600 + 6400 = T 1536 8 64 1536 3
Vrms =
1 T
575 T 3
=
575 = 13.84 V 3
2 Vrms = 12.78 W 15 [c] From Problem 16.16,
[b] P =
b1 =
80 (2 cos 45◦ + π sin 90◦ − 3) = 12.61 V π2
vg ∼ = 12.61 sin ω0 t V √ (19.57/ 2)2 = 5.30 W P = 15
5.30 [d] % error = − 1 (100) = −61.71% 13.84 P 16.39 Figure P16.39(b): ta = 0.2 s;
tb = 0.6 s
v = 50t 0 ≤ t ≤ 0.2 v = −50t + 20 v = 25t − 25
0.2 ≤ t ≤ 0.6 0.6 ≤ t ≤ 1.0
Area 1 under v = A1 = 2
Area 2 = A2 = Area 3 = A3 =
0.6 0.2
1.0
A1 + A2 + A3 =
Vrms =
0.6
0.2 0
20 3
100(4 − 20t + 25t2 ) dt = 625(t2 − 2t + 1) dt =
100 3
2500t2 dt =
1 100 10 = √ V. 1 3 3
40 3
40 3
Problems Figure P16.39(c): ta = tb = 0.4 s v(t) = 25t 0 ≤ t ≤ 0.4 v(t) = A1 = A2 =
50 (t − 1) 3
0.4 0
1.0 0.4
625t2 dt =
Vrms =
40 3
2500 2 60 (t − 2t + 1) dt = 9 3 100 3
A1 + A2 =
0.4 ≤ t ≤ 1
1 (A1 + A2 ) = T
1 100 10 = √ V. 1 3 3
Figure P16.39 (d): ta = tb = 1 v = 10t 0 ≤ t ≤ 1 A1 =
1 0
100t2 dt =
Vrms =
100 3
1 100 10 = √ V. 1 3 3
1 T /4 Vm e−jnωo t −jnωo t Vm e dt = P 16.40 cn = T 0 T −jnωo
T /4 0
Vm Vm nπ nπ Vm = [j(e−jnπ/2 − 1)] = cos sin +j −1 T nωo 2πn 2 2πn 2
Vm nπ nπ − j 1 − cos = sin 2πn 2 2 v(t) =
∞
cn ejnωo t
n=−∞
1 T /4 Vm c o = av = Vm dt = T 0 4
16–39
CHAPTER 16. Fourier Series
16–40 or
sin(nπ/2) Vm 1 − cos(nπ/2) co = lim −j n→0 2π n n
(π/2) cos(nπ/2) Vm (π/2) sin(nπ/2) = lim −j 2π n→0 1 1
Vm π Vm − j0 = = 2π 2 4 Note it is much easier to use co = av than to use L’Hopital’s rule to find the limit of 0/0. P 16.41 co = av = 1 cn = T
Vm Vm T 1 · = 2 T 2 T
Vm −jnωo t te dt 0 T
T
Vm e−jnω0 t = 2 (−jnω0 t − 1) T −n2 ω02 Vm = 2 T
e−jn2πT /T
−n2 ω02
0
1 Vm 1 = 2 (1 + jn2π) − 2 T n2 ω0 n2 ω02 =j
Vm , 2nπ
P 16.42 [a] Vrms =
=
2π 1 −jn T − 1 − (−1) T −n2 ω02
n = ±1, ±2, ±3, . . . 1T 2 v dt = T 0
1 T Vm 2 2 t dt T 0 T
Vm2 t3 T T 3 3 0
Vm Vm2 =√ = 3 3 √ 2 (120/ 3) P = = 480 W 10 [b] From the solution to Problem 16.41 15 120 120 = 60 V; c4 = j =j c0 = 2 8π π c1 = j
60 120 =j ; 2π π
c5 = j
12 120 =j 10π π
Problems c2 = j
120 30 =j ; 4π π
c6 = j
120 10 =j 12π π
c3 = j
120 20 =j ; 6π π
c7 = j
120 8.57 =j 14π π
Vrms =
c2
o +2
∞
|cn |2
n=1
=
602 +
2 (602 π2
+ 302 + 202 + 152 + 122 + 102 + 8.572 )
= 68.58 V [c] P =
(68.58)2 = 470.29 W 10
470.29 % error = − 1 (100) = −2.02% 480 P 16.43 [a] Co = av = Cn =
Vm (1/2)(T /2)Vm = T 4
1 T /2 2Vm −jnωo t te dt T 0 T
T /2
2Vm e−jnωo t (−jnωo t − 1) = 2 T −n2 ωo2 =
0
Vm −jnπ [e (jnπ + 1) − 1] 2n2 π 2
Since e−jnπ = cos nπ we can write Vm Vm cos nπ Cn = 2 2 (cos nπ − 1) + j 2π n 2nπ 54 [b] Co = = 13.5 V 4 C−1 =
−54 27 = 10.19/122.48◦ V +j 2 π π
C1 = 10.19/− 122.48◦ V C−2 = −j
13.5 = 4.30/− 90◦ V π
C2 = 4.30/90◦ V C−3 =
−6 9 + j = 2.93/101.98◦ V π2 π
C3 = 2.93/− 101.98◦ V
16–41
16–42
CHAPTER 16. Fourier Series C−4 = −j
6.75 = 2.15/− 90◦ V π
C4 = 2.15/90◦ V [c]
Vo Vo Vo − Vg + + Vo sC + =0 250 sL 62.5 × 103 .·.
(250LCs2 + 1.004sL + 250)Vo = 0.004sLVg
Vo (1/62, 500C)s = H(s) = 2 Vg s + 1/249C + 1/LC H(s) = ωo =
16s s2 + 1/249Cs + 4 × 1010
2π 2π = × 106 = 2 × 105 rad/s T 10π
H(j0) = 0 H(j2 × 105 k) =
jk 12, 500(1 − k 2 ) + j251k
Therefore, H−1 = 0.0398/0◦ ; H−2 = H−3 = H−4 =
H1 = 0.0398/0◦
−j2 = 5.33 × 10−5 /86.23◦ ; −37, 500 − j20 −3j = 3.00 × 10−5 /89.57◦ ; − j753
−10−5
H2 = 5.33 × 10−5 /− 89.23◦ H2 = 3.00 × 10−5 /− 89.57◦
−4j = 2.13 × 10−5 /89.69◦ ; −187, 500 − j1004
The output voltage coefficients: C0 = 0 C−1 = (10.19/122.48◦ )(0.00398/0◦ ) = 0.0406/122.48◦ V
H2 = 2.13 × 10−5 /− 89.69◦
Problems
16–43
C1 = 0.0406/− 122.48◦ V C−2 = (4.30/− 90◦ )(5.33 × 10−5 /86.23◦ ) = 2.29 × 10−4 /− 3.77◦ V C2 = 2.29 × 10−4 /3.77◦ V C−3 = (2.93/101.98◦ )(3.00 × 10−5 /89.57◦ ) = 8.79 × 10−5 /191.55◦ V C3 = 8.79 × 10−5 /− 191.55◦ V C−4 = (2.15/− 90◦ )(2.13 × 10−5 /89.69◦ ) = 4.58 × 10−5 /− 0.31◦ V C4 = 4.58 × 10−5 /0.31◦ V [d] Vrms ∼ =
C 2 o
+2
4
|Cn |2 ∼ =
4
2 |C |2
n=1
∼ = P =
n
n=1
2(0.04062 + (2.29 × 10−4 )2 + (8.79 × 10−5 )2 + (4.58 × 10−5 )2 ∼ = 0.0574 V
(0.0574)2 = 13.2 µW 250
P 16.44 [a] Vrms = =
=
1 T /2 2Vm 2 t dt T 0 T
1 4V 2 t3 T /2 m
T2 3
T
0
Vm 4Vm2 =√ (3)(8) 6
54 Vrms = √ = 22.05 V 6 [b] From the solution to Problem 16.43 C0 = 13.5;
|C3 | = 2.93
|C1 | = 10.19;
|C4 | = 2.15
|C2 | = 4.30 Vg (rms) ∼ =
13.52 + 2(10.192 + 4.302 + 2.932 + 2.152 ) ∼ = 21.29 V
21.29 [c] % Error = − 1 (100) = −3.44% 22.05
16–44
CHAPTER 16. Fourier Series
P 16.45 [a] From Example 16.3 we have: 40 av = = 10 V, 4
kπ 40 bk = 1 − cos πk 2
[b] Cn =
Ak /− θk◦ = ak − jbk
,
θ3 = −135◦ ,
A6 = 4.24 V,
an − jbn , 2
θ1 = −45◦ ,
A1 = 18.01 V A3 = 6 V,
kπ 40 ak = sin πk 2
A4 = 0,
θ6 = −90◦ ,
C−n =
A2 = 12.73 V,
θ2 = −90◦
A5 = 3.6 V,
A7 = 2.57 V,
θ5 = −45◦
θ7 = −135◦
an + jbn = Cn∗ 2
C0 = av = 10 V
C3 = 3/135◦ V
C1 = 9/45◦ V
C−3 = 3/− 135◦ V C−6 = 2.12/− 90◦ V
C−1 = 9/− 45◦ V
C4 = C−4 = 0
C7 = 1.29/135◦ V
C2 = 6.37/90◦ V
C5 = 1.8/45◦ V
C−7 = 1.29/− 135◦ V
C−2 = 6.37/− 90◦ V C−5 = 1.8/− 45◦ V
C6 = 2.12/90◦ V
Problems P 16.46 [a] From the solution to Problem 16.29 we have Ak = ak − jbk =
Im Im (cos kπ − 1) + j π2k2 πk
A0 = 0.75Im = 180 mA A1 =
240 240 = 90.56/122.48◦ mA (−2) + j 2 π π
A2 = j A3 =
240 240 = 26.03/101.98◦ mA (−2) + j 2 9π 3π
A4 = j A5 =
240 = 38.20/90◦ mA 2π
240 = 19.10/90◦ mA 4π
240 240 = 15.40/97.26◦ mA (−2) + j 2 25π 5π
A6 = j
240 = 12.73/90◦ mA 6π
[b] C0 = A0 = 180 mA 1 C1 = A1 /− θ1 = 45.28/122.48◦ mA 2 C−1 = 45.28/− 122.48◦ mA 1 C2 = A2 /− θ2 = 19.1/90◦ mA 2 C−2 = 19.1/− 90◦ mA
16–45
16–46
CHAPTER 16. Fourier Series 1 C3 = A3 /− θ3 = 13.02/101.98◦ mA 2 C−3 = 13.02/− 101.98◦ mA 1 C4 = A4 /− θ4 = 9.55/90◦ mA 2 C−4 = 9.55/− 90◦ mA 1 C5 = A5 /− θ5 = 7.70/97.26◦ mA 2 C−5 = 7.70/− 97.26◦ mA 1 C6 = A6 /− θ6 = 6.37/90◦ mA 2 C−6 = 6.37/− 90◦ mA
P 16.47 [a] v = A1 cos(ωo t + 90◦ ) + A3 cos(3ωo t − 90◦ ) +A5 cos(5ωo t + 90◦ ) + A7 cos(7ωo t − 90◦ ) v = −A1 sin ωo t + A3 sin 3ωo t − A5 sin 5ωo t + A7 sin 7ωo t [b] v(−t) = A1 sin ωo t − A3 sin 3ωo t + A5 sin 5ωo t − A7 sin 7ωo t .·. v(−t) = −v(t);
odd function
[c] v(t − T /2) = −A1 sin(ωo t − π) + A3 sin(3ωo t − 3π) −A5 sin(5ωo t − 5π) + A7 sin(7ωo t − 7π) = A1 sin ωo t − A3 sin 3ωo t + A5 sin 5ωo t − A7 sin 7ωo t .·. v(t − T /2) = −v(t), yes, the function has half-wave symmetry
Problems
16–47
[d] Since the function is odd, with hws, we test to see if f (T /2 − t) = f (t) f (T /2 − t) = −A1 sin(π − ωo t) + A3 sin(3π − 3ωo t) A5 sin(5π − 5ωo t) + A7 sin(7π − 7ωo t) = −A1 sin ωo t + A3 sin 3ωo t − A5 sin 5ωo t + A7 sin 7ωo t .·. f (T /2 − t) = f (t) and the voltage has quarter-wave symmetry P 16.48 [a] i = 11,025 cos 10,000t + 1225 cos(30,000t − 180◦ ) + 441 cos(50,000t − 180◦ ) + 225 cos 70,000t µA = 11,025 cos 10,000t − 1225 cos 30,000t − 441 cos 50,000t + 225 cos 70,000t µA [b] i(t) = i(−t), [c] Yes,
A0 = 0,
Function is even An = 0 for n even
11,0252 + 12252 + 4412 + 2252 = 7.85 mA 2 [e] A1 = 11,025/0◦ µA; C1 = 5512.50/0◦ µA
[d] Irms =
A3 = 1225/180◦ µA; A5 = 441/180◦ µA; A7 = 225/0◦ µA;
C3 = 612.5/180◦ µA C5 = 220.5/180◦ µA C7 = 112.50/0◦ µA
C−1 = 5512.50/0◦ µA;
C−3 = 612.5/− 180◦ µA
C−5 = 220.5/− 180◦ µA;
C−7 = 112.50/0◦ µA ◦
◦
i = 112.5e−j70,000t + 220.5e−j180 e−j50,000t + 612.5e−j180 e−j30,000t ◦
+ 5512.5e−j10,000t + 5512.5ej10,000t + 612.5ej180 ej30,000t ◦
+ 220.5ej180 ej50,000t + 112.5ej70,000t µA
16–48
CHAPTER 16. Fourier Series
[f]
θn 180˚ −70 −50 −30 −10 10
30
−180˚
P 16.49 From Table 15.1 we have H(s) =
1 (s + 1)(s2 + s + 1)
After scaling we get H (s) =
ωo =
106 (s + 100)(s2 + 100s + 104 )
2π 2π = × 103 = 400 rad/s T 5π
.·. H (jnωo ) = It follows that H(j0) = 1/0◦
1 (1 + j4n)[(1 − 16n2 ) + j4n]
50
70
(krad/s)
Problems H(jωo ) =
16–49
1 = 0.0156/− 241.03◦ (1 + j4)(−15 + j4)
H(j2ωo ) =
1 = 0.00195/− 255.64◦ (1 + j8)(−63 + j8)
∞ cos nωo t A A 2A vg (t) = + sin ωo t − π 2 π n=2,4,6, n2 − 1
= 54 + 27π sin ωo t − 36 cos 2ωo t − · · · V .·. vo = 54 + 1.33 sin(400t − 241.03◦ ) − 0.07 cos(800t − 255.64◦ ) − · · · V P 16.50 Using the technique outlined in Problem 16.17 we can derive the Fourier series for vg (t). We get vg (t) = 100 +
∞ 1 800 cos nωo t 2 π n=1,3,5, n2
The transfer function of the prototype second-order low pass Butterworth filter is H(s) =
s2
1 √ , + 2s + 1
where ωc = 1 rad/s
Now frequency scale using kf = 2000 to get ωc = 2 krad/s: H(s) =
4 × 106 √ s2 + 2000 2s + 4 × 106
H(j0) = 1 H(j5000) =
4 × 106 √ = 0.1580/− 146.04◦ (j5000)2 + 2000 2(j5000)2 + 4 × 106
H(j15,000) =
4 × 106 √ = 0.0178/− 169.13◦ (j15,000)2 + 2000 2(j15,000)2 + 4 × 106
Vdc = 100 V Vg1 =
800 ◦ /0 V π2
Vg3 =
800 ◦ /0 V 9π 2
16–50
CHAPTER 16. Fourier Series
Vodc = 100(1) = 100 V Vo1 =
800 (0.1580/− 146.04◦ ) = 12.81/− 146.04◦ V π2
Vo3 =
800 (0.0178/− 169.13◦ ) = 0.16/− 169.13◦ V 9π 2
vo (t) = 100 + 12.81 cos(5000t − 146.04◦ ) + 0.16 cos(15,000t − 169.13◦ ) + · · · V P 16.51 [a] Let Va represent the node voltage across R2 , then the node-voltage equations are Va Va − Vg + + Va sC2 + (Va − Vo )sC1 = 0 R1 R2 (0 − Va )sC2 +
0 − Vo =0 R3
Solving for Vo in terms of Vg yields Vo = H(s) = Vg s2 +
1 R3
−1 s R1 C1 1 + C12 s C1
+
R1 +R2 R1 R2 R3 C1 C2
It follows that R1 + R2 ωo2 = R1 R2 R3 C1 C2 β=
1 R3
R3 Ko = R1
1 1 + C1 C2
C2 C1 + C2
Note that H(s) =
3 −R R1
s2 +
C2 1 C1 +C2 R3 1 1 1 + s R3 C1 C2
1 C1
+
+
1 s C2 R1 +R2 R1 R2 R3 C1 C2
[b] For the given values of R1 , R2 , R3 , C1 , and C2 we have R3 − R1 1 R3
C2 C1 + C2
1 1 + C1 C2
=−
R3 400 =− 2R1 313
= 2000
R1 + R2 = 0.16 × 1010 = 16 × 108 R1 R2 R3 C1 C2
Problems H(s) = ωo =
−(400/313)(2000)s + 2000s + 16 × 108
s2
2π 2π = × 106 = 4 × 104 rad/s T 50π
H(jnωo ) = = H(jωo ) =
−(400/313)(2000)jnωo 16 × 108 − n2 ωo2 + j2000nωo −j(20/313)n (1 − n2 ) + j0.05n
400 −j(20/313) =− = −1.28 j(0.050) 313
H(j3ωo ) =
−j(20/313)(3) = 0.0240/91.07◦ −8 + j0.15
H(j5ωo ) =
−j(100/313) = 0.0133/90.60◦ −24 + j0.25
vg (t) =
∞ 1 4A sin(nπ/2) cos nωo t π n=1,3,5 n
A = 15.65π V vg (t) = 62.60 cos ωo t − 20.87 cos 3ωo t + 12.52 cos 5ωo t − · · · vo (t) = −80 cos ωo t − 0.50 cos(3ωo t + 91.07◦ ) + 0.17 cos(5ωo t + 90.60◦ ) − · · · V
16–51
The Fourier Transform
Assessment Problems AP 17.1 [a] F (ω) =
0
−jωt
−τ /2
(−Ae
) dt +
τ /2 0
Ae−jωt dt
A [2 − ejωτ /2 − e−jωτ /2 ] jω ejωτ /2 + e−jωτ /2 2A 1− = jω 2 ωτ −j2A [1 − cos ] = ω 2 =
[b] F (ω) = AP 17.2 f (t) = =
1 2π
∞ 0
−at −jωt
te
−2 −3
e
dt =
4ejtω dω +
2 −2
∞ 0
te−(a+jω)t dt =
ejtω dω +
3 2
4ejtω dω
1 {4e−j2t − 4e−j3t + ej2t − e−j2t + 4ej3t − 4ej2t } j2πt
1 3e−j2t − 3ej2t 4ej3t − 4e−j3t + = πt j2 j2 =
1 (a + jω)2
1 (4 sin 3t − 3 sin 2t) πt
AP 17.3 [a] F (ω) = F (s) |s=jω = L{e−at sin ω0 t}s=jω =
ω0 ω0 = (s + a)2 + ω02 s=jω (a + jω)2 + ω02
[b] F (ω) = L{f (−t)}s=−jω
1 = (s + a)2
17–1
= s=−jω
1 (a − jω)2
17
17–2
CHAPTER 17. The Fourier Transform [c] f + (t) = te−at ,
f − (t) = teat 1 , (s + a)2
L{f + (t)} =
.·.
2A , τ
−τ < t < 0; 2
f (t) =
2A 2A [u(t + τ /2) − u(t)] − [u(t) − u(t − τ /2)] τ τ
f (t) =
[b] F{f (t)} =
f (t) =
−2A , τ
2A 2A τ 4A τ δ t+ δ(t) + δ t− − τ 2 τ τ 2
2A jωτ /2 4A 2A −jωτ /2 e + e − τ τ τ
[c] F{f (t)} = (jω)2 F (ω) = −ω 2 F (ω); 1 F (ω) = − 2 ω
Thus we have AP 17.5 v(t) = Vm u t +
τ F u t+ 2 τ F u t− 2
τ τ −u t− 2 2
4A ωτ cos τ 2
−1
F (ω) = −
therefore
ωτ 4A ejωτ /2 + e−jωτ /2 4A −1 = = cos τ 2 τ 2
τ 2
0 0 2 from a table of definite integrals −π · = −1 t < 0 2
Therefore f (t) = sgn t P 17.9
From Problem 17.4[c] we have F (ω) =
2
+ 2 2 + (ω − ω0 ) + (ω + ω0 )2
Note that as → 0, F (ω) → 0 everywhere except at ω = ±ω0 . At ω = ±ω0 , F (ω) = 1/, therefore F (ω) → ∞ at ω = ±ω0 as → 0. The area under each bell-shaped curve is independent of , that is ∞ −∞
∞ dω dω = =π 2 + (ω − ω0 )2 −∞ 2 + (ω + ω0 )2
F (ω) → πδ(ω − ω0 ) + πδ(ω + ω0 )
Therefore as → 0, P 17.10 A(ω) = =
∞
−∞
0 −∞
∞
=2
0
f (t) cos ωt dt f (t) cos ωt dt +
f (t) cos ωt dt,
∞ 0
f (t) cos ωt dt
since f (t) cos ωt is also even.
B(ω) = 0, since f (t) sin ωt is an odd function and 0 −∞
f (t) sin ωt dt = −
P 17.11 A(ω) =
0 −∞
∞ 0
f (t) sin ωt dt
f (t) cos ωt dt +
∞ 0
f (t) cos ωt dt = 0
since f (t) cos ωt is an odd function. B(ω) = −2
∞ 0
f (t) sin ωt dt,
since f (t) sin ωt is an even function.
∞ df (t) df (t) −jωt e P 17.12 [a] F = dt dt −∞ dt Let u = e−jωt , then du = −jωe−jωt ;
Therefore F
df (t) dt
=
let dv = [df (t)/dt] dt, then v = f (t).
∞ −jωt f (t)e
−∞
= 0 + jωF (ω)
∞
−
−∞
f (t)[−jωe−jωt dt]
17–12
CHAPTER 17. The Fourier Transform
[b] Fourier transform of f (t) exists, i.e., f (∞) = f (−∞) = 0.
F
[c] To find
F
Then
d2 f (t) , dt2
d2 f (t) dt2
=F
G(ω) = F
But
df (t) dt
dg(t) dt
= jωG(ω)
Therefore we have F
df (t) dt
g(t) =
let
= jωF (ω) d2 f (t) dt2
= (jω)2 F (ω)
Repeated application of this thought process gives
F P 17.13 [a] F
dn f (t) dtn
t
= (jω)n F (ω).
−∞
∞ t
f (x) dx =
Now let Let
u=
t −∞
−∞
f (x) dx e−jωt dt
−∞
f (x) dx,
dv = e−jωt dt,
then v=
then
du = f (t)dt
e−jωt −jω
Therefore, F
t
e−jωt f (x) dx = −jω −∞ = 0+ ∞
[b] We require
−∞ ∞
[c] No, because P 17.14 [a] F{f (at)} = Let
−∞
u = at,
−∞
−
−∞
−∞
−jω
F (ω) jω
e−ax u(x) dx =
1 = 0 a
f (at)e−jωt dt
du = a dt,
u = ±∞ when
Therefore, F{f (at)} =
f (x) dx
∞ −jωt e
∞
f (x) dx = 0
−∞
∞
t
∞ −∞
−jωu/a
f (u)e
du a
1 = F a
t = ±∞
ω , a
a>0
f (t) dt
Problems [b] F{e−|t| } =
17–13
1 1 2 + = 1 + jω 1 − jω 1 + ω2
Therefore F{e−a|t| } =
(1/a)2 (ω/a)2 + 1
Therefore F{e−0.5|t| } =
4 , +1
F{e−|t| } =
4ω 2
ω2
2 +1
F{e−2|t| } = 1/[0.25ω 2 + 1], yes as “a” increases, the sketches show that f (t) approaches zero faster and F (ω) flattens out over the frequency spectrum.
P 17.15 [a] F{f (t − a)} =
∞ −∞
f (t − a)e−jωt dt
Let u = t − a, then du = dt, t = u + a, and u = ±∞ when t = ±∞. Therefore, F{f (t − a)} =
∞
−∞
f (u)e−jω(u+a) du
= e−jωa [b] F{ejω0 t f (t)} =
∞ −∞
∞ −∞
f (u)e−jωu du = e−jωa F (ω)
f (t)e−j(ω−ω0 )t dt = F (ω − ω0 )
ejω0 t + e−jω0 t [c] F{f (t) cos ω0 t} = F f (t) 2
P 17.16 Y (ω) = =
∞ ∞ −∞
∞ −∞
−∞
x(λ)
1 1 = F (ω − ω0 ) + F (ω + ω0 ) 2 2
x(λ)h(t − λ) dλ e−jωt dt
∞ −∞
h(t − λ)e−jωt dt dλ
17–14
CHAPTER 17. The Fourier Transform
Let u = t − λ, du = dt, and u = ±∞, when t = ±∞. Therefore Y (ω) = = = P 17.17 F{f1 (t)f2 (t)} =
∞ −∞
−∞
∞ −∞
−∞
1 = 2π
−∞
1 = 2π
P 17.18 [a] F (ω) = dF = dω
∞
−∞
∞ −∞
∞ −∞
−∞
−∞ ∞ −∞
∞ −∞
du dλ
−jωu
h(u)e
du dλ
x(λ)e−jωλ H(ω) dλ = H(ω)X(ω)
1 2π
∞
h(u)e
−jωλ
x(λ) e
∞ ∞
1 = 2π
−jω(u+λ)
−∞
∞
∞
x(λ)
∞
F1 (u)ejtu du f2 (t)e−jωt dt −jωt jtu
F1 (u)f2 (t)e ∞
F1 (u)
−∞
e
−j(ω−u)t
f2 (t)e
du dt
dt du
F1 (u)F2 (ω − u) du
f (t)e−jωt dt ∞ d −jωt f (t)e dt = −j tf (t)e−jωt dt = −jF{tf (t)} dω −∞
Therefore j
dF (ω) = F{tf (t)} dω
d2 F (ω) ∞ = (−jt)(−jt)f (t)e−jωt dt = (−j)2 F{t2 f (t)} 2 dω −∞ Note that
(−j)n =
Thus we have [b] (i)
jn
F{e−at u(t)} =
1 jn
dn F (ω) = F{tn f (t)} dω n
1 = F (ω); a + jω
dF (ω) 1 = Therefore j dω (a + jω)2 Therefore F{te−at u(t)} =
1 (a + jω)2
dF (ω) −j = dω (a + jω)2
Problems (ii)
F{|t|e−a|t| } = F{te−at u(t)} − F{teat u(−t)} 1 d = −j 2 (a + jω) dω
1 a − jω
F{te−a|t| } = F{te−at u(t)} + F{teat u(−t)} 1 d = +j 2 (a + jω) dω =
P 17.19 [a] f1 (t) = cos ω0 t, f2 (t) = 1, Thus
1 1 + (a + jω)2 (a − jω)2
= (iii)
17–15
1 a − jω
1 1 − 2 (a + jω) (a − jω)2 F1 (u) = π[δ(u + ω0 ) + δ(u − ω0 )]
−τ /2 < t < τ /2,
and f2 (t) = 0 elsewhere
τ sin(uτ /2) uτ /2
F2 (u) =
Using convolution, 1 F (ω) = 2π =
1 2π
τ = 2 + =
∞ −∞ ∞ −∞
∞ −∞
τ 2
F1 (u)F2 (ω − u) du π[δ(u + ω0 ) + δ(u − ω0 )]τ
δ(u + ω0 )
∞ −∞
sin[(ω − u)τ /2] du (ω − u)(τ /2)
sin[(ω − u)τ /2] du (ω − u)(τ /2)
δ(u − ω0 )
sin[(ω − u)τ /2] du (ω − u)(τ /2)
τ sin[(ω + ω0 )τ /2] τ sin[(ω − ω0 )τ /2] · + · 2 (ω + ω0 )(τ /2) 2 (ω − ω0 )τ /2
[b] As τ increases, the amplitude of F (ω) increases at ω = ±ω0 and at the same time the duration of F (ω) approaches zero as ω deviates from ±ω0 . The area under the [sin x]/x function is independent of τ, that is ∞ τ ∞ sin[(ω − ω0 )(τ /2)] sin[(ω − ω0 )(τ /2)] dω = [(τ /2) dω] = π 2 −∞ (ω − ω0 )(τ /2) (ω − ω0 )(τ /2) −∞
Therefore as t → ∞, f1 (t)f2 (t) → cos ω0 t and
F (ω) → π[δ(ω − ω0 ) + δ(ω + ω0 )]
17–16
CHAPTER 17. The Fourier Transform
P 17.20 [a] vg = 100u(t)
1 Vg (ω) = 100 πδ(ω) + jω H(s) =
10 2 = 5s + 10 s+2
H(ω) =
2 jω + 2
Vo (ω) = H(ω)Vg (ω) =
200 200πδ(ω) + jω + 2 jω(jω + 2)
= V1 (ω) + V2 (ω) v1 (t) =
1 200π 1 ∞ 200πejtω δ(ω) dω = = 50 (sifting property) 2π −∞ jω + 2 2π 2
V2 (ω) =
K2 100 100 K1 + = − jω jω + 2 jω jω + 2
v2 (t) = 50sgn(t) − 100e−2t u(t) vo (t) = v1 (t) + v2 (t) = 50 + 50sgn(t) − 100e−2t u(t) = 100u(t) − 100e−2t u(t) vo (t) = 100(1 − e−2t )u(t) V [b]
Problems
17–17
P 17.21 [a] From the solution to Problem 17.20 H(ω) =
2 jω + 2
Now, Vg (ω) =
200 jω
Then, Vo (ω) = H(ω)Vg (ω) = .·.
K1 K2 200 200 400 = + = − jω(jω + 2) jω jω + 2 jω jω + 2
vo (t) = 100sgn(t) − 200e−2t u(t) V
[b]
P 17.22 [a] Find the Thévenin equivalent with respect to the terminals of the capacitor:
5 vTh = vg ; 6 Io =
RTh = 6012 = 10 kΩ
2sVTh VTh = 6 10,000 + 10 /2s 20,000s + 106
H(s) =
Io 10−4 s ; = VTh s + 50
H(ω) =
jω × 10−4 jω + 50
17–18
CHAPTER 17. The Fourier Transform 5 vTh = vg = 30 sgn(t); 6
Io = H(ω)VTh (ω) =
60 jω
VTh =
60 jω
jω × 10−4 jω + 50
=
6 × 10−3 jω + 50
io (t) = 6e−50t u(t) mA [b] At t = 0− the circuit is
At t = 0+ the circuit is
ig (0+ ) =
30 + 36 = 5.5 mA 12
i60k (0+ ) =
30 = 0.5 mA 60
io (0+ ) = 5.5 + 0.5 = 6 mA which agrees with our solution. We also know io (∞) = 0, which agrees with our solution. The time constant with respect to the terminals of the capacitor is RTh C Thus, τ = (10,000)(2 × 10−6 ) = 20 ms;
.·.
1 = 50, τ
which also agrees with our solution. Thus our solution makes sense in terms of known circuit behavior.
Problems
17–19
P 17.23 [a] From the solution of Problem 17.22 we have
Vo =
106 VTh · 104 + (106 /2s) 2s
H(s) =
50 Vo = VTh s + 50
H(jω) =
50 jω + 50
VTh (ω) =
60 jω
Vo (ω) = H(jω)VTh (ω) = =
60 jω
50 jω + 50
60 60 3000 = − (jω)(jω + 50) jω jω + 50
vo (t) = 30sgn(t) − 60e−50t u(t) V [b] vo (0− ) = −30 V vo (0+ ) = 30 − 60 = −30 V This makes sense because there cannot be an instantaneous change in the voltage across a capacitor. vo (∞) = 30 V This agrees with vTh (∞) = 30 V. As in Problem 17.22 we know the time constant is 20 ms. P 17.24 [a]
Vo 4/s = H(s) = Vg 0.5 + 0.01s + 4/s H(s) =
s2
H(jω) = Vg (ω) =
400 400 = + 50s + 400 (s + 10)(s + 40)
400 (jω + 10)(jω + 40) 6 jω
17–20
CHAPTER 17. The Fourier Transform Vo (ω) = Vg (ω)H(jω) = Vo (ω) =
K2 K3 K1 + + jω jω + 10 jω + 40
K1 =
2400 = 6; 400
K3 =
2400 =2 (−40)(−30)
Vo (ω) =
2400 jω(jω + 10)(jω + 40)
K2 =
2400 = −8 (−10)(30)
8 2 6 − + jω jω + 10 jω + 40
vo (t) = 3sgn(t) − 8e−10t u(t) + 2e−40t u(t) V [b] vo (0− ) = −3 V [c] vo (0+ ) = 3 − 8 + 2 = −3 V [d] For t ≥ 0+ :
(Vo + 3/s)s Vo − 3/s + =0 0.5 + 0.01s 4
Vo
s 100 300 + − 0.75 = s + 50 4 s(s + 50)
Vo =
K1 K2 K3 1200 − 3s2 − 150s = + + s(s + 10)(s + 40) s s + 10 s + 40
K1 =
1200 = 3; 400
K3 =
1200 − 4800 + 6000 =2 (−40)(−30)
K2 =
1200 − 300 + 1500 = −8 (−10)(30)
vo (t) = (3 − 8e−10t + 2e−40t )u(t) V [e] Yes.
Problems P 17.25 [a] Io =
Vg 0.5 + 0.01s + 4/s
H(s) =
100s 100s Io = = 2 Vg s + 50s + 400 (s + 10)(s + 40)
H(ω) =
100(jω) (jω + 10)(jω + 40)
Vg (ω) =
6 jω
Io (ω) = H(ω)Vg (ω) = =
600 (jω + 10)(jω + 40)
20 20 − jω + 10 jω + 40
io (t) = (20e−10t − 20e−40t )u(t) A [b] io (0− ) = 0 [c] io (0+ ) = 0 [d]
Io =
6/s 600 = 2 0.5 + 0.01s + 4/s s + 50s + 400
=
600 20 20 = − (s + 10)(s + 40) s + 10 s + 40
io (t) = (20e−10t − 20e−40t )u(t) A [e] Yes. P 17.26 [a] Io =
RCsIg Ig R = ; R + 1/sC RCs + 1
106 1 = = 40; RC 25 × 103
H(s) = H(ω) =
Io s = Ig s + 1/RC
jω jω + 40
ig = 200sgn(t) µA;
2 Ig = (200 × 10 ) jω −6
=
400 × 10−6 jω
17–21
17–22
CHAPTER 17. The Fourier Transform Io = Ig [H(ω)] =
jω 400 × 10−6 400 × 10−6 · = jω jω + 40 jω + 40
io (t) = 400e−40t u(t) µA [b] Yes, at the time the source current jumps from −200 µA to +200 µA the capacitor is charged to (200)(50) × 10−3 = 10 V, positive at the lower terminal. The circuit at t = 0− is
At t = 0+ the circuit is
The time constant is (50 × 103 )(0.5 × 10−6 ) = 25 ms. .·. P 17.27 [a] Vo =
1 = 40 τ
.·.
io = 400e−40t µA
for t > 0,
Ig R Ig R(1/sC) = R + (1/sC) RCs + 1
H(s) =
1/C Vo 2 × 106 = = Ig s + (1/RC) s + 40
H(ω) =
2 × 106 ; 40 + jω
Vo (ω) = H(ω)Ig (ω) = =
Ig (ω) =
400 × 10−6 jω
400 × 10−6 jω
2 × 106 40 + jω
20 20 800 = − jω(40 + jω) jω 40 + jω
vo (t) = 10sgn(t) − 20e−40t u(t) V [b] Yes, at the time the current source jumps from −200 to +200 µA the capacitor is charged to −10 V. That is, at t = 0− , vo (0− ) = (50 × 103 )(−200 × 10−6 ) = −10 V.
Problems At t = ∞ the capacitor will be charged to +10 V. That is, vo (∞) = (50 × 103 )(200 × 10−6 ) = 10 V The time constant of the circuit is (50 × 103 )(0.5 × 10−6 ) = 25 ms, so 1/τ = 40. The function vo (t) is plotted below:
P 17.28 [a] ig = 3e−5|t| .·.
Ig (ω) =
3 30 3 + = jω + 5 −jω + 5 (jω + 5)(−jω + 5)
Vo Vo s + = Ig 10 10 10 Vo ; = H(s) = Ig s+1
.·.
Vo (ω) = Ig (ω)H(ω) = =
H(ω) =
10 jω + 1
300 (jω + 1)(jω + 5)(−jω + 5)
K1 K2 K3 + + jω + 1 jω + 5 −jω + 5
K1 =
300 = 12.5 (4)(6)
K2 =
300 = −7.5 (−4)(10)
K3 =
300 =5 (6)(10)
Vo (ω) =
7.5 5 12.5 − + jω + 1 jω + 5 −jω + 5
vo (t) = [12.5e−t − 7.5e−5t ]u(t) + 5e5t u(−t) V
17–23
17–24
CHAPTER 17. The Fourier Transform
[b] vo (0− ) = 5 V [c] vo (0+ ) = 12.5 − 7.5 = 5 V [d] ig = 3e−5t u(t), Ig =
3 ; s+5
t ≥ 0+ H(s) =
vo (0+ ) = 5 V;
10 s+1
γC = 0.5
Vo Vo s + = Ig + 0.5 10 10 Vo (s + 1) = Vo =
5 30 + (s + 5)(s + 1) s + 1 =
.·.
30 +5 s+5
−7.5 7.5 5 12.5 7.5 + + = − s+5 s+1 s+1 s+1 s+5
vo (t) = (12.5e−t − 7.5e−5t )u(t) V
[e] Yes, for t ≥ 0+ the solution in part (a) is also vo (t) = (12.5e−t − 7.5e−5t )u(t) V P 17.29 [a]
Vo − Vg Vo Vo =0 + + sL1 sL2 R .·. Vo =
RVg
1 1 s+R + L1 L2
L1
Problems Io =
Vo sL2
R/L1 L2 Io = H(s) = Vg s(s + R[(1/L1 ) + (1/L2 )])
.·.
R = 12 × 105 L1 L2
1 1 R + L1 L2 .·. H(s) = H(ω) =
= 3 × 104
12 × 105 s(s + 3 × 104 )
12 × 105 jω(jω + 3 × 104 )
Vg (ω) = 125π[δ(ω + 4 × 104 ) + δ(ω − 4 × 104 )] Io (ω) = H(ω)Vg (ω) =
1500π × 105 [δ(ω + 4 × 104 ) + δ(ω − 4 × 104 )] jω(jω + 3 × 104 )
1500π × 105 ∞ [δ(ω + 4 × 104 ) + δ(ω − 4 × 104 )]ejtω dω io (t) = 2π jω(jω + 3 × 104 ) −∞
io (t) = 750 × 105
e−j40,000t −j40,000(30,000 − j40,000)
ej40,000t + j40,000(30,000 + j40,000) 75 × 106 = 4 × 108 75 = 400
ej40,000t e−j40,000t + −j(3 − j4) j(3 + j4)
e−j40,000t ej40,000t + 5/− 143.13◦ 5/143.13◦
= 0.075 cos(40,000t − 143.13◦ ) A io (t) = 75 cos(40,000t − 143.13◦ ) mA
17–25
17–26
CHAPTER 17. The Fourier Transform
[b] In the phasor domain:
Vo − 125 Vo Vo + + =0 j200 j800 120 12Vo − 1500 + 3Vo + j20Vo = 0 Vo = Io =
1500 = 60/− 53.13◦ V 15 + j20 Vo = 75 × 10−3 /− 143.13◦ A j800
io (t) = 75 cos(40,000t − 143.13◦ ) mA P 17.30 [a]
Vg s Vg s2 Vo = = 2 25 + (100/s) + s s + 25s + 100 H(s) =
Vo s2 = ; Vg (s + 5)(s + 20)
H(ω) =
(jω)2 (jω + 5)(jω + 20)
vg = 25ig = −450e10t u(−t) − 450e−10t u(t) V Vg = −
450 450 − −jω + 10 jω + 10
Vo (ω) = H(ω)Vg =
−450(jω)2 (−jω + 10)(jω + 5)(jω + 20)
+
−450(jω)2 (jω + 10)(jω + 5)(jω + 20)
=
K2 K3 K4 K5 K6 K1 + + + + + −jω + 10 jω + 5 jω + 20 jω + 5 jω + 10 jω + 20
Problems K1 =
450(100) = −100 (15)(30)
K2 =
450(25) = −50 (15)(15)
K3 =
450(400) = 400 (30)(−15)
Vo (ω) =
K4 = K5 =
17–27
−450(25) = −150 (5)(15)
−450(100) = 900 (−5)(10)
K6 =
−450(400) = −1200 (−15)(−10)
−200 −800 900 −100 + + + −jω + 10 jω + 5 jω + 20 jω + 10
vo = −100e10t u(−t) + [900e−10t − 200e−5t − 800e−20t ]u(t) V [b] vo (0− ) = −100 V [c] vo (0+ ) = 900 − 200 − 800 = −100 V [d] At t = 0− the circuit is
Therefore, the solution predicts v1 (0− ) will be −350 V. Now v1 (0+ ) = v1 (0− ) because the inductor will not let the current in the 25 Ω resistor change instantaneously, and the capacitor will not let the voltage across the 0.01 F capacitor change instantaneously. At t = 0+ the circuit is
From the circuit at t = 0+ we see that vo must be −100 V, which is consistent with the solution for vo obtained in part (c).
CHAPTER 17. The Fourier Transform
17–28 P 17.31
Vo s Vo − Vg 100Vo + =0 + 25 s 100s + 125 × 104 .·. Vo =
Io =
s(100s + 125 × 104 )Vg 125(s2 + 12,000s + 25 × 106 )
sVo 100s + 125 × 104
Io s2 H(s) = = Vg 125(s2 + 12,000s + 25 × 106 ) −8 × 10−3 ω 2 H(ω) = (25 × 106 − ω 2 ) + j12,000ω Vg (ω) = 300π[δ(ω + 5000) + δ(ω − 5000)] Io (ω) = H(ω)Vg (ω) =
io (t) =
−2.4π 2π
−2.4πω 2 [δ(ω + 5000) + δ(ω − 5000)] (25 × 106 − ω 2 ) + j12,000ω
ω 2 [δ(ω + 5000) + δ(ω − 5000)] jtω e dω −∞ (25 × 106 − ω 2 ) + j12,000ω ∞
25 × 106 ej5000t 25 × 106 e−j5000t + = −1.2 −j(12,000)(5000) j(12,000)(5000) 6 = 12
e−j5000t ej5000t + −j j ◦
◦
= 0.5[e−j(5000t+90 ) + ej(5000t+90 ) ] io (t) = 1 cos(5000t + 90◦ ) A
Problems
17–29
P 17.32 [a]
From the plot of vg note that vg is −10 V for an infinitely long time before t = 0. Therefore .·. vo (0− ) = −10 V There cannot be an instantaneous change in the voltage across a capacitor, so vo (0+ ) = −10 V [b] io (0− ) = 0 A At t = 0+ the circuit is
30 − (−10) 40 = = 8A 5 5 [c] The s-domain circuit is io (0+ ) =
Vg Vo = 5 + (10/s)
Vo 2 = H(s) = Vg s+2
10 2Vg = s s+2
17–30
CHAPTER 17. The Fourier Transform H(ω) =
2 jω + 2
2 Vg (ω) = 5 jω
− 5[2πδ(ω)] +
10 30 30 = − 10πδ(ω) + jω + 5 jω jω + 5
10 2 30 Vo (ω) = H(ω)Vg (ω) = − 10πδ(ω) + jω + 2 jω jω + 5 =
20πδ(ω) 60 20 − + jω(jω + 2) jω + 2 (jω + 2)(jω + 5)
=
K0 K1 K2 K3 20πδ(ω) + + + − jω jω + 2 jω + 2 jω + 5 jω + 2
K0 =
20 = 10; 2
Vo (ω) =
K1 =
20 = −10; −2
K2 =
60 = 20; 3
K3 =
60 = −20 −3
10 20 20πδ(ω) 10 10 20 10 + − − = + + − 10πδ(ω) jω jω + 2 jω + 5 jω + 2 jω jω + 2 jω + 5
vo (t) = 5sgn(t) + [10e−2t − 20e−5t ]u(t) − 5 V P 17.33 [a]
Vo (Vo − Vg )s Vo + =0 + 6 10 4s 800 .·. Vo =
s2 Vg s2 + 1250s + 25 × 104
Vo s2 = H(s) = Vg (s + 250)(s + 1000) H(ω) =
(jω)2 (jω + 250)(jω + 1000)
vg = 45e−500|t| ;
Vg (ω) =
.·. Vo (ω) = H(ω)Vg (ω) = =
45,000 (jω + 500)(−jω + 500)
45,000(jω)2 (jω + 250)(jω + 500)(jω + 1000)(−jω + 500)
K2 K3 K4 K1 + + + jω + 250 jω + 500 jω + 1000 −jω + 500
Problems K1 =
45,000(−250)2 = 20 (250)(750)(750)
K2 =
45,000(−500)2 = −90 (−250)(500)(1000)
K3 =
45,000(−1000)2 = 80 (−750)(−500)(1500)
K4 =
45,000(500)2 = 10 (750)(1000)(1500)
.·. vo (t) = [20e−250t − 90e−500t + 80e−1000t ]u(t) + 10e500t u(−t) V [b] vo (0− ) = 10 V;
Vo (0+ ) = 20 − 90 + 80 = 10 V
vo (∞) = 0 V [c] IL =
0.25sVg Vo = 4s (s + 250)(s + 1000)
H(s) =
0.25s IL = Vg (s + 250)(s + 1000)
H(ω) =
0.25(jω) (jω + 250)(jω + 1000)
IL (ω) =
0.25(jω)(45,000) (jω + 250)(jω + 500)(jω + 1000)(−jω + 500)
= K4 =
K1 K2 K3 K4 + + + jω + 250 jω + 500 jω + 1000 −jω + 500
(0.25)(500)(45,000) = 5 mA (750)(1000)(1500)
iL (t) = 5e500t u(−t);
.·. iL (0− ) = 5 mA
K1 =
(0.25)(−250)(45,000) = −20 mA (250)(750)(750)
K2 =
(0.25)(−500)(45,000) = 45 mA (−250)(500)(1000)
K3 =
(0.25)(−1000)(45,000) = −20 mA (−750)(−500)(1500)
.·. iL (0+ ) = K1 + K2 + K3 = −20 + 45 − 20 = 5 mA Checks, i.e.,
iL (0+ ) = iL (0− ) = 5 mA
17–31
17–32
CHAPTER 17. The Fourier Transform At t = 0− : vC (0− ) = 45 − 10 = 35 V At t = 0+ : vC (0+ ) = 45 − 10 = 35 V
[d] We can check the correctness of our solution for t ≥ 0+ by using the Laplace transform. Our circuit becomes
Vo (Vo − Vg )s Vo 5 × 10−3 −6 + =0 + + 35 × 10 + 106 s 800 4s .·. (s2 + 1250s + 25 × 104 )Vo = s2 Vg − (35s + 5000) vg (t) = 45e−500t u(t) V;
Vg =
.·. (s + 250)(s + 1000)Vo =
45 s + 500
45s2 − (35s + 5000)(s + 500) (s + 500)
10s2 − 22,500s − 250 × 104 .·. Vo = (s + 250)(s + 500)(s + 1000) =
90 80 20 − + s + 250 s + 500 s + 1000
.·. vo (t) = [20e−250t − 90e−500t + 80e−1000t ]u(t) V This agrees with our solution for vo (t) for t ≥ 0+ . P 17.34 [a]
Vg (ω) =
36 36 72jω − = 4 − jω 4 + jω (4 − jω)(4 + jω)
Problems Vo (s) =
(16/s) Vg (s) 10 + s + (16/s)
H(s) =
16 16 Vo (s) = 2 = Vg (s) s + 10s + 16 (s + 2)(s + 8)
H(ω) =
16 (jω + 2)(jω + 8)
Vo (ω) = H(ω) · Vg (ω) = =
17–33
1152jω (4 − jω)(4 + jω)(2 + jω)(8 + jω)
K2 K3 K4 K1 + + + 4 − jω 4 + jω 2 + jω 8 + jω
K1 =
1152(4) =8 (8)(6)(12)
K2 =
1152(−4) = 72 (8)(−2)(4)
K3 =
1152(−2) = −32 (6)(2)(6)
K4 =
1152(−8) = −32 (12)(−4)(−6)
.·. Vo (jω) =
8 72 32 32 + − − 4 − jω 4 + jω 2 + jω 8 + jω
.·. vo (t) = 8e4t u(−t) + [72e−4t − 32e−2t − 32e−8t ]u(t)V [b] vo (0− ) = 8 V [c] vo (0+ ) = 72 − 32 − 32 = 8 V The voltages at 0− and 0+ must be the same since the voltage cannot change instantaneously across a capacitor. P 17.35 Vo (s) =
30 40 600(s + 10) 10 + − = s s + 20 s + 30 s(s + 20)(s + 30)
Vo (s) = H(s) ·
15 s
.·.
H(s) =
40(s + 10) (s + 20)(s + 30)
.·.
H(ω) =
40(jω + 10) (jω + 20)(jω + 30)
17–34 .·.
CHAPTER 17. The Fourier Transform 40(jω + 10) 1200(jω + 10) 30 · = jω (jω + 20)(jω + 30) jω(jω + 20)(jω + 30)
Vo (ω) =
vo (ω) =
60 80 20 + − jω jω + 20 jω + 30
vo (t) = 10sgn(t) + [60e−20t − 80e−30t ]u(t) V 1 P 17.36 [a] f (t) = 2π [b] W = 2 [c] W =
1 π
0
∞ 0
∞ 0
ω jtω
−∞
e e
dω +
∞
2
(1/π) 2 dt = 2 2 2 (1 + t ) π e−2ω dω =
−2ω
1e π −2
0
e
∞ 0
−ω jtω
dω =
1/π 1 + t2
dt 1 J = 2 2 (1 + t ) 2π
∞ = 0
e
1 J 2π
1 ω1 −2ω 0.9 [d] , 1 − e−2ω1 = 0.9, e dω = π 0 2π ω1 = (1/2) ln 10 ∼ = 1.15 rad/s
e2ω1 = 10
P 17.37
Io =
Ig R RCsIg = R + (1/sC) RCs + 1
H(s) =
s Io = Ig s + (1/RC)
RC = (100 × 103 )(1.25 × 10−6 ) = 125 × 10−3 ; H(s) =
s ; s+8
Ig (ω) =
30 × 10−6 jω + 2
H(ω) =
Io (ω) = H(ω)Ig (ω) =
jω jω + 8
30 × 10−6 jω (jω + 2)(jω + 8)
1 1 = =8 RC 0.125
Problems
17–35
ω(30 × 10−6 ) √ |Io (ω)| = √ 2 ( ω + 4)( ω 2 + 64) |Io (ω)|2 =
K1 K2 900 × 10−12 ω 2 = 2 + 2 2 2 (ω + 4)(ω + 64) ω + 4 ω + 64
K1 =
(900 × 10−12 )(−4) = −60 × 10−12 (60)
K2 =
(900 × 10−12 )(−64) = 960 × 10−12 (−60)
|Io (ω)|2 = W1Ω =
960 × 10−12 60 × 10−12 − ω 2 + 64 ω2 + 4
60 × 10−12 ∞ dω 1∞ 960 × 10−12 ∞ dω − |Io (ω)|2 dω = π 0 π ω 2 + 64 π ω2 + 4 0 0
= =
120 × 10−12 ω ∞ 30 × 10−12 ω ∞ tan−1 tan−1 − 8 0 π 2 0 π
120 π 30 π × 10−12 = (60 − 15) × 10−12 = 45 pJ · − · π 2 π 2
Between 0 and 4 rad/s W1Ω = %=
120 1 30 tan−1 − tan−1 2 × 10−12 = 7.14 pJ π 2 π
7.14 (100) = 15.86% 45
P 17.38 [a] Vg (ω) =
60 (jω + 1)(−jω + 1) 0.4 (jω + 0.5)
H(s) =
Vo 0.4 ; = Vg s + 0.5
Vo (ω) =
24 (jω + 1)(jω + 0.5)(−jω + 1)
Vo (ω) =
32 8 −24 + + jω + 1 jω + 0.5 −jω + 1
H(ω) =
vo (t) = [−24e−t + 32e−t/2 ]u(t) + 8et u(−t) V
17–36
CHAPTER 17. The Fourier Transform
[b] |Vg (ω)| =
[c] |Vo (ω)| =
[d] Wi = 2 [e] Wo =
60 + 1)
(ω 2
24 √ (ω 2 + 1) ω 2 + 0.25
∞ 0
0
−∞
= 32 +
−2t
900e
64e2t dt + ∞ 0
∞
e−2t = 900 J dt = 1800 −2 0 ∞ 0
(−24e−t + 32e−t/2 )2 dt
[576e−2t − 1536e−3t/2 + 1024e−t ] dt
= 32 + 288 − 1024 + 1024 = 320 J
Problems [f] |Vg (ω)| = Wg =
60 , +1
3600 (ω 2 + 1)2
dω 3600 2 π 0 (ω 2 + 1)2
3600 = π =
|Vg2 (ω)| =
ω2
2 1 ω −1 + tan ω 2 2 ω +1 0
1800 2 + tan−1 2 = 863.53 J π 5
.·. % =
[g] |Vo (ω)|2 = =
863.53 × 100 = 95.95% 900 576 (ω 2
+
1)2 (ω 2
+ 0.25)
768 1024 1024 − − ω 2 + 0.25 (ω 2 + 1)2 (ω 2 + 1)
2 2 1 ω 1 −1 + tan 1024 · 2 · tan−1 2ω −768 ω Wo = π 2 ω2 + 1 0 0 −1
−1024 tan =
ω
2 0
= 319.2 J %= P 17.39 Io =
319.2 × 100 = 99.75% 320
sIg 0.5sIg = 0.5s + 25 s + 50
H(s) =
s Io = Ig s + 50
H(ω) =
jω jω + 50
I(ω) =
2048 384 2 1024 tan−1 4 − + tan−1 2 − tan−1 2 π π 5 π
12 jω + 10
Io (ω) = H(ω)I(ω) =
12(jω) (jω + 10)(jω + 50)
17–37
17–38
CHAPTER 17. The Fourier Transform
|Io (ω)| =
|Io (ω)|2 = =
12ω (ω 2 + 100)(ω 2 + 2500)
144ω 2 (ω 2 + 100)(ω 2 + 2500) ω2
Wo (total) = =
150 −6 + 2 + 100 ω + 2500
1 π
∞ 0
1 150dω − ω 2 + 2500 π
∞ 0 ω2
6dω + 100
∞ ∞ 3 ω ω 0.6 −1 tan−1 tan − π 50 0 π 10 0
= 1.5 − 0.3 = 1.2 J Wo (0–100 rad/s) =
3 0.6 tan−1 (2) − tan−1 (10) π π
= 1.06 − 0.28 = 0.78 J Therefore, the percent between 0 and 100 rad/s is 0.78 (100) = 64.69% 1.2 P 17.40 [a] |Vi (ω)|2 =
4 × 104 ; ω2
|Vi (100)|2 =
4 × 104 = 4; 1002
|Vi (200)|2 =
4 × 104 =1 2002
Problems [b] Vo =
RCVi Vi R = R + (1/sC) RCs + 1
H(s) =
s Vo ; = Vi s + (1/RC)
H(ω) =
jω jω + 100
|Vo (ω)| =
4 × 104 , ω 2 + 104
|Vo (100)|2 =
[c] W1Ω =
1 106 10−3 1000 = = = 100 RC (0.5)(20) 10
|ω| 200 200 √ ·√ 2 = |ω| ω + 104 ω 2 + 104
|Vo (ω)|2 =
100 ≤ ω ≤ 200 rad/s;
4 × 104 = 2; 104 + 104
|Vo (200)|2 =
|Vo (ω)|2 = 0, 4 × 104 = 0.8 5 × 104
1 200 4 × 104 4 × 104 1 200 dω = − π 100 ω2 π ω 100
1 4 × 104 1 200 ∼ − = = = 63.66 J π 100 200 π [d] W1Ω
17–39
200 1 200 4 × 104 4 × 104 −1 ω · tan = dω = π 100 ω 2 + 104 π 100 100
=
400 [tan−1 2 − tan−1 1] ∼ = 40.97 J π
elsewhere
17–40
CHAPTER 17. The Fourier Transform A ; a + jω
P 17.41 [a] Vi (ω) = H(s) =
|Vi (ω)| = √
s ; s+α
H(ω) =
A + ω2
a2
jω ; α + jω
|H(ω)| = √
ω α2 + ω 2
ωA
Therefore |Vo (ω)| =
(a2 + ω 2 )(α2 + ω 2 )
Therefore |Vo (ω)|2 = WIN =
∞ 0
ω 2 A2 (a2 + ω 2 )(α2 + ω 2 )
A2 e−2at dt =
A2 ; 2a
when α = a we have
a a2 dω A2 a ω 2 dω A2 a dω WOUT (a) = = − π 0 (ω 2 + a2 )2 π 0 a2 + ω 2 0 (a2 + ω 2 )2
=
A2 4aπ
π −1 2
ω2 A2 ∞ A2 WOUT (total) = dω = π 0 (a2 + ω 2 )2 4a
Therefore
1 WOUT (a) = 0.5 − = 0.1817 or WOUT (total) π
18.17%
[b] When α = a we have WOUT (α) =
ω 2 A2 dω 1 α π 0 (a2 + ω 2 )(α2 + ω 2 )
A2 = π where K1 = Therefore
α 0
K1 K2 + 2 dω 2 2 a +ω α + ω2
2
a a2 − α2
and
K2 =
−α2 a2 − α2
α A2 απ a tan−1 − WOUT (α) = 2 2 π(a − α ) a 4
π A2 π A2 − α WOUT (total) = a = π(a2 − α2 ) 2 2 2(a + α)
Therefore
α απ 2 WOUT (α) · a tan−1 − = WOUT (total) π(a − α) a 4
√ For α = a 3, this ratio is 0.2723, √ or 27.23% of the output energy lies in the frequency band between 0 and a 3. √ [c] For α = a/ 3, the ratio is 0.1057, √ or 10.57% of the output energy lies in the frequency band between 0 and a/ 3.
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