circuits nilsson 7th solution manual
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1Circuit Variables
Assessment Problems
AP 1.1 To solve this problem we use a product of ratios to change units from dollars/year todollars/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 ofratios:
1 year365.25 days
· 1 day24 hours
· 1 hour60 mins
· 1 min60 secs
· 1 sec1000 ms
=1 year
31.5576 × 109 ms
Now we can convert from dollars/year to dollars/millisecond, again with a productof ratios:
$100 × 109
1 year· 1 year31.5576 × 109 ms
=100
31.5576= $3.17/ms
AP 1.2 First, we recognize that 1 ns = 10−9 s. The question then asks how far a signal willtravel in 10−9 s if it is traveling at 80% of the speed of light. Remember that thespeed 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 toinches/nanosecond:
2.4 × 108 m1s
· 1 s109 ns
· 100 cm1 m
· 1 in2.54 cm
=(2.4 × 108)(100)
(109)(2.54)=
9.45 in1 ns
Thus, a signal traveling at 80% of the speed of light will travel 9.45′′ in ananosecond.
1–1
1–2 CHAPTER 1. Circuit Variables
AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or i = dqdt
Inthis 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
0i(x) dx
We are given the expression for current, i, which can be substituted into the aboveexpression. To find the total charge, we let t → ∞ in the integral. Thus we have
qtotal =∫ ∞
020e−5000x dx =
20−5000
e−5000x
∣∣∣∣∞
0=
20−5000
(e∞ − e0)
=20
−5000(0 − 1) =
205000
= 0.004 C = 4000µC
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or i = dqdt
. Inthis problem we are given an expression for the charge, and asked to find themaximum current. First we will find an expression for the current using Eq. (1.2):
i =dq
dt=
d
dt
[ 1α2 −
(t
α+
1α2
)e−αt
]
=d
dt
( 1α2
)− d
dt
(t
αe−αt
)− d
dt
( 1α2 e−αt
)
= 0 −( 1
αe−αt − α
t
αe−αt
)−(−α
1α2 e−αt
)
=(− 1
α+ t +
1α
)e−αt
= te−αt
Now that we have an expression for the current, we can find the maximum value ofthe current by setting the first derivative of the current to zero and solving for t:
di
dt=
d
dt(te−αt) = e−αt + t(−α)eαt = (1 − αt)e−αt = 0
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 valueof t, the current is
i =1α
e−α/α =1α
e−1
Remember in the problem statement, α = 0.03679. Using this value for α,
i =1
0.03679e−1 ∼= 10 A
Problems 1–3
AP 1.5 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 thepolarities shown in Fig. 1.6. Remember that 4A of current entering Terminal 2is the same as 4A of current leaving Terminal 1. We get
(a) v = −20 V, i = −4 A; (b) v = −20 V, i = 4 A
(c) v = 20 V, i = −4 A; (d) v = 20 V, i = 4 A
[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 isabsorbing 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 andcurrent polarities shown in Fig. 1.5, p = vi. From Eq. (1.3), we know that power isthe time rate of change of energy, or p = dw
dt. If we know the power, we can find the
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 × 105e−10,000t W
Now find the expression for energy by integrating Eq. (1.3):
w(t) =∫ t
0p(x) dx
1–4 CHAPTER 1. Circuit Variables
Substitute the expression for power, p, above. Note that to find the total energy, welet t → ∞ in the integral. Thus we have
w =∫ ∞
02 × 105e−10,000x dx =
2 × 105
−10,000e−10,000x
∣∣∣∣∣∞
0
=2 × 105
−10,000(e−∞ − e0) =
2 × 105
−10,000(0 − 1) =
2 × 105
10,000= 20 J
AP 1.7 At the Oregon end of the line the current is leaving the upper terminal, and thusentering the lower terminal where the polarity marking of the voltage is negative.Thus, using the passive sign convention, p = −vi. Substituting the values of voltageand 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 tobe 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 pixelrequires 3 bytes of information. We can calculate the number of bytes ofinformation required for the display by multiplying the number of pixels in thedisplay by 3 bytes per pixel:
nbytes =1,310,720 pixels
1 display· 3 bytes
1 pixel= 3,932,160 bytes/display
Finally, we use the fact that there are 106 bytes per MB:
3,932,160 bytes1 display
· 1 MB106bytes
= 3.93 MB/display
Problems 1–5
P 1.2 c = 3 × 108 m/s so12c = 1.5 × 108 m/s
1.5 × 108 m1 s
=5 × 106 m
x sso x =
5 × 106
1.5 × 108 = 33.3 ms
P 1.3 We can set up a ratio to determine how long it takes the bamboo to grow 10 µm First,recall that 1 mm = 103µm. Let’s also express the rate of growth of bamboo using theunits mm/s instead of mm/day. Use a product of ratios to perform this conversion:
250 mm1 day
· 1 day24 hours
· 1 hour60 min
· 1 min60 sec
=250
(24)(60)(60)=
103456
mm/s
Use a ratio to determine the time it takes for the bamboo to grow 10 µm:
10/3456 × 10−3 m1 s
=10 × 10−6 m
x sso x =
10 × 10−6
10/3456 × 10−3 = 3.456 s
P 1.4 Volume = area × thickness
106 = (10 × 106)(thickness)
⇒ thickness =106
10 × 106 = 0.10 mm
P 1.5300 × 109 dollars
1 year· 100 pennies
1 dollar· 1 year365.25 days
· 1 day24 hr
· 1 hr3600 s
· 1.5 mm1 penny
· 1 m1000 mm
= 1426 m/s
P 1.6 Our approach is as follows: To determine the area of a bit on a track, we need toknow the height and width of the space needed to store the bit. The height of thespace used to store the bit can be determined from the width of each track on thedisk. The width of the space used to store the bit can be determined by calculatingthe number of bits per track, calculating the circumference of the inner track, anddividing the number of bits per track by the circumference of the track. Thecalculations are shown below.
Width of track =1 in
77 tracks25,400µm
in= 329.87µm/track
Bits on a track =1.4 MB2 sides
8 bitsbyte
1 side77 tracks
= 72,727.273 bits/track
Circumference of inner track = 2π(1/2′′)(25,400µm/in) = 79,796.453µm
Width of bit on inner track =79,796.453µm72,727.273 bits
= 1.0972µm/bit
Area of bit on inner track = (1.0972)(329.87) = 361.934µm2
1–6 CHAPTER 1. Circuit Variables
P 1.7 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)Cm
× ave vel(
m
s
)= i
Thus, average velocity =14008.652
× 10−6 = 161.81 µm/s
P 1.8 [a] n =20 × 10−6 C/s
1.6022 × 10−19 C/elec= 1.25 × 1014 elec/s
[b] m = 3303 mi · 5280 ft1 mi
· 12 in1 ft
· 2.54 cm1 in
· 104 µm1 cm
= 5.32 × 1012 µm
Therefore,n
m= 23.5
The number of electrons/second is approximately 23.5 times the number ofmicrometers between Sydney and San Francisco.
P 1.9 First we use Eq. (1.2) to relate current and charge:
i =dq
dt= 20 cos 5000t
Therefore, dq = 20 cos 5000t dt
To find the charge, we can integrate both sides of the last equation. Note that wesubstitute x for q on the left side of the integral, and y for t on the right side of theintegral:
∫ q(t)
q(0)dx = 20
∫ t
0cos 5000y dy
We solve the integral and make the substitutions for the limits of the integral,remembering that sin 0 = 0:
q(t) − q(0) = 20sin 5000y
5000
∣∣∣∣t
0=
205000
sin 5000t − 205000
sin 5000(0) =20
5000sin 5000t
But q(0) = 0 by hypothesis, i.e., the current passes through its maximum value att = 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 1–7
P 1.11 p = (6)(100 × 10−3) = 0.6 W; 3 hr · 3600 s1 hr
= 10,800 s
w(t) =∫ t
0p dt w(10,800) =
∫ 10,800
00.6 dt = 0.6(10,800) = 6480 J
P 1.12 Assume we are standing at box A looking toward box B. Then, using the passivesign convention p = vi, since the current i is flowing into the + terminal of thevoltage 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 beflowing from A to B. If the power is negative, B is generating power so the powermust be flowing from B to A.
[a] p = (20)(15) = 300 W 300 W from A to B
[b] p = (100)(−5) = −500 W 500 W from B to A
[c] p = (−50)(4) = −200 W 200 W from B to A
[d] p = (−25)(−16) = 400 W 400 W from A to B
P 1.13 [a]
p = vi = (−20)(5) = −100 WPower 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 Vbattery(the current i flows into the + terminal of the battery of Car A).Therefore using the passive sign convention, p = vi = (−40)(12) = −480W.Since the power is negative, the battery in Car A is generating power, so Car Bmust have the ”dead” battery.
1–8 CHAPTER 1. Circuit Variables
[b] w(t) =∫ t
0p dx; 1.5 min = 1.5 · 60 s
1 min= 90 s
w(90) =∫ 90
0480 dx
w = 480(90 − 0) = 480(90) = 43,200 J = 43.2 kJ
P 1.16 p = vi; w =∫ t
0p 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
w = (60 × 10−3)(216 × 103) +12(90 × 10−3 − 60 × 10−3)(216 × 103) = 16.2 kJ
P 1.17 [a] To find the power at 625 µs, we substitute this value of time into both theequations 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(625×10−6) = 18.394 − 38.94 = −20.546 V
i(625 µs) = 5 × 10−3e−1600(625×10−6) − 5 × 10−3e−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 to625 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
=14(e−3200t − 2e−2000t + e−800t)
Now we integrate this expression for p(t) to get an expression for w(t). Notewe substitute x for t on the right side of the integral.
w(t) =14
∫ t
0(e−3200x − 2e−2000x + e−800x)dx
=14
[e−3200x
−3200+
e−2000x
1000− e−800x
800
]∣∣∣∣∣t
0
=14
[e−3200t
−3200+
e−2000t
1000− e−800t
800−( 1
−3200+
11000
− 1800
)]
=14
[e−3200t
−3200+
e−2000t
1000− e−800t
800+ 5.625 × 10−4
]
Finally, substitute t = 625 µs into the equation for w(t):
w(625 µs) =14[−4.2292 × 10−5 + 2.865 × 10−4 − 7.5816 × 10−4 + 5.625 × 10−4]
= 12.137 µJ
[c] To find the total energy, we let t → ∞ in the above equation for w(t). Note thatthis will cause all expressions of the form e−nt to go to zero, leaving only theconstant term 5.625 × 10−4. Thus,
wtotal =14[5.625 × 10−4] = 140.625 µJ
P 1.18 [a] v(20 ms) = 100e−1 sin 3 = 5.19 Vi(20 ms) = 20e−1 sin 3 = 1.04 A p(20 ms) = vi = 5.39 W
[b] p = vi = 2000e−100t sin.2 150t
= 2000e−100t[12
− 12
cos 300t]
= 1000e−100t − 1000e−100t cos 300t
w =∫ ∞
01000e−100t dt −
∫ ∞
01000e−100t cos 300t dt
= 1000e−100t
−100
∣∣∣∣∣∞
0
−1000
e−100t
(100)2 + (300)2 [−100 cos 300t + 300 sin 300t]∣∣∣∣∣
∞
0
= 10 − 1000[ 1001 × 104 + 9 × 104
]= 10 − 1
w = 9 J
1–10 CHAPTER 1. Circuit Variables
P 1.19 [a] 0 s ≤ t < 1 s:
v = 5 V; i = 20t A; p = 100t W
1 s < t ≤ 2 s:
v = 0 V; i = 20 A; p = 0 W
2 s ≤ t < 3 s:
v = 0 V; i = 20 A; p = 0 W
3 s < t ≤ 4 s:
v = −5 V; i = 80 − 20t A; p = −400 + 100t W
4 s ≤ t < 5 s:
v = −5 V; i = 80 − 20t A; p = −400 + 100t W
5 s < t ≤ 6 s:
v = 5 V; i = −120 + 20t A; p = −600 + 100t W
6 s ≤ t < 7 s:
v = 5 V; i = −120 + 20t A; p = −600 + 100t W
t > 7 s:
v = 0 V; i = 20 A; p = 0 W
[b] Calculate the area under the curve from zero up to the desired time:
w(1) = 12(1)(100) = 50 J
w(6) = 12(1)(100) − 1
2(1)(100) + 12(1)(100) − 1
2(1)(100) = 0 J
w(10) = 12(1)(100) − 1
2(1)(100) + 12(1)(100) − 1
2(1)(100) + 12(1)(100) = 50 J
P 1.20 [a] p = vi = (100e−500t)(0.02 − 0.02e−500t) = (2e−500t − 2e−1000t) W
dp
dt= −1000e−500t + 2000e−1000t = 0 so 2e−1000t = e−500t
2 = e500t so ln 2 = 500t thus p is maximum at t = 1.4 ms
Problems 1–11
pmax = p(1.4 ms) = 0.5 W
[b] w =∫ ∞
0[2e−500t − 2e−1000t] dt =
[ 2−500
e−500t − 2−1000
e−1000t
∣∣∣∣∞
0
]
=4
1000− 2
1000= 2 mJ
P 1.21 [a] p = vi = 200 cos(500πt)4.5 sin(500πt) = 450 sin(1000πt) WTherefore, pmax = 450 W
[b] pmax(extracting) = 450 W
[c]
pavg =1
4 × 10−3
∫ 4×10−3
0450 sin(1000πx) dx =
4504 × 10−3
[− cos 1000πt
1000π
]4×10−3
0
=−4504π
[cos 4π − cos 0] =−4504π
[1 − 1] = 0 W
[d] pavg =−4504π
[cos 15π − cos 0] =−4504π
[−1 − 1] =9004π
= 71.62 W
P 1.22 [a] q = area under i vs. t plot
=6(5000)
2+ 6(5000) +
6(10,000)2
+ 8(15,000) +8(5000)
2= 15,000 + 30,000 + 30,000 + 120,000 + 20,000 = 215,000 C
[b] w =∫
pdt =∫
vi dt
v = 0.2 × 10−3t + 8 0 ≤ t ≤ 20 ks0 ≤ t ≤ 5000s
i = 20 − 1.2 × 10−3t
p = (8 + 0.2 × 10−3t)(20 − 1.2 × 10−3t)
= 160 − 5.6 × 10−3t − 2.4 × 10−7t2
w1 =∫ 5000
0(160 − 5.6 × 10−3t − 2.4 × 10−7t2) dt
=
(160t − 5.6 × 10−3
2t2 − 2.4 × 10−7
3t3)∣∣∣∣∣
5000
0
= 720 kJ
5000 ≤ t ≤ 15,000s
i = 17 − 0.6 × 10−3t
p = (8 + 0.2 × 10−3t)(17 − 0.6 × 10−3t)
= 136 − 1.4 × 10−3t − 1.2 × 10−7t2
w2 =∫ 15,000
5000(136 − 1.4 × 10−3t − 1.2 × 10−7t2) dt
=
(136t − 1.4 × 10−3
2t2 − 1.2 × 10−7
3t3)∣∣∣∣∣
15,000
5000
= 1090 kJ
1–12 CHAPTER 1. Circuit Variables
15,000 ≤ t ≤ 20,000s
i = 32 − 1.6 × 10−3t
p = (8 + 0.2 × 10−3t)(32 − 1.6 × 10−3t)
= 256 − 6.4 × 10−3t − 3.2 × 10−7t2
w3 =∫ 20,000
15,000(256 − 6.4 × 10−3t − 3.2 × 10−7t2) dt
=
(256t − 6.4 × 10−3
2t2 − 3.2 × 10−7
3t3)∣∣∣∣∣
20,000
15,000
= 226,666.67 J
wT = w1 + w2 + w3 = 720,000 + 1,090,000 + 226,666.67 = 2036.67 kJ
P 1.23 [a] p = vi = [104t + 5)e−400t][(40t + 0.05)e−400t]
= 400 × 103t2e−800t + 700te−800t + 0.25e−800t
= e−800t[400,000t2 + 700t + 0.25]dp
dt= e−800t[800 × 103t + 700] − 800e−800t[400,000t2 + 700t + 0.25]= [−3,200,000t2 + 2400t + 5]100e−800t
Therefore,dp
dt= 0 when 3,200,000t2 − 2400t − 5 = 0
so pmax occurs at t = 1.68 ms.
[b] pmax = [400,000(.00168)2 + 700(.00168) + 0.25]e−800(.00168)
= 666.34 mW
[c] w =∫ t
0pdx
w =∫ t
0400,000x2e−800x dx +
∫ t
0700xe−800x dx +
∫ t
00.25e−800x dx
=400,000e−800x
−512 × 106 [64 × 104x2 + 1600x + 2]∣∣∣∣∣t
0
+
700e−800x
64 × 104 (−800x − 1)∣∣∣∣∣t
0
+ 0.25e−800x
−800
∣∣∣∣∣t
0When t → ∞ all the upper limits evaluate to zero, hence
w =(400,000)(2)512 × 106 +
70064 × 104 +
0.25800
= 2.97 mJ.
P 1.24 [a] We can find the time at which the power is a maximum by writing an expressionfor p(t) = v(t)i(t), taking the first derivativeof 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 = vi = (t − 0.025t2)(4 − 0.2t) = 4t − 0.3t2 + 0.005t3 W 0 < t < 40 sdp
dt= 4 − 0.6t + 0.015t2 = 0
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 bysubstituting 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 whichthe power is maximum: pmax = 15.396 W (delivered)
[c] As we saw in part (a), the other “maximum” power is actually a minimum, orthe maximum negative power. As we calculated in part (a), maximum power isbeing extracted at t = 31.547 s.
[d] This maximum extracted power was calculated in part (a) to determine the timeat which power is maximum: pmaxext = 15.396 W (extracted)
[e] w =∫ t
0pdx =
∫ t
0(4x − 0.3x2 + 0.005x3)dx = 2t2 − 0.1t3 + 0.00125t4
w(0) = 0 J w(30) = 112.50 J
w(10) = 112.50 J w(40) = 0 J
w(20) = 200 JTo give you a feel for the quantities of voltage, current, power, and energy andtheir relationships among one another, they are plotted below:
1–14 CHAPTER 1. Circuit Variables
Problems 1–15
P 1.25 [a] p = vi = (8 × 104te−500t)(15te−500t) = 12 × 105t2e−1000t Wdp
dt= 12 × 105[t2(−1000)e−1000t + e−1000t(2t)]
= 12 × 105e−1000t[t(2 − 1000t)]dp
dt= 0 at t = 0, t = 2 ms
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)2e−2 = 649.61 mW
[c] w = 12 × 105∫ ∞
0t2e−1000t dt
= 12 × 105
e−1000t
(−1000)3 [106t2 + 2,000t + 2]∣∣∣∣∣∞
0
= 2400µJ
P 1.26 We use the passive sign convention to determine whether the power equation isp = vi or p = −vi and substitute into the power equation the values for v and i, asshown below:
pa = −vaia = −(−18)(−51) = −918 W
pb = vbib = (−18)(45) = −810 W
pc = vcic = (2)(−6) = −12 W
pd = −vdid = −(20)(−20) = 400 W
pe = −veie = −(16)(−14) = 224 W
pf = vfif = (36)(31) = 1116 WRemember 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 canadd the positive powers together and the negative powers together — if the powerbalances, 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.
1–16 CHAPTER 1. Circuit Variables
P 1.27 [a] From the diagram and the table we have
pa = −vaia = −(900)(−22.5) = 20,250 W
pb = −vbib = −(105)(−52.5) = 5512.5 W
pc = −vcic = −(−600)(−30) = −18,000 W
pd = vdid = (585)(−52.5) = −30,712.5 W
pe = −veie = −(−120)(30) = 3600 W
pf = vfif = (300)(60) = 18,000 W
pg = −vgig = −(585)(82.5) = −48,262.5 W
ph = −vhih = −(−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 powerabsorbed 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 canrecalculate 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 balancesfor the circuit.
P 1.28 pa = vaia = (9)(1.8) = 16.2 W
pb = −vbib = −(−15)(1.5) = 22.5 W
pc = −vcic = −(45)(−0.3) = 13.5 W
pd = −vdid = −(54)(−2.7) = 145.8 W
pe = veie = (−30)(−1) = 30 W
pf = −vfif = −(−240)(4) = 960 W
pg = −vgig = −(294)(4.5) = −1323 W
ph = vhih = (−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 = vaia = (−160)(−10) = 1600 W
pb = vbib = (−100)(−20) = 2000 W
pc = −vcic = −(−60)(6) = 360 W
pd = vdid = (800)(−50) = −40,000 W
pe = −veie = −(800)(−20) = 16,000 W
pf = −vfif = −(−700)(14) = 9800 W
pg = −vgig = −(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 thepassive 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.
2Circuit Elements
Assessment Problems
AP 2.1
[a] To find vg write a KVL equation clockwise around the left loop, starting belowthe dependent source:
−ib
4+ vg = 0 so vg =
ib
4To find ib write a KCL equation at the upper right node. Sum the currentsleaving the node:
ib + 8 A = 0 so ib = −8 A
Thus,
vg =−84
= −2 V
[b] To find the power associated with the 8 A source, we need to find the voltagedrop across the source, vi. To do this, write a KVL equation clockwise aroundthe left loop, starting below the voltage source:
−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 thetop left node, summing the currents leaving:
15 A + αvx = 0
Substituting for vx,
15 A + α(−25 V) = 0 so α(25 V) = 15 A
Thus α =15 A25 V
= 0.6 A/V
[b] To find the power associated with the voltage source we need to know thecurrent, 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. UseOhm’s law to calculate the value of the resistor:
R =vR
ig
=1 kV5 mA
= 200 kΩ
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 sourceis thus
psource = −vgig so vg = −psource
ig
= −(−3 W)75 mA
= 40 V
Since we now have the value of both the voltage and the current for theresistor, we can use Ohm’s law to calculate the resistor value:
R =vg
ig
=40 V
75 mA= 533.33 Ω
The power absorbed by the resistor must equal the power generated by thesource. Thus,
pR = −psource = −(−3 W) = 3 W
[c] Again, note the iR = ig. The power dissipated by the resistor can be determinedfrom the resistor’s current:
pR = R(iR)2 = R(ig)2
Solving for ig,
i2g =pr
R=
480 mW300 Ω
= 0.0016 so ig =√
0.0016 = 0.04 A = 40 mA
Then, since vR = vg
vR = RiR = Rig = (300 Ω)(40 mA) = 12 V so vg = 12 V
AP 2.4
[a] Note from the circuit that the current throught the conductance G is ig, flowingfrom top to bottom (from KCL), and the voltage drop across the current sourceis vg, positive at the top (from KVL). From a version of Ohm’s law,
vg =ig
G=
0.5 A50 mS
= 10 V
Now that we know the voltage drop across the current source, we can find thepower delivered by this source:
psource = −vgig = −(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 thecurrent using Ohm’s law and the conductance value:
pg = Gv2g so G =
pg
v2g
=9
152 = 0.04 S = 40 mS
ig = Gvg = (40 mS)(15 V) = 0.6 A
[c] We can find the voltage from the power and the conductance, and then use thevoltage value in Ohm’s law to find the current:
pg = Gv2g so v2
g =pg
G=
8 W200 µS
= 40,000
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 circuitelement.
Write a KVL equation clockwise around the circuit, starting below the voltagesource:
−24 V + v2 + v5 − v1 = 0
Next, use Ohm’s law to calculate the three unknown voltages from the threecurrents:
v2 = 3i2; v5 = 7i5; v1 = 2i1
A KCL equation at the upper right node gives i2 = i5; a KCL equation at thebottom right node gives i5 = −i1; a KCL equation at the upper left node givesis = −i2. Now replace the currents i1 and i2 in the Ohm’s law equations withi5:
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 voltageand 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 =1208
= 15 A
Also use Ohm’s law to calculate the value of the current through the 24 Ω resistor:
i2 =120 V24 Ω
= 5 A
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 voltagesource:
−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 unknownresistor, we can use Ohm’s law to calculate the resistance:
R =v1
i1=
8020
= 4 Ω
AP 2.7 [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 ofresistance:
R =∆v
∆i=
25 − 00.25 − 0
=25
0.25= 100 Ω
A circuit model having the same v − i characteristic is a 25 V source in serieswith 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 currentthrough the 25 Ω resistor. Do this by first using KCL to recognize that thecurrent in each of the components is it, flowing in a clockwise direction. Writea 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 thedirection of the current it flowing through the resistors:
−25 V + 100it + 25it = 0 so 125it = 25 so it =25125
= 0.2 A
Thus, the power delivered to the 25 Ω resistor is
p25 = (25)i2t = (25)(0.2)2 = 1 W.
Problems 2–7
AP 2.8 [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 astraight line, its slope can be used to calculate the value of resistance:
R =∆v
∆i=
25 − 00.25 − 0
=25
0.25= 100 Ω
A circuit model having the same v − i characteristic is a 0.25 A current sourcein parallel with a 100Ω resistor, as shown below:
[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 thevoltage drop across both resistors is vt. Write a KCL equation at the top centernode, summing the currents leaving the node. Use Ohm’s law to specify thecurrents through the resistors in terms of the voltage drop across the resistorsand the value of the resistors.
−0.25 +vt
100+
vt
25= 0, so 5vt = 25, thus vt = 5 V
p25 =v2
t
25= 1 W.
AP 2.9 First note that we know the current through all elements in the circuit except the 6kΩ resistor (the current in the three elements to the left of the 6 kΩ resistor is i1; thecurrent in the three elements to the right of the 6 kΩ resistor is 30i1). To find thecurrent 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 ina 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
240,000= 25 µA
[b] Now that we have the value of i1, we can calculate the voltage for eachcomponent except the dependent source. Then we can write a KVL equationfor the right-hand loop to find the voltage v of the dependent source. Sum thevoltages 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’sconstruct a power table:
Problems 2–9
Element Current Voltage Power Power
(µA) (V) Equation (µW)
5 V 25 5 p = −vi −125
54 kΩ 25 1.35 p = Ri2 33.75
1 V 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
8 V 750 8 p = −vi −6000
[c] The total power generated in the circuit is the sum of the negative power valuesin 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 inthe power table:
33.75 µW + 3603.75 µW + 1500µW + 1012.5 µW = 6150µW
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. Wecan 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 thecircuit below.
2–10 CHAPTER 2. Circuit Elements
[a] To find vs, write a KVL equation, summing the voltages counter-clockwisearound 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 theright-hand node. Sum the currents leaving the node
−4 A + 1 A + iv = 0 so iv = 4 A − 1 A = 3 A
The current in the voltage source is 3 A, flowing top to bottom. The powerassociated 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 writinga 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 = −v5Ai = −(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 KVLequation 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 = v4Ai = (−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
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
P 2.2 Since we know the device is a resistor, we can use Ohm’s law to calculate theresistance. From Fig. P2.2(a),
v = Ri so R =v
i
Using the values in the table of Fig. P2.2(b),
R =−160−0.02
=−80
−0.01=
800.01
=1600.02
=2400.03
= 8kΩ
P 2.3 The resistor value is the ratio of the power to the square of the current:
50012 =
200022 =
450032 =
800042 =
12,50052 =
18,00062 = 500 Ω
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
Rso R =
v2
p
Using the values in the table of Fig. P2.4(b)
R =(−8)2
3.2=
(−4)2
0.8=
(4)2
0.8=
(8)2
3.2=
(12)2
7.2=
(16)2
12.8= 20 Ω
2–12 CHAPTER 2. Circuit Elements
P 2.5 [a] Yes, independent voltage sources can carry whatever current is required by theconnection; independent current source can support any voltage required bythe connection.
[b] 18 V source: absorbing
5 mA source: delivering
7 V source: absorbing
[c] P18V = (5 × 10−3)(18) = 90 mW (abs)
P5mA = −(5 × 10−3)(25) = −125 mW (del)
P7V = (5 × 10−3)(7) = 35 mW (abs)∑Pabs =
∑Pdel = 125 mW
[d] Yes; 18 V source is delivering, the 5 mA source is absorbing, and the 7 V sourceis absorbing
P18V = −(5 × 10−3)(18) = −90 mW (del)
P5mA = (5 × 10−3)(11) = 55 mW (abs)
P7V = (5 × 10−3)(7) = 35 mW (abs)∑Pabs =
∑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: − v25 + v20 = 0
KVL, right loop: 60V − 100V − v5 − v20 = 0
Problems 2–13
KVL, outer loop: 60V − 100V − v5 − v25 = 0
Note that since v5, v20, and v25 are not specified, we can choose values that satisfythe equations. For example, let v5 = −80V, v20 = 40V, and v25 = 40V. There aremany other voltage values that will satisfy the equations, too.Thus, the interconnection is valid because it does not violate Kirchhoff’s laws. Wecan 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, ordeveloping −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 usethe values of i8, v30 and v10 stated above to calculate the power associated with eachsource:
p30A = −(30)(−8) = 240 W p16V = −(30)(16) = −480 W
2–14 CHAPTER 2. Circuit Elements
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 so v5A = 20 V (KVL)
15 + 5 + i50V = 0 so i50V = −20 A (KCL)
p15A = −(15)(50) = −750 W p50V = (20)(50) = 1000 W
p5A = −(5)(20) = −100 W p10V = (5)(10) = 50 W
p40V = −(5)(40) = −200 W∑
Pdev =∑
Pabs = 1050 W
P 2.9 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: − v75 − 50V + vdep − 20V = 0
KVL, right loop: 20V − vdep + v∆ = 0
KVL, outer loop: − v75 − 50V + v∆ = 0
2–16 CHAPTER 2. Circuit Elements
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. Thesevalues 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 inthe 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
∑Pdev = 3750 + 1000 = 4750 W =
∑Pabs
P 2.12 [a] Yes, Kirchhoff’s laws are not violated. (Note that i∆ = −8 A.)
[b] No, because the voltages across the independent and dependent current sourcesare 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 willbe 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 2–17
P 2.13 First, 10va = 5 V, so va = 0.5 VKVL for the outer loop: 5 − 20 + v9A = 0 so v9A = 15 VKVL for the right loop: 5 − 0.5 + vg = 0 so vg = −4.5 VKCL at the top node: 9 + 6 + ids = 9 so ids = −15 AThus,
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 Ωresistor:
−va + vb = −0 so va = vb
Using Ohm’s law,
va = 300ia and vb = 75ib
Substituting,
300ia = 75ib so 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 thevoltage source:
−200 V + v40 + va = 0
From Ohm’s law,
v40 = 40ig and va = 300ia
2–18 CHAPTER 2. Circuit Elements
Substituting,
−200 V + 40ig + 300ia = 0
Subsituting for ig:
−200 V + 40(5ia) + 300ia = −200 V + 200ia + 300ia = −200 V + 500ia = 0
Thus,
500ia = 200 V so ia =200 V500
= 0.4 A
[b] From part (a), ib = 4ia = 4(0.4 A) = 1.6 A.
[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:∑Pdis = 160 + 48 + 192 = 400 W
∑Pdel = 400 W
P 2.15 [a] vo = 8ia + 14ia + 18ia = 40(20) = 800 V
800 = 10io
io = 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 so 120io = 240 V
Thus io =240 V120
= 2 A
Now write a KCL equatiohn at the top middle node, summing the currentsleaving:
−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 = −voig = −(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
2–20 CHAPTER 2. Circuit Elements
P 2.17 [a]
v2 = 2(20) = 40 Vv8Ω = 80 − 40 = 40 Vi2 = 40 V/8 Ω = 5 Ai3 = io − i2 = 2 − 5 = −3 Av4Ω = (−3)(4) = −12 Vv1 = 4i3 + v2 = −12 + 40 = 28 Vi1 = 28 V/4 Ω = 7 A
[b] i4 = i1 + i3 = 7 − 3 = 4 A
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
[c]∑
Pdis = 208 + 200 + 196 + 36 + 80 = 720 W
ig = i4 + i2 = 4 + 5 = 9 A
Pdev = (9)(80) = 720 W
Problems 2–21
P 2.18 [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 = (25)2(2) + (20)2(8) + (5)2(4) + (15)216 + (20)22 + (5)2(80)
= 1250 + 3200 + 100 + 3600 + 800 + 2000 = 10,950 W∑Psup = 5750 + 5200 = 10,950 W
Therefore,∑
Pdis =∑
Psup = 10,950 W
P 2.19 [a]
v2 = 100 + 4(15) = 160 V; v1 = 160 − 30(2) = 100 V
i1 =v1
20=
10020
= 5 A; i3 = i1 − 2 = 5 − 2 = 3 A
2–22 CHAPTER 2. Circuit Elements
vg = v1 + 30i3 = 100 + 30(3) = 190 V
vg − 5i4 = v2 so 5i4 = vg − v2 = 190 − 160 = 30 V
Thus i4 =305
= 6 A
ig = i3 + i4 = 3 + 6 = 9 A
[b] Calculate power using the formula p = Ri2:
p9 Ω = (9)(2)2 = 36 W; p11 Ω = (11)(2)2 = 44 W
p10 Ω = (10)(2)2 = 40 W; p30 Ω = (30)(3)2 = 270 W
p5 Ω = (5)(6)2 = 180 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 = −vgig = −(190 V)(9 A) = −1710 W
Thus the total power dissipated is 1310 + 400 = 1710 W and the total powerdeveloped is 1710 W, so the power balances.
P 2.20 [a] Plot the v − i characteristic
From the plot:
R =∆v
∆i=
(125 − 50)(15 − 0)
= 5 Ω
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:
R =∆v
∆i=
130 − (−30)8 − 0
= 20 Ω
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 a20 Ω resistor, as shown below:
2–24 CHAPTER 2. Circuit Elements
[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 toexpress the voltage drop across the resistors in terms of the current it throughthe resistors:
−(−30 V) − 20it − 40it = 0 so − 60it = −30 V
Thus it =−30 V−60
= +0.5 A
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 =∆v
∆i=
50 V5 mA
= 10 kΩ
Problems 2–25
[c]
10,000i1 = 2500is so i1 = 0.25is
0.02 = i1 + is = 0.25is + is = 1.25is
Thus, is =0.021.25
= 0.016 A = 16 mA
[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 cancalculate R from the plotted values as follows:
R =∆v
∆i=
24 − 1824 − 0
=624
= 0.25 Ω
We can determine the value of the ideal voltage source by considering thevalue of vs when is = 0. When there is no current, there is no voltage dropacross the resistor, so all of the voltage drop at the output is due to the voltagesource. Thus the value of the voltage source must be 24 V. The model, validfor 0 ≤ is ≤ 24 A, is shown below:
2–26 CHAPTER 2. Circuit Elements
[c] The circuit is shown below:
Write a KVL equation in the clockwise direction, starting below the voltagesource. Use Ohm’s law to express the voltage drop across the resistors in termsof the current i:
−24 V + 0.25i + 1i = 0 so 1.25i = 24 V
Thus, i =24 V
1.25 Ω= 19.2 A
[d] The circuit is shown below:
Write a KVL equation in the clockwise direction, starting below the voltagesource. Use Ohm’s law to express the voltage drop across the resistors in termsof the current i:
−24 V + 0.25i = 0 so 0.25i = 24 V
Thus, i =24 V
0.25 Ω= 96 A
[e] The short circuit current can be found in the table of values (or from the plot) asthe 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 ispiecewise linear, but not linear for all values of is). Since the proposed circuitmodel is a linear model, it cannot be used to predict the nonlinear behaviorexhibited by the plotted data.
Problems 2–27
P 2.24
vab = 240 − 180 = 60 V; therefore, ie = 60/15 = 4 Aic = ie − 1 = 4 − 1 = 3 A; therefore, vbc = 10ic = 30 Vvcd = 180 − vbc = 180 − 30 = 150 V;
therefore, id = vcd/(12 + 18) = 150/30 = 5 Aib = id − ic = 5 − 3 = 2 Avac = vab + vbc = 60 + 30 = 90 VR = vac/ib = 90/2 = 45 Ω
CHECK: 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)
P 2.25 [a]
ib = 60 V/30 Ω = 2 Ava = (30 + 60)(2) = 180 V−500 + va + vb = 0 so vb = 500 − va = 500 − 180 = 320 Vie = vb/(60 + 36) = 320/96 = (10/3) Aid = ie − ib = (10/3) − 2 = (4/3) Avc = 30id + vb = 40 + 320 = 360 Vic = vc/180 = 360/180 = 2 Avd = 500 − vc = 500 − 360 = 140 Via = id + ic = 4/3 + 2 = (10/3) AR = vd/ia = 140/(10/3) = 42 Ω
[b] ig = ia + ib = (10/3) + 2 = (16/3) Apg (supplied) = (500)(16/3) = 2666.67 W
2–28 CHAPTER 2. Circuit Elements
P 2.26 [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:
i22.5 Ω =90 V
22.5 Ω= 4 A
Next we can calculate the voltage drop across the 15 Ω resistor by writing aKVL 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’slaw to find the current in this resistor:
i15 Ω =150 V15 Ω
= 10 A
Write a KCL equation at the middle right node to find the current through the5 Ω resistor. Sum the currents entering:
4 A − 10 A + i5 Ω = 0 so i5 Ω = 10 A − 4 A = 6 A
Write a KVL equation clockwise around the upper right loop, starting belowthe 4 Ω resistor. Use Ohm’s law to express the voltage drop across the resistorsin 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:
i4 Ω =60 V4 Ω
= 15 A
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 equationat 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 cancalculate the power supplied by the source:
pg = −240(19) = −4560 W thus pg (supplied) = 4560 W
[c]∑
Pdis = 900 + 1620 + 180 + 360 + 1500 = 4560 WTherefore,∑
Psupp =∑
Pdis
P 2.27 iE − iB − iC = 0
iC = βiB therefore iE = (1 + β)iB
i2 = −iB + i1
Vo + iERE − (i1 − iB)R2 = 0
−i1R1 + VCC − (i1 − iB)R2 = 0 or i1 =VCC + iBR2
R1 + R2
Vo + iERE + iBR2 − VCC + iBR2
R1 + R2R2 = 0
Now replace iE by (1 + β)iB and solve for iB. Thus
iB =[VCCR2/(R1 + R2)] − Vo
(1 + β)RE + R1R2/(R1 + R2)
P 2.28 [a] io = 0 because no current can exist in a single conductor connecting two partsof a circuit.
[b]
−200 + 8000ig + 12,000ig = 0 so ig = 200/20,000 = 10 mAv∆ = (12 × 103)(10 × 10−3) = 120 V5 × 10−3v∆ = 0.6 A9000i1 = 3000i2 so i2 = 3i10.6 + 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 the200 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 findthe 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 voltagesin 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
65,000= 100 µA
Now that we have the value of iβ , we can calculate the voltage for eachcomponent except the dependent source. Then we can write a KVL equationfor the right-hand loop to find the voltage vy of the dependent source. Sum thevoltages 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’sconstruct a power table:
Element Current Voltage Power Power
(µA) (V) 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
9 V 4900 9 p = −vi −44,100
The total power generated in the circuit is the sum of the negative power valuesin 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 powerabsorbed in the circuit is the sum of the positive power values in the powertable:
550 µW + 70 µW + 5000µW + 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 thecircuit 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 sourcevd = 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 = 2iσig + 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,∑Pgen =
∑Pdiss = 356 W
P 2.31 40i2 +540
+510
= 0 so i2 = −15.625 mA
v1 = 80i2 = −1.25 V
25i1 +−1.25
20+ (−15.625 × 10−3) = 0 so i1 = 3.125 mA
vg = 60i1 + 260i1 = 320i1
Therefore, vg = 1 V
P 2.32VCCR2
R1 + R2=
(10)(60 × 103)100 × 103 = 6V
R1R2
R1 + R2=
(40 × 103)(60 × 103)100 × 103 = 24 kΩ
iB =6 − 0.6
24 × 103 + 50(120)=
5.4(24 + 6) × 103 = 0.18 mA
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
R2=
1.6860
× 10−3 = 28 µA
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 2–33
P 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 warningsign should be posted at the 250 V source.
P 2.35
P 2.36 [a] p = i2R
parm =(250
650
)2
(400) = 59.17 W
pleg =(250
650
)2
(200) = 29.59 W
ptrunk =(250
650
)2
(50) = 7.40 W
2–34 CHAPTER 2. Circuit Elements
[b]
(dT
dt
)arm
=2.39 × 10−4parm
4= 35.36 × 10−4 C/s
tarm =5
35.36× 104 = 1414.23 s or 23.57 min
(dT
dt
)leg
=2.39 × 10−4
10Pleg = 7.07 × 10−4 C/s
tleg =5 × 104
7.07= 7,071.13 s or 117.85 min
(dT
dt
)trunk
=2.39 × 10−4(7.4)
25= 0.71 × 10−4 C/s
ttrunk =5 × 104
0.71= 70,677.37 s or 1,177.96 min
[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 = ispaceRspace = 3000 V.
3Simple Resistive Circuits
Assessment Problems
AP 3.1
Start from the right hand side of the circuit and make series and parallelcombinations of the resistors until one equivalent resistor remains. Begin bycombining 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 Ω‖64 Ω =(16)(64)16 + 64
=102480
= 12.8 Ω
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 Ω‖30 Ω =(20)(30)20 + 30
=60050
= 12 Ω
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 boththe 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, wecan 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 Ωresistor, so we can use Ohm’s law to calculate the current through this resistor:
iA =60 V30 Ω
= 2 A
Now write a KCL equation at the upper left node to find the current iB:
−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’slaw to express the voltage drop across the resistors in terms of the currentthrough the resistors:
−v + 7.2iB + 6iC + 10iC = 0
So 16iC = v − 7.2iB = 60 V − (7.2)(3) = 38.4 V
Thus iC =38.416
= 2.4 A
Now that we have the current through the 10 Ω resistor we can use the formulap = 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
75,000 + 25,000(200 V) = 150 V
Problems 3–3
[b] When we have a load resistance of 150 kΩ then the voltage vo is across theparallel combination of the 75 kΩ resistor and the 150 kΩ resistor. First,calculate the equivalent resistance of the parallel combination:
75 kΩ‖150 kΩ =(75,000)(150,000)75,000 + 150,000
= 50,000 Ω = 50 kΩ
Now use voltage division to find vo across this equivalent resistance:
vo =50,000
50,000 + 25,000(200 V) = 133.3 V
[c] If the load terminals are short-circuited, the 75 kΩ resistor is effectivelyremoved from the circuit, leaving only the voltage source and the 25 kΩresistor. We can calculate the current in the resistor using Ohm’s law:
i =200 V25 kΩ
= 8 mA
Now we can use the formula p = Ri2 to find the power dissipated in the 25 kΩresistor:
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 becan use the formula p = v2/R to calculate the power:
p75k(max) =(150)2
75,000= 0.3 W
AP 3.3
[a] We will write a current division equation for the current throught the 80Ωresistor and use this equation to solve for R:
i80Ω =R
R + 40 Ω + 80 Ω(20 A) = 4 A so 20R = 4(R + 120)
Thus 16R = 480 and R =48016
= 30 Ω
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 formulap = Ri2:
iR =40 + 80
40 + 80 + 30(20 A) = 16 A so pR = (30)(16)2 = 7680 W
[c] Write a KVL equation around the outer loop to solve for the voltage v, and thenuse the formula p = −vi to calculate the power delivered by the current source:
−v + (60 Ω)(20 A) + (30 Ω)(16 A) = 0 so v = 1200 + 480 = 1680 V
Thus, 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 and70 Ω resistors:
Req = 20 Ω‖30 Ω‖(50 Ω + 10 Ω) so1
Req=
120 Ω
+1
30 Ω+
160 Ω
=1
10 Ω
Thus, Req = 10 Ω
Now we can use voltage division to find the voltage vo:
vo =40
40 + 10 + 70(60 V) = 20 V
[b] The current through the 40 Ω resistor can be found using Ohm’s law:
i40Ω =vo
40=
20 V40 Ω
= 0.5 A
This current flows from left to right through the 40 Ω resistor. To use currentdivision, we need to find the equivalent resistance of the two parallel branchescontaining the 20 Ω resistor and the 50 Ω and 10 Ω resistors:
20 Ω‖(50 Ω + 10 Ω) =(20)(60)20 + 60
= 15 Ω
Now we use current division to find the current in the 30 Ω branch:
i30Ω =15
15 + 30(0.5 A) = 0.16667 A = 166.67 mA
Problems 3–5
[c] We can find the power dissipated by the 50 Ω resistor if we can find the currentin this resistor. We can use current division to find this current from the currentin the 40 Ω resistor, but first we need to calculate the equivalent resistance ofthe 20 Ω branch and the 30 Ω branch:
20 Ω‖30 Ω =(20)(30)20 + 30
= 12 Ω
Current division gives:
i50Ω =12
12 + 50 + 10(0.5 A) = 0.08333 A
Thus, p50Ω = (50)(0.08333)2 = 0.34722 W = 347.22 mW
AP 3.5 [a]
We can find the current i using Ohm’s law:
i =1 V
100 Ω= 0.01 A = 10 mA
[b]
Rm = 50 Ω‖5.555 Ω = 5 Ω
We can use the meter resistance to find the current using Ohm’s law:
imeas =1 V
100 Ω + 5 Ω= 0.009524 = 9.524 mA
AP 3.6 [a]
3–6 CHAPTER 3. Simple Resistive Circuits
Use voltage division to find the voltage v:
v =75,000
75,000 + 15,000(60 V) = 50 V
[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 equivalentresistance of the parallel combination of the 75 kΩ resistor and the voltmeter:
75,000 Ω‖150,000 Ω =(75,000)(150,000)75,000 + 150,000
= 50 kΩ
Thus, vmeas =50,000
50,000 + 15,000(60 V) = 46.15 V
AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite resistorsmust be equal. Therefore,
100Rx = (1000)(150) so Rx =(1000)(150)
100= 1500 Ω = 1.5 kΩ
[b] When the bridge is balanced, there is no current flowing through the meter, sothe meter acts like an open circuit. This places the following branches inparallel: The branch with the voltage source, the branch with the seriescombination R1 and R3 and the branch with the series combination of R2 andRx. We can find the current in the latter two branches using Ohm’s law:
iR1,R3 =5 V
100 Ω + 150 Ω= 20 mA; iR2,Rx =
5 V1000 + 1500
= 2 mA
We can calculate the power dissipated by each resistor using the formulap = Ri2:
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 beleft in the balanced state without exceeding the power-dissipating capacity ofthe resistors.
Problems 3–7
AP 3.8 Convert the three Y-connected resistors, 20 Ω, 10 Ω, and 5 Ω to three ∆-connectedresistors Ra, Rb, and Rc. To assist you the figure below has both the Y-connectedresistors and the ∆-connected resistors
Ra =(5)(10) + (5)(20) + (10)(20)
20= 17.5 Ω
Rb =(5)(10) + (5)(20) + (10)(20)
10= 35 Ω
Rc =(5)(10) + (5)(20) + (10)(20)
5= 70 Ω
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 Ω‖28 Ω =(70)(28)70 + 28
= 20 Ω
Also, the 17.5 Ω resistor is parallel to the 105 Ω resistor:
17.5 Ω‖105 Ω =(17.5)(105)17.5 + 105
= 15 Ω
Once the parallel combinations are made, we can see that the equivalent 20 Ωresistor 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 theother 35 Ω resistor:
35 Ω‖35 Ω =(35)(35)35 + 35
= 17.5 Ω
Thus, the resistance seen by the 2 A source is 17.5 Ω, and the voltage can becalculated 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. Thesimplified circuit is shown below:
[b] The 3 kΩ, 5 kΩ, and 7 kΩ resistors are in series. The simplified circuit is shownbelow:
[c] The 300 Ω, 400 Ω, and 500 Ω resistors are in series. The simplified circuit isshown 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 isshown below:
[c] The 600 Ω, 200 Ω, and 300 Ω resistors are in series. The simplified circuit isshown below:
P 3.3 [a] p4Ω = i2s4 = (12)24 = 576 W p18Ω = (4)218 = 288 W
p3Ω = (8)23 = 192 W p6Ω = (8)26 = 384 W
[b] p120V(delivered) = 120is = 120(12) = 1440 W
[c] pdiss = 576 + 288 + 192 + 384 = 1440 W
P 3.4 [a] From Ex. 3-1: i1 = 4 A, i2 = 8 A, is = 12 Aat node x: −12 + 4 + 8 = 0, at node y: 12 − 4 − 8 = 0
Problems 3–11
[b] v1 = 4is = 48 V v3 = 3i2 = 24 V
v2 = 18i1 = 72 V v4 = 6i2 = 48 Vloop 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 thatthe current in all series-connected circuits is the same, and that the voltage dropacross all parallel-connected resistors is the same.
[a] Req = 6 + 12 + [4‖(9 + 7)] = 18 + (4‖16) = 18 + 3.2 = 21.2 Ω
[b] Req = 4 k + [10 k‖(3 k + 5 k + 7 k)] = 4 k + (10 k‖15 k) = 4 k + 6 k = 10 kΩ
[c] Req = (300 + 400 + 500) + (600‖1200) = 1200 + 400 = 1600 Ω
P 3.6 Always work from the side of the circuit furthest from the source. Remember thatthe current in all series-connected circuits is the same, and that the voltage dropacross all parallel-connected resistors is the same.
[a] Req = 18 + (100‖25‖(22 + (10‖40))) = 18 + (20‖(22 + 8) = 18 + 12 = 30 Ω
[b] Req = 10 k‖(5 k + 2 k + (9 k‖18 k‖6 k)) = 10 k‖(7 k + 3 k) = 10 k‖10 k = 5 kΩ
[c] Req = 600‖200‖300‖(250 + 150) = 600‖200‖300‖400 = 80 Ω
P 3.7 [a] Req = 12 + (24‖(30 + 18)) + 10 = 12 + (24‖48) + 10 = 12 + 16 + 10 = 38 Ω
[b] Req = 4 k‖30 k‖60 k‖(1.2 k + (7.2 k‖2.4 k) + 2 k) = 4 k‖30 k‖60 k‖(3.2 k + 1.8 k)
= 4 k‖30 k‖60 k‖5 k = 2 kΩ
P 3.8 [a] 5‖20 = 100/25 = 4 Ω 5‖20 + 9‖18 + 10 = 20 Ω
9‖18 = 162/27 = 6 Ω 20‖30 = 600/50 = 12 Ω
Rab = 5 + 12 + 3 = 20 Ω
3–12 CHAPTER 3. Simple Resistive Circuits
[b] 5 + 15 = 20 Ω 30‖20 = 600/50 = 12 Ω
20‖60 = 1200/80 = 15 Ω 3‖6 = 18/9 = 2 Ω
15 + 10 = 25 Ω 3‖6 + 30‖20 = 2 + 12 = 14 Ω
25‖75 = 1875/100 = 18.75 Ω 26‖14 = 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 60‖40 = 2400/100 = 24 Ω
8‖12 = 96/20 = 4.8 Ω 24 + 6 = 30 Ω
4.8 + 5.2 = 10 Ω 30‖10 = 300/40 = 7.5 Ω
45 + 15 = 60 Ω Rab = 1.5 + 7.5 + 1.0 = 10 Ω
P 3.9 [a] For circuit (a)
Rab = 360‖(90 + 120‖(160 + 200)) = 360‖(90 + (120‖360)) = 360‖(90 + 90)
= 360‖180 = 120 Ω
For circuit (b)
1Re
=120
+115
+120
+14
+112
=3060
=12
Re = 2 Ω
Re + 16 = 18 Ω
18‖18 = 9 Ω
Rab = 10 + 8 + 9 = 27 Ω
For circuit (c)
15‖30 = 10 Ω
10 + 20 = 30 Ω
60‖30 = 20 Ω
20 + 10 = 30 Ω
30‖80‖(40 + 20) = 30‖80‖60 = 16 Ω
Rab = 16 + 24 + 10 = 50 Ω
Problems 3–13
[b] Pa = (0.032)(120) = 108 mW
Pb =1442
27= 768 W
Pc =0.082
50= 128 µ W
P 3.10 The equivalent resistance to the right of the 10 Ω resistor is
(6 + 5‖(8 + 12)) = 6 + 5‖20 = 6 + 4 = 10 Ω.
We can use current division to see that the current then splits equally between thetwo 10 Ω branches. Thus the current through the 6 Ω branch in the original circuit is5 A. This 5 A current splits between the branch with the 5 Ω resistor and the branchwith the 8 + 12 = 20 Ω resistor and we use current division to determine the currentin the 5 Ω resistor:
i5Ω =20
20 + 5(5) = 4 A
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 CHAPTER 3. Simple Resistive Circuits
P 3.12 [a] Req = R‖R =R2
2R=
R
2
[b] Req = R‖R‖R‖ · · · ‖R (n R’s)
= R‖ R
n − 1
=R2/(n − 1)
R + R/(n − 1)=
R2
nR=
R
n
[c] One solution:
400 =2000
nso n =
2000400
= 5
You can place 5 identical 2 kΩ resistors in parallel to get an equivalentresistance of 400 Ω.
[d] One solution:
12,500 =100,000
nso n =
100,00012,500
= 8
You can place 8 identical 100 kΩ resistors in parallel to get an equivalentresistance of 12.5 kΩ.
P 3.13 [a] We can calculate the no-load voltage using voltage division to determine thevoltage drop across the 500 Ω resistor:
vo =500
(2000 + 500)(75 V) = 15 V
[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 lawto calculate the current they share:
i =75 V
2000 Ω + 500 Ω= 0.03 A = 30 mA
Now use the formula p = Ri2 to calculate the power dissipated by eachresistor:
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-loadvoltage requirement, first pick R1 to meet the 1 W specification
iR1 =75 − 15
R1, Therefore,
( 60R1
)2
R1 ≤ 1
Thus, R1 ≥ 602
1or R1 ≥ 3600 Ω
Problems 3–15
Now use the voltage specification:
R2
R2 + 3600(75) = 15
Thus, R2 = 900 Ω
R1 = 1600 Ω and R2 = 400 Ω are the smallest values of resistors that satisfythe 1 W specification.
P 3.14 Use voltage division to determine R2 from the no-load voltage specification:
6 V =R2
(R2 + 40)(18 V); so 18R2 = 6(R2 + 40)
Thus, 12R2 = 240 so R2 =24012
= 20 Ω
Now use voltage division again, this time to determine the value of Re, the parallelcombination of R2 and RL. We use the loaded voltage specification:
4 V =Re
(40 + Re)(18 V) so 18Re = 4(40 + Re)
Thus, 14Re = 160 so Re =16014
= 11.43 Ω
Now use the definition Re to calculate the value of RL given that R2 = 20 Ω:
Re =20RL
20 + RL
= 11.43 so 20RL = 11.43(RL + 20)
Therefore, 8.57RL = 228.6 and RL =226.88.57
= 26.67 Ω
P 3.15 [a] From the constraint on the no-load voltage,
R2
R1 + R2(40) = 8 so R1 = 4R2
From the constraint on the loaded voltage divider:
7.5 =
3600R2
3600 + R2
R1 +3600R2
3600 + R2
(40)
=
3600R2
3600 + R2
4R2 +3600R2
3600 + R2
(40)
3–16 CHAPTER 3. Simple Resistive Circuits
=3600R2
4R2(3600 + R2) + 3600R2(40) =
144,000R2
4R22 + 18,000R2
So,144,000
4R2 + 18,000= 7.5 ·. . R2 = 300 Ω and R1 = 4R2 = 1200 Ω
[b] Power dissipated in R1 will be maximum when the voltage across R1 ismaximum. This will occur under load conditions.
vR1 = 40 − 7.5 = 32.5 V; PR1 =(32.5)2
1200= 880.2 mW
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.
PR2 =(8)2
300= 213.3 m W
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 limitwhen the power dissipated in R1 equals 1 W
So (v2R1
/1200) = 1; vR1 = 34.641 V vo = 40 − 34.641 = 5.359 V
Therefore,Re
1200 + Re(40) = 5.359, and Re = 185.641 Ω
1200RL
1200 + RL= 185.641 ·. . RL = 219.62 Ω
P 3.17 [a]
120 kΩ + 30 kΩ = 150 kΩ
75 kΩ‖150 kΩ = 50 kΩ
vo1 =240
(25,000 + 50,000)(50,000) = 160 V
vo =120,000
(150,000)(vo1) = 128 V, vo = 128 V
Problems 3–17
[b]
i =240
100,000= 2.4 mA
75,000i = 180 V
vo =120,000150,000
(180) = 144 V; 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
v′o1 =
75,000(100,000)
(240) = 180 V
Now note this is the input voltage to the second voltage divider when thecurrent controlled voltage source is used.
P 3.18(24)2
R1 + R2 + R3= 36, Therefore, R1 + R2 + R3 = 16 Ω
(R1 + R2)24(R1 + R2 + R3)
= 12
Therefore, 2(R1 + R2) = R1 + R2 + R3
Thus, R1 + R2 = R3; 2R3 = 16; R3 = 8 Ω
R2(24)R1 + R2 + R3
= 6
4R2 = R1 + R2 + R3 so R2 = R3/2 = 4 Ω
R2 = 4 Ω; R1 = 16 − 8 − 4 = 4 Ω
P 3.19 Note – in the problem description, the first equation defines R1 not RL.
[a] At no load: vo = kvs =R2
R1 + R2vs.
At full load: vo = αvs =Re
R1 + Revs, where Re =
RoR2
Ro + R2
Therefore k =R2
R1 + R2and R1 =
(1 − k)k
R2
α =Re
R1 + Reand R1 =
(1 − α)α
Re
3–18 CHAPTER 3. Simple Resistive Circuits
Thus(1 − α
α
) [R2Ro
Ro + R2
]=
(1 − k)k
R2
Solving for R2 yields R2 =(k − α)α(1 − k)
Ro
Also, R1 =(1 − k)
kR2
·. . R1 =(k − α)
αkRo
[b] R1 =(0.05
0.68
)Ro = 2.5 kΩ
R2 =(0.05
0.12
)Ro = 14.167 kΩ
[c]
Maximum dissipation in R2 occurs at no load, therefore,
PR2(max) =[(60)(0.85)]2
14,167= 183.6 mW
Maximum dissipation in R1 occurs at full load.
PR1(max) =[60 − 0.80(60)]2
2500= 57.60 mW
[d ]
PR1 =(60)2
2500= 1.44 W = 1440 mW
PR2 =(0)2
14,167= 0 W
P 3.20 [a] Let vo be the voltage across the parallel branches, positive at the upper terminal,then
ig = voG1 + voG2 + · · · + voGN = vo(G1 + G2 + · · · + GN)
It follows that vo =ig
(G1 + G2 + · · · + GN)
Problems 3–19
The current in the kth branch is ik = voGk; Thus,
ik =igGk
[G1 + G2 + · + GN ]
[b] io =120(0.00125)
[0.0025 + 0.0004167 + 0.00125 + 0.000625 + 0.0002083]= 30 mA
P 3.21 Begin by using the relationships among the branch currents to express all branchcurrents in terms of i4:
i1 = 2i2 = 2(10i3) = 20i4
i2 = 10i3 = 10i4
i3 = i4
Now use KCL at the top node to relate the branch currents to the current supplied bythe 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 =0.00832
= 0.00025 = 0.25 mA
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
i4=
4 V0.25 mA
= 16 kΩ
Calculate i3 from i4 and use Ohm’s law as above to find R3:
i3 = i4 = 0.25 mA ·. . R3 =vg
i3=
4 V0.25 mA
= 16 kΩ
Calculate i2 from i4 and use Ohm’s law as above to find R2:
i2 = 10i4 = 10(0.25 mA) = 2.5 mA ·. . R2 =vg
i2=
4 V2.5 mA
= 1.6 kΩ
Calculate i1 from i4 and use Ohm’s law as above to find R1:
i1 = 20i4 = 20(0.25 mA) = 5 mA ·. . R1 =vg
i1=
4 V5 mA
= 800 Ω
The resulting circuit is shown below:
3–20 CHAPTER 3. Simple Resistive Circuits
P 3.22 [a]
Using voltage division,
v18Ω =18
18 + 30(40) = 15 V positive at the top
[b]
Using current division,
i30Ω =24
24 + 30 + 18(60 × 10−3) = 20 mA flowing from right to left
[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 kΩ‖60 kΩ = 3.75 kΩ
Using current division,
i30 kΩ =3.75 k
30 k + 3.75 k(9 × 10−3) = 1 m A, flowing bottom to top
[d]
Problems 3–21
The voltage drop across the 4 kΩ resistor is the same as the voltage drop acrossthe series combination of the 1.2 kΩ, the (7.2 k‖2.4 k)Ω combined resistor,and the 2 kΩ resistor. Note that
7.2 k‖2.4 k =(7200)(2400)
9600= 1.8 kΩ
Using voltage division,
vo =1800
1200 + 1800 + 2000(50) = 18 V positive at the top
P 3.23 [a]
First, note the following: 18‖9 = 6 Ω; 20‖5 = 4 Ω; and the voltage drop acrossthe 18 Ω resistor is the same as the voltage drop across the parallel combinationof the 18 Ω and 9 Ω resistors. Using voltage division,
vo =6
6 + 4 + 10(0.1 V) = 30 mV positive at the left
[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,
vRe =15
15 + 10(10) = 6 V
Using voltage division again,
vo =15
5 + 15(6) = 4.5 V positive at the top
3–22 CHAPTER 3. Simple Resistive Circuits
[c]
Find equivalent resistance on the right side
Rr = 5.2 +(12)(5 + 3)(12 + 3 + 5)
= 10 Ω
Find voltage bottom to top across Rr
(10)(3) = 30 V
Find the equivalent resistance on the left side
Rl = 6 +(40)(45 + 15)(40 + 45 + 15)
= 30 Ω
The current in the 6 Ω is
i6 Ω =3030
= 1 A left to right
Use current division to find io
io = (1)( 40
40 + 15 + 45
)= 0.4 A bottom to top
P 3.24 [a] v20k =20
20 + 5(45) = 36 V
v90k =90
90 + 60(45) = 27 V
vx = v20k − v90k = 36 − 27 = 9 V
[b] v20k =2025
(Vs) = 0.8Vs
v90k =90150
(Vs) = 0.6Vs
vx = 0.8Vs − 0.6Vs = 0.2Vs
P 3.25 150‖75 = 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
4590 + 45
(3) = 1 V
Use voltage division to find v1, which is the voltage drop across the parallelcombination whose equivalent resistance is 50 Ω:
v1 =50
50 + 40(1) = 5/9 V
Use voltage division to find v2:
v2 =30
30 + 60(1) = 1/3 V
P 3.26 i300Ω =1000 + 200
1000 + 200 + 300 + 300(15 × 10−3) = 10 mA
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 Ω‖20 Ω = 4 Ω; 4 Ω + 6 Ω = 10 Ω; 10‖40 = 8 Ω;
Therefore, ig =125
8 + 2= 12.5 A
i6Ω =(40)(12.5)
50= 10 A; io =
(5)(10)25
= 2 A
P 3.28 [a] Combine resistors in series and parallel to find the equivalent resistance seen bythe source. Use this equivalent resistance to find the current through thesource, and use current division to find io:
80 + 70 = 150 Ω 100‖150‖90 = 36 Ω 36 + 24 = 60 Ω
i24Ω =60 V60Ω
= 1 A
io =100‖90‖150
150(1) =
36150
= 0.24 A
3–24 CHAPTER 3. Simple Resistive Circuits
[b] Use current division to find the current through the 90 Ω resistor from the sourcecurrent found in part (a), and use the calculated current to find the power in the90 Ω resistor:
i90Ω =100‖90‖150
90(1) =
3690
= 0.4 A
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
i5‖20Ω = i40Ω + i25Ω + i4Ω = 1 + 1 + 4 = 6 A
v5‖20Ω = (4)(6) = 24 V
v32Ω = v40Ω + v5‖20Ω = 40 + 24 = 64 V
i32Ω = 64/32 = 2 A;
i10Ω = i32Ω + i5‖20Ω = 2 + 6 = 8 A
vg = 10(8) + v32Ω = 80 + 64 = 144 V.
[b] P20Ω =(v5‖20Ω)2
20=
242
20= 28.8 W
P 3.30
40‖10 = 8 Ω 15‖60 = 12 Ω
Problems 3–25
i1 =(3)(40)(60)
= 2 A; vx = 8i1 = 16 V
vg = 20i1 = 40 V
v60 = vg − vx = 24 V
Pdevice =242
60+
162
10+
402
40= 75.2 W
P 3.31 [a] The model of the ammeter is an ideal ammeter in parallel with a resistor whoseresistance is given by
Rs =100 µV10 µA
= 10 Ω.
We can calculate the current through the real meter using current division:
im =(10/99)
10 + (10/99)(imeas) =
10990 + 10
(imeas) =1
100imeas
[b] Rs =100 µV10 µA
= 10 Ω.
im =(100/999,990)
10 + (100/999,990)(imeas) =
1100,000
(imeas)
[c] Yes
P 3.32 Measured value: 60‖20.1 = 15.056 Ω
ig =50
(15.056 + 10)= 1.9955 A; imeas = (1.9955)
6080.1
= 1.495 A
True value: 60‖20 = 15 Ω
ig =50
(15 + 10)=
5025
= 2.0 A; itrue = (2)(60
80
)= 1.5 A
% error =[1.495
1.5− 1
]× 100 = −0.3488%
3–26 CHAPTER 3. Simple Resistive Circuits
P 3.33 Begin by using current division to find the actual value of the current io:
itrue =15
15 + 45(50 mA) = 12.5 mA
imeas =15
15 + 45 + 0.1(50 mA) = 12.48 mA
% error =[12.48
12.5− 1
]100 = −0.1664%
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 mV1 mA
= 20 Ω
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)[15‖45‖(4980 + 20)] = 0.5612 V
[b] vtrue = (50 × 10−3)(15‖45) = 0.5625 V
% error =(0.5612
0.5625− 1
)× 100 = −0.23%
P 3.36
Original meter: Re =50 × 10−3
5= 0.01 Ω
Modified meter: Re =(0.02)(0.01)
0.03= 0.00667 Ω
·. . (Ifs)(0.00667) = 50 × 10−3
·. . Ifs = 7.5 A
Problems 3–27
P 3.37 At full scale the voltage across the shunt resistor will be 100 mV; therefore thepower dissipated will be
PA =(100 × 10−3)2
RA
Therefore RA ≥ (100 × 10−3)2
0.25= 40 mΩ
Otherwise the power dissipated in RA will exceed its power rating of 0.25 WWhen RA = 40 mΩ, the shunt current will be
iA =100 × 10−3
40 × 10−3 = 2.5 A
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
ifullscale − 2 × 10−3 =10
1000ifs − 0.02
[a] RA =10
10,000 − 0.02= 1 mΩ
[b] RA =10
1000 − 0.02= 10 mΩ
[c] RA =10
100 − 0.02= 1 Ω
[d] RA =10
0.1 − 0.02= 125 Ω
P 3.39 [a]
3–28 CHAPTER 3. Simple Resistive Circuits
10 × 103i1 + 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 × 103iB + 0.4 + 30iB(300)
50i1 − 61iB = 0.4 × 10−3
Calculator solution yields iB = 496.6 µA
[c] % error =( 496.6
553.85− 1
)100 = −10.34%
P 3.40 [a] vmeter = 100 V
[b] Rmeter = (100 Ω/V)(100 V) = 10 kΩ
10 k‖60 k = 8.57 kΩ
vmeter =8.57 k23.57 k
(100) = 36.36 V
[c] 10 k‖1 k = 6 kΩ
vmeter =666
(100) = 9.09 V
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 =300300
× 10−3 = 1 mA; i2 max =150120
× 10−3 = 1.25 mA
·. . 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
[c] im =399
420 × 103 = 0.95 mA
vm1 = (0.95)(300) = 285 V vm2 = (0.95)(120) = 114 V
P 3.42 The current in the series-connected voltmeters is
im =288300
= 0.96 mA
v80 kΩ = (0.96)(80) = 76.8 V
Vpower supply = 288 + 115.2 + 76.8 = 480 V
P 3.43 Rmeter = Rm + Rmovement =750 V1.5 mA
= 500 kΩ
vmeas = (25 kΩ‖125 kΩ‖500 kΩ)(30 mA) = (20 kΩ)(30 mA) = 600 V
vtrue = (25 kΩ‖125 kΩ)(30 mA) = (20.833 kΩ)(30 mA) = 625 V
% error =(600
625− 1
)100 = −4%
3–30 CHAPTER 3. Simple Resistive Circuits
P 3.44 Note – the upper terminal of the voltmeter should be labeled 820 V, not 300 V.
[a] Rmeter = 360 kΩ + 200 kΩ‖50 kΩ = 400 kΩ
400‖600 = 240 kΩ
Vmeter =240300
(300) = 240 V
[b] What is the percent error in the measured voltage?
True value =600660
(300) = 272.73 V
% error =( 240
272.73− 1
)100 = −12%
P 3.45 [a] R1 =100 V2 mA
= 50 kΩ
R2 =10 V2 mA
= 5 kΩ
R3 =1 V
2 mA= 500 Ω
[b] Let ia = actual current in the movement
id = design current in the movement
Then % error =(
iaid
− 1)
100
For the 100 V scale:
ia =100
50,000 + 25=
10050,025
, id =100
50,000
iaid
=50,00050,025
= 0.9995 % error = (0.9995 − 1)100 = −0.05%
For the 10 V scale:iaid
=50005025
= 0.995 % error = (0.995 − 1.0)100 = −0.5%
For the 1 V scale:iaid
=500525
= 0.9524 % error = (0.9524 − 1.0)100 = −4.76%
P 3.46 [a] Rmovement = 50 Ω
R1 + Rmovement =30
1 × 10−3 = 30 kΩ ·. . R1 = 29,950 Ω
R2 + R1 + Rmovement =150
1 × 10−3 = 150 kΩ ·. . R2 = 120 kΩ
Problems 3–31
R3 + R2 + R1 + Rmovement =300
1 × 10−3 = 300 kΩ
·. . R3 = 150 kΩ
3–32 CHAPTER 3. Simple Resistive Circuits
[b]
imove =288300
(1) = 0.96 mA
v1 = (0.96 m)(150 k) = 144 V
i1 =144
750 k= 0.192 mA
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
50 =Vs(10)10 + Rs
(1) Vs in mV; Rs in MΩ
48.75 =Vs(6)6 + Rs
(2)
[a] From Eq (1) 10 + Rs = 0.2Vs
·. . Rs = 0.2Vs − 10
Substituting into Eq (2) yields
48.75 =6Vs
0.2Vs − 6or Vs = 52 mV
[b] From Eq (1)
50 =520
10 + Rs
or 50Rs = 20
So Rs = 400 kΩ
Problems 3–33
P 3.48 Since the bridge is balanced, we can remove the detector without disturbing thevoltages and currents in the circuit.
It follows that
i1 =ig(R2 + Rx)
R1 + R2 + R3 + Rx
=ig(R2 + Rx)∑
R
i2 =ig(R1 + R3)
R1 + R2 + R3 + Rx
=ig(R1 + R3)∑
R
v3 = R3i1 = vx = i2Rx
·. .R3ig(R2 + Rx)∑
R=
Rxig(R1 + R3)∑R
·. . R3(R2 + Rx) = Rx(R1 + R3)
From which Rx =R2R3
R1
P 3.49 [a]
The condition for a balanced bridge is that the product of the opposite resistorsmust be equal:
(200)(Rx) = (500)(800) so Rx =(500)(800)
200= 2000 Ω
3–34 CHAPTER 3. Simple Resistive Circuits
[b] The source current is the sum of the two branch currents. Each branch currentcan be determined using Ohm’s law, since the resistors in each branch are inseries and the voltage drop across each branch is 6 V:
is =6 V
200 Ω + 800 Ω+
6 V500 Ω + 2000 Ω
= 8.4 mA
[c] We can use current division to find the current in each branch:
ileft =500 + 2000
500 + 2000 + 200 + 800(8.4 mA) = 6 mA
iright = 8.4 mA − 6 mA = 2.4 mA
Now we can use the formula p = Ri2 to find the power dissipated by eachresistor:
p200 = (200)(0.006)2 = 7.2 mW p800 = (800)(0.006)2 = 28.8 mW
p500 = (500)(0.0024)2 = 2.88 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; itabsorbs 2.88 mW of power.
P 3.50 Redraw the circuit, replacing the detector branch with a short circuit.
6 kΩ‖30 kΩ = 5 kΩ
12 kΩ‖20 kΩ = 7.5 kΩ
ig =75
5000 + 7500= 6 mA
v1 = 6 mA(5000) = 30 V
v2 = 6 mA(7500) = 45 V
Problems 3–35
i1 =30 V
6000 Ω= 5 mA
i2 =45 V
12,000 Ω= 3.75 mA
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 nocurrent 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 = 163048
= 10 mA
·. . 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 tothe 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 ∆-connectedresistors. Both the Y-connected and ∆-connected resistors are shown below to assistin using Eqs. 3.44 – 3.46:
Now use Eqs. 3.44 – 3.46 to calculate the values of the ∆-connected resistors:
R1 =(40)(10)
10 + 40 + 50= 4 Ω; R2 =
(50)(10)10 + 40 + 50
= 5 Ω; R3 =(40)(50)
10 + 40 + 50= 20 Ω
The transformed circuit is shown below:
3–36 CHAPTER 3. Simple Resistive Circuits
The equivalent resistance seen by the 24 V source can be calculated by makingseries and parallel combinations of the resistors to the right of the 24 V source:
Req = (15 + 5)‖(4 + 1) + 20 = 20‖5 + 20 = 4 + 20 = 24 Ω
Therefore, the current i in the 24 V source is given by
i =24 V24 Ω
= 1 A
Use current division to calculate the currents i1 and i2. Note that the current i1 flowsin the branch containing the 15 Ω and 5 Ω series connected resistors, while thecurrent i2 flows in the parallel branch that contains the series connection of the 1 Ωand 4 Ω resistors:
i1 =1 + 4
1 + 4 + 15 + 5(i) =
525
(1 A) = 0.2 A, 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 voltagedrop 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 thevoltage drop across the 5 Ω resistor, 5i1, and the voltage drop across the 20 Ωresistor, 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 the10 Ω, 40 Ω, and 50Ω ∆-connected resistors:
RX =(10)(50)
10 + 40 + 50= 5 Ω; RY =
(40)(50)10 + 40 + 50
= 20 Ω;
RZ =(10)(40)
10 + 40 + 50= 4 Ω
Replacing the R2—R3—R4 delta with its equivalent Y gives
Problems 3–37
Now calculate the equivalent resistance Rab by making series and parallelcombinations 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 the8 Ω, 10 Ω, and 40 Ω Y-connected resistors:
RX =(10)(40) + (40)(8) + (8)(10)
8=
8008
= 100 Ω
RY =(10)(40) + (40)(8) + (8)(10)
10=
80010
= 80 Ω
RZ =(10)(40) + (40)(8) + (8)(10)
40=
80040
= 20 Ω
Replacing the R2, R4, R5 wye with its equivalent ∆ gives
Make series and parallel combinations of the resistors to find the equivalentresistance Rab:
100 Ω‖50 Ω = 33.33 Ω; 80 Ω‖4 Ω = 3.81 Ω
·. . 100‖50 + 80‖4 = 33.33 + 3.81 = 37.14 Ω
·. . 37.14‖20 =(37.14)(20)
57.14= 13 Ω
·. . 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)
100= 5 Ω; R2U =
(50)(40)100
= 20 Ω; R3U =(40)(10)
100= 4 Ω
R1L =(60)(10)
100= 6 Ω; R2L =
(60)(30)100
= 18 Ω; R3L =(30)(10)
100= 3 Ω
The resulting circuit is shown below:
Now make series and parallel combinations of the resistors:
(4 + 6)‖(20 + 32 + 20 + 18) = 10‖90 = 9 Ω
Rab = 33 + 5 + 9 + 3 + 40 = 90 Ω
P 3.56 18 + 2 = 20 Ω
20‖80 = 16 Ω
16 + 4 = 20 Ω
20‖30 = 12 Ω
12 + 8 = 20 Ω
20‖60 = 15 Ω
15 + 5 = 20 Ω
ig =240 V20 Ω
= 12 A
Problems 3–39
io =60
60 + 20(12 A) = 9 A
i30Ω =20
20 + 30(9 A) = 3.6 A
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)
27= 6 Ω
The lower left and right deltas can be replaced by wyes. Each resistance in the wyeequals 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)
200= 12.5 Ω
R2 =(50)(100)
200= 25 Ω
R3 =(50)(100)
200= 25 Ω
Convert the lower delta to a wye.
R4 =(60)(80)
200= 24 Ω
R5 =(60)(60)
200= 18 Ω
3–40 CHAPTER 3. Simple Resistive Circuits
R6 =(60)(80)
200= 24 Ω
Now redraw the circuit using the wye equivalents.
Rab = 1.5 + 12.5 + (25 + 71 + 24)‖(25 + 31 + 24) + 18
= 1.5 + 12.5 + (120‖85) + 18 = 1.5 + 12.5 + 48 + 18 = 80 Ω
[b] When vab = 400 V
ig =40080
= 5 A
io =120
120 + 80(5) = 3 A
p31Ω = (31)(3)2 = 279 W
P 3.59 [a] After the 20 Ω—80 Ω—40 Ω wye is replaced by its equivalent delta, the circuitreduces to
Problems 3–41
Now the circuit can be reduced to
Req = 44 + 280‖92.5 = 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
3–42 CHAPTER 3. Simple Resistive Circuits
P 3.60 [a] After the 30 Ω—60 Ω—10 Ω delta is replaced by its equivalent wye, the circuitreduces to
Use current division to calculate i1:
i1 =22 + 18
22 + 18 + 4 + 6(5 A) =
4050
(5 A) = 4 A
[b] Return to the original circuit and write a KVL equation around the upper leftloop:
(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:
i2 = i1 +v
60= 4 +
−660
= 3.9 A
[d] Write a KVL equation around the bottom loop of the original circuit:
−v5A + (4 Ω)(4 A) + (10 Ω)(3.9 A) + (1 Ω)(5 A) = 0
So, v5A = (4)(4) + (10)(3.9) + (1)(5) = 60 V
Thus, p5A = (5 A)(60 V) = 300 W
P 3.61 Subtracting Eq. 3.42 from Eq. 3.43 gives
R1 − R2 = (RcRb − RcRa)/(Ra + Rb + Rc).
Adding this expression to Eq. 3.41 and solving for R1 gives
R1 = RcRb/(Ra + Rb + Rc).
To find R2, subtract Eq. 3.43 from Eq. 3.41 and add this result to Eq. 3.42. To findR3, 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]
(R2/R1)Rb + Rb + (R2/R3)Rb=
Rb(R1 + R3)R2
(R1R2 + R2R3 + R3R1)
Problems 3–43
Solving for Rb gives Rb = (R1R2 + R2R3 + R3R1)/R2. To find Ra: First useEqs. 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 eliminateRb and Rc from Eq. 3.42. To find Rc, use Eqs. 3.44–3.46 to obtain the ratiosRb = (R3/R2)Rc and Ra = (R3/R1)Rc. Now use the relationships to eliminate Rb
and Ra from Eq. 3.41.
P 3.62 Ga =1Ra
=R1
R1R2 + R2R3 + R3R1
=1/G1
(1/G1)(1/G2) + (1/G2)(1/G3) + (1/G3)(1/G1)
=(1/G1)(G1G2G3)G1 + G2 + G3
=G2G3
G1 + G2 + G3Similar manipulations generate the expressions for Gb and Gc.
P 3.63 [a] Rab = 2R1 +R2(2R1 + RL)2R1 + R2 + RL
= RL
Therefore 2R1 − RL +R2(2R1 + RL)2R1 + R2 + RL
= 0
Thus R2L = 4R2
1 + 4R1R2 = 4R1(R1 + R2)
When Rab = RL, the current into terminal a of the attenuator will be vi/RL
Using current division, the current in the RL branch will be
vi
RL· R2
2R1 + R2 + RL
Therefore vo =vi
RL· R2
2R1 + R2 + RLRL
andvo
vi
=R2
2R1 + R2 + RL
[b] (600)2 = 4(R1 + R2)R1
9 × 104 = R21 + R1R2
vo
vi
= 0.6 =R2
2R1 + R2 + 600
·. . 1.2R1 + 0.6R2 + 360 = R2
0.4R2 = 1.2R1 + 360
R2 = 3R1 + 900
·. . 9 × 104 = R21 + R1(3R1 + 900) = 4R2
1 + 900R1
·. . R21 + 225R1 − 22,500 = 0
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 +3RRL
3R + RL=
2.25R2L + 3.75RRL
3R + RL
Therefore Rab =3R
(2.25R2
L + 3.75RRL
3R + RL
)
3R +(
2.25R2L + 3.75RRL
3R + RL
) =3R(3R + 5RL)
15R + 9RL
When Rab = RL, we have 15RRL + 9R2L = 9R2 + 15RRL
Therefore R2L = R2 or RL = R
Problems 3–45
[b] When R = RL, the circuit reduces to
io =ii(3RL)4.5RL
=1
1.5ii =
11.5
vi
RL, vo = 0.75RLio =
12vi,
Thereforevo
vi
= 0.5
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
3(180)2 − (300)2 = 4500 Ω
[b]
vo =vi
3.5=
423.5
= 12 V
io =12300
= 40 mA
i1 =42 − 124500
=30
4500= 6.67 mA
ig =42300
= 140 mA
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
3–46 CHAPTER 3. Simple Resistive Circuits
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 =vinR4
Ro + R4 + ∆R
vb =R3
R2 + R3vin
vo = va − vb =R4vin
Ro + R4 + ∆R− R3
R2 + R3vin
When the bridge is balanced,
R4
Ro + R4vin =
R3
R2 + R3vin
·. .R4
Ro + R4=
R3
R2 + R3
Thus, vo =R4vin
Ro + R4 + ∆R− R4vin
Ro + R4
= R4vin
[ 1Ro + R4 + ∆R
− 1Ro + R4
]
=R4vin(−∆R)
(Ro + R4 + ∆R)(Ro + R4)
≈ −(∆R)R4vin
(Ro + R4)2
Problems 3–47
[b] ∆R = 0.03Ro
Ro =R2R4
R3=
(1000)(5000)500
= 10,000 Ω
∆R = (0.03)(104) = 300 Ω
·. . vo ≈ −300(5000)(6)(15,000)2 = −40 mV
[c] vo =−(∆R)R4vin
(Ro + R4 + ∆R)(Ro + R4)
=−300(5000)(6)
(15,300)(15,000)
= −39.2157 mV
P 3.67 [a] approx value =−(∆R)R4vin
(Ro + R4)2
true value =−(∆R)R4vin
(Ro + R4 + ∆R)(Ro + R4)
·. .approx value
true value=
(Ro + R4 + ∆R)(Ro + R4)
·. . % error =[Ro + R4 + ∆R
Ro + R4− 1
]× 100 =
∆R
Ro + R4× 100
But Ro =R2R4
R3
·. . % error =R3∆R
R4(R2 + R3)
[b] % error =(500)(300)
(5000)(1500)× 100 = 2%
P 3.68∆R(R3)(100)(R2 + R3)R4
= 0.5
∆R(500)(100)(1500)(5000)
= 0.5
·. . ∆R = 75 Ω
% change =75
10,000× 100 = 0.75%
3–48 CHAPTER 3. Simple Resistive Circuits
P 3.69 [a] From Eq 3.64 we have(i1i2
)2
=R2
2
R21(1 + 2σ)2
Substituting into Eq 3.63 yields
R2 =R2
2
R21(1 + 2σ)2R1
Solving for R2 yields
R2 = (1 + 2σ)2R1
[b] From Eq 3.63 we have
i1ib
=R2
R1 + R2 + 2Ra
But R2 = (1 + 2σ)2R1 and Ra = σR1 therefore
i1lb
=(1 + 2σ)2R1
R1 + (1 + 2σ)2R1 + 2σR1=
(1 + 2σ)2
(1 + 2σ) + (1 + 2σ)2
=1 + 2σ
2(1 + σ)
It follows that(i1ib
)2
=(1 + 2σ)2
4(1 + σ)2
Substituting into Eq 3.66 gives
Rb =(1 + 2σ)2Ra
4(1 + σ)2 =(1 + 2σ)2σR1
4(1 + σ)2
P 3.70 From Eq 3.69
i1i3
=R2R3
D
But D = (R1 + 2Ra)(R2 + 2Rb) + 2RbR2
where Ra = σR1; R2 = (1 + 2σ)2R1 and Rb =(1 + 2σ)2σR1
4(1 + σ)2
Therefore D can be written as
Problems 3–49
D = (R1 + 2σR1)[(1 + 2σ)2R1 +
2(1 + 2σ)2σR1
4(1 + σ)2
]+
2(1 + 2σ)2R1
[(1 + 2σ)2σR1
4(1 + σ)2
]
= (1 + 2σ)3R21
[1 +
σ
2(1 + σ)2 +(1 + 2σ)σ2(1 + σ)2
]
=(1 + 2σ)3R2
1
2(1 + σ)2 2(1 + σ)2 + σ + (1 + 2σ)σ
=(1 + 2σ)3R2
1
(1 + σ)2 1 + 3σ + 2σ2
D =(1 + 2σ)4R2
1
(1 + σ)
·. .i1i3
=R2R3(1 + σ)(1 + 2σ)4R2
1
=(1 + 2σ)2R1R3(1 + σ)
(1 + 2σ)4R21
=(1 + σ)R3
(1 + 2σ)2R1When this result is substituted into Eq 3.69 we get
R3 =(1 + σ)2R2
3R1
(1 + 2σ)4R21
Solving for R3 gives
R3 =(1 + 2σ)4R1
(1 + σ)2
P 3.71 From the dimensional specifications, calculate σ and R3:
σ =y
x=
0.0251
= 0.025; R3 =V 2
dc
p=
122
120= 1.2 Ω
Calculate R1 from R3 and σ:
R1 =(1 + σ)2
(1 + 2σ)4R3 = 1.0372 Ω
Calculate Ra, Rb, and R2:
Ra = σR1 = 0.0259 Ω Rb =(1 + 2σ)2σR1
4(1 + σ)2 = 0.0068 Ω
3–50 CHAPTER 3. Simple Resistive Circuits
R2 = (1 + 2σ)2R1 = 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) + 2R2Rb = 1.2758
ib =Vdc(R1 + R2 + 2Ra)
D= 21 A
i1 =VdcR2
D= 10.7561 A i2 =
Vdc(R1 + 2Ra)D
= 10.2439 A
It follows that i2bRb = 3 W and the power dissipation per meter is 3/0.025 = 120W/m. The value of i21R1 = 120 W/m. The value of i22R2 = 120 W/m. Finally,i21Ra = 3 W/m.
P 3.72 From the solution to Problem 3.71 we have ib = 21 A and i3 = 10 A. By symmetryic = 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 totalpower delivered to the vertical resistors is pV = (8)(3) = 24 W. The total powerdelivered to the five horizontal resistors is pH = 5(120) = 600 W.
·. .∑
pdiss = pH + pV = 624 W =∑
pdel
P 3.73 [a] σ = 0.03/1.5 = 0.02Since the power dissipation is 150 W/m the power dissipated in R3 must be200(1.5) or 300 W. Therefore
R3 =122
300= 0.48 Ω
From Table 3.1 we have
R1 =(1 + σ)2R3
(1 + 2σ)4 = 0.4269 Ω
Ra = σR1 = 0.0085 Ω
R2 = (1 + 2σ)2R1 = 0.4617 Ω
Problems 3–51
Rb =(1 + 2σ)2σR1
4(1 + σ)2 = 0.0022 Ω
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 =VdcR2
D= 26.51 A
i21R1 = 300 W or 200 W/m
i2 =R1 + 2Ra
DVdc = 25.49 A
i22R2 = 300 W or 200 W/m
i21Ra = 6 W or 200 W/m
ib =R1 + R2 + 2Ra
DVdc = 52 A
i2bRb = 6 W or 200 W/m
isource = 52 + 52 +12
0.48= 129 A
pdel = 12(129) = 1548 W
pH = 5(300) = 1500 W
pV = 8(6) = 48 W∑
pdel =∑
pdiss = 1548 W
4Techniques 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
−15 +v1
60+
v1
15+
v1 − v2
5= 0
5 +v2
2+
v2 − v1
5= 0
Place these equations in standard form:
v1
( 160
+115
+15
)+ v2
(−1
5
)= 15
v1
(−1
5
)+ v2
(12
+15
)= −5
Solving, v1 = 60 V and v2 = 10 V;Therefore, i1 = (v1 − v2)/5 = 10 A
[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:
−4.5 +v1
1+
v1 − v2
6 + 2= 0
v2
12+
v2 − v1
6 + 2+
v2 − 304
= 0
Place these equations in standard form:
v1
(1 +
18
)+ v2
(−1
8
)= 4.5
v1
(−1
8
)+ v2
( 112
+18
+14
)= 7.5
Solving, v1 = 6 V v2 = 18 VTo find the voltage v, first find the current i through the series-connected 6 Ω and 2 Ωresistors:
i =v1 − v2
6 + 2=
6 − 188
= −1.5 A
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
6+
v1
8+
v1 − v2
2− 3i1 = 0
−5 +v2
4+
v2 − v1
2+ 3i1 = 0
Problems 4–3
The dependent source requires the following constraint equation:
i1 =50 − v1
6Place these equations in standard form:
v1
(16
+18
+12
)+ v2
(−1
2
)+ i1(−3) =
506
v1
(−1
2
)+ v2
(14
+12
)+ i1(3) = 5
v1
(16
)+ v2(0) + i1(1) =
506
Solving, v1 = 32 V; v2 = 16 V; i1 = 3 AUsing these values to calculate the power associated with each source:
p50V = −50i1 = −150 W
p5A = −5(v2) = −80 W
p3i1 = 3i1(v2 − v1) = −144 W
[b] All three sources are delivering power to the circuit because the powercomputed 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 nodevoltage equation will be written:
The node voltage equation is
vo
40+
vo − 1010
+vo + 20i∆
20= 0
The constraint equation required by the dependent source is
i∆ = i10 Ω + i30 Ω =10 − vo
10+
10 + 20i∆30
Place these equations in standard form:
4–4 CHAPTER 4. Techniques of Circuit Analysis
vo
( 140
+110
+120
)+ i∆(1) = 1
vo
( 110
)+ i∆
(1 − 20
30
)= 1 +
1030
Solving, vo = 24 V 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 asupernode. The v1 node voltage equation is
v1
7.5+
v1 − v
2.5− 4.8 = 0
The supernode equation is
v − v1
2.5+
v
10+
v2
2.5+
v2 − 121
= 0
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:
v1
( 17.5
+1
2.5
)+ v
(− 1
2.5
)+ v2(0) + ix(0) = 4.8
v1
(− 1
2.5
)+ v
( 12.5
+110
)+ v2
( 12.5
+ 1)
+ ix(0) = 12
v1
(− 1
7.5
)+ v(0) + v2(0) + ix(1) = 0
v1(0) + v(1) + v2(−1) + ix(1) = 0
Solving this set of equations for v gives v = 8 V
v1 = 15 V, v2 = 10 V, ix = 2 A
Problems 4–5
AP 4.6 Redraw the circuit identifying the reference node and the two unknown nodevoltages. Note that the right-most node voltage is the sum of the 60 V source and thedependent source voltage.
The node voltage equation at v1 is
v1 − 602
+v1
24+
v1 − (60 + 6iφ)3
= 0
The constraint equation due to the dependent source is
iφ =60 + 6iφ − v1
3
Place these two equations in standard form:
v1
(12
+124
+13
)+ iφ(−2) = 30 + 20
v1
(13
)+ iφ(1 − 2) = 20
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 = 0Solving,
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) = 0The 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 = 0Solving
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 simpleexpression for iφ into the third mesh equation and place the equations in standardform:
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 thati3 = −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 AThe current in the 2 Ω resistor is i1 − i2 = 6 A ·. . p2 Ω = (6)2(2) = 72 WThus, the 2 Ω resistors dissipates 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, sowe 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; ib = −10 A; ic = 10 A; vφ = 25 VThus, ia = 15 A.
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 thesemeshes 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
ia − ib = 2
Place these three equations in standard form:
2ia + 4ib − 4ic = 10
−2ia − 2ib + 5ic = 6
ia − ib + 0ic = 2
Solving, ia = 7 A; ib = 5 A; ic = 6 AThus, 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 − 2015
− 2 +v1 − 25
10= 0
4–10 CHAPTER 4. Techniques of Circuit Analysis
Rearranging and solving,
v1
( 115
+110
)= 2 +
2015
+2510
·. . 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 twomesh 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
ix = i1 − i3 = i1 − 4
Substitute the constraint equation into the second mesh equation and place theresulting two mesh equations in standard form:
12i1 − 6i2 = 144
24i1 + 14i2 = 132
Solving,
i1 = 9 A; i2 = −6 A; i3 = 4 A; ix = 9 − 4 = 5 A
·. . v4A = 3(i3 − i2) − 4ix = 10 V
p4A = −v4A(4) = −(10)(4) = −40 W
Thus, the 2 A current source delivers 40 W.
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 inparallel with the 20 Ω resistor. Also transform the −60 V source in series withthe 5 Ω resistor into a −12 A source in parallel with the 5 Ω resistor. The resultis the following circuit:
Combine the three current sources into a single current source, using KCL, andcombine the 20 Ω, 5 Ω, and 6 Ω resistors in parallel. The resulting circuit isshown on the left. To simplify the circuit further, transform the resulting 30 Asource in parallel with the 2.4 Ω resistor into a 72 V source in series with the2.4 Ω resistor. Combine the 2.4 Ω resistor in series with the 1.6 Ω resisor to geta very simple circuit that still maintains the voltage v. The resulting circuit ison the right.
Use voltage division in the circuit on the right to calculate v as follows:
v =812
(72) = 48 V
[b] Calculate i in the circuit on the right using Ohm’s law:
i =v
8=
488
= 6 A
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 thevoltage drop across the series combination of the 120 V source and 20 Ωresistor. Use this fact to calculate the current in the 120 V source, ia:
ia =120 − va
20=
120 − 57.620
= 3.12 A
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 of5‖20 = 4 Ω. The equivalent 4 Ω resistance is in series with the 8 Ω resistor for anequivalent resistance of 4 + 8 = 12 Ω. Finally, the 12 Ω equivalent resistance is inparallel with the 12 Ω resistor, so RTh = 12‖12 = 6 Ω.
Use node voltage analysis to find vTh. Begin by redrawing the circuit and labelingthe node voltages:
The node voltage equations arev1 − 72
5+
v1
20+
v1 − vTh
8= 0
vTh − v1
8+
vTh − 7212
= 0
Problems 4–13
Place these equations in standard form:
v1
(15
+120
+18
)+ vTh
(−1
8
)=
725
v1
(−1
8
)+ vTh
(18
+112
)= 6
Solving, v1 = 60 V and vTh = 64.8 V. Therefore, the Thévenin equivalent circuit isa 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 ofthe 15 A source and 8 Ω resistor into a series combination of a 120 V source and an8 Ω resistor, as shown in the figure on the left. Next, combine the 2 Ω, 8 Ω and 10 Ωresistors in series to give an equivalent 20 Ω resistance. Then transform the seriescombination of the 120 V source and the 20 Ω equivalent resistance into a parallelcombination 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 = 20‖12 = 7.5 Ω.Thus, the Norton equivalent circuit is the parallel combination of a 6 A source and a7.5 Ω resistor.
AP 4.18 Find the Thévenin equivalent with respect to A, B using source transformations. Tobegin, convert the series combination of the −36 V source and 12 kΩ resistor into aparallel combination of a −3 mA source and 12 kΩ resistor. The resulting circuit isshown 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 k‖60 k= 10 kΩ resistor. Thentransform the 15 mA source in parallel with the 10 kΩ resistor into a 150 V source inseries with a 10 kΩ resistor, and combine this 10 kΩ resistor in series with the 15 kΩresistor. 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 equivalentand use voltage division to calculate the meter reading vAB:
vAB =100,000125,000
(150) = 120 V
AP 4.19 Begin by calculating the open circuit voltage, which is also vTh, from the circuitbelow:
Summing the currents away from the node labeled vTh We have
vTh
8+ 4 + 3ix +
vTh − 242
= 0
Also, using Ohm’s law for the 8 Ω resistor,
ix =vTh
8Substituting 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 voltagesource with a short circuit, the current source with an open circuit, and apply the testvoltage vT, as shown in the circuit below:
Problems 4–15
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 nodevoltage method in the circuit below:
The node voltage equations are
v
60+
v − (vTh + 160i∆)20
− 4 = 0,
vTh
40+
vTh
80+
vTh + 160i∆ − v
20= 0
The dependent source constraint equation is
i∆ =vTh
40
4–16 CHAPTER 4. Techniques of Circuit Analysis
Substitute the constraint equation into the node voltage equations and put the twoequations in standard form:
v( 1
60+
120
)+ vTh
(− 5
20
)= 4
v(− 1
20
)+ vTh
( 140
+180
+520
)= 0
Solving, v = 172.5 V and vTh = 30 V.
Now use the test source method to calculate the test current and thus RTh. Replacethe current source with a short circuit and apply the test source to get the followingcircuit:
Write a KCL equation at the rightmost node:
iT =vT
80+
vT
40+
vT + 160i∆80
The dependent source constraint equation is
i∆ =vT
40
Substitute the constraint equation into the KCL equation and simplify the right-handside:
iT =vT
10
Therefore,
RTh =vT
iT= 10 Ω
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 circuitbetween 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)4
+vTh − v1
4= 0
v1 − 1004
+v1 − 20
4+
v1 − vTh
4= 0
The dependent source constraint equation is
vφ = v1 − 20
Place these three equations in standard form:
vTh
(14
+14
)+ v1
(−1
4
)+ vφ
(−1
4
)= 25
vTh
(−1
4
)+ v1
(14
+14
+14
)+ vφ (0) = 30
vTh (0) + v1 (1) + vφ (−1) = 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 methodwith the circuit below:
The mesh current equations are
−100 + 4(i1 − i2) + vφ + 20 = 0
−vφ + 4i2 + 4(i2 − isc) + 4(i2 − i1) = 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
isc=
12040
= 3 Ω
[a] For maximum power transfer, R = RTh = 3 Ω
[b] The Thévenin voltage, vTh = 120 V, splits equally between the Théveninresistance and the load resistance, so
vload =1202
= 60 V
Therefore,
pmax =v2
load
Rload=
602
3= 1200 W
AP 4.22 Sustituting the value R = 3 Ω into the circuit and identifying three mesh currents wehave 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 isdelivering 3000 W.
[b] pdepsource = −vφi2 = −(40)(20) = −800 W. Thus, the dependent source isdelivering 800 W.
[c] From Assessment Problem 4.21(b), the power delivered to the load resistor is1200 W, so the load power is (1200/3800)100 = 31.58% of the combinedpower 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, 1branch with a dependent source
[b] The current is unknown in every branch except the one containing the 8 Acurrent source, so the current is unknown in 10 branches.
[c] 9 essential branches – R4 − R5 forms an essential branch as does R8 − 10 V. Theremaining 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, twoidentified with single black dots, and one identified by a triangle.
[f] There are 4 essential nodes, three identified with rectangular boxes and oneidentified with a triangle
[g] A mesh is like a window pane, and as can be seen from the figure there are 6window 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 voltagesource 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 nowonly a single part in the circuit. There would now be 4 nodes, because the twolower nodes are now joined as a single node. The number of branches remainsat 7, where each branch contains one of the seven individual circuitcomponents.
P 4.3 [a] From Problem 4.1(d) there are 8 essential branches were the current isunknown, 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 sourceconstraint equation.
[c] The remaining 4 equations needed to describe the circuit will be derived fromKVL equations.
[d] We must avoid using the topmost mesh and the leftmost mesh. Each of thesemeshes contains a current source, and we have no way of determining thevoltage drop across a current source.
P 4.4 [a] There are six circuit components, five resistors and the current source. Since thecurrent 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 inFig. P4.4. At three of these nodes you can write KCL equations that will beindependent of one another. A KCL equation at the fourth node would bedependent on the first three. Therefore there are three independent KCLequations.
[c]
Sum the currents at any three of the four essential nodes a, b, c, and d. Usingnodes a, b, and c we get
−ig + i1 + i2 = 0
4–22 CHAPTER 4. Techniques of Circuit Analysis
−i1 + i4 + i3 = 0
i5 − i2 − 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 onthe bottom right with components R3, R4, and R5. We cannot write a KVLequation for the left mesh because we don’t know the voltage drop across thecurrent source. Therefore, we can write KVL equations for the two meshes onthe right, giving a total of two independent KVL equations.
[e] Sum the voltages around two independent closed paths, avoiding a path thatcontains the independent current source since the voltage across the currentsource is not known. Using the upper and lower meshes formed by the fiveresistors gives
R1i1 + R3i3 − R2i2 = 0
R3i3 + R5i5 − R4i4 = 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 1for 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 voltageat 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 + 25120 + 5
+vo
25+ 0.04 = 0
Solve by multiplying both sides of the KCL equation by 125 and collecting theterms involving vo on one side of the equation and the constants on the other side ofthe 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; thereforep40mA = (−5)(0.04) = −0.2 WThe 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 ofProblem 4.6 isig = (−5 + 25)/125 = 160 mAThe power in the 25 V source isp25V = −(25)(0.16) = −4 WThe power developed by the 25 V source is 4 W
[c] p5Ω = (0.16)2(5) = 128 mW
p120Ω = (0.16)2(120) = 3.072 W
p25Ω = (−5)2/25 = 1 W∑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
125+
vo
25+ 0.04 = 0
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 Vp40mA = (−9)(0.04) = −360 mWThe 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 mAp25V = −(25)(0.16) = −4 WThe power developed by the 25 V source is 4 W
[d] p5Ω = (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∑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 mAcurrent source
P 4.9
The two node voltage equations are:
−6 +v1
40+
v1 − v2
8= 0
v2 − v1
8+
v2
80+
v2
120+ 1 = 0
Place these equations in standard form:
v1
( 140
+18
)+ v2
(−1
8
)= 6
v1
(−1
8
)+ v2
(18
+180
+1
120
)= −1
Solving, v1 = 120 V and v2 = 96 V.
Check this result by calculating the power associated with each component:
Problems 4–25
Component Power Delivered (W) Power Absorbed (W)
6A −(6 A)(120 V) = −720
40 Ω1202
40= 360
8 Ω(120 − 96)2
8= 72
80 Ω962
80= 115.2
120 Ω962
120= 76.8
1 A (96 V)(1 A) = 96
Total −720 720
P 4.10 [a]
The two node voltage equations are:v1
6+
v1 − 444
+v1 − v2
1= 0
v2
3+
v2 − v1
1+
v2 + 22
= 0
Place these equations in standard form:
v1
(16
+14
+ 1)
+ v2(−1) =444
v1(−1) + v2
(13
+ 1 +12
)= −2
2Solving, v1 = 12 V; v2 = 6 VNow calculate the branch currents from the node voltage values:
ia =44 − 12
4= 8 A
ib =126
= 2 A
ic =12 − 6
1= 6 A
id =63
= 2 A
ie =6 + 2
2= 4 A
4–26 CHAPTER 4. Techniques of Circuit Analysis
[b] psources = p44V + p2V = −(44)ia − (2)ie = −(44)(8) − (2)(4) = −352 − 8 = −360 WThus, the power developed in the circuit is 360 W. Note that the resistorscannot develop power!
P 4.11 [a]
v1 − 1102
+v1 − v2
8+
v1 − v3
16= 0 so 11v1 − 2v2 − v3 = 880
v2 − v1
8+
v2
3+
v2 − v3
24= 0 so −3v1 + 12v2 − v3 = 0
v3 + 1102
+v3 − v2
24+
v3 − v1
16= 0 so −3v1 − 2v2 + 29v3 = −2640
Solving, v1 = 74.64 V; v2 = 11.79 V; v3 = −82.5 V
Thus, i1 =110 − v1
2= 17.68 A i4 =
v1 − v2
8= 7.86 A
i2 =v2
3= 3.93 A i5 =
v2 − v3
24= 3.93 A
i3 =v3 + 110
2= 13.75 A i6 =
v1 − v3
16= 9.82 A
[b]∑
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 − 150
20+
v1
80+
v1 − v2
40= 0
v2 − v1
40− 11.25 +
v2
4= 0
Problems 4–27
Place these equations in standard form:
v1
( 120
+180
+140
)+ v2(− 1
40) =
15020
v1
(− 1
40
)+ v2
( 140
+14
)= 11.25
Solving, v1 = 100 V; v2 = 50 V
P 4.13
At vo : −2 +vo
50+
vo − 454 + 1
= 0
Solving, vo = 50 V
p2A = −(50)(2) = −100 W
Thus, the 2 A current source delivers 100 W, or the current source extracts −100 Wfrom the circuit.
P 4.14
The three node voltage equations are:v1 − 40
4+
v1
40+
v1 − v2
2= 0
v2 − v1
2+
v2 − v3
4− 28 = 0
v3
2+
v3 − v2
4+ 28 = 0
Place these equations in standard form:
v1
(14
+140
+12
)+ v2
(−1
2
)+ v3(0) =
404
v1
(−1
2
)+ v2
(12
+14
)+ v3
(−1
4
)= 28
v1(0) + v2
(−1
4
)+ v3
(12
+14
)= −28
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
12+
v1
25+ 5 +
v1 − v2
20= 0
v2 − v1
20+
v2 − v3
40− 7.5 − 5 = 0
v3
40+
v3 − v2
40+ 7.5 = 0
Place these equations in standard form:
v1
( 112
+125
+120
)+ v2
(− 1
20
)+ v3(0) = −40
12− 5
v1
(− 1
20
)+ v2
( 120
+140
)+ v3
(− 1
40
)= 12.5
v1(0) + v2
(− 1
40
)+ v3
( 140
+140
)= −7.5
Solving, v1 = −10 V; v2 = 132 V; v3 = −84 V.Find the power:
i40V = (−10 + 40)/12 = 2.5 A
p40V = −(2.5)(40) = −100 W (del)
p5A = (5)(−10 − 132) = −710 W (del)
p7.5A = (7.5)(−84 − 132) = −1620 W (del)
p12Ω = (−10 + 40)2/12 = 75 W (abs)
p25Ω = (−10)2/25 = 4 W (abs)
Problems 4–29
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 (abs)
∑pdiss = 75 + 4 + 1008.2 + 1166.4 + 176.4 = 2430 W
∑pdev = 100 + 710 + 1620 W = 2430 W (CHECKS)
P 4.16 [a]vo − v1
R+
vo − v2
R+
vo − v3
R+ · · · +
vo − vn
R= 0
·. . nvo = v1 + v2 + v3 + · · · + vn
·. . vo =1n
[v1 + v2 + v3 + · · · + vn] =1n
∑n
k=1vk
[b] vo =13(120 + 60 − 30) = 50 V
P 4.17 [a]
The node voltage equation is:
−0.45 +vo
100+
vo − 6.25i∆5
+vo − 45
25= 0
The dependent source constraint equation is:
i∆ =45 − vo
25Place these equations in standard form:
vo
( 1100
+15
+125
)+ i∆
(−6.25
5
)=
4525
+ 0.45
vo
( 125
)+ i∆(1) =
4525
Solving, vo = 15 V; i∆ = 1.2 A
[b] ids =vo − 6.25i∆
5=
15 − 7.55
= 1.5 A
pds = [6.25(1.2)](1.5) = 11.25 WThus, the dependent source absorbs 11.25 W
4–30 CHAPTER 4. Techniques of Circuit Analysis
[c] p450mA = −(0.45)(15) = −6.75 Wp45V = −(1.2)(45) = −54 W∑
pdev = 6.75 + 54 = 60.75 WThus the independent sources develop 60.75 WAlso,∑
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:
−5io +v1
20+
v1 − v2
5= 0
v2 − v1
5+
v2
40+
v2 − v3
10= 0
v3 − v2
10+
v3 − 11.5io
5+
v3 − 964
= 0
The dependent source constraint equation is:io = v2/40Place these equations in standard form:
v1
( 120
+15
)+ v2
(−1
5
)+ v3(0) + io(−5) = 0
v1
(−1
5
)+ v2
(15
+140
+110
)+ v3
(− 1
10
)+ io(0) = 0
v1(0) + v2
(− 1
10
)+ v3
( 110
+15
+14
)+ io
(−11.5
5
)=
964
v1(0) + v2
(− 1
40
)+ v3(0) + io(1) = 0
Solving, v1 = 156 V; v2 = 120 V; v3 = 78 V; io = 3 A
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 − 16010
+vo
100+
vo − 150iσ
30 + 20= 0
The dependent source constraint equation is:
iσ = − vo
100Place these equations in standard form:
vo
( 110
+1
100+
150
)+ iσ
(−150
50
)=
16010
vo
( 1100
)+ iσ(1) = 0
Solving, vo = 100 V; iσ = −1 ANow find the power:
io =160 − 100
10− 1 = 5 A
pds = [150(−1)](5) = −750 W.Thus, the dependent source delivers 750 W
4–32 CHAPTER 4. Techniques of Circuit Analysis
P 4.20 [a]
The node voltage equations are:
−25 +v1
40+
v1
160+
v1 − v2
10= 0
v2 − v1
10+
v2
20+
v2 − 84i∆8
= 0
The dependent source constraint equation is:i∆ = v1/160Place these three equations in standard form:
v1
( 140
+1
160+
110
)+ v2
(− 1
10
)+ i∆(0) = 25
v1
(− 1
10
)+ v2
( 110
+120
+18
)+ i∆
(−84
8
)= 0
v1
(− 1
160
)+ v2(0) + i∆(1) = 0
Solving, v1 = 352 V; v2 = 212 V; i∆ = 2.2 ANow calculate the power. Only the two sources can develop power, so focus onthe 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 inthe circuit is 8800 W
[b] The dependent source and all of the resistors dissipate the power developed bythe current source. Check that the power developed equals the powerdissipated:
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 4–33
P 4.21
The two node voltage equations are:v1 − 40
5+
v1
50+
v1 − v2
10= 0
v2 − v1
10− 10 +
v2
40+
v2 − 408
= 0
Place these equations in standard form:
v1
(15
+150
+110
)+ v2
(− 1
10
)=
405
v1
(− 1
10
)+ v2
( 110
+140
+18
)= 10 +
408
Solving, v1 = 50 V; v2 = 80 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 (abs)
p5Ω = (50 − 40)2/5 = 20 W (abs)
p8Ω = (80 − 40)2/8 = 200 W (abs)
p10Ω = (80 − 50)2/10 = 90 W (abs)
p50Ω = 502/50 = 50 W (abs)
p40Ω = 802/40 = 160 W (abs)
p10A = −(80)(10) = −800 W (del)∑pabs = 280 + 20 + 200 + 90 + 50 + 160 = 800 W =
∑pdel
4–34 CHAPTER 4. Techniques of Circuit Analysis
P 4.22
The node voltage equations are:v1 − 2.26
20+
v1 − v2
50+
v1
25= 0
v2 − 2.2640
+v2 − v1
50+
v2
100= 0
Place these equations in standard form:
v1
( 120
+150
+125
)+ v2
(− 1
50
)=
2.2620
v1
(− 1
50
)+ v2
( 140
+150
+1
100
)=
2.2640
Solving, v1 = 1.3 V; v2 = 1.5 V.
Thus, io =v1 − v2
50=
1.3 − 1.550
= −4 mA
P 4.23 [a]
The node voltage equations are:
Problems 4–35
v2 − 2301
+v2 − v4
1+
v2 − v3
1= 0
v3 − v2
1+
v3 − v5
1+
v3
1= 0
v4 − 2305 + 1
+v4 − v2
1+
v4 − v5
2= 0
v5 − v4
2+
v5 − v3
1+
v5
5 + 1= 0
Place these equations in standard form:
v2(1 + 1 + 1) + v3(−1) + v4(−1) + v5(0) = 230
v2(−1) + v3(1 + 1 + 1) + v4(0) + v5(−1) = 0
v2(−1) + v3(0) + v4
(16
+ 1 +12
)+ v5
(−1
2
)=
2306
v2(0) + v3(−1) + v4
(−1
2
)+ v5
(12
+ 1 +16
)= 0
Solving, v2 = 150 V; v3 = 80 V; v4 = 140 V; v5 = 90 VFind the power dissipated by the 2 Ω resistor:
i2Ω =v4 − v5
2=
140 − 902
= 25 A
p2Ω = (25)2(2) = 1250 W
[b] Find the power developed by the 230 V source:
i230V =v2 − 230
1+
v4 − 2306
= −80 − 15 = −95 A
p230V = (230)(−95) = −21,850 W, so the source supplies 21,850 WCheck:
∑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)
4–36 CHAPTER 4. Techniques of Circuit Analysis
P 4.24 [a]
There is only one node voltage equation:va + 305000
+va
500+
va − 801000
+ 0.01 = 0
Solving,va + 30 + 10va + 5va − 400 + 50 = 0 so 16va = 320·. . 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
i3 + i4 + i5 − 10 mA = 0 so i3 = 0.01 − 0.02 − 0.06 = −0.07 = −70 mA
[b] p30V = (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∑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:
7 +vb
3+
vb − vc
1= 0
−2vx +vc − vb
1+
vc − 42
= 0
The constraint equation for the dependent source is:vx = vc − 4
Place these equations in standard form:
vb
(13
+ 1)
+ vc(−1) + vx(0) = −7
vb(−1) + vc
(1 +
12
)+ vx(−2) =
42
vb(0) + vc(1) + vx(−1) = 4
Solving, vo = vb = 1.5 V 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. Thesupernode equation is
2 +v1
50+
v2
150+
v2
20 + 55= 0
The supernode constraint equation is
v2 + 25 = v1
Place these two equations in standard form:
4–38 CHAPTER 4. Techniques of Circuit Analysis
v1
( 150
)+ v2
( 1150
+175
)= −2
v1(1) + v2(−1) = 25
Solving, v1 = −37.5 V and v2 = −62.5 V.
p25V = (25)i25
i25 = −2 A − i50 = −2 A − v1
50= 2 A − −37.5
50= −2 A + 0.75 A = −1.25 A
Thus, p25V = (25)(−1.25) = −31.25 W
The 25 V source delivers 31.25 W.
P 4.27
The supernode equation is:
v1 − 10010
+v1
60+
v1 − 4v∆
20+
v1 − 4v∆
30= 0
The constraint equation for the dependent source is:4v∆ = v1 − v∆
Place these equations in standard form:
v1
( 110
+160
+120
+130
)+ v∆
(− 4
20− 4
30
)=
10010
v1(1) + v∆(−5) = 0
Solving, v1 = 75 V; v∆ = 15 VThus, vo = 100 − v1 = 25 V
P 4.28 Calculate currents and voltages needed to calculate the power for the variouscomponents:
iφ =v4 − v3
8=
81.6 − 1088
= −3.3 A
403
iφ =403
(−3.3) = −44 V
v1 = v4 +403
iφ = 81.6 − 44 = 37.6 V
Problems 4–39
v3 + v∆ = 120 ·. . v∆ = 120 − 108 = 12 V
1.75v∆ = (1.75)(12) = 21 A
i120V =v1 − 120
4+
v3 − 1202
=37.6 − 120
4+
108 − 1202
= −26.6 A
iccvs =0 − v1
20+
v2 − v1
4=
−37.620
+120 − 37.6
4= 18.72 A
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∑
pabs =∑
pdel = 4015.6 W
pccvs = (18.72)(−44) = −823.68 W
Now sum the powers:∑ptotal = 70.688 + 1697.44 − 3192 + 72 + 291.6 + 87.12
+83.232 + 1712.6 − 823.68 = 0 WThus, the power balances and the staff analyst has correctly calculated the voltagevalues
P 4.29
The supernode equation is:v1
100+
v1 − 6010
+v3 − v2
20+
v3
400− 0.625v∆ = 0
The node voltage equation at v2 is:
4–40 CHAPTER 4. Techniques of Circuit Analysis
v2 − 605
+v2
200+
v2 − v3
20= 0
The supernode constraint equation is:v3 − v1 = 175iφ
The two dependent source constraint equations are:v∆ = v2 − 60iφ = −v2/200Place the four equations above in standard form:
v1
( 1100
+110
)+ v2
(− 1
20
)+ v3
( 1400
+120
)+ iφ(0) + v∆(−0.625) =
6010
v1(0) + v2
(15
+1
200+
120
)+ v3
(− 1
20
)+ iφ(0) + v∆(0) =
605
v1(1) + v2(0) + v3(−1) + iφ(175) + v∆(0) = 0
v1(0) + v2(1) + v3(0) + iφ(0) + v∆(−1) = 60
v1(0) + v2
( 1200
)+ v3(0) + iφ(1) + v∆(0) = 0
Solving,v1 = −60.75 V v2 = 30 V; v3 = −87 V; iφ = −0.15 A; v∆ = −30 VCalculate the power for the 60 V source:
i60V =v1 − 60
10+
v2 − 605
=−60.75 − 60
10+
30 − 605
= −18.075 A
p60V = (60)(−18.075) = −1084.5 W
Thus, the 60 V source delivers 1084.5 W
P 4.30 From Eq. 4.16, iB = vc/(1 + β)RE
From Eq. 4.17, iB = (vb − Vo)/(1 + β)RE
From Eq. 4.19,
iB =1
(1 + β)RE
[VCC(1 + β)RER2 + VoR1R2
R1R2 + (1 + β)RE(R1 + R2)− Vo
]
=VCCR2 − Vo(R1 + R2)
R1R2 + (1 + β)RE(R1 + R2)=
[VCCR2/(R1 + R2)] − Vo
[R1R2/(R1 + R2)] + (1 + β)RE
Problems 4–41
P 4.31 [a]
The mesh current equations are:
−60 + 4i1 + 10(i1 − i2) + 1i1 = 0
20 + 3i2 + 10(i2 − i1) + 2i2 = 0Place the equations in standard form:
i1(4 + 10 + 1) + i2(−10) = 60
i1(−10) + i2(3 + 10 + 2) = −20Solving, i1 = 5.6 A; i2 = 2.4 A
Now solve for the requested currents:ia = i1 = 5.6 A; ib = i1 − i2 = 3.2 A; ic = −i2 = −2.4 A
[b] If the polarity of the 60 V source is reversed, we have the following meshcurrent equations in standard form:
i1(4 + 10 + 1) + i2(−10) = −60
i1(−10) + i2(3 + 10 + 2) = −20Solving, i1 = −8.8 A; i2 = −7.2 A
Now solve for the requested currents:ia = i1 = −8.8 A; ib = i1 − i2 = −1.6 A; ic = −i2 = 7.2 A
P 4.32 [a]
The mesh current equations are:
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
460 + 5i3 − 115 + 2(i3 − i1) + 3(i3 − i2) = 0Place 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) = −345Solving, i1 = 4.4 A; i2 = −10.6 A; i3 = −36.8 AThe 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 thepower 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 4–43
−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:iσ = i2 − i1Place 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; iσ = −25.8 ACalculate the power:p20Ω = 20(68.4 − 64.8)2 = 259.2 WThus 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:iφ = i2 − 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
4–44 CHAPTER 4. Techniques of Circuit Analysis
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 WThus, 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
Place these equations in standard form:
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
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 4–45
∑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
4–46 CHAPTER 4. Techniques of Circuit Analysis
[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
[c] p4Ω = (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
∑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) = 0Solving, 9250i1 = 27.75 ·. . i1 = 3 mAThen, i∆ = 0.005 − i1 = 0.005 − 0.003 = 0.002 = 2 mA
[b] vo = (0.005)(10,000) + (0.002)(5400) = 60.8 Vp5mA = −(60.8)(0.005) = −304 mWThus, the 5 mA source delivers 304 mW
Problems 4–47
[c] 150i∆ = 150(0.002) = 0.3 Vpdep source = 150i∆i1 = −(0.3)(0.003) = −0.9 mWThe 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:i3 = −0.5v∆; v∆ = 2(i1 − i2)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 VSolve the outer loop KVL equation to find vcs:−125 + 7i1 + vcs = 0; ·. . vcs = 125 − 7(6) = 83 VCalculate 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
4–48 CHAPTER 4. Techniques of Circuit Analysis
·. .∑
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 currentequations:
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
Solving, i1 = 16 A; i2 = 15 AFind v:−v + 5(20 − i1) + 20(20 − i2) = 0 ·. . v = 5(4) + 20(5) = 120 VCalculate 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
∑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) = 0The 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) = 0Solving, 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 WThe 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:
28885245
× 100 = 55.06%
4–50 CHAPTER 4. Techniques of Circuit Analysis
P 4.42 [a]
The mesh current equations are:
75 + 6i1 + 12(i1 − i2) − 7i∆ = 0
15i2 + 60(i2 − i3) + 7i∆ + 12(i2 − i1) = 0The two constraint equations are:
i∆ = −i2
i3 = 1.6v∆ = 1.6(6i1) = 9.6i1Place 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) = 0Solving, i1 = 4 A; i2 = 29.4 A; i3 = 38.4 A; i∆ = −29.4 ACalculate 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
The two dependent sources are generating a total of5227.32 + 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 4–51
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 = 0The supermesh constraint equation is :i1 − i2 = 6Place these equations in standard form:
i1(4 + 1) + i2(9 + 6) = 20 + 90
i1(1) + i2(−1) = 6
Solving, i1 = 10 A; i2 = 4 ANow find the power:
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
4–52 CHAPTER 4. Techniques of Circuit Analysis
P 4.44
The supermesh equation is:−120 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0The supermesh constraint equation is :i1 − i2 = 6Place these equations in standard form:
i1(4 + 1) + i2(9 + 6) = 120 + 90
i1(1) + i2(−1) = 6
Solving, i1 = 15 A; i2 = 9 ANow find the power:
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 (note that the 6 A source is
now dissipating power!)∑pdiss = 1800 + 810 = 2610 W
Thus the power dissipated in the circuit is 2610 W
P 4.45 [a]
Problems 4–53
The supermesh equation is:−60 + 4i1 + 9i2 − 90 + 6i2 + 1i1 = 0The supermesh constraint equation is :i1 − i2 = 6Place these equations in standard form:
i1(4 + 1) + i2(9 + 6) = 60 + 90
i1(1) + i2(−1) = 6Solving, i1 = 12 A; i2 = 6 ANow find the power:
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 equationsare:−60 + 4i1 + 1i1 = 0
−90 + 6i2 + 9i2 = 0Since these equations are uncoupled, each can be solved separately:
4–54 CHAPTER 4. Techniques of Circuit Analysis
5i1 = 60 ·. . i1 = 60/5 = 12 A
15i2 = 90 ·. . i2 = 90/15 = 6 A
Since the currents are the same as in part (a), the power will be the same ascalculated 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 ashort 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 = 0The i2 — i3 supermesh equationa:2i2 + 20i3 + 10(i3 − i1) + 5(i2 − i1) = 0The supermesh constraint:i3 − i2 = 1.2ib = 1.2i1Place 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) = 0Solving, i1 = 7.4 A; i2 = −4.2 A; i3 = 4.68 ASolve 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 ·. . vcs = −2(−4.2) − 5(−4.2 − 7.4) = 66.4 V
Problems 4–55
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) = 0The i3 — i4 supermesh equation:40(i3 − 19) + 5(i3 − i2) + 10(i4 − i2) − 240 = 0The 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) = 0Solving, i2 = 18 A; i3 = 26 A; i4 = 10 ASolve for the requested currents:
4–56 CHAPTER 4. Techniques of Circuit Analysis
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 + 50i3Solving, i1 = 23.93 A; i2 = 17.79 A; i3 = 11.40 A
Problems 4–57
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]∑
pdev = 125(i1 + i2) = 5213.99 W
∑pload = 5021.15 W
% delivered =5021.155213.99
× 100 = 96.3%
[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 notfunction properly, and could be damaged.
4–58 CHAPTER 4. Techniques of Circuit Analysis
P 4.49
125 = (R1 + 0.6)ia − 0.4ib − R1ic
125 = −0.4ia + (R2 + 0.6)ib − R2ic
0 = −R1ia − R2ib + (R1 + R2 + 21.2)ic
∆ =
∣∣∣∣∣∣∣∣∣∣∣
(R1 + 0.6) −0.4 −R1
−0.4 (R2 + 0.6) −R2
−R1 −R2 (R1 + R2 + 21.2)
∣∣∣∣∣∣∣∣∣∣∣When R1 = R2, ∆ reduces to
∆ = 21.6R21 + 25.84R1 + 4.24.
Na =
∣∣∣∣∣∣∣125 −0.4 −R1
125 (R2 + 0.6) −R2
0 −R2 (R1 + R2 + 21.2)
∣∣∣∣∣∣∣= 125 [2R1R2 + R1 + 22.2R2 + 21.2]
Nb =
∣∣∣∣∣∣∣(R1 + 0.6) 125 −R1
−0.4 125 −R2
−R1 0 (R1 + R2 + 21.2)
∣∣∣∣∣∣∣= 125 [2R1R2 + 22.2R1 + R2 + 21.2]
ia =Na
∆, ib =
Nb
∆
ineutral = ia − ib =Na − Nb
∆=
125[(R1 − R2) + 22.2(R2 − R1)]∆
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 thesource must be zero. This means that the voltage drop across the 15 Ω resistor is alsozero, 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 ANow 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
4–60 CHAPTER 4. Techniques of Circuit Analysis
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) = 0Place the equations in standard form:
i2(−5) + i3(−10) + Vdc(0) = −23
i2(30 + 15 + 5) + i3(−15) + Vdc(0) = 0
i2(−15) + i3(25 + 10 + 15) + Vdc(1) = 46Solving, i2 = 0.6 A; i3 = 2 A; Vdc = −45 VThus, 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 simultaneousequations.
[b]
The mesh current equations:
2i1 + 10(i1 − i2) + 8(i1 − 4) = 0
4i2 + 1(i2 − 4) + 10(i2 − i1) = 0Place the equations in standard form:
i1(2 + 10 + 8) + i2(−10) = 32
i1(−10) + i2(4 + 1 + 10) = 4Solving, i1 = 2.6 A; i2 = 2 AFind the power in the 10 Ω resistor:i10Ω = i1 − i2 = 0.6 Ap10Ω = (0.6)2(10) = 3.6 W
[c] No, the voltage across the 4 A current source is readily available from the meshcurrents, and solving two simultaneous mesh-current equations is less workthan solving three node voltage equations.
[d] vg = 2i1 + 4i2 = 2(2.6) + 4(2) = 13.2 Vp4A = −(13.2)(4) = −52.8 WThus the 4 A source develops 52.8 W.
P 4.53 [a] There are three unknown node voltages and three unknown mesh currents, sothe number of simultaneous equations required is the same for both methods.The node voltage method has the advantage of having to solve the threesimultaneous equations for one unknown voltage provided the connection ateither the top or bottom of the circuit is used as the reference node. Thereforerecommend the node voltage method.
[b]
4–62 CHAPTER 4. Techniques of Circuit Analysis
The node voltage equations are:v1
1+
v1 − v2
8+
v1 − v3
10= 0
−4 +v2
20+
v2 − v1
8+
v2 − v3
2= 0
v3 − v1
10+
v3 − v2
2+
v3
4= 0
Put the equations in standard form:
v1
(1 +
18
+110
)+ v2
(−1
8
)+ v3
(− 1
10
)= 0
v1
(−1
8
)+ v2
( 120
+18
+12
)+ v3
(−1
2
)= 4
v1
(− 1
10
)+ v2
(−1
2
)+ v3
(12
+110
+14
)= 0
Solving, v1 = 1.72 V; v2 = 11.33 V; v3 = 6.87 Vp4A = −(11.33)(4) = −45.32 WTherefore, the 4 A source is developing 45.32 W
P 4.54 [a] The node voltage method requires summing the currents at two supernodes interms of four node voltages and using two constraint equations to reduce thesystem of equations to two unknowns. If the connection at the bottom of thecircuit is used as the reference node, then the voltages controlling thedependent sources are node voltages. This makes it easy to formulate theconstraint equations. The current in the 20 V source is obtained by summingthe currents at either terminal of the source.
The mesh current method requires summing the voltages around the twomeshes not containing current sources in terms of four mesh currents. Inaddition the voltages controlling the dependent sources must be expressed interms of the mesh currents. Thus the constraint equations are morecomplicated, and the reduction to two equations and two unknowns involvesmore algebraic manipulation. The current in the 20 V source is found bysubtracting two mesh currents.
Because the constraint equations are easier to formulate in the node voltagemethod, it is the preferred approach.
Problems 4–63
[b]
Node voltage equations:v1
100+
v2
250− 0.2 + 3 × 10−3v3 = 0
v3
500+
v4
200− 3 × 10−3v3 + 0.2 = 0
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)
2700 + 2300 + 1000= −1 mA
[b]
The node voltage equations:
4–64 CHAPTER 4. Techniques of Circuit Analysis
−2 × 10−3 +v1
2700+
v1 − v2
2300= 0
v2
1000+
v2 − v1
2300+ 0.6 × 10−3 = 0
Place these equations in standard form:
v1
( 12700
+1
2300
)+ v2
(− 1
2300
)= 2 × 10−3
v1
(− 1
2300
)+ v2
( 11000
+1
2300
)= −0.6 × 10−3
Solving, v1 = 2.7 V; v2 = 0.4 V
·. . io =v2 − v1
2300= −1 mA
P 4.56 [a]
Problems 4–65
io = −135/30,000 = −4.5 mA
[b]
va = (7500)(−0.0045) = −33.75 V
ia =va
90,000=
−33.7590,000
= −0.375 mA
ib = −8.4 × 10−3 + 0.375 × 10−3 + 4.5 × 10−3 = −3.525 mA
vb = (6000)(3.525 × 10−3) − 33.75 = −12.6 V
ig =−12.6 − 120
40,000= −3.315 mA
p120V = (120)(−3.315 × 10−3) = −397.8 mW
Check:p8.4mA = (−33.75)(8.4 × 10−3) = −283.5 mW∑
Pdev = 397.8 + 283.5 = 681.3 mW∑Pdis = (40,000)(−3.315 × 10−3)2 +
(−12.6)2
60,000+
(−33.75)2
90,000+(6000)(−3.525 × 10−3)2 + (7500)(−4.5 × 10−3)2
= 681.3 mW
P 4.57 [a]
4–66 CHAPTER 4. Techniques of Circuit Analysis
·. . vo =250300
(480) = 400 V; io =400250
= 1.6 A
[b]
p520V = −(520)(3.6) = −1872 WTherefore, the 520 V source is developing 1872 kW.
[c] v = −(16)(1) − 40(2.6) = −120 Vp1A = (−120)(1) = −120 WTherefore 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 the6 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 =8.510
(−1) = −0.85 A
[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) = 0Put these equations in standard form:
ia(6 + 6 + 5 + 17) + io(−17) = 12 + 5 + 34
ia(−17) + io(1.5 + 17) = −34Solving, ia = 1.075 A; io = −0.85 A
P 4.59 VTh =30
30 + 10(80) = 60 V
RTh = 10‖30 + 2.5 = 10 Ω
P 4.60
Write and solve the node voltage equation at v1:v1 − 60
10+
v1
40− 4 = 0
4v1 − 240 + v1 − 160 = 0 ·. . v1 = 400/5 = 80 VCalculate VTh:VTh = v1 + (8)(4) = 80 + 32 = 112 VCalculate RTh by removing the independent sources and making series and parallelcombinations of the resistors:RTh = 8 + 40‖10 = 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
Solving, i1 = 30 A; i2 = 12.5 AVTh = 5.2i2 + 12i1 = 425 VRTh = (8‖12 + 5.2)‖30 = 7.5 Ω
P 4.62 First we make the observation that the 10 mA current source and the 10 kΩ resistorwill 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 ourcircuit can be simplified to
4–70 CHAPTER 4. Techniques of Circuit Analysis
or
Therefore the Norton equivalent is determined by adding the current sources andcombining the resistors in parallel:
P 4.63 [a] First, find the Thévenin equivalent with respect to a,b using a succession ofsource transformations.
Problems 4–71
·. . VTh = 54 V RTh = 4.5 kΩ
vmeas =85.590
(54) = 51.3 V
[b] %error =(51.3 − 54
54
)× 100 = −5%
P 4.64 [a] Open circuit:
The node voltage equations are:v1 − 17.4
40+
v1
15+ 0.1 = 0
−0.1 +v2
14+
v2 − 17.426
= 0
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
voc =10
10 + 4(7) = 5 V
Short circuit:
4–72 CHAPTER 4. Techniques of Circuit Analysis
Write a node voltage equation at v2:
−0.1 +v2 − 17.4
26+
v2
4= 0
Solving,−5.2 + 2v2 − 34.8 + 13v2 = 0 ·. . v2 = 40/15 VCalculate the short circuit current:isc = (40/15)/4 = 2/3 ATherefore, RTh = 5/(2/3) = 7.5 Ω
[b]
RTh = 10‖(26 + 4) = 7.5 Ω (CHECKS)
Problems 4–73
P 4.65
OPEN CIRCUITUse Ohm’s law to solve for v2 on the right hand side of the circuit:v2 = −80ib(50,000) = −40 × 105ibUse this value of v2 to express the value of the dependent voltage source in terms ofib:4 × 10−5v2 = 4 × 10−5(−40 × 105ib) = −160ibWrite the mesh current equation for the ib mesh:1310ib − 160ib + 100(ib − 500 × 10−6) = 0Solving,1250ib = 0.05 ·. . ib = 0.05/1250 = 40µ AThus,VTh = v2 = −40 × 105ib = −40 × 105(40 × 10−6) = −160 VSHORT CIRCUIT
v2 = 0; isc = −80ib
Calculate ib using current division on the left hand side of the circuit:
ib =100
100 + 1310500 × 10−6 = 35.461 µ A
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
−2.8369 × 10−3 = 56.4 kΩ
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, RTh = 40 mΩ
·. . IN = 320 A, RN = 40 mΩ
P 4.67 First, find the Thévenin equivalent with respect to Ro.
Problems 4–75
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
P 4.68
The node voltage equations are:v1 − 9015,000
+v1
10,000+
v1 − v2
4000= 0
v2 − v1
4000+
v2
40,000+
v2 − v3
5000− 19i∆ = 0
v3 − v2
5000+
v3
89,000+ 19i∆ = 0
The dependent source constraint equation is:
i∆ =v1 − v2
4000Substitute the constraint equation into the node voltage equations and put the threeremaining equations in standard form:
v1
(1
15,000+
110,000
+1
4000
)+ v2
(− 1
4000
)+ v3(0) =
9015,000
v1
(− 1
4000− 19
4000
)+ v2
(1
4000+
140,000
+1
5000+
194000
)+ v3
(− 1
5000
)= 0
v1
( 194000
)+ v2
(− 1
5000− 19
4000
)+ v3
(1
5000+
189,000
)= 0
Solving, v1 = 32.75 V; v2 = 30.58 V; v3 = −19.8 VVTh = 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 Aisc = −444.94 µ ARTh = −19.8/ − 444.94 × 10−6 = 44.5 kΩ
P 4.69 [a] Use source transformations to simplify the left side of the circuit.
ib =7.7 − 5.522,000
= 0.1 mA
Problems 4–77
Let Ro = Rmeter‖1.3 kΩ = 5.5/4.4 × 10−3 = 1250 Ω
·. .(Rmeter)(1300)Rmeter + 1300
= 1250; Rmeter =(1250)(1300)
50= 32.5 kΩ
[b] Actual value of ve:
ib =7.7
22,000 + 44(1300)= 97.22 µ A
ve = 44ib(1300) = 5.56 V
% error =(5.5 − 5.56
5.56
)× 100 = −1.10%
P 4.70 [a] Find the Thévenin equivalent with respect to the terminals of the ammeter. Thisis most easily done by first finding the Thévenin with respect to the terminalsof the 4.8 Ω resistor.Thévenin voltage: note iφ is zero.
VTh − 242
+VTh
100+
VTh
25+
VTh
20= 0
50VTh + VTh + 4VTh + 5VTh = 50(24) ·. . VTh = 50(24)/60 = 20 V
Short-circuit current:
isc = 12 + 2isc, ·. . isc = −12 A
RTh =20
−12= −1.67 Ω
4–78 CHAPTER 4. Techniques of Circuit Analysis
Rtotal =206
= 3.333 Ω
Rmeter = 3.333 − 3.133 = 0.20 Ω
[b] Actual current:
iactual =20
3.133= 6.383 A
% error =6 − 6.383
6.383× 100 = −6%
P 4.71
i1 = 100/20,000 = 5 mA
100 = VTh − 0.005RTh, VTh = 100 + 0.005RTh
Problems 4–79
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:
v1 =501
501 + 100(5) = 4.168053 V
v2 =5000
5000 + 1000(5) = 4.1666667 V
Now calculate VTh:VTh = v1 − v2 = 4.168053 − 4.1666667 = 1.3866 mVCalculate RTh by removing the voltage source and creating series and parallelcombinations of the resisitors:
4–80 CHAPTER 4. Techniques of Circuit Analysis
RTh = 100‖501 + 1000‖5000 =(100)(501)
601+
(1000)(5000)6000
= 916.69 ΩThe resulting Thévenin equivalent circuit is shown below:
Use KVL to calculate igal:
igal =1.3866 × 10−3
916.69 + 50= 1.43 µA
P 4.73 VTh = 0, since circuit contains no independent sources.
iT =vT − v1
20+
vT − 40i∆60
v1 − 40i∆16
+v1
80+
v1 − vT
20= 0
·. . 10v1 − 200i∆ = 4vT i∆ =−v1
80, 200i∆ = −2.5v1
·. . 12.5v1 = 4vT; v1 = 0.32vT
60iT = 4vT − 2.5v1 = 3.2vT
·. .vT
iT=
603.2
= 18.75 Ω
RTh = 18.75 Ω
Problems 4–81
P 4.74 VTh = 0 since there are no independent sources in the circuit. To find RTh, apply a 1A test source and calculate the voltage drop across the test source. Use the meshcurrent 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) = 0
ix(10 − 5) + iy(20 + 10 + 5) = 20
Solving, ix = 80 mA; iy = 560 mAFind 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 acouple 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
30+
vT − 30(−vT/30)20
iTvT
=130
+110
=430
=215
RTh =vT
iT=
152
= 7.5 Ω
Thus our problem is reduced to analyzing the circuit shown below.
p =( 100
7.5 + Ro
)2
Ro = 250
104
R2o + 15Ro + 56.25
Ro = 250
104Ro
250= R2
o + 15Ro + 56.25
40Ro = R2o + 15Ro + 56.25
R2o − 25Ro + 56.25 = 0
Ro = 12.5 ± √156.25 − 56.25 = 12.5 ± 10
Ro = 22.5 Ω
Ro = 2.5 Ω
Problems 4–83
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) = 0The dependent source constraint equation is:iβ = i2 − 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) = 0Solving, i1 = 99.6 A; i2 = 78 A; i3 = 100.8 A; iβ = −21.6 AVTh = 20(i1 − i3) = −24 VShort-circuit current:
The mesh current equations are:
4–84 CHAPTER 4. Techniques of Circuit Analysis
−240 + 3(i1 − i2) + 2i1 = 0
2i2 + 4(i2 − i3) + 3(i2 − i1) = 0
10iβ + 1i3 + 4(i3 − i2) = 0The dependent source constraint equation is:iβ = i2 − 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) = 0Solving, i1 = 92 A; i2 = 73.33 A; i3 = 96 A; iβ = −18.67 A
isc = i1 − i3 = −4 A; RTh =VTh
isc=
−24−4
= 6 Ω
RL = RTh = 6 Ω
[b] pmax =122
6= 24 W
P 4.77 [a]
v − 1212,000
+v − 1020,000
+v
12,500= 0
Solving, v = 7.03125 V
v10k =10,00012,500
(7.03125) = 5.625 V
·. . VTh = v − 10 = −4.375 V
Problems 4–85
RTh = [(12,000‖20,000) + 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, andsolve:
At v1: − 3 × 10−3 +v1
4000+
v1 − v2
8000= 0
At v2:v2 − v1
8000+
v2 − 1020,000
+v2 − v3
2500= 0
At v3:v3 − v2
2500+
v3
10,000+
v3 − 105000
= 0
4–86 CHAPTER 4. Techniques of Circuit Analysis
Standard form:
v1
( 14000
+1
8000
)+ v2
(− 1
8000
)+ v3(0) = 0.003
v1
(− 1
8000
)+ v2
(1
8000+
120,000
+1
2500
)+ v3
(− 1
2500
)=
1020,000
v1(0) + v2
(− 1
2500
)+ v3
(1
2500+
110,000
+1
5000
)=
105000
Calculator solution:
v1 = 10.890625 V v2 = 8.671875 V v3 = 7.8125 V
Calculate currents:
i2 =10 − v2
20,000= 66.40625µ A i3 =
10 − v3
5000= 437.5 µ A
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:0.9579312537.7109375
(100) = 2.54%
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
44,500= 2.2 mW
Problems 4–87
[c]
The node voltage equations are:v1 − 9015,000
+v1
10,000+
v1 − v2
4000= 0
v2 − v1
4000+
v2
40,000+
v2 − v3
5000− 19i∆ = 0
v3 − v2
5000+
v3
89,000+ 19i∆ +
v3
44,500= 0
The dependent source constraint equation is:
i∆ =v1 − v2
4000Place these equations in standard form:
v1
(1
15,000+
110,000
+1
4000
)+ v2
(− 1
4000
)+ v3(0) + i∆(0) =
9015,000
v1
(− 1
4000
)+ v2
(1
4000+
140,000
+1
5000
)+ v3
(− 1
5000
)+ i∆(−19) = 0
v1(0) + v2
(− 1
5000
)+ v3
(1
5000+
189,000
+1
44,500
)+ i∆(19) = 0
v1
( 14000
)+ v2
(− 1
4000
)+ v3(0) + i∆(−1) = 0
Solving,v1 = 33.2818 V; v2 = 31.4697 V; v3 = −9.9 V; i∆ = 453 µACalculate the power:
ig =90 + 33.2818
15,000= 3.78 mA
p90V = −(90)(3.78 × 10−3) = −340.31 mWpdep source = (v3 − v2)(19i∆) = −356.07 mW∑
pdev = 340.31 + 356.07 = 696.38 mW
% delivered =2.2 × 10−3
696.38 × 10−3 × 100 = 0.316%
P 4.80 [a] From the solution to Problem 4.67 we have
4–88 CHAPTER 4. Techniques of Circuit Analysis
Ro(Ω) Po(W) Ro(Ω) Po(W)
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
[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); ic = ia − ib
Problems 4–89
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) ·. . ic = ia − isc
Solving, isc = 60 A; ia = 80 A; ic = 20 A; v∆ = 40 V
RTh = VTh/isc = 184.8/60 = 3.08 Ω
Ro = 3.08 Ω
pRo =(92.4)2
3.08= 2772 W
With Ro equal to 3.08 Ω the circuit becomes
4–90 CHAPTER 4. Techniques of Circuit Analysis
220 = 5i1 − 3(0.5)(2)(i1 − i3) − 2i3 = 2i1 + i3
·. . 2i1 = 220 − i3 = 220 − 43.2 = 176.8 ·. . i1 = 88.4 A
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
Problems 4–91
Therefore, the only source developing power is the 440 V source.
p440V = −(440)(88.4) = −38,896 W Power delivered is 38,896 W
% delivered =2772
38,896(100) = 7.13%
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) = 0The constraint equation is:i∆ = i3 − i1Place 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) = 0Solving, i1 = 4.7 A; i2 = 10.5 A; i3 = 4.1 A; i∆ = −0.6 AAlso, VTh = vab = −80i∆ = 48 VNow find the short-circuit current.
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) = 0Place 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) = −50Solving, i1 = 5 A; i2 = 0 A; i3 = −2.5 AThen, isc = i1 − i3 = 7.5 ARTh = 48/7.5 = 6.4 Ω
For maximum power transfer Ro = RTh = 6.4 Ω
[b] pmax =242
6.4= 90 W
P 4.83 From the solution of Problem 4.82 we know that when Ro is 6.4 Ω, the voltageacross Ro is 24 V, positive at the upper terminal. Therefore our problem reduces tothe analysis of the following circuit. In constructing the circuit we have used the factthat i∆ is −0.3 A, and hence 124i∆ is −37.2 V.
Problems 4–93
Using the node voltage method to find v1 and v2 yields
4.05 +24 − v1
4+
24 − v2
8= 0
2v1 + v2 = 104.4; v1 + 37.2 = v2
Solving, v1 = 22.4 V; v2 = 59.6 V.It follows that
ig1 =22.4 − 100
16= −4.85 A
ig2 =59.6 − 50
12= 0.8 A
i2 =59.6 − 24
8= 4.45 A
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
680.3(100) = 13.23%
·. . 13.23% of developed power is delivered to load
4–94 CHAPTER 4. Techniques of Circuit Analysis
P 4.84 [a] Open circuit voltage
Node voltage equations:v1 − 60
2+
v1 − 4i∆5
+v1 − v2
4= 0
v2 − v1
4+ 2v∆ = 0
Constraint equations:
v∆ = 60 − v1
i∆ =v1 − v2
4Place the equations in standard form:
v1
(12
+15
+14
)+ v2
(−1
4
)+ i∆
(−4
5
)+ v∆(0) = 30
v1
(−1
4
)+ v2
(14
)+ i∆(0) + v∆(2) = 0
v1(1) + v2(0) + i∆(0) + v∆(1) = 60
v1(1) + v2(−1) + i∆(−4) + v∆(0) = 0
Solving, v1 = 20 V; v2 = −300 V; i∆ = 80 A; v∆ = 40 VShort circuit current:
The node voltage equation:v1 − 60
2+
v1 − 4i∆5
+v1
4= 0
The constraint equation:
Problems 4–95
i∆ = v1/4Place these equations in standard form:
v1
(12
+15
+14
)+ i∆
(−4
5
)= 30
v1
(14
)+ i∆(−1) = 0
Solving, v1 = 40 V; i∆ = 10 AThen, v∆ = 60 − 40 = 20 Vand isc = i∆ − 2v∆ = 10 − 40 = −30 AThus, RTh = −300/ − 30 = 10 Ω
[b]
pmax =(150)2
10= 2250 W
[c]
The node voltage equation:va − 60
2+
va − 4i∆5
+va + 150
4= 0
The constraint equation is:
i∆ =va + 150
4Place the equations in standard form:
va
(12
+15
+14
)+ i∆
(−4
5
)= 30 − 150
4
va
(−1
4
)+ i∆(1) =
1504
Solving, va = 30 V; i∆ = 45 ACalculate the power:
4–96 CHAPTER 4. Techniques of Circuit Analysis
i60V =va − 60
2= −15 A
p60V = (60)(−15) = −900 W
iccvs =va − 4i∆
5= −30 A
pccvs = 4(45)(−30) = −5400 W
pvccs = (−150)[2(30)] = −9000 W
∑pdev = 900 + 5400 + 9000 = 15,300 W
% delivered =2250
15,3000× 100 = 14.7%
P 4.85 [a] First find the Thévenin equivalent with respect to Ro.Open circuit voltage: iφ = 0; 50iφ = 0
v1
100+
v1 − 28010
+v1 − 280
25+
v1
400+ 0.5125v∆ = 0
v∆ =(280 − v1)
255 = 56 − 0.2v1
v1 = 210 V; v∆ = 14 V
VTh = 280 − v∆ = 280 − 14 = 266 V
Short circuit current
Problems 4–97
v1
100+
v1 − 28010
+v2
20+
v2
400+ 0.5125(280) = 0
v∆ = 280 V
v2 + 50iφ = v1
iφ =2805
+v2
20= 56 + 0.05v2
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
4–98 CHAPTER 4. Techniques of Circuit Analysis
[c]
v1
100+
v1 − 28010
+v2 − 133
20+
v2
400+ 0.5125(280 − 133) = 0
v2 + 50iφ = v1; iφ = 133/35 = 3.8 A
Therefore, v1 = −189 V and v2 = −379 V; thus,
ig =280 − 133
5+
280 + 18910
= 76.30 A
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 whenRo = 0.
[b] P =242
8= 72 W
P 4.87 [a] 110 V source acting alone:
Re =10(14)
24=
356
Ω
i′ =110
5 + 35/6=
13213
A
v′ =(35
6
)(13213
)=
77013
V
4 A source acting alone:
Problems 4–99
5 Ω‖10 Ω = 50/15 = 10/3 Ω
10/3 + 2 = 16/3 Ω
16/3‖12 = 48/13 Ω
Hence our circuit reduces to:
It follows that
v′′a = 4(48/13) = (192/13) V
and
v′′ =−v′′
a
(16/3)(10/3) = −5
8v′′
a = −(120/13) V
·. . v = v′ + v′′ =77013
− 12013
= 50 V
[b] p =v2
o
10= 250 W
4–100 CHAPTER 4. Techniques of Circuit Analysis
P 4.88 70-V source acting alone:
v′ = 70 − 4i′b
i′s =v′
b
2+
v′
10= i′a + i′b
70 = 20i′a + v′b
i′a =70 − v′
b
20
·. . i′b =v′
b
2+
v′
10− 70 − v′
b
20=
1120
v′b +
v′
10− 3.5
v′ = v′b + 2i′b
·. . v′b = v′ − 2i′b
·. . i′b =1120
(v′ − 2i′b) +v′
10− 3.5 or i′b =
1342
v′ − 7042
·. . v′ = 70 − 4(13
42v′ − 70
42
)or v′ =
322094
=161047
V
50-V source acting alone:
v′′ = −4i′′b
Problems 4–101
v′′ = v′′b + 2i′′b
v′′ = −50 + 10i′′d
·. . i′′d =v′′ + 50
10
i′′s =v′′
b
2+
v′′ + 5010
i′′b =v′′
b
20+ i′′s =
v′′b
20+
v′′b
2+
v′′ + 5010
=1120
v′′b +
v′′ + 5010
v′′b = v′′ − 2i′′b
·. . i′′b =1120
(v′′ − 2i′′b ) +v′′ + 50
10or i′′b =
1342
v′′ +10042
Thus, v′′ = −4(13
42v′′ +
10042
)or v′′ = −200
47V
Hence, v = v′ + v′′ =161047
− 20047
=141047
= 30 V
P 4.89 10 V source acting alone:
vo1 =20
20 + 5 + 15(10) = 5 V
20 V source acting alone:
vo2 =13.333
13.333 + 10 + 30(20) = 5 V
4–102 CHAPTER 4. Techniques of Circuit Analysis
6 A current source acting alone:
Node voltage equations:v1
15+
v1 − v2
5− 6 = 0
v2 − v1
5+
v2
40+
v2 − v3
10= 0
v3 − v2
10+
v3
30+ 6 = 0
In standard form:
v1
( 115
+15
)+ v2
(−1
5
)+ v3(0) = 6
v1
(−1
5
)+ v2
(15
+140
+110
)+ v3
(− 1
10
)= 0
v1(0) + v2
(− 1
10
)+ v3
( 110
+130
)= −6
Solving, v1 = 22.5 V; v2 = 0 V; v3 = −45 VNote that vo3 = v2 = 0 VFinally, vo = vo1 + vo2 + vo3 = 5 + 5 + 0 = 10 V
P 4.90 Voltage source acting alone:
vo1 − 254000
+vo1
20,000− 2.2
(vo1 − 25
4000
)= 0
Problems 4–103
Simplifying 5vo1 − 125 + vo1 − 11vo1 + 275 = 0
·. . vo1 = 30 V
Current source acting alone:
vo2
4000+
vo2
20,000+ 0.005 − 2.2
(vo2
4000
)= 0
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
40 + 100‖25= −2.25 A
vo1 =6090
(−135) = −90 V
Current source acting alone:
4–104 CHAPTER 4. Techniques of Circuit Analysis
v1
30+
v1
60+ 18 = 0 ·. . v1 = −360 V; vo2 = 360 V
−18 +v2
80+
v2 − v3
20= 0
v3 − v2
20+
v3
25+
v3
40= 0
·. . 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 Ω‖5 Ω‖60 Ω = 4 Ω
·. . io1 =20
20 + 5(6) = 4.8 A
10 A source:
Problems 4–105
io2 =425
(10) = 1.6 A
75 V source:
io3 = − 425
(15) = −2.4 A
io = io1 + io2 + io3 = 4.8 + 1.6 − 2.4 = 4 A
P 4.93 [a] By hypothesis i′o + i′′o = 3.5 mA.
i′′′o =20008000
(−0.005) = −1.25 mA; ·. . io = 3.5 − 1.25 = 2.25 mA
[b] With all three sources in the circuit write a single node voltage equation.
vb − 82000
+vb
6000+ 0.005 − 0.010 = 0
4–106 CHAPTER 4. Techniques of Circuit Analysis
·. . vb = 13.5 V
io =vb
6000=
13.56000
= 2.25 mA
P 4.94 [a]
voc = VTh = 75 V; iL =6020
= 3 A; iL =75 − 60
RTh=
15RTh
Therefore RTh =153
= 5 Ω
[b] iL =vo
RL
=VTh − vo
RTh
Therefore RTh =VTh − vo
vo/RL
=(
VTh
vo
− 1)
RL
P 4.95 [a]
v − v1
2xr+
v
R+
v − v2
2r(L − x)= 0
v
[1
2xr+
1R
+1
2r(L − x)
]=
v1
2xr+
v2
2r(L − x)
v =v1RL + xR(v2 − v1)RL + 2rLx − 2rx2
[b] Let D = RL + 2rLx − 2rx2
dv
dx=
(RL + 2rLx − 2rx2)R(v2 − v1) − [v1RL + xR(v2 − v1)]2r(L − 2x)D2
dv
dx= 0 when numerator is zero.
The numerator simplifies to
x2 +2L − v1
(v2 − v1)x +
RL(v2 − v1) − 2rv1L2
2r(v2 − v1)= 0
Problems 4–107
Solving for the roots of the quadratic yields
x =L
v2 − v1
−v1 ±
√v1v2 − R
2rL(v2 − v1)2
[c] x =L
v2 − v1
−v1 ±
√v1v2 − R
2rL(v1 − v2)2
v2 = 1200 V, v1 = 1000 V, L = 16 km
r = 5 × 10−5 Ω/m; R = 3.9 Ω
L
v2 − v1=
16,0001200 − 1000
= 80; v1v2 = 1.2 × 106
R
2rL(v1 − v2)2 =
3.9(−200)2
(10 × 10−5)(16 × 103)= 0.975 × 105
x = 80−1000 ±√
1.2 × 106 − 0.0975 × 106= 80−1000 ± 1050 = 80(50) = 4000 m
[d]
vmin =v1RL + R(v2 − v1)xRL + 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 sixessential nodes. Further study shows that by replacing the parallel resistorswith their equivalent values the circuit reduces to four meshes and fouressential nodes as shown in the following diagram.
The node Voltage approach will require solving three node Voltage equationsalong with equations involving v∆ and iβ .
The mesh-current approach will require writing one supermesh equation plusthree constraint equations involving the three current sources. Thus at theoutset we know the supermesh equation can be reduced to a single unknowncurrent. Since we are interested in the power developed by the 1 V source, wewill 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
−9iβ +43ia + 0.75ib + 1 + 5ib + 7(ic − id) + 8ic = 0
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
id = −6v∆ = −6(−4
3ia
)= 8ia = −16 − 16ib
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
B–C supernode:vB − 3vx
4+
vB − vE
7− 0.1 = 0
At node E:vE
6+
vE − 3vx
5+
vE − vB
7+ 5 = 0
At node D:vD + 13v∆
3− 5 + 0.1 +
vD
2= 0
Constraint: v∆ = vB − vE
Constraint: vx = −v∆ + 5i∆ − 0.9Constraint: i∆ = (3vx − vB)/4In standard form:
vB
(14
+17
)+ vD(0) + vE
(−1
7
)+ v∆(0) + vx
(−3
4
)+ i∆(0) = 0.1
vB(0) + vD
(12
+13
)+ vE(0) + v∆
(133
)+ vx(0) + i∆(0) = 4.9
vB
(−1
7
)+ vD(0) + vE
(16
+15
+17
)+ v∆(0) + vx
(−3
5
)+ i∆(0) = −5
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
Solving, 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 WThe 5 A source absorbs 23.1 W
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; ic = 26.33 A; id = 23.27 A; ie = 20.14 AFind the requested voltages:v1 = 18.4(ic − ie) = 113.90 Vv2 = 38.4(id − ie) = 120.19 Vv3 = 11.6ie = 233.62 V
Problems 4–111
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 Wp1L = (ia − ib)2 = (ia − ig)2 = 400 Wp2C = 2(ib − ic)2 = (ig − if )2 = 50 Wp3 = 3(ic − id)2 = 3(ie − if )2 = 300 Wp4 = 4(id − ie)2 = 0 Wp8 = 8(ic)2 = 8(if )2 = 200 Wp2L = 2(ia − id)2 = 2(ia − ie)2 = 450 W
4–112 CHAPTER 4. Techniques of Circuit Analysis
∑pabs = 100 + 400 + 50 + 200 + 300 + 450 + 0 + 450 + 300+
200 + 50 + 400 + 100 = 3000 W∑pgen = 100ia = 100(30) = 3000 W (CHECKS)
P 4.100dv1
dIg1=
−R1[R2(R3 + R4) + R3R4](R1 + R2)(R3 + R4) + R3R4
dv1
dIg2=
R1R3R4
(R1 + R2)(R3 + R4) + R3R4
dv2
dIg1+
−R1R3R4
(R1 + R2)(R3 + R4) + R3R4
dv2
dIg2=
R3R4(R1 + R2)(R1 + R2)(R3 + R4) + R3R4
P 4.101 From the solution to Problem 4.100 we have
dv1
dIg1=
−25[5(125) + 3750]30(125) + 3750
= −17512
V/A
and
dv2
dIg1=
−(25)(50)(75)30(125) + 3750
= −12.5 V/A
By hypothesis, ∆Ig1 = 11 − 12 = −1 A
·. . ∆v1 = (−17512
)(−1) =17512
= 14.5833 V
Thus, v1 = 25 + 14.5833 = 39.5833 VAlso,
∆v2 = (−12.5)(−1) = 12.5 V
Thus, v2 = 90 + 12.5 = 102.5 VThe PSpice solution is
v1 = 39.5830 V
and
v2 = 102.5000 V
These values are in agreement with our predicted values.
Problems 4–113
P 4.102 From the solution to Problem 4.100 we have
dv1
dIg2=
(25)(50)(75)30(125) + 3750
= 12.5 V/A
and
dv2
dIg2=
(50)(75)(30)30(125) + 3750
= 15 V/A
By hypothesis, ∆Ig2 = 17 − 16 = 1 A
·. . ∆v1 = (12.5)(1) = 12.5 V
Thus, v1 = 25 + 12.5 = 37.5 VAlso,
∆v2 = (15)(1) = 15 V
Thus, v2 = 90 + 15 = 105 VThe 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
dv1
dIg1= −175
12V/A;
dv1
dIg2= 12.5 V/A
dv2
dIg1= −12.5 V/A;
dv2
dIg2= 15 V/A
By hypothesis,
∆Ig1 = 11 − 12 = −1 A
∆Ig2 = 17 − 16 = 1 A
Therefore,
∆v1 =17512
+ 12.5 = 27.0833 V
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.
5The Operational Amplifier
Assessment Problems
AP 5.1 [a] This is an inverting amplifier, so
vo = (−Rf/Ri)vs = (−80/16)vs, so vo = −5vs
vs( V) 0.4 2.0 3.5 −0.6 −1.6 −2.4
vo( V) −2.0 −10.0 −15.0 3.0 8.0 10.0Two 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, vs = −2 V
Therefore − 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 = 4×10−5Rx
Use the negative power supply value to determine one limit on the value of Rx:
4×10−5Rx = −15 so Rx = −15/4×10−5 = −375 kΩ
Since we cannot have negative resistor values, the lower limit for Rx is 0. Now usethe positive power supply value to determine the upper limit on the value of Rx:
4×10−5Rx = 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 thenegative 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 thenegative 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 eachequation 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, thecurrent into the op amp at the inputs is zero:
vn
4500+
vn − vo
63,000= 0
Solve for vo in terms of vn by multiplying both sides by 63,000 and collectingterms:
14vn + vn − vo = 0 so vo = 15vn
Now use voltage division to calculate vp. We can use voltage division becausethe op amp is ideal, so no current flows into the non-inverting input terminaland the 400 mV divides between the 15 kΩ resistor and the Rx resistor:
vp =Rx
15,000 + Rx
(0.400)
Problems 5–3
Now substitute the value Rx = 60 kΩ:
vp =60,000
15,000 + 60,000(0.400) = 0.32 V
Finally, remember that for an ideal op amp, vn = vp, so substitute the value ofvp 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 resultingequation equal to the positive power supply value:
vo = 15(
0.4Rx
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 outputvoltage in terms of the input voltages and the resistor values given in Eq. 5.22:
vo =20(60)10(24)
vb − 5010
va
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)10(12)
vb − 5va = 4vb − 5va
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
500,000+
vn − vg
5000+
vn − vo
100,000= 0
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)5000
+vo − vn
100,000= 0
Multiply through by 100,000 and simplify:
20vo + 6 × 106vn + vo − vn = 0 so 6 × 106vn + 21vo = 0
Use Cramer’s method to solve for vo:
∆ =
∣∣∣∣∣∣∣21.2 −1
6 × 106 21
∣∣∣∣∣∣∣ = 6,000,445.2
No =
∣∣∣∣∣∣∣21.2 20vg
6 × 106 0
∣∣∣∣∣∣∣ = −120 × 106vg
vo =No
∆= −19.9985vg; so
vo
vg
= −19.9985
[b] Use Cramer’s method again to solve for vn:
N1 =
∣∣∣∣∣∣∣20vg −1
0 21
∣∣∣∣∣∣∣ = 420vg
vn =N1
∆= 6.9995 × 10−5vg
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 tothe input current, so calculate the input current as a function of the inputvoltage:
ig =vg − vn
5000=
vg − 6.9995 × 10−5vg
5000Solve for the ratio of vg to ig to get the input resistance:
Rg =vg
ig
=5000
1 − 6.9995 × 10−5 = 5000.35 Ω
[d] This is a simple inverting amplifier configuration, so the voltage gain is the ratioof the feedback resistance to the input resistance:
vo
vg
= −100,0005000
= −20
Since this is now an ideal op amp, the voltage difference between the two inputterminals is zero; since vp = 0, vn = 0Since there is no current into the inputs of an ideal op amp, the resistance seenby 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 ofthe input currents to 0. Thus, in = 0 A.
[c] The open-loop voltage gain of an ideal op amp is infinite, which constrains thedifference between the voltage at the two input terminals to 0. Thus,(vp − vn) = 0.
[d] Write a node voltage equation at vn:
vn − 2.510,000
+vn − vo
40,000= 0
But vp = 0 and vn = vp = 0. Thus,
−2.510,000
− vo
40,000= 0 so vo = −10 V
P 5.2vb − va
20+
vb − vo
100= 0, therefore vo = 6vb − 5va
[a] va = 4 V, vb = 0 V, vo = −15 V (sat)
[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, vo = 9.6 − 5va = ±15
·. . −1.08 ≤ va ≤ 4.92 V
P 5.3 vo = −(0.5 × 10−3)(10 × 103) = −5 V
·. . io =−5
5000= −1 mA
P 5.4 Since the current into the inverting input terminal of an ideal op-amp is zero, thevoltage across the 2.2 MΩ resistor is (2.2 × 106)(3.5 × 10−6) or 7.7 V. Thereforethe voltmeter reads 7.7 V.
Problems 5–7
P 5.5 [a] ia =25 × 10−3
5000= 5 µA
va = −50 × 103ia = −250 mV
[b]va
50,000+
va
10,000+
va − vo
40,000= 0
·. . 4va + 20va + 5va − 5vo = 0
·. . vo = 29va/5 = −1.45 V
[c] ia = 5 µA
[d] io =−vo
30,000+
va − vo
40,000= 78.33 µ A
P 5.6 [a] The gain of an inverting amplifier is the ratio of the feedback resistor to theinput resistor. If the gain of the inverting amplifier is to be 6, the feedbackresistor must be 6 times as large as the input resistor. There are many possibledesigns that use only 20 kΩ resistors. We present two here. Use a single 20 kΩresistor as the input resistor, and use six 20 kΩ resistors in series as thefeedback resistor to give a total of 120 kΩ.
Alternately, Use a single 20 kΩ resistor as the feedback resistor and use six 20kΩ 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 supplyvoltages must be greater than or equal to the product of the input voltage andthe 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 atvn:
340,000
+3 − vo
80,000= 0
Solving,
vo = 9 V.
P 5.8 vp =1824
(12) = 9 V = vn
vn − 2430
+vn − vo
20= 0
vo = (45 − 48)/3 = −1.0 V
iL =vo
5× 10−3 = − 1
5× 10−3 = −200 × 10−6
iL = −200 µA
P 5.9 [a] First, note that vn = vp = 2.5 VLet vo1 equal the voltage output of the op-amp. Then
2.5 − vg
5000+
2.5 − vo1
10,000= 0, ·. . vo1 = 7.5 − 2vg
Also note that vo1 − 2.5 = vo, ·. . vo = 5 − 2vg
[b] Yes, the circuit designer is correct!
Problems 5–9
P 5.10 [a] Let v∆ be the voltage from the potentiometer contact to ground. Then
0 − vg
2000+
0 − v∆
50,000= 0
−25vg − v∆ = 0, ·. . v∆ = −25(40 × 10−3) = −1 V
v∆
αR∆+
v∆ − 050,000
+v∆ − vo
(1 − α)R∆= 0
v∆
α+ 2v∆ +
v∆ − vo
1 − α= 0
v∆
( 1α
+ 2 +1
1 − α
)=
vo
1 − α
·. . vo = −1[1 + 2(1 − α) +
(1 − α)α
]
When α = 0.2, vo = −1(1 + 1.6 + 4) = −6.6 V
When α = 1, vo = −1(1 + 0 + 0) = −1 V
·. . −6.6 V ≤ vo ≤ −1 V
[b] −1[1 + 2(1 − α) +
(1 − α)α
]= −7
α + 2α(1 − α) + (1 − α) = 7α
α + 2α − 2α2 + 1 − α = 7α
·. . 2α2 + 5α − 1 = 0 so α ∼= 0.186
P 5.11 [a] Replace the combination of vg, 1.6 kΩ, and the 6.4 kΩ resistors with itsThévenin equivalent.
Then vo =−[12 + σ50]
1.28(0.2)
At saturation vo = −5 V; therefore
−(12 + σ50
1.28
)(0.2) = −5, or σ = 0.4
Thus for 0 ≤ σ < 0.40 the operational amplifier will not saturate.
5–10 CHAPTER 5. The Operational Amplifier
[b] When σ = 0.272, vo =−(12 + 13.6)
1.28(0.2) = −4 V
Alsovo
10+
vo
25.6+ io = 0
·. . io = − vo
10− vo
25.6=
410
+4
25.6mA = 556.25 µA
P 5.12 [a]
vn − va
R+
vn − vo
R= 0
2vn − va = vo
va
Ra+
va − vn
R+
va − vo
R= 0
va
[ 1Ra
+2R
]− vn
R=
vo
R
va
(2 +
R
Ra
)− vn = vo
vn = vp = va + vg
·. . 2vn − va = 2va + 2vg − va = va + 2vg
·. . va − vo = −2vg (1)
2va + va
(R
Ra
)− va − vg = vo
·. . va
(1 +
R
Ra
)− vo = vg (2)
Now combining equations (1) and (2) yields
−vaR
Ra= −3vg
Problems 5–11
or va = 3vgRa
R
Hence ia =va
Ra=
3vg
RQ.E.D.
[b] At saturation Vo = ± Vcc
·. . va = ± Vcc − 2vg (3)
and
·. . va
(1 +
R
Ra
)= ± Vcc + vg (4)
Dividing Eq (4) by Eq (3) gives
1 +R
Ra=
± Vcc + vg
± Vcc − 2vg
·. .R
Ra=
± Vcc + vg
± Vcc − 2vg
− 1 =3vg
± Vcc − 2vg
or Ra =(± Vcc − 2vg)
3vg
R Q.E.D.
P 5.13 [a] Assume the op-amp is operating within its linear range, then
iL =8
4000= 2 mA
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 asRL ≤ 6 kΩ. Therefore when RL = 16 kΩ the op-amp is saturated. We canestimate the value of iL by assuming ip = in iL. TheniL = 20/(4,0000 + 16,000) = 1 mA. To justify neglecting the current into theop-amp assume the drop across the 50 kΩ resistor is negligible, and the inputresistance to the op-amp is at least 500 kΩ. Thenip = in = (8 − 4)/(500 × 103) = 8 µA. But 8 µA 1 mA, hence ourassumption is reasonable.
5–12 CHAPTER 5. The Operational Amplifier
[d]
P 5.14 [a] Let vo1 = output voltage of the amplifier on the left. Let vo2 = output voltage ofthe amplifier on the right. Then
vo1 =−4710
(1) = −4.7 V; vo2 =−22033
(−0.15) = 1.0 V
ia =vo2 − vo1
1000= 5.7 mA
[b] ia = 0 when vo1 = v02 so from (a) vo2 = 1 V
Thus−4710
(vL) = 1
vL = −1047
= −212.77 mV
P 5.15 [a] p600Ω =(60 × 10−3)2
(600)= 6 µW
[b] v600Ω =600
30,000(60 × 10−3) = 1.2 mV
p600Ω =(1.2 × 10−3)2
(600)= 2.4 nW
[c]pa
pb=
6 × 10−6
2.4 × 10−9 = 2500
[d] Yes, the operational amplifier serves several useful purposes:
Problems 5–13
• First, it enables the source to control 2, 500 times as much power deliveredto the load resistor. When a small amount of power controls a largeramount 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 functionof the operational amplifier.
• Third, it allows the load resistor voltage (and thus its current) to be setwithout drawing any current from the input voltage source. This is thecurrent amplification function of the circuit.
P 5.16 [a] This circuit is an example of an inverting summing amplifier.
[b] vo = −22033
va − 22022
vb − 22080
vc = −8 + 15 − 11 = −4 V
[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 − va
40,000+
vd − vb
22,000+
vd − vc
100,000+
vd
352,000+
vd − vo
220,000= 0
Multiply through by 220,000, plug in the values of input voltages, andrearrange to solve for vo:
vo = 220,000(
440,000
+−1
22,000+
−5100,000
+8
352,000+
8220,000
)= 14 V
[b] Write a KCL equation at the inverting input to the op amp. Use the given valuesof input voltages in the equation:
8 − va
40,000+
8 − 922,000
+8 − 13100,000
+8
352,000+
8 − vo
220,000= 0
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
5–14 CHAPTER 5. The Operational Amplifier
Therefore,
3.818 V ≤ va ≤ 9.273 V
P 5.18 [a]8 − 440,000
+8 − 922,000
+8 − 13100,000
+8
352,000+
8 − v0
Rf
= 0
8 − vo
Rf
= −2.7272 × 10−5 so Rf =8 − vo
−2.727 × 10−5
For vo = 15 V, Rf = 256.7 kΩ
For vo = −15 V, Rf < 0 so this solution is not possible.
[b] io = −(if + i10k) = −[
15 − 8256.7 × 103 +
1510,000
]= −1.527 mA
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 gainthat is the ratio of the feedback resistance to the resistance in the forward path forthe input voltage:
vo = −[ 48Ra
va +48Rb
vb +48Rc
vc +48Rd
vd
]
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 =R1vo
R1 + R2, vn = vp
Therefore vo =(
R1 + R2
R1
)vs =
(1 +
R2
R1
)vs
Problems 5–15
[b] vo = vs
[c] Because vo = vs, thus the output voltage follows the signal voltage.
P 5.21 vo = −[
Rf
3000(0.15) +
Rf
5000(0.1) +
Rf
25,000(0.25)
]
−6 = −8 × 10−5Rf ; Rf = 75 kΩ; ·. . 0 ≤ Rf ≤ 75 kΩ
P 5.22 [a] This circuit is an example of the non-inverting amplifier.
[b] Use voltage division to calculate vp:
vp =10,000
10,000 + 30,000vs =
vs
4
Write a KCL equation at vn = vp = vs/4:
vs/44000
+vs/4 − vo
28,000= 0
Solving,
vo = 7vs/4 + vs/4 = 2vs
[c] 2vs = 8 so vs = 4 V
2vs = −12 so vs = −6 V
Thus, −6 V ≤ vs ≤ 4 V.
P 5.23 [a] vp = vn =68,00080,000
vg = 0.85vg
·. .0.85vg
30,000+
0.85vg − vo
63,000= 0;
·. . vo = 2.635vg = 2.635(4), vo = 10.54 V
[b] vo = 2.635vg = ±12
vg = ±4.554 V, −4.554 ≤ vg ≤ 4.554 V
[c]0.85vg
30,000+
0.85vg − vo
Rf= 0
(0.85Rf
30,000+ 0.85
)vg = vo = ±12
·. . 1.7×10−3Rf + 51 = ±360; 1.7×10−3Rf = 360 − 51; Rf = 181.76 kΩ
P 5.24 [a] This circuit is an example of a non-inverting summing amplifier.
5–16 CHAPTER 5. The Operational Amplifier
[b] Write a KCL equation at vp and solve for vp in terms of vs:
vp − vs
15,000+
vp − 630,000
= 0
2vp − 2vs + vp − 6 = 0 so vp = 2vs/3 + 2
Now write a KCL equation at vn and solve for vo:
vn
20,000+
vn − vo
60,000= 0 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 so vs = 3 V
8vs/3 + 8 = −12 so vs = −7.5 V
Thus, −7.5 V ≤ vs ≤ 3 V.
P 5.25 [a] The circuit is a non-inverting summing amplifier.
[b]vp − va
3.3 × 103 +vp − vb
4.7 × 103 = 0
·. . vp = 0.5875va + 0.4125vb
vn
10,000+
vn − vo
100,000= 0
·. . vo = 11vn = 11vp = 6.4625va + 4.5375vb = 8.03 V
[c] vp = vn =vo
11= 730 mV
ia =va − vp
3.3 × 103 = −100 µA
ib =vb − vp
4.7 × 103 = 100 µA
[d] 6.4625 for va
4.5375 for vb
P 5.26 [a]vp − va
Ra+
vp − vb
Rb+
vp − vc
Rc+
vp
Rg
= 0
·. . vp =RbRcRg
Dva +
RaRcRg
Dvb +
RaRbRg
Dvc
where D = RbRcRg + RaRcRg + RaRbRg + RaRbRc
Problems 5–17
vn
Rs
+vn − vo
Rf= 0
vn
( 1Rs
+1Rf
)=
vo
Rf
·. . vo =(1 +
Rf
Rs
)vn = kvn
where k =(1 +
Rf
Rs
)
vp = vn
·. . vo = kvp
or
vo =kRgRbRc
Dva +
kRgRaRc
Dvb +
kRgRaRb
Dvc
kRgRbRc
D= 3 ·. .
Rb
Ra= 1.5
kRgRaRc
D= 2 ·. .
Rc
Rb= 2
kRgRaRb
D= 1 ·. .
Rc
Ra= 3
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
180 × 109 = 3
k =540 × 109
72 × 109 = 7.5
·. . 7.5 = 1 +Rf
Rs
Rf
Rs
= 6.5
Rf = (6.5)(12,000) = 78 kΩ
[b] vo = 3(0.8) + 2(1.5) + 2.10 = 7.5 V
vn = vp =7.57.5
= 1.0 V
ia =0.8 − 12000
=−0.22000
= −0.1 mA = −100 µA
5–18 CHAPTER 5. The Operational Amplifier
ib =1.5 − 1.0
3000=
0.53000
= 166.67 µA
ic =2.10 − 1.0
6000=
1.16000
= 183.33 µA
ig =1
4000= 250 µA
is =vn
12,000=
112,000
= 83.33 µA
P 5.27 [a]vp − va
Ra+
vp − vb
Rb+
vp − vc
Rc= 0
·. . vp =RbRc
Dva +
RaRc
Dvb +
RaRb
Dvc
where D = RbRc + RaRc + RaRb
vn
10,000+
vn − vo
Rf= 0
(Rf
10,000+ 1
)vn = vo
LetRf
10,000+ 1 = k
vo = kvn = kvp
·. . vo =kRbRc
Dva +
kRaRc
Dvb +
kRaRb
Dvc
·. .kRbRc
D= 5 ·. .
Rc
Ra= 5
kRaRc
D= 4
kRaRb
D= 1 ·. .
Rc
Rb= 4
·. . Rc = 5Ra = 5 kΩ
Rb = Rc/4 = 1.25 kΩ
·. . D = (1.25)(5) + (1)(5) + (1.25)(1) = 12.5 × 106
·. . k =5D
RbRc=
(5)(12.5) × 106
(1.25)(5) × 106 = 10
·. .Rf
10,000+ 1 = 10, Rf = 90 kΩ
Problems 5–19
[b] vo = 5(0.5) + 4(1) + 1.5 = 8 V
vn = vo/10 = 0.8 V = vp
ia =va − vp
1000=
0.5 − 0.81000
= −300 µA
ib =vb − vp
1250=
1 − 0.81250
= 160 µA
ic =vc − vp
5000=
1.5 − 0.85000
= 140 µA
P 5.28 [a] Assume va is acting alone. Replacing vb with a short circuit yields vp = 0,therefore vn = 0 and we have
0 − va
Ra+
0 − v′o
Rb+ in = 0, in = 0
Therefore
v′o
Rb= − va
Ra, v′
o = −Rb
Rava
Assume vb is acting alone. Replace va with a short circuit. Now
vp = vn =vbRd
Rc + Rd
vn
Ra+
vn − v′′o
Rb+ in = 0, in = 0
( 1Ra
+1
Rb
)(Rd
Rc + Rd
)vb − v′′
o
Rb= 0
v′′o =
(Rb
Ra+ 1
)(Rd
Rc + Rd
)vb =
Rd
Ra
(Ra + Rb
Rc + Rd
)vb
vo = v′o + v′′
o =Rd
Ra
(Ra + Rb
Rc + Rd
)vb − Rb
Rava
[b]Rd
Ra
(Ra + Rb
Rc + Rd
)=
Rb
Ra, therefore Rd(Ra + Rb) = Rb(Rc + Rd)
RdRa = RbRc, thereforeRa
Rb=
Rc
Rd
WhenRd
Ra
(Ra + Rb
Rc + Rd
)=
Rb
Ra
Eq. (5.22) reduces to vo =Rb
Ravb − Rb
Rava =
Rb
Ra(vb − va).
5–20 CHAPTER 5. The Operational Amplifier
P 5.29 Use voltage division to find vp:
vp =2000
2000 + 8000(5) = 1 V
Write a KCL equation at vn and solve it for vo:
vn − va
5000+
vn − vo
Rf
= 0 so(
Rf
5000+ 1
)vn − Rf
5000va = vo
Since the op amp is ideal, vn = vp = 1V, so
vo =(
Rf
5000+ 1
)− Rf
5000va
To satisfy the equation,
(Rf
5000+ 1
)= 5 and
Rf
5000= 4
Thus, Rf = 20 kΩ.
P 5.30 [a]
vp
72,000+
vp − vc
9,000+
vp − vd
24,000= 0
·. . vp = (2/3)vc + 0.25vd = vn
vn − va
12,000+
vn − vb
18,000+
vn − vo
144,000= 0
·. . 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 5–21
[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)Ra(Rc + Rd)
vb − Rb
Rava =
47(110)10(80)
(0.80) − 10(0.67)
vo = 5.17 − 6.70 = −1.53 V
[b] vn = vp =(800)(47)
80= 470 mV
ia =(670 − 470)10−3
10 × 103 = 20 µA
Ra =va
ia=
670 × 10−3
20 × 10−6 = 33.5 kΩ
[c] Rin b = Rc + Rd = 80 kΩ
P 5.32 vp =vbRb
Ra + Rb= vn
vn − va
4700+
vn − vo
Rf= 0
vn
(Rf
4700+ 1
)− vaRf
4700= vo
·. .(
Rf
4700+ 1
)Rb
Ra + Rbvb − Rf
4700va = vo
·. .Rf
4700= 10; Rf = 47 kΩ
·. .Rf
4700+ 1 = 11
·. . 11(
Rb
Ra + Rb
)= 10
11Rb = 10Rb + 10Ra Rb = 10Ra
5–22 CHAPTER 5. The Operational Amplifier
Ra + Rb = 220 kΩ
11Ra = 220 kΩ
Ra = 20 kΩ
Rb = 220 − 20 = 200 kΩ
P 5.33 vp = vn = Rbib
Rbib − 3000ia3000
+Rbib − vo
Rf= 0
(Rb
3000+
Rb
Rf
)ib − ia =
vo
Rf
vo =[RbRf
3000+ Rb
]ib − Rfia
·. . Rf = 2000 Ω
(2/3)Rb + Rb = 2000
·. . Rb = 1200 Ω
P 5.34 vo =Rd(Ra + Rb)Ra(Rc + Rd)
vb − Rb
Rava
By hypothesis: Rb/Ra = 4; Rc + Rd = 470 kΩ;Rd(Ra + Rb)Ra(Rc + Rd)
= 3
·. .Rd
Ra
(Ra + 4Ra)470,000
= 3 so Rd = 282 kΩ; Rc = 188 kΩ
Also, when vo = 0 we have
vn − va
Ra+
vn
Rb= 0
·. . vn
(1 +
Ra
Rb
)= va; vn = 0.8va
ia =va − 0.8va
Ra= 0.2
va
Ra; Rin =
va
ia= 5Ra = 22 kΩ
·. . Ra = 4.4 kΩ; Rb = 17.6 kΩ
Problems 5–23
P 5.35 [a] vn = vp = αvg vo = (αvg − vg)4 + αvg
vn − vg
R1+
vn − vo
Rf= 0 = [(α − 1)4 + α]vg
(vn − vg)Rf
R1+ vn − vo = 0 = (5α − 4)vg
= (5α − 4)(2) = 10α − 8
α vo α vo α 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
vo =(
Rf
R1+ 1
)vgα + −
(Rf
R1
)vg
Therefore,
slope =(
Rf
R1+ 1
)vg; intercept = −
(Rf
R1
)vg
[c] Using the equations from (b),
−6 =(
Rf
R1+ 1
)vg; 4 = −
(Rf
R1
)vg
Solving,
vg = −2 V;Rf
R1= 2
P 5.36 vp =15009000
(−18) = −3 V = vn
18 − 31600
+−3 − vo
Rf= 0
5–24 CHAPTER 5. The Operational Amplifier
·. . vo =15
1600Rf − 3
vo = 9 V; Rf = 1280 Ω
vo = −9 V; Rf = −640 Ω
But Rf ≥ 0, ·. . Rf = 1280 Ω
P 5.37 [a] Adm =(24)(26) + (25)(25)
(2)(1)(25)= 24.98
[b] Acm =(1)(24) − 25(1)
1(25)= −0.04
[c] CMRR =∣∣∣∣24.980.04
∣∣∣∣ = 624.50
P 5.38 Acm =(20)(50) − (50)Rx
20(50 + Rx)
Adm =50(20 + 50) + 50(50 + Rx)
2(20)(50 + Rx)
Adm
Acm=
Rx + 1202(20 − Rx)
·. .Rx + 120
2(20 − Rx)= ±1000 for the limits on the value of 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 rightnode:vn − vg
5000+
vn − vo
100,000+
vn
500,000= 0
vo + 3 × 105vn
5000+
vo − vn
100,000+
vo
1000= 0
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 − vn
5000=
vg − 403.2 × 10−6vg
5000
Rg =vg
ig
=5000
1 − 403.2 × 10−6 = 5002.02 Ω
[d] For an ideal op amp, the voltage gain is the ratio between the feedback resistorand the input resistor:
vo
vg
= −100,0005000
= −20
For an ideal op amp, the difference between the voltages at the input terminalsis zero, and the input resistance of the op amp is infinite. Therefore,
vn = vp = 0 V; Rg = 5000 Ω
P 5.40 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 voltageequation at the inverting input:
vn
40,000+
vn − vg
500,000+
vn − vo
80,000= 0
Simplify:
12.5vn + vn − vg + 6.25vn − 6.25vo = 0
The node voltage equation at the op amp output:
vo
4000+
vo − 20,000(vp − vn)5000
+vo − vn
80,000= 0
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 andsimplify into two simultaneous equations:
19.75vn − 6.25vo = 1
255,999vn + 37vo = 256,000
These equations are in standard form, so solve them to yieldvo = 2.9986 V; vn = 999.571 mVThus,
vo
vg
=2.9986
1= 2.9986
[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
[d] ig =vg − vp
100,000=
1 − 999.914 × 10−3
100,000= 859 pA
[e] For an ideal op amp, vn = vp = vg, so the KVL equation at the inverting node is
vo
40,000+
vg − vo
80,000= 0
Then,
vo = 3vg sovo
vg
= 3
Also,
vn = vp = 1 V; vp − vn = 0 V; ig = 0 A
Problems 5–27
P 5.41 [a]
vn − 0.881600
+vn
500,000+
vn − vTh
24,000= 0
vTh + 105vn
2000+
vTh − vn
24,000= 0
Solving, vTh = −13.198 V
Short-circuit current calculation:
vn
500,000+
vn − 0.881600
+vn − 024,000
= 0
·. . vn = 0.823 V
isc =vn
24,000− 105
2000vn = −41.13 A
RTh =vTh
isc= 320.90 mΩ
5–28 CHAPTER 5. The Operational Amplifier
[b] The output resistance of the inverting amplifier is the same as the Théveninresistance, i.e.,
Ro = RTh = 320.90 mΩ
[c]
vo =( 330
330.32
)(−13.198) = −13.185 V
vn − 0.881600
+vn
500,000+
vn + 13.18524,000
= 0
·. . vn = 941.92 µV
ig =0.88 − 941.92 × 10−6
1600= 549.41 µA
Rg =0.88
0.88 − 941.92 × 10−6 (1600) = 1601.7 Ω
P 5.42 [a] vTh =−241.6
(0.88) = −13.2 V
RTh = 0, since op-amp is ideal
Problems 5–29
[b] Ro = RTh = 0 Ω
[c] Rg = 1.6 kΩ since vn = 0
P 5.43 [a]
vn − vg
15,000+
vn − vo
135,000= 0
·. . vo = 10vn − 9vg
Also vo = A(vp − vn) = −Avn
·. . vn =−vo
A
·. . vo
(1 +
10A
)= −9vg
vo =−9A
(10 + A)vg
[b] vo =−9(90)(0.4)(10 + 90)
= −3.24 V
[c] vo = −9(0.4) = −3.60 V
[d] −3.42 =−9(0.4)A10 + A
·. . A = 190
P 5.44 From Eq. 5.57,
vref
R + ∆R= vn
(1
R + ∆R+
1R − ∆R
+1
Rf
)− vo
Rf
Substituting Eq. 5.59 for vp = vn:
vref
R + ∆R=
vref
(1
R+∆R+ 1
R−∆R+ 1
Rf
)(R − ∆R)
(1
R+∆R+ 1
R−∆R+ 1
Rf
) − vo
Rf
5–30 CHAPTER 5. The Operational Amplifier
Rearranging,
vo
Rf
= vref
( 1R − ∆R
− 1R + ∆R
)
Thus,
vo = vref
( 2∆R
R2 − ∆R2
)Rf
P 5.45
i1 =15 − 105000
= 1 mA
i2 + i1 + 0 = 10 mA; i2 = 9 mA
vo2 = 10 + (400)(9) × 10−3 = 13.6 V
i3 =15 − 13.6
2000= 0.7 mA
i4 = i3 + i1 = 1.7 mA
vo1 = 15 + 1.7(0.5) = 15.85 V
Problems 5–31
P 5.46 vp =5.68.0
vg = 0.7vg = 7 sin(π/3)t V
vn
15,000+
vn − vo
75,000= 0
6vn = vo; vn = vp
·. . vo = 42 sin(π/3)t V 0 ≤ t ≤ ∞
vo = 0 t ≤ 0
At saturation
42 sin(
π
3
)t = ±21; sin
π
3t = ±0.5
·. .π
3t =
π
6,
5π6
,7π6
,11π6
, etc.
t = 0.50 s, 2.50 s, 3.50 s, 5.50 s, etc.
P 5.47 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
vg = t − 4 3.5 ≤ t ≤ 4.5, etc.Therefore
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 thepeak 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:
Rf =voR
2∆vref=
(5)(120)(2)(0.01)(15)
= 2 kΩ
[b] Now solve for ∆ given vo = 50 mV:
∆ =voR
2Rfvref=
(0.05)(120)2(2000)(15)
= 100 × 10−6
The change in strain gage resistance that corresponds to a 50 mV change inoutput voltage is thus
∆R = ∆R = (100 × 10−6)(120) = 12 mΩ
Problems 5–33
P 5.49 [a]
Let R1 = R + ∆R
vp
Rf
+vp
R+
vp − vin
R1= 0
·. . vp
[1
Rf
+1R
+1R1
]=
vin
R1
·. . vp =RRfvin
RR1 + RfR1 + RfR= vn
vn
R+
vn − vin
R+
vn − vo
Rf
= 0
vn
[1R
+1R
+1
Rf
]− vo
Rf
=vin
R
·. . vn
[R + 2Rf
RRf
]− vin
R=
vo
Rf
·. .vo
Rf
=[R + 2Rf
RRf
]RRfvin
[RR1 + RfR1 + RfR]− vin
R
·. .vo
Rf
=[
R + 2Rf
RR1 + RfR1 + RfR− 1
R
]vin
·. . vo =[R2 + 2RRf − R1(R + Rf ) − RRf ]Rf
R[R1(R + Rf ) + RRf ]vin
Now substitute R1 = R + ∆R and get
vo =−∆R(R + Rf )Rfvin
R[(R + ∆R)(R + Rf ) + RRf ]
If ∆R R
vo ≈ (R + Rf )Rf (−∆R)vin
R2(R + 2Rf )
[b] vo ≈ 47 × 104(48 × 104)(−95)15108(95 × 104)
≈ −3.384 V
5–34 CHAPTER 5. The Operational Amplifier
[c] vo =−95(48 × 104)(47 × 104)15
104[(1.0095)104(48 × 104) + 47 × 108]= −3.368 V
P 5.50 [a] vo ≈ (R + Rf )Rf (−∆R)vin
R2(R + 2Rf )
vo =(R + Rf )(−∆R)Rfvin
R[(R + ∆R)(R + Rf ) + RRf ]
·. .approx value
true value=
R[(R + ∆R)(R + Rf ) + RRf ]R2(R + 2Rf )
·. . Error =R[(R + ∆R)(R + Rf ) + RRf ] − R2(R + 2Rf )
R2(R + 2Rf )
=∆R
R
(R + Rf )(R + 2Rf )
·. . % error =∆R(R + Rf )R(R + 2Rf )
× 100
[b] % error =95(48 × 104) × 100
104(95 × 104)= 0.48%
P 5.51 1 =∆R(48 × 104)104(95 × 104)
× 100
·. . ∆R =950048
= 197.91667 Ω
·. . % change in R =197.19667
104 × 100 ≈ 1.98%
P 5.52 [a] It follows directly from the solution to Problem 5.49 that
vo =[R2 + 2RRf − R1(R + Rf ) − RRf ]Rfvin
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 5–35
[b] It follows directly from the solution to Problem 5.49 that
·. .approx value
true value=
R[(R − ∆R)(R + Rf ) + RRf ]R2(R + 2Rf )
·. . Error =(R − ∆R)(R + Rf ) + RRf − R(R + 2Rf )
R(R + 2Rf )
=−∆R(R + Rf )R(R + 2Rf )
·. . % error =−∆R(R + Rf )R(R + 2Rf )
× 100
[c] R − ∆R = 9810 Ω ·. . ∆R = 10,000 − 9810 = 190 Ω
·. . vo ≈ (48 × 104)(47 × 104)(190)(15)108(95 × 104)
≈ 6.768 V
[d] % error =−190(48 × 104)(100)
104(95 × 104)= −0.96%
6Inductance, Capacitance, and Mutual
Inductance
Assessment Problems
AP 6.1 [a] ig = 8e−300t − 8e−1200tA
v = Ldig
dt= −9.6e−300t + 38.4e−1200tV, 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
[d]dp
dt= 0 when e1800t − 12.5e900t + 16 = 0
Let x = e900t and solve the quadratic x2 − 12.5x + 16 = 0
x = 1.45, t =ln 1.45900
= 411.05 µs
x = 11.05, t =ln 11.05
900= 2.67 ms
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 = Cdv
dt= 24 × 10−6 d
dt[e−15,000t sin 30,000t]
= [0.72 cos 30,000t − 0.36 sin 30,000t]e−15,000t A, i(0+) = 0.72 A
[b] i(
π
80ms)
= −31.66 mA, v(
π
80ms)
= 20.505 V,
p = vi = −649.23 mW
[c] w =(1
2
)Cv2 = 126.13 µJ
AP 6.3 [a] v =( 1
C
) ∫ t
0−i dx + v(0−)
=1
0.6 × 10−6
∫ t
0−3 cos 50,000x dx = 100 sin 50,000t V
[b] p(t) = vi = [300 cos 50,000t] sin 50,000t
= 150 sin 100,000t W, p(max) = 150 W
[c] w(max) =(1
2
)Cv2
max = 0.30(100)2 = 3000µJ = 3 mJ
AP 6.4 [a] Leq =60(240)
300= 48 mH
[b] i(0+) = 3 + −5 = −2 A
[c] i =1256
∫ t
0+(−0.03e−5x) dx − 2 = 0.125e−5t − 2.125 A
[d] i1 =503
∫ t
0+(−0.03e−5x) dx + 3 = 0.1e−5t + 2.9 A
i2 =256
∫ t
0+(−0.03e−5x) dx − 5 = 0.025e−5t − 5.025 A
i1 + i2 = i
AP 6.5 v1 = 0.5 × 106∫ t
0+240 × 10−6e−10x dx − 10 = −12e−10t + 2 V
v2 = 0.125 × 106∫ t
0+240 × 10−6e−10x dx − 5 = −3e−10t − 2 V
v1(∞) = 2 V, v2(∞) = −2 V
W =[12(2)(4) +
12(8)(4)
]× 10−6 = 20 µJ
Problems 6–3
AP 6.6 [a] Summing the voltages around mesh 1 yields
4di1dt
+ 8d(i2 + ig)
dt+ 20(i1 − i2) + 5(i1 + ig) = 0
or
4di1dt
+ 25i1 + 8di2dt
− 20i2 = −(
5ig + 8dig
dt
)
Summing the voltages around mesh 2 yields
16d(i2 + ig)
dt+ 8
di1dt
+ 20(i2 − i1) + 780i2 = 0
or
8di1dt
− 20i1 + 16di2dt
+ 800i2 = −16dig
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
20+
7.8780
)= −0.4A; i2(∞) = − 7.8
780= −0.01A
From the solutions for i1 and i2 we have
di1dt
= 46.40e−4t − 60e−5t
di2dt
= 3.96e−4t − 5e−5t
Also,dig
dt= 7.84e−4t
Thus
4di1dt
= 185.60e−4t − 240e−5t
6–4 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
25i1 = −10 − 290e−4t + 300e−5t
8di2dt
= 31.68e−4t − 40e−5t
20i2 = −0.20 − 19.80e−4t + 20e−5t
5ig = 9.8 − 9.8e−4t
8dig
dt= 62.72e−4t
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,
8di1dt
= 371.20e−4t − 480e−5t
20i1 = −8 − 232e−4t + 240e−5t
16di2dt
= 63.36e−4t − 80e−5t
800i2 = −8 − 792e−4t + 800e−5t
16dig
dt= 125.44e−4t
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 6–5
Problems
P 6.1 [a] i = 0 t < 0
i = 50t A 0 ≤ t ≤ 5 ms
i = 0.5 − 50t A 5 ≤ t ≤ 10 ms
i = 0 10 ms < t
[b] v = Ldi
dt= 20 × 10−3(50) = 1 V 0 ≤ t ≤ 5 ms
v = 20 × 10−3(−50) = −1 V 5 ≤ t ≤ 10 ms
v = 0 t < 0
v = 1 V 0 < t < 5 ms
v = −1 V 5 < t < 10 ms
v = 0 10 ms < t
p = vi
p = 0 t < 0
p = (50t)(1) = 50t W 0 < t < 5 ms
p = (0.5 − 50t)(−1) = 50t − 0.5 W 5 < t < 10 ms
p = 0 10 ms < t
w = 0 t < 0
w =∫ t
0(50x) dx = 50
x2
2
∣∣∣∣t0= 25t2 J 0 < t < 5 ms
w =∫ t
0.005(50x − 0.5) dx + 0.625 × 10−3
= 25x2 − 0.5x∣∣∣∣t0.005
+0.625 × 10−3
= 25t2 − 0.5t + 2.5 × 10−3 J 5 < t < 10 ms
w = 0 10 ms < t
P 6.2 [a] 0 ≤ t ≤ 2 ms :
i =1L
∫ t
0vs dx + i(0) =
1200 × 10−6
∫ t
05 × 10−3 dx + 0
6–6 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
= 25x∣∣∣∣t0= 25t A
2 ms ≤ t < ∞ :
i =1
200 × 10−6
∫ t
2×10−3(0) dx + 25(2 × 10−3) = 50 mA
[b]
P 6.3 Note – the initial current should be 1 A.
0 ≤ t ≤ 2 s
iL =1
2.5 × 10−4
∫ t
03 × 10−3e−4x dx + 0 = 1.2
e−4x
−4
∣∣∣∣t0
+ 0
= 0.3 − 0.3e−4t A, 0 ≤ t ≤ 2 s
iL(2) = 0.3A
2 s < t < ∞
iL = −1.2(
e−4(x−2)
−4
∣∣∣∣t2
+ 0.3)
= 0.3e−4(t−2) A, 2 s ≤ t < ∞
Problems 6–7
P 6.4 [a] v = Ldi
dt
di
dt= 18[t(−10e−10t) + e−10t] = 18e−10t(1 − 10t)
v = (50 × 10−6)(18)e−10t(1 − 10t)
= 0.9e−10t(1 − 10t) mV, t > 0
[b] p = vi
v(200 ms) = 0.9e−2(1 − 2) = −121.8 µV
i(200 ms) = 18(0.2)e−2 = 487.2 mA
p(200 ms) = (−121.8 × 10−6)(487.2 × 10−3) = −59.34 µW
[c] delivering
[d] w =12Li2 =
12(50 × 10−6)(487.2 × 10−3)2 = 5.93 µJ
[e] The energy is a maximum where the current is a maximum:
diL
dt= 18[t(−10)e−10t + e−10t) = 18e−10t(1 − 10t)
diL
dt= 0 when t = 0.1 s
6–8 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
imax = 18(0.1)e−1 = 662.2 mA
wmax =12(50 × 10−6)(662.2 × 10−3)2 = 10.96 µJ
P 6.5 [a] 0 ≤ t ≤ 1 s :
v = −100t
i =15
∫ t
0−100x dx + 0 = −20
x2
2
∣∣∣∣t0
i = −10t2 A
1 s ≤ t ≤ 3 s :
v = −200 + 100t
i(1) = −10 A
·. . i =15
∫ t
1(100x − 200) dx − 10
= 20∫ t
1x dx − 40
∫ t
1dx − 10
= 10(t2 − 1) − 40(t − 1) − 10
= 10t2 − 40t + 20 A3 s ≤ t ≤ 5 s :
v = 100
i(3) = 10(9) − 120 + 20 = −10 A
i =15
∫ t
3100 dx − 10
= 20t − 60 − 10 = 20t − 70 A5 s ≤ t ≤ 6 s :
v = −100t + 600
i(5) = 100 − 70 = 30
i =15
∫ t
5(−100x + 600) dx + 30
= −20∫ t
5x dx + 120
∫ t
5dx + 30
= −10(t2 − 25) + 120(t − 5) + 30
= −10t2 + 120t − 320 A
Problems 6–9
[b] i(6) = −10(36) + 120(6) − 320 = 720 − 680 = 40 A, 6 ≤ t < ∞[c]
P 6.6 [a] vL = Ldi
dt= [125 sin 400t]e−200t V
·. .dvL
dt= 25,000(2 cos 400t − sin 400t)e−200t V/s
dvL
dt= 0 when tan 400t = 2
·. . t = 2.77 ms
Also 400t = 1.107 + π etc.
Because of the decaying exponential vL will be maximum the first time thederivative is zero.
[b] vL(max) = [125 sin 1.107]e−0.554 = 64.27 V
vL max = 64.27 V
Note: When t = (1.107 + π)/400; vL = −13.36 V
P 6.7 [a] i =1
15 × 10−3
∫ t
030 sin 500x dx − 4
= 2000∫ t
0sin 500x dx − 4
= 2000[− cos 500x
500
∣∣∣∣t
0− 4
= 4(1 − cos 500t) − 4
i = −4 cos 500t A
6–10 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[b] p = vi = (30 sin 500t)(−4 cos 500t)
= −120 sin 500t cos 500t
p = −60 sin 1000t W
w =12Li2
=12(15 × 10−3)16 cos2 500t
= 120 cos2 500t mJ
w = [60 + 60 cos 1000t] mJ.
Problems 6–11
[c] Absorbing power: Delivering power:
π ≤ t ≤ 2π ms 0 ≤ t ≤ π ms
3π ≤ t ≤ 4π ms 2π ≤ t ≤ 3π ms
P 6.8 [a] i(0) = A1 + A2 = 0.04
di
dt= −10,000A1e
−10,000t − 40,000A2e−40,000t
v = −200A1e−10,000t − 800A2e
−40,000t V
v(0) = −200A1 − 800A2 = 28
Solving, A1 = 0.1 and A2 = −0.06
Thus,
i = 0.1e−10,000t − 0.06e−40,000t A, t ≥ 0
v = −20e−10,000t + 48e−40,000t V, t ≥ 0
[b] If p = 0 then either i = 0 or v = o. Suppose i = 0:
i = 0.1e−10,000t − 0.06e−40,000t = 0
·. . 0.1e30,000t = 0.06 so t = −17.03 µs
This answer is impossible! So assume that v = 0:
v = −20e−10,000t + 48e−40,000t = 0
Then, − 20e30,000t = −48 ·. . t = 29.18 µs
This answer makes sense; therefore, the power is 0 at t = 29.18 µs.
6–12 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P 6.9 [a] From Problem 6.8 we have
A1 + A2 = 0.04
Now, we add the second equation for the coefficients:
−200A1 − 800A2 = −68
Solving, A1 = −0.06; A2 = 0.1Thus,
i = −0.06e−10,000t + 0.1e−40,000tA t ≥ 0
v = 12e−10,000t − 80e−40,000tA t ≥ 0
[b] i = 0 when 0.06e−10,000t = 0.1e−40,000t
·. . e30,0000t = 5/3 so t = 17.03 µs
Thus,
i > 0 for 0 ≤ t ≤ 17.03 µs and i < 0 for 17.03 µs ≤ t < ∞v = 0 when 12e−10,000t = 80e−40,000t
·. . e30,0000t = 20/3 so t = 63.24 µs
Thus,
v < 0 for 0 ≤ t ≤ 63.24 µs and v > 0 for 63.24 µs ≤ t < ∞Therefore,
p < 0 for 0 ≤ t ≤ 17.03 µs and 63.24 µs ≤ t < ∞(inductor delivers energy)
p > 0 for 17.03 µs ≤ t ≤ 63.24 µs (inductor stores energy)
[c] The energy stored at t = 0 is
w(0) =12L[i(0)]2 =
12(20 × 10−3)(40 × 10−3)2 = 16 µJ
The power for t > 0 is
p = vi = 6e−50,000t − 8e−80,000t − 0.72e−20,000t
The energy for t > 0 is
w =∫ ∞
0p dt =
∫ ∞
06e−50,000x dx −
∫ ∞
08e−80,000x dx −
∫ ∞
00.72e−20,000x dx
=6
50,000− 8
80,000− 0.72
20,000= −16 µJ
Thus, the energy stored at t = 0 equals the energy extracted for t > 0.
Problems 6–13
P 6.10 i = (B1 cos 1.6t + B2 sin 1.6t)e−0.4t
i(0) = B1 = 5 A
di
dt= (B1 cos 1.6t + B2 sin 1.6t)(−0.4e−0.4t) + e−0.4t(−1.6B1 sin 1.6t + 1.6B2 cos 1.6t)
= [(1.6B2 − 0.4B1) cos 1.6t − (1.6B1 + 0.4B2) sin 1.6t]e−0.4t
v = 2di
dt= [(3.2B2 − 0.8B1) cos 1.6t − (3.2B1 + 0.8B2) sin 1.6t]e−0.4t
v(0) = 28 = 3.2B2 − 0.8B1 = 3.2B2 − 4 ·. . B2 = 32/3.2 = 10 A
Thus,
i = (5 cos 1.6t + 10 sin 1.6t)e−0.4t A, t ≥ 0
v = (28 cos 1.6t − 24 sin 1.6t)e−0.4t V, t ≥ 0
i(5) = 1.24 A; v(5) = −3.76 V
p(5) = (1.24)(−3.76) = −4.67 W
The power delivered is 4.67 W.
P 6.11 For 0 ≤ t ≤ 1.6 s:
iL =15
∫ t
03 × 10−3 dx + 0 = 0.6 × 10−3t
iL(1.6 s) = (0.6 × 10−3)(1.6) = 0.96 mA
Rm = (20)(1000) = 20 kΩ
vm(1.6 s) = (0.96 × 10−3)(20 × 103) = 19.2 V
P 6.12 p = vi = 40t[e−10t − 10te−20t − e−20t]
W =∫ ∞
0p dx =
∫ ∞
040x[e−10x − 10xe−20x − e−20x] dx = 0.2 J
This is energy stored in the inductor at t = ∞.
6–14 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P 6.13 [a] v(20 µs) = 12.5 × 109(20 × 10−6)2 = 5 V (end of first interval)
v(20 µs) = 106(20 × 10−6) − (12.5)(400) × 10−3 − 10
= 5 V (start of second interval)
v(40 µs) = 106(40 × 10−6) − (12.5)(1600) × 10−3 − 10
= 10 V (end of second interval)
[b] p(10µs) = 62.5 × 1012(10−5)3 = 62.5 mW, v(10 µs) = 1.25 V,
i(10µs) = 50 mA, p(10 µs) = vi = 62.5 mW,
p(30 µs) = 437.50 mW, v(30 µs) = 8.75 V, i(30 µs) = 0.05 A
[c] w(10 µs) = 15.625 × 1012(10 × 10−6)4 = 0.15625µJ
w = 0.5Cv2 = 0.5(0.2 × 10−6)(1.25)2 = 0.15625µJ
w(30 µs) = 7.65625µJ
w(30 µs) = 0.5(0.2 × 10−6)(8.75)2 = 7.65625µJ
P 6.14 iC = C(dv/dt)
0 < t < 0.5 :
vc = 30t2 V
iC = 20 × 10−6(60)t = 1.2t mA
0.5 < t < 1 :
vc = 30(t − 1)2 V
iC = 20 × 10−6(60)(t − 1) = 1.2(t − 1) mA
Problems 6–15
P 6.15 [a] 0 ≤ t ≤ 5 µs
C = 5 µF1C
= 2 × 105
v = 2 × 105∫ t
04 dx + 12
v = 8 × 105t + 12 V 0 ≤ t ≤ 5 µs
v(5 µs) = 4 + 12 = 16 V
[b] 5 µs ≤ t ≤ 20 µs
v = 2 × 105∫ t
5×10−6−2 dx + 16 = −4 × 105t + 2 + 16
v = −4 × 105t + 18V 5 ≤ t ≤ 20 µs
v(20 µs) = −4 × 105(20 × 10−6) + 18 = 10 V
[c] 20 µs ≤ t ≤ 25 µs
v = 2 × 105∫ t
20×10−66 dx + 10 = 12 × 105t − 24 + 10
v = 12 × 105t − 14 V, 20 µs ≤ t ≤ 25 µs
v(25 µs) = 12 × 105(25 × 10−6) − 14 = 16 V
[d] 25 µs ≤ t ≤ 35 µs
v = 2 × 105∫ t
25×10−64 dx + 16 = 8 × 105t − 20 + 16
v = 8 × 105t − 4 V, 25 µs ≤ t ≤ 35 µs
v(35 µs) = 8 × 105(35 × 10−6) − 4 = 24 V
[e] 35 µs ≤ t < ∞
v = 2 × 105∫ t
35×10−60 dx + 24 = 24
v = 24 V, 35 µs ≤ t < ∞[f]
6–16 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P 6.16 v = −10 V, t ≤ 0; C = 0.8 µF
v = 40 − e−1000t(50 cos 500t + 20 sin 500t)V, t ≥ 0
[a] i = 0, t < 0
[b]dv
dt= 1000e−1000t(50 cos 500t + 20 sin 500t)
−e−1000t(−25,000 sin 500t + 10,000 cos 500t)
= e−1000t(50,000 cos 500t + 20,000 sin 500t
+25,000 sin 500t − 10,000 cos 500t)
= (40,000 cos 500t + 45,000 sin 500t)e−1000t
i = Cdv
dt= (32 cos 500t + 36 sin 500t)e−1000t mA
[c] no
[d] yes, from 0 to 32 mA
[e] v(∞) = 40 V
w =12Cv2 =
12(0.8 × 10−6)(40)2 = 640 µJ
P 6.17 [a] i =400 × 10−3
5 × 10−6 t = 8 × 104t 0 ≤ t ≤ 5 µs
i = 400 × 10−3 5 ≤ t ≤ 20 µs
q =∫ 5×10−6
08 × 104t dt +
∫ 15×10−6
5×10−6400 × 10−3 dt
= 8 × 104 t2
2
∣∣∣∣5×10−6
0+400 × 10−3(10 × 10−6)
= 8 × 104(12)(25 × 10−12) + 4 × 10−6
= 5 µC
[b] v =1
0.25 × 10−6
[∫ 5 µs
08 × 104x dx +
∫ 20 µs
5 µs0.4x dx +
∫ 30 µs
20 µs(104x − 0.5) dx
]
=1
0.25 × 10−6
[4 × 104t2
∣∣∣∣5 µs
0+0.4t
∣∣∣∣20 µs
5 µs+(5000t2 − 0.5t)
∣∣∣∣30 µs
20 µs
]
=1
0.25 × 10−6 [1 × 10−6 + 6 × 10−6 − 10.5 × 10−6 + 8 × 10−6] = 18V
Problems 6–17
[c] v(50 µs) = 18 +1
0.25 × 10−6 (5000t2 − 0.5t)∣∣∣∣50 µs
30 µs
= 18 +1
0.25 × 10−6 (−12.5 × 10−6 + 10.5 × 10−6) = 10V
w =12Cv2 =
12(0.25 × 10−6)(10)2 = 12.5 µJ
P 6.18 [a] v =1
0.5 × 10−6
∫ 500×10−6
050 × 10−3e−2000t dt − 20
= 100 × 103 e−2000t
−2000
∣∣∣∣500×10−6
0−20
= 50(1 − e−1) − 20 = 11.61 V
w = 12Cv2 = 1
2(0.5)(10−6)(11.61)2 = 33.7 µJ
[b] v(∞) = 50 − 20 = 30V
w(∞) =12(0.5 × 10−6)(30)2 = 225 µJ
P 6.19 [a] w(0) =12C[v(0)]2 =
12(0.25) × 10−6(50)2 = 312.5 µJ
[b] v = (A1t + A2)e−4000t
v(0) = A2 = 50 V
dv
dt= −4000e−4000t(A1t + A2) + e−4000t(A1)
= (−4000A1t − 4000A2 + A1)e−4000t
dv
dt(0) = A1 − 4000A2
i = Cdv
dt, i(0) = C
dv(0)dt
·. .dv(0)
dt=
i(0)C
=400 × 10−3
0.25 × 10−6 = 16 × 105
·. . 16 × 105 = A1 − 4000(50)
Thus, A1 = 16 × 105 + 2 × 105 = 18 × 105 Vs
6–18 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[c] v = (18 × 105t + 50)e−4000t
i = Cdv
dt= 0.25 × 10−6 d
dt(18 × 105t + 50)e−4000t
i =d
dt[(0.45t + 12.5 × 10−6)e−4000t]
= (0.45t + 12.5 × 10−6)(−4000)e−4000t + e−4000t(0.45)
= (−1800t − 0.05 + 0.45)e−4000t
= (0.40 − 1800t)e−4000t A, t ≥ 0
P 6.20 5‖(12 + 8) = 4 H
4‖4 = 2 H
15‖(8 + 2) = 6 H
3‖6 = 2 H
6 + 2 = 8 H
P 6.21 30‖20 = 12 H
80‖(8 + 12) = 16 H
60‖(14 + 16) = 20 H
15‖(20 + 10) = 20 H
Lab = 5 + 10 = 15 H
P 6.22 [a]
i(t) = −12
∫ t
012e−x dx + 6
= 6e−x
∣∣∣∣t0
+6
= 6e−t − 6 + 6
i(t) = 6e−t A, t ≥ 0
Problems 6–19
[b] i1(t) = −13
∫ t
012e−x dx + 2
= 4e−x
∣∣∣∣t0
+2
= 4(e−t − 1) + 2
i1(t) = 4e−t − 2 A, t ≥ 0
[c] i2(t) = −16
∫ t
012e−x dx + 4
= 2e−x
∣∣∣∣t0
+4
= 2(e−t − 1) + 4
i2(t) = 2e−t + 2 A, t ≥ 0
[d] p = vi = (12e−t)(6e−t) = 72e−2t W
w =∫ ∞
0p dt =
∫ ∞
072e−2t dt
= 72e−2t
−2
∣∣∣∣∞0
= 36 J
[e] w =12(3)(2)2 +
12(6)(4)2 = 54 J
[f] wtrapped =12(3)(−2)2 +
12(6)(2)2 = 18 J
wtrapped = 54 − 36 = 18 J checks
[g] Yes, they agree.
P 6.23 [a] io(0) = i1(0) + i2(0) = 4 A
[b]
io = − 110
∫ t
0160e−4x dx + 4 = −16
[e−4x
−4
∣∣∣∣∣t
0
+ 4
= 4(e−4t − 1) + 4 = 4e−4t A, t ≥ 0
6–20 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[c]
va = 8d
dt(4e−4t) = −128e−4t V
vc = va + vb = −128e−4t + 160e−4t
= 32e−4t V
i1 = − 13
∫ t
032e−4x dx + 1
= 2.67e−4t − 2.67 + 1
i1 = 2.67e−4t − 1.67 A, t ≥ 0
[d] i2 = −16
∫ t
032e−4x dx + 3
= 1.33e−4t − 1.33 + 3
i2 = 1.33e−4t + 1.67 A, t ≥ 0
[e] w(0) =12(3)(1)2 +
12(6)(3)2 +
12(8)(4)2 = 92.5 J
[f] wdel =12(10)(4)2 = 80 J
[g] wtrapped = 92.5 − 80 = 12.5 J
P 6.24 vb = 160e−4t V
io = 4e−4t A
p = 640e−8t W
w =∫ t
0640e−8x dx = 640
e−8x
−8
∣∣∣∣t0= 80(1 − e−8t) W
wtotal = 80 J
w(0.2) = 80(1 − e−1.6) = 63.85 J
Thus,
% delivered =63.8580
(100) = 79.8%
Problems 6–21
P 6.2514
+16
=512
·. . Ceq = 2.4 µF
14
+112
=412
·. . Ceq = 3 µF
124
+18
=424
·. . Ceq = 6 µF
P 6.26 Work from the right hand side of the circuit, simplifying step by step:
1. 48 µF in series with 16 µF : 1/C = 1/16 µ + 1/48 µ ·. . C = 12 µFThe voltages add in series, so the 12 µF capacitor has a voltage of 20 V,negative at the top.
2. Previous 12 µF in parallel with 3 µF : C = 12 µ + 3 µ = 15 µFThe voltage is 20 V, negative at the top.
3. Previous 15 µF in series with 30 µF :1/C = 1/15 µ + 1/30 µ ·. . C = 10 µF
The voltages add in series, so the 10 µF capacitor has a voltage of 10 V,positive at the right.
6–22 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
4. Previous 10 µF in parallel with 10 µF : C = 10 µ + 10 µ = 20 µFThe voltage is 10 V, negative at the top.
5. Previous 20 µF in series with 5 µF and 4 µF :1/C = 1/20 µ + 1/5 µ + 1/4 µ ·. . C = 2 µF
The voltages in series add: 5V − 10V + 30V = 25V positive at the top.
The equivalent capacitance is 2 µF with a voltage of 25 V, positive at the top.
P 6.27 [a]
vo = − 12 × 10−6
∫ t
020 × 10−6e−x dx + 10
= 10e−x
∣∣∣∣t0
+10
= 10e−t V, t ≥ 0
[b] v1 = − 13 × 10−6 (20 × 10−6)e−x
∣∣∣∣t0
+4
= 6.67e−t − 2.67 V, t ≥ 0
[c] v2 = − 16 × 10−6 (20 × 10−6)e−x
∣∣∣∣t0
+6
= 3.33e−t + 2.67 V, t ≥ 0
[d] p = vi = (10e−t)(20 × 10−6)e−t
= 200 × 10−6e−2t
w =∫ ∞
0200 × 10−6e−2t dt
= 200 × 10−6 e−2t
−2
∣∣∣∣∞0
= −100 × 10−6(0 − 1) = 100µJ
[e] w = 12(3 × 10−6)(4)2 + 1
2(6 × 10−9)(6)2
= 132 µJ
[f] wtrapped = 12(3 × 10−6)(8/3)2 + 1
2(6 × 10−6)(8/3)2
= 32 µJ
Problems 6–23
CHECK: 100 + 32 = 132µJ
[g] Yes, they agree.
P 6.28 C1 = 10 + 2 = 12µF
1C2
=1
12 µ+
18 µ
·. . C2 = 4.8 µF
vo(0) + v1(0) = −5 + 25 = 20 V
[a]
v2 = − 14.8 × 10−6
∫ t
01.92 × 10−3e−20x dx + 20
= −400e−20x
−20
∣∣∣∣t0
+20
= 20(e−20t − 1) + 20
= 20e−20t V, t ≥ 0
[b] vo = − 18 × 10−6
∫ t
01.92 × 10−3e−20x dx − 5
= −240e−20x
−20
∣∣∣∣t0
−5
= 12(e−20t − 1) − 5
= 12e−20t − 17 V, t ≥ 0
[c] v1 = − 112 × 10−6
∫ t
01.92 × 10−3e−20x dx + 25
= −160e−20x
−20
∣∣∣∣t0
+25
= 8(e−20t − 1) + 25
= 8e−20t + 17 V, t ≥ 0
6–24 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
[d] i1 = −10 × 10−6 d
dt[8e−20t + 17]
= −10 × 10−6(−20)8e−20t
= 1.6e−20t mA, t > 0
[e] i2 = −2 × 10−6 d
dt[8e−20t + 17]
= −2 × 10−6(−20)8e−20t
= 0.32e−20t mA, t > 0CHECK: i1 + i2 = 1.92e−20t mA = io
P 6.29 [a] w(0) = [12(8 × 10−6)(−5)2 + 12(10 × 10−6)(25)2 + 1
2(2 × 10−6)(25)2]
= 3850 µJ
[b] vo(∞) = −17 V
v1(∞) = 17 V
w(∞) = [12(8 × 10−6)(−17)2 + 12(12 × 10−6)(17)2]
= 2890 µJ
[c] w =∫ ∞
0(20e−20t)(1.92 × 10−3e−20t) dt = 960 µJ
CHECK: 3850 − 2890 = 960µJ
[d] % delivered =9603850
× 100 = 24.9%
[e] w(40 ms) =∫ 0.04
0(20e−20t)(1.92 × 10−3e−20t) dt
= 0.0384e−40t
−40
∣∣∣∣0.04
0
= 960 × 10−6(1 − e−1.6) = 766.2 µJ
% delivered =766.2960
(100) = 79.8%
P 6.30 From Figure 6.17(a) we have
v =1C1
∫ t
0i + v1(0) +
1C2
∫ t
0i dx + v2(0) + · · ·
v =[ 1C1
+1C2
+ · · ·] ∫ t
0i dx + v1(0) + v2(0) + · · ·
Problems 6–25
Therefore1
Ceq=[ 1C1
+1C2
+ · · ·], veq(0) = v1(0) + v2(0) + · · ·
P 6.31 From Fig. 6.18(a)
i = C1dv
dt+ C2
dv
dt+ · · · = [C1 + C2 + · · ·]dv
dt
Therefore Ceq = C1 + C2 + · · ·. Because the capacitors are in parallel, the initialvoltage on every capacitor must be the same. This initial voltage would appear onCeq.
P 6.32dio
dt= 5e−2000t[−8000 sin 4000t + 4000 cos 4000t]
−2000e−2000t[2 cos 4000t + sin 4000t]dio
dt(0+) = 5[1(4000) + (−2000)(2)] = 0
v2(0+) = 10 × 10−3dio
dt(0+) = 0
v1(0+) = 40io(0+) + v2(0+) = 40(10) + 0 = 400V
P 6.33 vc = − 10.625 × 10−6
(∫ t
01.5e−16,000x dx −
∫ t
00.5e−4000x dx
)− 50
= 150(e−16,000t − 1) − 200(e−4000t − 1) − 50
= 150e−16,000t − 200e−4000t V
vL = 25 × 10−3dio
dt
= 25 × 10−3(−24,000e−16,000t + 2000e−4000t)
= −600e−16,000t + 50e−4000t V
vo = vc − vL
= (150e−16,000t − 200e−4000t) − (−600e−16,000t + 50e−4000t)
= 750e−16,000t − 250e−4000t V, t > 0
P 6.34 [a] −2dig
dt+ 16
di2dt
+ 32i2 = 0
16di2dt
+ 32i2 = 2dig
dt
[b] i2 = e−t − e−2t A
di2dt
= −e−t + 2e−2t A/s
6–26 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
ig = 8 − 8e−t A
dig
dt= 8e−t A/s
·. . −16e−t + 32e−2t + 32e−t − 32e−2t = 16e−t
[c] v1 = 4dig
dt− 2
di2dt
= 4(8e−t) − 2(−e−t + 2e−2t)
= 34e−t − 4e−2t V, t > 0
[d] v1(0) = 34 − 4 = 30 V; Also
v1(0) = 4dig
dt(0) − 2
di2dt
(0)
= 4(8) − 2(−1 + 2) = 32 − 2 = 30 VYes, 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 = L2d
dt(ig − i2) + M
di1dt
= 16d
dt(15 + 36e−5t − 51e−4t) + 8
d
dt(4 + 64e−5t − 68e−4t)
= −2880e−5t + 3264e−4t − 2560e−5t + 2176e−4t
vo = −5440e−5t + 5440e−4t V
P 6.36 [a] vg = 5(ig − i1) + 20(i2 − i1) + 60i2
= 5(16 − 16e−5t − 4 − 64e−5t + 68e−4t)+
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 6–27
[c] pdev = vgig
= 960 + 92,480e−4t − 94,400e−5t − 92,480e−9t+
93,440e−10tW
[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∑
pabs = 720 + 180 + 60 = 960 W
·. .∑
pdev =∑
pabs = 960 W
P 6.37 [a] Rearrange by organizing the equations by di1/dt, i1, di2/dt, i2 and transfer theig terms to the right hand side of the equations. We get
4di1dt
+ 25i1 − 8di2dt
− 20i2 = 5ig − 8dig
dt
−8di1dt
− 20i1 + 16di2dt
+ 80i2 = 16dig
dt
[b] From the given solutions we have
di1dt
= −320e−5t + 272e−4t
di2dt
= 260e−5t − 204e−4t
Thus,
4di1dt
= −1280e−5t + 1088e−4t
25i1 = 100 + 1600e−5t − 1700e−4t
8di2dt
= 2080e−5t − 1632e−4t
20i2 = 20 − 1040e−5t + 1020e−4t
5ig = 80 − 80e−5t
8dig
dt= 640e−5t
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 (OK)
8di1dt
= −2560e−5t + 2176e−4t
20i1 = 80 + 1280e−5t − 1360e−4t
16di2dt
= 4160e−5t − 3264e−4t
80i2 = 80 − 4160e−5t + 4080e−4t
16dig
dt= 1280e−5t
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 (OK)
P 6.38 [a] L2 =(
M2
k2L1
)=
(0.09)2
(0.75)2(0.288)= 50 mH
N1
N2=
√L1
L2=
√28850
= 2.4
[b] P1 =L1
N21
=0.288
(1200)2 = 0.2 × 10−6 Wb/A
P2 =L2
N22
=0.05
(500)2 = 0.2 × 10−6 Wb/A
P 6.39 P1 =L1
N21
= 2 nWb/A; P2 =L2
N22
= 2 nWb/A; M = k√
L1L2 = 180 µH
P12 = P21 =M
N1N2= 1.2 nWb/A
P11 = P1 − P21 = 0.8 nWb/A
Problems 6–29
P 6.40 [a] k =M√L1L2
=7.2√81
= 0.8
[b] M =√
81 = 9 mH
[c]L1
L2=
N21 P1
N22 P2
=(
N1
N2
)2
·. .(
N1
N2
)2
=273
= 9
N1
N2= 3
P 6.41 [a] M = k√
L1L2 = 0.8√
324 = 14.4 mH
P1 =L1
N21
=36 × 10−3
(200)2 = 900 nWb/A
dφ11
dφ21=
P11
P21= 0.1; P21 = 10P11
P1 = P11 + P21 = 11P11
P11 =111
P1 = 81.82 nWb/A
P21 = 10P11 = 818.18 nWb/A
N2 =M
N1P21=
14.4 × 10−3
(200)(818.18 × 10−9)= 88 turns
[b] P2 =L2
N22
=9 × 10−3
(88)2 = 1162.19 nWb/A
[c] P11 = 81.82 nWb/A [see part (a)]
[d]φ22
φ12=
P22
P12
P12 = P21 = 818.18 nWb/A
P22 = P2 − P12 = 1162.19 × 10−9 − 818.18 × 10−9 = 344.01 nWb/A
φ22
φ12=
344.01818.18
= 0.4205
P 6.42 [a] Dot terminal 1; the flux is up in coil 1-2, and down in coil 3-4. Assign thecurrent 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 thecurrent into terminal 4; the flux is right-to-left in coil 3-4. Therefore, dotterminal 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 thecurrent into terminal 4; the flux is right-to-left in coil 3-4. Therefore, dotterminal 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 thecurrent 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]1k2 =
(1 +
P11
P12
)(1 +
P22
P12
)=(1 +
P11
P21
)(1 +
P22
P12
)Therefore
k2 =P12P21
(P21 + P11)(P12 + P22)
Now note that
φ1 = φ11 + φ21 = P11N1i1 + P21N1i1 = N1i1(P11 + P21)
and similarly
φ2 = N2i2(P22 + P12)
It follows that
(P11 + P21) =φ1
N1i1
and
(P22 + P12) =(
φ2
N2i2
)
Therefore
k2 =(φ12/N2i2)(φ21/N1i1)(φ1/N1i1)(φ2/N2i2)
=φ12φ21
φ1φ2
or
k =
√√√√(φ21
φ1
)(φ12
φ2
)
[b] The fractions (φ21/φ1) and (φ12/φ2) are by definition less than 1.0, thereforek < 1.
P 6.44 [a] vab = L1di
dt+ L2
di
dt+ M
di
dt+ M
di
dt= (L1 + L2 + 2M)
di
dt
It follows that Lab = (L1 + L2 + 2M)
[b] vab = L1di
dt− M
di
dt+ L2
di
dt− M
di
dt= (L1 + L2 − 2M)
di
dt
Therefore Lab = (L1 + L2 − 2M)
Problems 6–31
P 6.45 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 bepositive at its positive terminal. Therefore, the voltage is negative at the negativeterminal of the voltmeter.
Thus, the lower terminal of the unmarked coil has the same instantaneous polarity asthe dotted terminal. Therefore, place a dot on the lower terminal of the unmarkedcoil.
P 6.46 [a] vab = L1d(i1 − i2)
dt+ M
di2dt
0 = L1d(i2 − i1)
dt− M
di2dt
+ Md(i1 − i2)
dt+ L2
di2dt
Collecting coefficients of [di1/dt] and [di2/dt], the two mesh-current equationsbecome
vab = L1di1dt
+ (M − L1)di2dt
and
0 = (M − L1)di1dt
+ (L1 + L2 − 2M)di2dt
Solving for [di1/dt] gives
di1dt
=L1 + L2 − 2ML1L2 − M2 vab
from which we have
vab =(
L1L2 − M2
L1 + L2 − 2M
)(di1dt
)
·. . Lab =L1L2 − M2
L1 + L2 − 2M
[b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses, therefore
Lab =L1L2 − M2
L1 + L2 + 2M
P 6.47 [a] W = (0.5)L1i21 + (0.5)L2i
22 + Mi1i2
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, i22 + 9i2 + 20.25 = 0
Therefore i2 = −(9
2
)±√(9
2
)2
− 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 squareroot sign negative.
P 6.49 When the button is not pressed we have
C2dv
dt= C1
d
dt(vs − v)
or
(C1 + C2)dv
dt= C1
dvs
dt
dv
dt=
C1
(C1 + C2)dvs
dt
Assuming C1 = C2 = C
dv
dt= 0.5
dvs
dt
or
v = 0.5vs(t) + v(0)
When the button is pressed we have
Problems 6–33
C1dv
dt+ C3
dv
dt+ C2
d(v − vs)dt
= 0
·. .dv
dt=
C2
C1 + C2 + C3
dvs
dt
Assuming C1 = C2 = C3 = C
dv
dt=
13
dvs
dt
v =13vs(t) + v(0)
Therefore interchanging the fixed capacitor and the button has no effect on thechange in v(t).
P 6.50 With no finger touching and equal 10 pF capacitors
v(t) =1020
(vs(t)) + 0 = 0.5vs(t)
With a finger touching
Let Ce = equivalent capacitance of person touching lamp
Ce =(10)(100)
110= 9.091 pF
Then C + Ce = 10 + 9.091 = 19.091 pF
·. . v(t) =10
29.091vs = 0.344vs
·. . ∆v(t) = (0.5 − 0.344)vs = 0.156vs
6–34 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance
P 6.51 With no finger on the button the circuit is
C1dv
dt(v − vs) + C2
d
dt(v + vs) = 0
when C1 = C2 = C (2C)dv
dt= 0
With a finger on the button
C1d(v − vs)
dt+ C2
d(v + vs)dt
+ C3dv
dt= 0
(C1 + C2 + C3)dv
dt+ C2
dvs
dt− C1
dvs
dt= 0
when C1 = C2 = C3 = C (3C)dv
dt= 0
·. . there is no change in the output voltage of this circuit.
7Response 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 shortcircuit, effectively eliminating the 2 Ω resistor from the circuit.
First combine the 30 Ω and 6 Ω resistors in parallel:30‖6 = 5 ΩUse voltage division to find the voltage drop across the parallel resistors:
v =5
5 + 3(120) = 75 V
Now find the current using Ohm’s law:
i(0−) = −v
6= −75
6= −12.5 A
[b] w(0) =12Li2(0) =
12(8 × 10−3)(12.5)2 = 625 mJ
[c] To find the time constant, we need to find the equivalent resistance seen by theinductor for t > 0. When the switch opens, only the 2 Ω resistor remainsconnected to the inductor. Thus,
τ =L
R=
8 × 10−3
2= 4 ms
[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) = 12Li2(5 ms) = 1
2(8) × 10−3(3.58)2 = 51.3 mJw (dis) = 625 − 51.3 = 573.7 mJ
% dissipated =(573.7
625
)100 = 91.8%
AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor:
Using current division,
i(0−) =10
10 + 6(6.4) = 4 A
Now use the circuit for t > 0 to find the equivalent resistance seen by theinductor, and use this value to find the time constant:
Req = 4‖(6 + 10) = 3.2 Ω, ·. . τ =L
Req=
0.323.2
= 0.1 s
Use the initial inductor current and the time constant to find the current in theinductor:i(t) = i(0−)e−t/τ = 4e−t/0.1 = 4e−10t A, t ≥ 0Use current division to find the current in the 10 Ω resistor:
io(t) =4
4 + 10 + 6(−i) =
420
(−4e−10t) = −0.8e−10t A, t ≥ 0+
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
w(0) =12Li2(0−) =
12(0.32)(4)2 = 2.56 J
Find the energy dissipated in the 4 Ω resistor by integrating the power over alltime:
v4Ω(t) = Ldi
dt= 0.32(−10)(4e−10t) = −12.8e−10t V, t ≥ 0+
Problems 7–3
p4Ω(t) =v2
4Ω
4= 40.96e−20t W, t ≥ 0+
w4Ω(t) =∫ ∞
040.96e−20tdt = 2.048 J
Find the percentage of the initial energy in the inductor dissipated in the 4 Ωresistor:
% dissipated =(2.048
2.56
)100 = 80%
AP 7.3 [a] The circuit for t < 0 is shown below. Note that the capacitor behaves like anopen circuit.
Find the voltage drop across the open circuit by finding the voltage drop acrossthe 50 kΩ resistor. First use current division to find the current through the50 kΩ resistor:
i50k =80 × 103
80 × 103 + 20 × 103 + 50 × 103 (7.5 × 10−3) = 4 mA
Use Ohm’s law to find the voltage drop:v(0−) = (50 × 103)i50k = (50 × 103)(0.004) = 200 V
[b] To find the time constant, we need to find the equivalent resistance seen by thecapacitor for t > 0. When the switch opens, only the 50 kΩ resistor remainsconnected 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
[d] w(0) =12Cv2 =
12(0.4 × 10−6)(200)2 = 8 mJ
[e] w(t) =12Cv2(t) =
12(0.4 × 10−6)(200e−50t)2 = 8e−100t mJ
The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains:
8 × 10−3e−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 isshown 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 thetwo RC circuits:
i =15
75,000= 0.2 mA, v5(0−) = 4 V, v1(0−) = 8 V
There are two time constants in the circuit, one for each RC subcircuit. τ5 isthe time constant for the 5 µF – 20 kΩ subcircuit, and τ1 is the time constant forthe 1 µF – 40 kΩ subcircuit:τ5 = (20 × 103)(5 × 10−6) = 100 ms; τ1 = (40 × 103)(1 × 10−6) = 40 msTherefore,v5(t) = v5(0−)e−t/τ5 = 4e−t/0.1 = 4e−10t V, t ≥ 0v1(t) = v1(0−)e−t/τ1 = 8e−t/0.04 = 8e−25t V, t ≥ 0Finally,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 tofind the energy at 60 ms:v1(60 ms) = 8e−25(0.06) ∼= 1.79 V, v5(60 ms) = 4e−10(0.06) ∼= 2.20 Vw1(60 ms) = 1
2Cv21(60 ms) = 1
2(1 × 10−6)(1.79)2 ∼= 1.59 µJw5(60 ms) = 1
2Cv25(60 ms) = 1
2(5 × 10−6)(2.20)2 ∼= 12.05 µJw(60 ms) = 1.59 + 12.05 = 13.64 µJFind the initial energy from the initial voltage:w(0) = w1(0) + w2(0) = 1
2(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 initialenergy: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 theinductor:
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 theinductor at 0+. Note that this is the same as the voltage drop across the 10 Ωresistor, which has current from two sources — 8 A from the current sourceand 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 theinductor for t > 0. Only the 10 Ω resistor is connected to the inductor fort > 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. Whenthe switch has been in position a for a long time, the circuit reduces to the onebelow:
Note that the inductor behaves as a short circuit and all of the current from the8 A source flows through the short circuit. Thus,if = −8 ANow,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) = Ldi(t)dt
= (200 × 10−3)(−50)(20e−50t) = −200e−50t V, 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 relationshipbetween vo and vA:vA − vo
8000+
vA
160,000+
vA + 7540,000
= 0
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 dropacross the capacitor:
i =(
40 × 103
125 × 103
)(10 × 10−3) = 3.2 mA
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 thatinterval:
Problems 7–7
For 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 thecapacitor voltage from the previous interval:vc(0.01) = 80e−40(0.01) = 53.63 VNow use the next circuit, valid for t ≥ 10 ms, to calculate vc(t) for that interval:
For t ≥ 10 ms :
Req = 25 kΩ‖100 kΩ = 20 kΩ
τ = ReqC = (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 powerabsorbed by the resistor over all time. Use the expression p = v2/R tocalculate the power absorbed by the resistor.
w25k =∫ 0.01
0
[80e−40t]2
25,000dt +
∫ ∞
0.01
[53.63e−50(t−0.01)]2
25,000dt = 2.91 mJ
[d] Repeat the process in part (c), but recognize that the voltage across this resistoris non-zero only for the second interval:
w100kΩ =∫ ∞
0.01
[53.63e−50(t−0.01)]2
100,000dt = 0.29 mJ
We can check our answers by calculating the initial energy stored in thecapacitor. All of this energy must eventually be dissipated by the 25 kΩresistor 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 the8 A current source should be 9 Ω.Prior to switch a closing at t = 0, there are no sources connected to theinductor; 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 + (3‖0.8). The currentleaving the 10 V source is
102 + (3‖0.8)
= 3.8 A
The final current in the inductor, which is equal to the current in the 0.8 Ωresistor is
i(∞) =3
3 + 0.8(3.8) = 3 A
The resistance seen by the inductor is calculated to find the time constant:
0.8 + (2‖3) = 2 Ω τ =L
R=
22
= 1 s
Therefore,
i = i(∞) + [i(0+) − i(∞)]e−t/τ = 3 − 3e−t A, 0 ≤ t ≤ 1 s
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:
i =9
9 + 6(−8) = −4.8 A
The equivalent resistance seen by the inductor is used to calculate the timeconstant:
3‖(9 + 6) = 2.5 Ω τ =L
R=
22.5
= 0.8 s
Problems 7–9
Therefore,
i = i(∞) + [i(1+) − i(∞)]e−(t−1)/τ
= −4.8 + 6.696e−1.25(t−1) A, t ≥ 1 s
AP 7.9 0 ≤ t ≤ 32 ms:
vo = − 1RCf
∫ 32×10−3
0−10 dt + 0 = − 1
RCf
(−10t)∣∣∣∣32×10−3
0= − 1
RCf
(−320 × 10−3)
RCf = (200 × 103)(0.2 × 10−6) = 40 × 10−3 so1
RCf
= 25
vo = −25(−320 × 10−3) = 8 V
t ≥ 32 ms:
vo = − 1RCf
∫ t
32×10−35 dy + 8 = − 1
RCf
(5y)∣∣∣∣t32×10−3
+8 = − 1RCf
5(t − 32 × 10−3) + 8
RCf = (250 × 103)(0.2 × 10−6) = 50 × 10−3 so1
RCf
= 20
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 thenon-inverting input:
vp = Vf + [Vo − Vf ]e−t/τ = −2 + (0 + 2)e−t/τ
τ = (160 × 103)(10 × 10−9) = 10−3; 1/τ = 625
vp = −2 + 2e−625t V; vn = vp
Write a KVL equation at the inverting input, and use it to determine vo:
vn
10,000+
vn − vo
40,000= 0
·. . 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 thenon-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 kΩ‖6 kΩ = 1.5kΩ
Find the current from the voltage source by combining the resistors in seriesand parallel and using Ohm’s law:
ig(0−) =40
(1500 + 500)= 20 mA
Find the branch currents using current division:
i1(0−) =20008000
(0.02) = 5 mA
i2(0−) =60008000
(0.02) = 15 mA
[b] The current in an inductor is continuous. Therefore,
i1(0+) = i1(0−) = 5 mA
i2(0+) = −i1(0+) = −5 mA (when switch is open)
[c] τ =L
R=
0.4 × 10−3
8 × 103 = 5 × 10−5 s;1τ
= 20,000
i1(t) = i1(0+)e−t/τ = 5e−20,000t mA, t ≥ 0
[d] i2(t) = −i1(t) when t ≥ 0+
·. . i2(t) = −5e−20,000t mA, t ≥ 0+
[e] The current in a resistor can change instantaneously. The switching operationforces 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
[b] τ =L
R=
445 + 5
= 80 ms
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 t ≥ 0+
v2 = Ldi
dt= (4)(−12.5)(4e−12.5t) = −200e−12.5t V t ≥ 0+
[d] pdiss = i2(45) = 720e−25t W
wdiss =∫ t
0720e−25x dx = 720
e−25x
−25
∣∣∣∣t0= 28.8 − 28.8e−25t J
wdiss(40 ms) = 28.8 − 28.8e−1 = 18.205 J
w(0) =12(4)(4)2 = 32 J
% dissipated =18.205
32(100) = 56.89%
P 7.3 [a] io(0−) = 0 since the switch is open for t < 0.
[b] For t = 0− the circuit is:
120 Ω‖60 Ω = 40 Ω
·. . ig =12
10 + 40= 0.24 A = 240 mA
iL(0−) =(120
180
)ig = 160 mA
[c] For t = 0+ the circuit is:
120 Ω‖40 Ω = 30 Ω
Problems 7–13
·. . ig =12
10 + 30= 0.30 A = 300 mA
ia =(120
160
)300 = 225 mA
·. . io(0+) = 225 − 160 = 65 mA
[d] iL(0+) = iL(0−) = 160 mA
[e] io(∞) = ia = 225 mA
[f] iL(∞) = 0, since the switch short circuits the branch containing the 20 Ωresistor and the 100 mH inductor.
[g] τ =L
R=
100 × 10−3
20= 5 ms;
1τ
= 200
·. . iL = 0 + (160 − 0)e−200t = 160e−200t mA, t ≥ 0
[h] vL(0−) = 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; ·. . vL(0+) = −3.2 V
[j] vL(∞) = 0, since the current in the inductor is a constant at t = ∞.
[k] vL(t) = 0 + (−3.2 − 0)e−200t = −3.2e−200t V, t ≥ 0+
[l] io = ia − iL = 225 − 160e−200t mA, t ≥ 0+
P 7.4 [a]v
i= R =
400e−5t
10e−5t= 40 Ω
[b] τ =15
= 200 ms
[c] τ =L
R= 200 × 10−3
L = (200 × 10−3)(40) = 8 H
[d] w(0) =12L[i(0)]2 =
12(8)(10)2 = 400 J
[e] wdiss =∫ t
04000e−10x dx = 400 − 400e−10t
0.8w(0) = (0.8)(400) = 320 J
400 − 400e−10t = 320 ·. . e10t = 5
Solving, t = 160.9 ms.
7–14 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.5 [a] iL(0) =126
= 2 A
io(0+) =122
− 2 = 6 − 2 = 4 A
io(∞) =122
= 6 A
[b] iL = 2e−t/τ ; τ =L
R=
14
s
iL = 2e−4t A
io = 6 − iL = 6 − 2e−4t A, t ≥ 0+
[c] 6 − 2e−4t = 5
1 = 2e−4t
e6t = 2 ·. . t = 173.3 ms
P 7.6 w(0) =12(30 × 10−3)(32) = 135 mJ
15w(0) = 27 mJ
iR = 3e−t/τ
pdiss = i2RR = 9Re−2t/τ
wdiss =∫ t
0R(9)e−2x/τ dx
wdiss = 9Re−2x/τ
−2/τ
∣∣∣∣to0= −4.5τR(e−2to/τ − 1) = 4.5L(1 − e−2to/τ )
4.5L = (4.5)(30) × 10−3 = 0.135; to = 15 µs
1 − e−2to/τ =15
e2to/τ = 1.25;2toτ
=2toRL
= ln 1.25
R =L ln 1.25
2to=
30 × 10−3 ln 1.2530 × 10−6 = 223.14 Ω
Problems 7–15
P 7.7 [a] w(0) =12LI2
g
wdiss =∫ to
0I2gRe−2t/τ dt = I2
gRe−2t/τ
(−2/τ)
∣∣∣∣to0
=12I2gRτ(1 − e−2to/τ ) =
12I2gL(1 − e−2to/τ )
wdiss = σw(0)
·. .12LI2
g (1 − e−2to/τ ) = τ(1
2LI2
g
)
1 − e−2to/τ = σ; e2to/τ =1
(1 − σ)
2toτ
= ln[
1(1 − σ)
];
R(2to)L
= ln[1/(1 − σ)]
R =L ln[1/(1 − σ)]
2to
[b] R =(30 × 10−3) ln[1/0.8]
30 × 10−6
R = 223.14 Ω
P 7.8 [a] t < 0
iL(0−) =150180
(12) = 10 A
t ≥ 0
τ =1.6 × 10−3
8= 200 × 10−6; 1/τ = 5000
io = −10e−5000t A t ≥ 0
7–16 CHAPTER 7. Response of First-Order RL and RC Circuits
[b] wdel =12(1.6 × 10−3)(10)2 = 80 mJ
[c] 0.95wdel = 76 mJ
·. . 76 × 10−3 =∫ to
08(100e−10,000t) dt
·. . 76 × 10−3 = −80 × 10−3e−10,000t
∣∣∣∣to0= 80 × 10−3(1 − e−10,000to)
·. . e−10,000to = 4 × 10−3 so to = 552.1 µs
·. .toτ
=552.1 × 10−6
200 × 10−6 = 2.76 so to ≈ 2.76τ
P 7.9 For t < 0+
ig =−48
6 + (18‖1.5)= −6.5 A
iL(0−) =18
18 + 1.5(−6.5) = −6 A = iL(0+)
For t > 0
iL(t) = iL(0+)e−t/τ A, t ≥ 0
τ =L
R=
0.510 + 12.45 + (54‖26)
= 0.0125 s;1τ
= 80
iL(t) = −6e−80t A, t ≥ 0
io(t) =5480
(−iL(t)) =5480
(6e−80t) = 4.05e−80t V, t ≥ 0+
Problems 7–17
P 7.10 From the solution to Problem 7.9,
i54Ω =2680
(−iL) = −1.95e−80t A
P54Ω = 54(i54Ω)2 = 205.335e−160tW
wdiss =∫ 0.0125
0205.335e−160t dt
=205.335−160
e−160t
∣∣∣∣0.0125
0
= 1.28(1 − e−2) = 1.11 J
wstored =12(0.5)(−6)2 = 9 mJ.
% diss =1.119
× 100 = 12.3%
P 7.11 [a] t < 0 :
iL(0−) = iL(0+) =70
70 + 4(11.84) = 11.2 A
i∆ =70160
iT = 0.4375iT
vT = 30i∆ + iT(90)(70)
160= 30(0.4375)iT +
(90)(70)160
iT = 52.5iT
vT
iT= RTh = 52.5 Ω
7–18 CHAPTER 7. Response of First-Order RL and RC Circuits
τ =L
R=
20 × 10−3
52.5= ·. .
1τ
= 2625
iL = 11.2e−2625t A, t ≥ 0
[b] vL = LdiLdt
= 20 × 10−3(−2625)(11.2e−2625t) = −588e−2625t V, t ≥ 0+
[c]
vL = 30i∆ + 90i∆ = 120i∆
i∆ =vL
120= −4.9e−2625t A t ≥ 0+
P 7.12 w(0) =12(20 × 10−3)(11.2)2 = 1254.4 mJ
p30i∆ = −30i∆iL = −30(−4.9e−2625t)(11.2e−2625t) = 1646.4e−5250t W
w30i∆ =∫ ∞
01646.4e−5250t dt = 1646.4
e−5250t
−5250
∣∣∣∣∞0
= 313.6mJ
% dissipated =313.61254.4
(100) = 25%
P 7.13 t < 0
iL(0−) = iL(0+) = 4 A
Problems 7–19
t > 0
Find Thévenin resistance seen by inductor
iT = 4vT ;vT
iT= RTh =
14
= 0.25 Ω
τ =L
R=
5 × 10−3
0.25= 20 ms; 1/τ = 50
io = 4e−50t A, t ≥ 0
vo = Ldiodt
= (5 × 10−3)(−200e−50t) = −e−50t V, t ≥ 0+
P 7.14 t < 0:
iL(0+) = 8 A
7–20 CHAPTER 7. Response of First-Order RL and RC Circuits
t > 0:
Re =(10)(40)
50+ 10 = 18 Ω
τ =L
Re
=0.07218
= 4 ms;1τ
= 250
·. . iL = 8e−250t A
·. . vo = −10iL − 0.072diLdt
= −80e−250t + 144e−250t
= 64e−250t A t ≥ 0+
P 7.15 w(0) =12(72 × 10−3)(8)2 = 2304 mJ
p40Ω =v2
o
40=
642
40e−500t = 102.4e−500t W
w40Ω =∫ ∞
0102.4e−500t dt = 204.8 mJ
%diss =204.82304
(100) = 8.89%
P 7.16 [a] vo(t) = vo(0+)e−t/τ
·. . vo(0+)e−1×10−3/τ = 0.5vo(0+)
·. . e1×10−3/τ = 2
·. . τ =L
R=
1 × 10−3
ln 2
·. . L =10 × 10−3
ln 2= 14.43 mH
Problems 7–21
[b] vo(0+) = −10iL(0+) = −10(1/10)30 × 10−3 = −30 mV
·. . vo = −0.03e−t/τ V, t ≥ 0+
p10Ω =v2
o
10= 9 × 10−5e−2t/τ
w10Ω(1 ms) =∫ 10−3
0+9 × 10−5e−2t/τ dt
= 4.5τ × 10−5(1 − e−2(0.001)/τ )
τ =1
1000 ln 2·. . w10Ω(1 ms) = 48.69 nJ
wL(0) =12Li2L(0) =
12(14.43 × 10−3)(3 × 10−3)2 = 64.92 nJ
%dissipated in 1 ms =48.6964.92
(100) = 75%
P 7.17 [a] t < 0 :
t = 0+:
33 = iab + 9 + 15, iab = 9 A, t = 0+
[b] At t = ∞:
iab = 165/5 = 33 A, t = ∞
7–22 CHAPTER 7. Response of First-Order RL and RC Circuits
[c] i1(0) = 9, τ1 =12.5 × 10−3
5= 2.5 ms
i2(0) = 15, τ2 =3.75 × 10−3
31.25 ms
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 x = 0.7116 = e−400t
·. . t =[ln(1/0.7116)]
400= 850.6 µs
P 7.18 [a] t < 0
1 kΩ‖4 kΩ = 0.8 kΩ
20 kΩ‖80 kΩ = 16 kΩ
(105 × 10−3)(0.8 × 103) = 84 V
Problems 7–23
iL(0−) =84
16,800= 5 mA
t > 0
τ =L
R=
624
× 10−3 = 250 µs;1τ
= 4000
iL(t) = 5e−4000t mA, t ≥ 0
p4k = 25 × 10−6e−8000t(4000) = 0.10e−8000t W
wdiss =∫ t
00.10e−8000x dx = 12.5 × 10−6[1 − e−8000t] J
w(0) =12(6)(25 × 10−6) = 75 µJ
0.10w(0) = 7.5 µJ
12.5(1 − e−8000t) = 7.5; ·. . e8000t = 2.5
t =ln 2.58000
= 114.54 µs
[b] wdiss(total) = 75(1 − e−8000t) µJ
wdiss(114.54 µs) = 45 µJ
% = (45/75)(100) = 60%
7–24 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.19 [a] t < 0:
t = 0+:
t > 0:
iR = −2e−t/τ A; τ =L
R=
57.5
= 666.67 ms ·. .1τ
= 1.5
iR = −2e−1.5t A
vR = (7.5)(−2e−1.5t) = −15e−1.5t V
Problems 7–25
v1 = 1.25[(−1.5)(−2e−1.5t)] = 3.75e−1.5t V,
vo = −v1 − vR = 11.25e−1.5t V t ≥ 0+
[b] io =16
∫ t
011.25e−1.5x dx + 0 = 1.25 − 1.25e−1.5t A t ≥ 0
P 7.20 [a] From the solution to Problem 7.19,
iR = −2e−1.5t A
pR = (−2e−1.5t)2(7.5) = 30e−3t W
wdiss =∫ ∞
030e−3t dt
= 30e−3t
−3
∣∣∣∣∞0
= 10 J
[b] wtrapped =12(10)(−1.25)2 +
12(6)(1.25)2 = 12.5 J
CHECK: w(0) = 12(1.25)(2)2 + 1
2(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;1τ
= 50
i =40
25,000e−50t = 1.6e−50t mA, t ≥ 0+
v1 =−1
10−6
∫ t
01.6 × 10−3e−50x dx + 40 = 32e−50t + 8 V, t ≥ 0
v2 =1
4 × 10−6
∫ t
01.6 × 10−3e−50x dx + 0 = −8e−50t + 8 V, t ≥ 0
7–26 CHAPTER 7. Response of First-Order RL and RC Circuits
[b] w(0) =12(10−6)(40)2 = 800 µJ
[c] wtrapped =12(10−6)(8)2 +
12(4 × 10−6)(8)2 = 160 µJ.
The energy dissipated by the 25 kΩ resistor is equal to the energy dissipated bythe two capacitors; it is easier to calculate the energy dissipated by thecapacitors (final voltage on the equivalent capacitor is zero):
wdiss =12(0.8 × 10−6)(40)2 = 640 µJ.
Check: wtrapped + wdiss = 160 + 640 = 800µJ; w(0) = 800µJ.
P 7.22 [a] Calculate the initial voltage drop across the capacitor:
v(0) = (2.7 k‖3.3 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 k‖6 k = 2 kΩ
τ = ReC = (2000)(0.5) × 10−6 = 1000µs;1τ
= 1000
v = v(0)e−t/τ = 59.4e−1000t V t ≥ 0
io =v
2.4 k + 3.6 k= 9.9e−1000t mA, t ≥ 0+
[b] w(0) =12(0.5 × 10−6)(59.4)2 = 882.09 µJ
i3k =59.4e−1000t
3000= 19.8e−1000t mA
p3k = [(19.8 × 10−3)e−1000t]2(3000) = 1.176e−2000t
w3k(500 µs) = 1.176e−2000x
−2000
∣∣∣∣500×10−6
0=
1.176−2000
(e−1 − 1) = 371.72 µJ
% =371.72882.09
× 100 = 42.14%
P 7.23 [a] R =v
i= 4 kΩ
[b]1τ
=1
RC= 25; C =
1(25)(4 × 103)
= 10 µF
[c] τ =125
= 40 ms
[d] w(0) =12(10 × 10−6)(48)2 = 11.52 mJ
Problems 7–27
[e] wdiss(60 ms) =∫ 0.06
0
v2
Rdt =
∫ 0.06
0
(48e−25t)2
(4 × 103)dt
= 0.576e−50t
−50
∣∣∣∣0.06
0= −5.74 × 10−4 + 0.01152 = 10.95 mJ
P 7.24 [a] t < 0:
i1(0−) = i2(0−) =3 V30 Ω
= 100 mA
[b] t > 0:
i1(0+) =0.22
= 100 mA
i2(0+) =−0.2
8= −25 mA
[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 resistivebranch. In this circuit
i2(0−) = 100 mA and i2(0+) = −25 mA
[e] vc = 0.2e−t/τ V, t ≥ 0 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
i1 =vc
2= 0.1e−312,500t A, t ≥ 0
[f] i2 =−vc
8= −25e−312,500t mA, t ≥ 0+
7–28 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.25 [a] t < 0:
Re = 12 k||68 k = 10.2 kΩ
vo(0) =10,200
10,200 + 1800(−120) = −102 V
t > 0:
τ = [(10/3) × 10−6)(12,000) = 40 ms;1τ
= 25
vo = −102e−25t V, t ≥ 0
p =v2
o
12,000= 867 × 10−3e−50t W
wdiss =∫ 12×10−3
0867 × 10−3e−50t dt
= 17.34 × 10−3(1 − e−50(12×10−3)) = 7.82 mJ
[b] w(0) =(1
2
)(103
)(102)2 × 10−6 = 17.34 mJ
0.75w(0) = 13 mJ∫ to
0867 × 10−3e−50x dx = 13 × 10−3
·. . 1 − e−50to = 0.75; e50to = 4; so to = 27.73 ms
Problems 7–29
P 7.26 [a]
vT = 20 × 103(iT + αv∆) + 5 × 103iT
v∆ = 5 × 103iT
vT = 25 × 103iT + 20 × 103α(5 × 103iT )
RTh = 25,000 + 100 × 106α
τ = RThC = 40 × 10−3 = RTh(0.8 × 10−6)
RTh = 50 kΩ = 25,000 + 100 × 106α
α =25,000
100 × 106 = 2.5 × 10−4 A/V
[b] vo(0) = (−5 × 10−3)(3600) = −18 V t < 0t > 0:
vo = −18e−25t V, t ≥ 0
v∆
5000+
v∆ − vo
20,000+ 2.5 × 10−4v∆ = 0
7–30 CHAPTER 7. Response of First-Order RL and RC Circuits
4v∆ + v∆ − vo + 5v∆ = 0
·. . v∆ =vo
10= −1.8e−25t V, t ≥ 0+
P 7.27 [a]
pds = (16.2e−25t)(−450 × 10−6e−25t) = −7290 × 10−6e−50t W
wds =∫ ∞
0pds dt = −145.8 µJ.
·. . dependent source is delivering 145.8 µJ
[b] w5k =∫ ∞
0(5000)(0.36 × 10−3e−25t)2 dt = 648 × 10−6
∫ ∞
0e−50t dt = 12.96 µJ
w20k =∫ ∞
0
(16.2e−25t)2
20,000dt = 13,122 × 10−6
∫ ∞
0e−50t dt = 262.44 µJ
wc(0) =12(0.8 × 10−6)(18)2 = 129.6 µJ
∑wdiss = 12.96 + 262.44 = 275.4 µJ
∑wdev = 145.8 + 129.6 = 275.4 µJ.
P 7.28 t < 0
Problems 7–31
t > 0
vT = −5io − 15io = −20io = 20iT ·. . RTh =vT
iT= 20 Ω
τ = RC = 40 µs;1τ
= 25,000
vo = 15e−25,000t V, t ≥ 0
io = − vo
20= −0.75e−25,000t A, t ≥ 0+
P 7.29 [a] The equivalent circuit for t > 0:
τ = 2 ms; 1/τ = 500
vo = 10e−500t V, t ≥ 0
io = e−500t mA, t ≥ 0+
i24kΩ = e−500t(16
40
)= 0.4e−500t mA, t ≥ 0+
p24kΩ = (0.16 × 10−6e−1000t)(24,000) = 3.84e−1000t mW
w24kΩ =∫ ∞
03.84 × 10−3e−1000t dt = −3.84 × 10−6(0 − 1) = 3.84 µJ
7–32 CHAPTER 7. Response of First-Order RL and RC Circuits
w(0) =12(0.25 × 10−6)(40)2 +
12(1 × 10−6)(50)2 = 1.45 mJ
% diss (24 kΩ) =3.84 × 10−6
1.45 × 10−3 × 100 = 0.26%
[b] p400Ω = 400(1 × 10−3e−500t)2 = 0.4 × 10−3e−1000t
w400Ω =∫ ∞
0p400 dt = 0.40 µJ
% diss (400Ω) =0.4 × 10−6
1.45 × 10−3 × 100 = 0.03%
i16kΩ = e−500t(24
40
)= 0.6e−500t mA, t ≥ 0+
p16kΩ = (0.6 × 10−3e−500t)2(16,000) = 5.76 × 10−3e−1000t W
w16kΩ =∫ ∞
05.76 × 10−3e−1000t dt = 5.76 µJ
% diss (16kΩ) = 0.4%
[c]∑
wdiss = 3.84 + 5.76 + 0.4 = 10 µJ
wtrapped = w(0) −∑wdiss = 1.45 × 10−3 − 10 × 10−6 = 1.44 mJ
% trapped =1.441.45
× 100 = 99.31%
Check: 0.26 + 0.03 + 0.4 + 99.31 = 100%
P 7.30 [a] Ce =(2 + 1)62 + 1 + 6
= 2 µF
vo(0) = −5 + 30 = 25 V
τ = (2 × 10−6)(250 × 103) = 0.5 s;1τ
= 2
vo = 25e−2t V, t > 0+
Problems 7–33
[b] wo =12(3 × 10−6)(30)2 +
12(6 × 10−6)(5)2 = 1425µJ
wdiss =12(2 × 10−6)(25)2 = 625 µJ
% diss =1425 − 625
1425× 100 = 56.14%
[c] io =vo
250 × 10−3 = 100e−2t µA
v1 = − 16 × 10−6
∫ t
0100 × 10−6e−2x dx − 5 = −16.67
∫ t
0e−2x dx − 5
= −16.67e−2x
−2
∣∣∣∣t0
−5 = 8.33e−2t − 13.33 V t ≥ 0
[d] v1 + v2 = vo
v2 = vo − v1 = 25e−2t − 8.33e−2t + 13.33 = 16.67e−2t + 13.33 V t ≥ 0
[e] wtrapped =12(6 × 10−6)(13.33)2 +
12(3 × 10−6)(13.33)2 = 800 µJ
wdiss + wtrapped = 625 + 800 = 1425µJ (check)
P 7.31 [a] At t = 0− the voltage on each capacitor will be 150 V(5 × 30), positive at theupper terminal. Hence at t ≥ 0+ we have
·. . isd(0+) = 5 +1500.2
+1500.5
= 1055 A
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−5×106t A, t ≥ 0+
i2(t) = 300e−20,000t A, t ≥ 0
·. . isd = 5 + 750e−5×106t + 300e−20,000t mA, t ≥ 0+
P 7.32 [a] t < 0:
io(0−) =6000
6000 + 4000(40 m) = 24 mA
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;1τ
= 5000
io(t) =24
1 × 103 e−t/τ = 24e−5000t mA, t ≥ 0+
[b]
Problems 7–35
vo =1
0.6 × 10−6
∫ t
024 × 10−3e−5000x dx + 72
= (40,000)e−5000x
−5000
∣∣∣∣t0
+72
= −8e−5000t + 8 + 72vo = [−8e−5000t + 80] V, t ≥ 0
[c] wtrapped = (1/2)(0.3 × 10−6)(80)2 + (1/2)(0.6 × 10−6)(80)2
wtrapped = 2880µJ.
Check:
wdiss =12(0.2 × 10−6)(24)2 = 57.6 µJ
w(0) =12(0.3 × 10−6)(96)2 +
12(0.6 × 10−6)(72)2 = 2937.6 µJ.
wtrapped + wdiss = w(0)
2880 + 57.6 = 2937.6 OK.
P 7.33 [a] t < 0
iL(0−) = −5 A
t > 0
iL(∞) =40 − 804 + 16
= −2 A
τ =L
R=
4 × 10−3
4 + 16= 200 µs;
1τ
= 5000
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, t ≥ 0+
[b] vL = LdiLdt
= 4 × 10−3(−5000)[−3e−5000t] = 60e−5000t V, t ≥ 0+
vL(0+) = 60 V
From part (a) vo(0+) = 0 V
Check: at t = 0+ the circuit is:
vL(0+) = 40 + (5 A)(4 Ω) = 60 V, vo(0+) = 80 − (16 Ω)(5 A) = 0 V
P 7.34 [a] t < 0
KVL equation at the top node:
50 =vo
8+
vo
40+
vo
10Multiply by 40 and solve:
2000 = (5 + 1 + 4)vo; vo = 200 V
·. . io(0−) =vo
10= 200/10 = 20 A
t > 0
Problems 7–37
Use voltage division to find the Thévenin voltage:
VTh = vo =40
40 + 120(800) = 200 V
Remove the voltage source and make series and parallel combinations ofresistors to find the equivalent resistance:
RTh = 10 + 120‖40 = 10 + 30 = 40 Ω
The simplified circuit is:
τ =L
R=
40 × 10−3
40= 1 ms;
1τ
= 1000
io(∞) =20040
= 5 A
·. . io = io(∞) + [io(0+) − io(∞)]e−t/τ
= 5 + (20 − 5)e−1000t = 5 + 15e−1000t A, t ≥ 0
[b] vo = 10io + Ldiodt
= 10(5 + 15e−1000t) + 0.04(−1000)(15e−1000t)
= 50 + 150e−1000t − 600e−1000t
vo = 50 − 450e−1000t V, t ≥ 0+
P 7.35 After making a Thévenin equivalent we have
For t < 0, the 15 Ω resistor is bypassed:
io(0−) = io(0+) = 50/5 = 10 A
7–38 CHAPTER 7. Response of First-Order RL and RC Circuits
τ =L
R=
16 × 10−3
5 + 15= 8 × 10−4;
1τ
= 1250
i(∞) =V
Req=
505 + 15
= 2.5 A
io = io(∞) + [io(0+) − io(∞)]e−t/τ = 2.5 + (10 − 2.5)e−1250t = 2.5 + 7.5e−1250t A, t ≥ 0
vo = Ldiodt
= 16 × 10−3(−1250)(7.5e−1250t) = −150e−1250t V, t ≥ 0+
P 7.36 [a] vo(0+) = −IgR2; τ =L
R1 + R2
vo(∞) = 0
vo(t) = −IgR2e−[(R1+R2)/L]t V, t ≥ 0+
[b] vo = −(10)(15)e− (5+15)0.016 t = −150e−1250t V, t ≥ 0+
[c] vo(0+) → ∞, and the duration of vo(t) → zero
[d] vsw = R2io; τ =L
R1 + R2
io(0+) = Ig; io(∞) = IgR1
R1 + R2
Therefore io(t) = IgR1R1+R2
+[Ig − IgR1
R1+R2
]e−[(R1+R2)/L]t
io(t) = R1Ig
(R1+R2) + R2Ig
(R1+R2)e−[(R1+R2)/L]t
Therefore vsw = R1Ig
(1+R1/R2) + R2Ig
(1+R1/R2)e−[(R1+R2)/L]t, t ≥ 0+
[e] |vsw(0+)| → ∞; duration → 0
P 7.37 Opening the inductive circuit causes a very large voltage to be induced across theinductor 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 damagesthe meter movement.
P 7.38 [a] From Eqs. (7.35) and (7.42)
i =Vs
R+(Io − Vs
R
)e−(R/L)t
v = (Vs − IoR)e−(R/L)t
·. .Vs
R= 4; Io − Vs
R= 4
Problems 7–39
Vs − IoR = −80;R
L= 40
·. . Io = 4 +Vs
R= 8 A
Now since Vs = 4R we have
4R − 8R = −80; R = 20 Ω
Vs = 80 V; L =R
40= 0.5 H
[b] i = 4 + 4e−40t; i2 = 16 + 32e−40t + 16e−80t
w =12Li2 =
12(0.5)[16 + 32e−40t + 16e−80t] = 4 + 8e−40t + 4e−80t
·. . 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; e40t = 2; t = 25 ln 2 = 17.33 ms
P 7.39 For t < 0
vx
15− 0.8vφ +
vx − 48021
= 0
vφ =vx − 480
21
vx
15− 0.8
(vx − 480
21
)+(
vx − 48021
)
=vx
15+ 0.2
(vs − 480
21
)= 21vx + 3(vx − 480) = 0
7–40 CHAPTER 7. Response of First-Order RL and RC Circuits
·. . 24vx = 1440 so vx = 60 V io(0−) =vx
15= 4 A
t > 0
Find Thévenin equivalent with respect to a, b
VTh − 3205
− 0.8(
VTh − 3205
)= 0 VTh = 320 V
vT = (iT + 0.8vφ)(5) =(iT + 0.8
vT
5
)(5)
Problems 7–41
vT = 5iT + 0.8vT·. . 0.2vT = 5iT
vT
iT= RTh = 25 Ω
io(∞) = 320/40 = 8 A
τ =80 × 10−3
40= 2 ms; 1/τ = 500
io = 8 + (4 − 8)e−500t = 8 − 4e−500t A, t ≥ 0
P 7.40 t > 0; calculate vo(0+)
va
15+
va − vo(0+)5
= 20 × 10−3
·. . va = 0.75vo(0+) + 75 × 10−3
15 × 10−3 +vo(0+) − va
5+
vo(0+)8
− 9i∆ + 50 × 10−3 = 0
13vo(0+) − 8va − 360i∆ = −2600 × 10−3
i∆ =vo(0+)
8− 9i∆ + 50 × 10−3
·. . i∆ =vo(0+)
80+ 5 × 10−3
·. . 360i∆ = 4.5vo(0+) + 1800 × 10−3
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
20+
vT
8− 9i∆
i∆ =vT
8− 9i∆ ·. . 10i∆ =
vT
8; i∆ =
vT
80
iT =vT
20+
10vT
80− 9vT
80
iTvT
=120
+180
=580
=116
S
·. . RTh = 16Ω
τ =4 × 10−3
16= 0.25 ms; 1/τ = 4000
·. . vo = 0 + (−80 − 0)e−4000t = −80e−4000t mV, t ≥ 0+
Problems 7–43
P 7.41 [a]
v
R+
1L
∫ t
0v dx =
Vs
R
1R
dv
dt+
v
L= 0
dv
dt+
R
Lv = 0
[b]dv
dt= −R
Lv
dv
dtdt = −R
Lv dt
·. .dv
v= −R
Ldt
∫ v(t)
v(0+)
dy
y= −R
L
∫ t
0+dx
ln y∣∣∣∣v(t)
v(0+)= −
(R
L
)t
ln[
v(t)v(0+)
]= −
(R
L
)t
v(t) = v(0+)e−(R/L)t; v(0+) =(
Vs
R− Io
)R = Vs − IoR
·. . v(t) = (Vs − IoR)e−(R/L)t
P 7.42 t > 0
τ =140
7–44 CHAPTER 7. Response of First-Order RL and RC Circuits
io = 5e−40t A, t ≥ 0
vo = 40io = 200e−40t V, t > 0+
200e−40t = 100; e40t = 2
·. . t =140
ln 2 = 17.33 ms
P 7.43 [a] wdiss =12Lei
2(0) =12(1)(5)2 = 12.5 J
[b] i3H =13
∫ t
0(200)e−40x dx − 5
= 1.67(1 − e−40t) − 5 = −1.67e−40t − 3.33 A
i1.5H =1
1.5
∫ t
0(200)e−40x dx + 0
= −3.33e−40t + 3.33 A
wtrapped =12(4.5)(3.33)2 = 25 J
[c] w(0) =12(3)(5)2 = 37.5 J
P 7.44 [a] t < 0
t > 0
iL(0−) = iL(0+) = 25 mA; τ =24 × 10−3
120= 0.2 ms;
1τ
= 5000
iL(∞) = −50 mA
iL = −50 + (25 + 50)e−5000t = −50 + 75e−5000t mA, t ≥ 0
vo = −120[75 × 10−3e−5000t] = −9e−5000t V, t ≥ 0+
Problems 7–45
[b] i1 =1
60 × 10−3
∫ t
0−9e−5000x dx + 10 × 10−3 = (30e−5000t − 20) mA, t ≥ 0
[c] i2 =1
40 × 10−3
∫ t
0−9e−5000x dx + 15 × 10−3 = (45e−5000t − 30) mA, t ≥ 0
P 7.45 [a] Let v be the voltage drop across the parallel branches, positive at the top node,then
−Ig +v
Rg
+1L1
∫ t
0v dx +
1L2
∫ t
0v dx = 0
v
Rg
+( 1
L1+
1L2
) ∫ t
0v dx = Ig
v
Rg
+1Le
∫ t
0v dx = Ig
1Rg
dv
dt+
v
Le
= 0
dv
dt+
Rg
Le
v = 0
Therefore v = IgRge−t/τ ; τ = Le/Rg
Thus
i1 =1L1
∫ t
0IgRge
−x/τ dx =IgRg
L1
e−x/τ
(−1/τ)
∣∣∣∣t0=
IgLe
L1(1 − e−t/τ )
i1 =IgL2
L1 + L2(1 − e−t/τ ) and i2 =
IgL1
L1 + L2(1 − e−t/τ )
[b] i1(∞) =L2
L1 + L2Ig; i2(∞) =
L1
L1 + L2Ig
P 7.46 For t < 0, i80mH(0) = 50 V/10 Ω = 5 AFor t > 0, after making a Thévenin equivalent we have
i =Vs
R+(Io − Vs
R
)e−t/τ
7–46 CHAPTER 7. Response of First-Order RL and RC Circuits
1τ
=R
L=
8100 × 10−3 = 80
Io = 5 A; If =Vs
R=
−808
= −10 A
i = −10 + (5 + 10)e−80t = −10 + 15e−80t A, t ≥ 0
vo = 0.08di
dt= 0.08(−1200e−80t) = −96e−80t V, t > 0+
P 7.47 For t < 0
Simplify the circuit:
80/10,000 = 8 mA, 10 kΩ‖40 kΩ‖24 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
Problems 7–47
Simplify the circuit:
8 mA + 2 mA = 10 mA
10 k‖40 k‖24 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;1τ
= 2000
vo = vo(∞) + [vo(0+) − vo(∞)]e−t/τ = −60 + [30 − (−60)]e−2000t
= −60 + 90e−2000t V t ≥ 0
P 7.48 [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
7–48 CHAPTER 7. Response of First-Order RL and RC Circuits
[b] ic = Cdvo
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
60 × 103 = 1 − 0.6e−100t mA, t ≥ 0+
[c] i1(t) = io + ic = 1 + 2.4e−100t mA t ≥ 0+
[d] i2(t) =v1
15 × 103 = 4 − 2.4e−100t mA t ≥ 0+
[e] i1(0+) = 1 + 2.4 = 3.4 mA
At t = 0+:
Re = 15 k‖60 k‖8 k = 4800 Ω
v1(0+) = (5 × 10−3)(4800) = 24 V
i1(0+) =v1(0+)60,000
+v1(0+)8000
= 0.4 m + 3 m = 3.4mA (checks)
P 7.49 [a] v = IsR + (Vo − IsR)e−t/RC i =(Is − Vo
R
)e−t/RC
·. . IsR = 40, Vo − IsR = −24
·. . Vo = 16 V
Is − Vo
R= 3 × 10−3; Is − 16
R= 3 × 10−3; R =
40Is
·. . Is − 0.4Is = 3 × 10−3; Is = 5 mA
R =405
× 103 = 8 kΩ
1RC
= 2500; C =1
2500R=
10−3
20 × 103 = 50 nF; τ = RC =1
2500= 400 µs
[b] v(∞) = 40 V
w(∞) =12(50 × 10−9)(1600) = 40µJ
0.81w(∞) = 32.4 µJ
v2(to) =32.4 × 10−6
25 × 10−9 = 1296; v(to) = 36 V
40 − 24e−2500to = 36; e2500to = 6; ·. . to = 716.70 µs
Problems 7–49
P 7.50 [a] For t > 0:
τ = RC = 250 × 103 × 8 × 10−9 = 2 ms;1τ
= 500
vo = 50e−500t V, t ≥ 0+
[b] io =vo
250,000=
50e−500t
250,000= 200e−500t µA
v1 =−1
40 × 10−9 × 200 × 10−6∫ t
0e−500x dx + 50 = 10e−500t + 40 V, t ≥ 0
P 7.51 [a] w =12Ceqv
2o =
12(8 × 10−9)(502) = 10 µJ
[b] wtrapped =12(40)2(50 × 10−9) = 40 µJ
[c] w(0) =12(40 × 10−9)(502) = 50 µJ
P 7.52 For t > 0
VTh = (−25)(16,000)ib = −400 × 103ib
ib =33,00080,000
(120 × 10−6) = 49.5 µA
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
7–50 CHAPTER 7. Response of First-Order RL and RC Circuits
vo = −19.8 + 19.8e−250t V, t ≥ 0
w(t) =12(0.25 × 10−6)v2
o = w(∞)(1 − e−250t)2 J
(1 − e−250t)2 =0.36w(∞)
w(∞)= 0.36
1 − e−250t = 0.6
e−250t = 0.4 ·. . t = 3.67 ms
P 7.53 [a]
io(0+) =−365000
= −7.2 mA
[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 × 10−3e−250t)] = −36 + 12.96e−250t V, t ≥ 0+
P 7.54 [a] vo(0−) = vo(0+) = 120 V
vo(∞) = −150 V; τ = 2 ms;1τ
= 500
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+
Problems 7–51
[c] vg = vo − 12.5 × 103io = −150 + 202.5e−500t V
[d] vg(0+) = −150 + 202.5 = 52.5 VChecks:
vg(0+) = io(0+)[37.5 × 103] − 150 = 202.5 − 150 = 52.5 V
i50k =vg
50k= −3 + 4.05e−500t mA
i150k =vg
150k= −1 + 1.35e−500t mA
-io + i50k + i150k + 4 = 0 (ok)
P 7.55 For t < 0, vo(0) = (−3 m)(15 k) = −45 Vt > 0:
VTh = −20 × 103i∆ +1050
(75) = −20 × 103( −75
50 × 103
)+ 15 = 45 V
vT = −20 × 103i∆ + 8 × 103iT = −20 × 103(0.2)iT + 8 × 103iT = 4 × 103iT
RTh =vT
iT= 4 kΩ
t > 0
vo = 45 + (−45 − 45)e−t/τ
7–52 CHAPTER 7. Response of First-Order RL and RC Circuits
τ = RC = (4000)( 1
16× 10−6
)= 250 µs;
1τ
= 4000
vo = 45 − 90e−4000t V, t ≥ 0
P 7.56 vo(0) = 45 V; vo(∞) = −45 V
RTh = 20 kΩ
τ = (20 × 103)( 1
16× 10−6
)= 1.25 × 10−3;
1τ
= 800
v = −45 + (45 + 45)e−800t = −45 + 90e−800t V, t ≥ 0
P 7.57 t < 0;
io(0−) =20100
(10 × 10−3) = 2 mA; vo(0−) = (2 × 10−3)(50,000) = 100 V
t = ∞:
io(∞) = −5 × 10−3( 20
100
)= −1 mA; vo(∞) = io(∞)(50,000) = −50 V
RTh = 50 kΩ‖50 kΩ = 25 kΩ; C = 16 nF
τ = (25,000)(16 × 10−9) = 0.4 ms;1τ
= 2500
·. . vo(t) = −50 + 150e−2500t V, t ≥ 0
ic = Cdvo
dt= −6e−2500t mA, t ≥ 0+
Problems 7–53
i50k =vo
50,000= −1 + 3e−2500t mA, t ≥ 0+
io = ic + i50k = −(1 + 3e−2500t) mA, t ≥ 0+
P 7.58 [a] Let i be the current in the clockwise direction around the circuit. Then
Vg = iRg +1C1
∫ t
0i dx +
1C2
∫ t
0i dx
= iRg +( 1
C1+
1C2
) ∫ t
0i dx = iRg +
1Ce
∫ t
0i dx
Now differentiate the equation
0 = Rgdi
dt+
i
Ce
ordi
dt+
1RgCe
i = 0
Therefore i =Vg
Rg
e−t/RgCe =Vg
Rg
e−t/τ ; τ = RgCe
v1(t) =1C1
∫ t
0
Vg
Rg
e−x/τ dx =Vg
RgC1
e−x/τ
−1/τ
∣∣∣∣t0
= − VgCe
C1(e−t/τ − 1)
v1(t) =VgC2
C1 + C2(1 − e−t/τ ); τ = RgCe
v2(t) =VgC1
C1 + C2(1 − e−t/τ ); τ = RgCe
[b] v1(∞) =C2
C1 + C2Vg; v2(∞) =
C1
C1 + C2Vg
P 7.59 [a]
IsR = Ri +1C
∫ t
0+i dx + Vo
0 = Rdi
dt+
i
C+ 0
·. .di
dt+
i
RC= 0
7–54 CHAPTER 7. Response of First-Order RL and RC Circuits
[b]di
dt= − i
RC;
di
i= − dt
RC∫ i(t)
i(0+)
dy
y= − 1
RC
∫ t
0+dx
lni(t)
i(0+)=
−t
RC
i(t) = i(0+)e−t/RC ; i(0+) =IsR − Vo
R=(Is − Vo
R
)
·. . i(t) =(Is − Vo
R
)e−t/RC
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; 1/τ = 1000
vo = 80 − 40e−1000t V, t ≥ 0
[b] io = −Cdvo
dt= −0.16 × 10−6[40,000e−1000t]
= −6.4e−1000t mA; t ≥ 0+
Problems 7–55
[c] v1 =−1
0.2 × 10−6
∫ t
0−6.4 × 10−3e−1000x dx + 32
= 64 − 32e−1000t V, t ≥ 0
[d] v2 =−1
0.8 × 10−6
∫ t
0−6.4 × 10−3e−1000x dx + 8
= 16 − 8e−1000t V, t ≥ 0
[e] wtrapped =12(0.2 × 10−6)(64)2 +
12(0.8 × 10−6)(16)2 = 512 µJ.
P 7.61 [a] vc(0+) = 50 V
[b] Use voltage division to find the final value of voltage:
vc(∞) =20
20 + 5(−30) = −24 V
[c] Find the Thévenin equivalent with respect to the terminals of the capacitor:
VTh = −24 V, RTh = 20‖5 = 4 Ω,
Therefore τ = ReqC = 4(25 × 10−9) = 0.1 µs
The simplified circuit for t > 0 is:
[d] i(0+) =−24 − 50
4= −18.5 A
[e] vc = vc(∞) + [vc(0+) − vc(∞)]e−t/τ
= −24 + [50 − (−24)]e−t/τ = −24 + 74e−107t V, t ≥ 0
[f] i = Cdvc
dt= (25 × 10−9)(−107)(74e−107t) = −18.5e−107t A, t ≥ 0+
P 7.62 [a] Use voltage division to find the initial value of the voltage:
vc(0+) = v9k =9 k
9 k + 3 k(120) = 90 V
[b] Use Ohm’s law to find the final value of voltage:
vc(∞) = v40k = −(1.5 × 10−3)(40 × 103) = −60 V
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 = RThC = 1 ms = 1000µs
[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 =ln(150/60)
1000= 916.3 µs
P 7.63 [a] For t < 0, calculate the Thévenin equivalent for the circuit to the left and rightof the 400-mH inductor. We get
i(0−) =−60 − 20015 k + 5 k
= −13 mA
i(0−) = i(0+) = −13 mA
[b] For t > 0, the circuit reduces to
Therefore i(∞) = −60/5,000 = −12 mA
[c] τ =L
R=
400 × 10−3
5000= 80 µs
[d] i(t) = i(∞) + [i(0+) − i(∞)]e−t/τ
= −12 + [−13 + 12]e−12,500t = −12 − e−12,500t mA, t ≥ 0
P 7.64 [a] From Example 7.10,
Leq =L1L2 − M2
L1 + L2 − 2M=
36 − 1620 − 8
=53
H
τ =Leq
R=
(5/3)(50/3)
=110
Problems 7–57
io =100
(50/3)− 100
(50/3)e−10t = 6 − 6e−10t A t ≥ 0
[b] vo = 100 − 503
io = 100 − 503
(6 − 6e−10t) = 100e−10t V, t ≥ 0+
[c] vo = 2di1dt
+ 4di2dt
io = i1 + i2
diodt
=di1dt
+di2dt
di2dt
=diodt
− di1dt
= 60e−10t − di1dt
·. . 100e−10t = 2di1dt
+ 4(
60e−10t − di1dt
)
·. .di1dt
= 70e−10t
di1 = 70e−10t dt
∫ i1
0dx = 70
∫ t
0e−10y dy
·. . i1 = 70e−10y
−10
∣∣∣∣t0= 7 − 7e−10t A, t ≥ 0
[d] i2 = io − i1
= 6 − 6e−10t − 7 + 7e−10t
= −1 + e−10t A, t ≥ 0
[e] vo = L2di2dt
+ Mdi1dt
= 18(−10e−10t) + 4(70e−10t)
= 100e−10t V, t ≥ 0+ (checks)Also,
vo = L1di1dt
+ Mdi2dt
= 2(70e−10t) + 4(−10e−10t)
= 100e−10t V, t ≥ 0+ CHECKS
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 ofWb-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 =(3)(15)3 + 15
= 2.5 H
τ =Leq
R=
2.57.5
=13
s
io(0) = 0; io(∞) =1207.5
= 16 A
·. . io = 16 − 16e−3t A, t ≥ 0
vo = 120 − 7.5io = 120e−3t V, t ≥ 0+
i1 =13
∫ t
0120e−3x dx =
403
− 403
e−3t A, t ≥ 0
i2 = io − i1 =83
− 83e−3t A, t ≥ 0
[b] io(0) = i1(0) = i2(0) = 0, consistent with initial conditions.vo(0+) = 120 V, consistent with io(0) = 0.
vo = 3di1dt
= 120e−3t V, t ≥ 0+
or
vo = 15di2dt
= 120e−3t V, t ≥ 0+
The voltage solution is consistent with the current solutions.
λ1 = 3i1 = 40 − 40e−3t Wb-turns
λ2 = 15i2 = 40 − 40e−3t Wb-turns
·. . λ1 = λ2 as it must, since
vo =dλ1
dt=
dλ2
dt
λ1(∞) = λ2(∞) = 40 Wb-turns
Problems 7–59
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 =L1L2 − M2
L1 + L2 + 2M=
50 − 2515 + 10
= 1 H
τ =L
R=
120
;1τ
= 20
·. . io(t) = 4 − 4e−20t A, t ≥ 0
[b] vo = 80 − 20io = 80 − 80 + 80e−20t = 80e−20t V, t ≥ 0+
[c] vo = 5di1dt
− 5di2dt
= 80e−20t V
io = i1 + i2
diodt
=di1dt
+di2dt
= 80e−20t A/s
·. .di2dt
= 80e−20t − di1dt
·. . 80e−20t = 5di1dt
− 400e−20t + 5di1dt
·. . 10di1dt
= 480e−20t; di1 = 48e−20t dt
∫ t1
0dx =
∫ t
048e−20y dy
i1 =48
−20e−20y
∣∣∣∣t0= 2.4 − 2.4e−20t A, 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 = Leqdiodt
= 1(80)e−20t = 80e−20t V, t ≥ 0+ (checks)
Also,
vo = 5di1dt
− 5di2dt
= 80e−20t V, t ≥ 0+ (checks)
7–60 CHAPTER 7. Response of First-Order RL and RC Circuits
vo = 10di2dt
− 5di1dt
= 80e−20t V, 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 offlux linkage. Hence, the answers make sense in terms of known circuitbehavior.
P 7.67 [a] Leq = 5 + 10 − 2.5(2) = 10 H
τ =L
R=
1040
=14;
1τ
= 4
i = 2 − 2e−4t A, t ≥ 0
[b] v1(t) = 5di1dt
− 2.5di
dt= 2.5
di
dt= 2.5(8e−4t) = 20e−4t V, t ≥ 0+
[c] v2(t) = 10di1dt
− 2.5di
dt= 7.5
di
dt= 7.5(8e−4t) = 60e−4t V, t ≥ 0+
[d] i(0) = 2 − 2 = 0, which agrees with initial conditions.
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, theanswers make sense in terms of known circuit behavior.
P 7.68 [a] Leq = 5 + 10 + 2.5(2) = 20 H
τ =L
R=
2040
=12;
1τ
= 2
i = 2 − 2e−2t A, t ≥ 0
[b] v1(t) = 5di1dt
+ 2.5di
dt= 7.5
di
dt= 7.5(4e−2t) = 30e−2t V, t ≥ 0+
[c] v2(t) = 10di1dt
+ 2.5di
dt= 12.5
di
dt= 12.5(4e−2t) = 50e−2t V, t ≥ 0+
[d] i(0) = 0, which agrees with initial conditions.
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, theanswers make sense in terms of known circuit behavior.
Problems 7–61
P 7.69 Use voltage division to find the initial voltage:
vo(0) =60
40 + 60(50) = 30 V
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;1τ
= 200
vo = vo(∞) + [vo(0+) − vo(∞)]e−t/τ
= −100 + (30 + 100)e−200t = −100 + 130e−200t V, t ≥ 0
P 7.70 [a] t < 0:
Using Ohm’s law,
ig =800
40 + 60‖40= 12.5 A
Using current division,
i(0−) =60
60 + 40(12.5) = 7.5 A = i(0+)
[b] 0 ≤ t ≤ 1 ms:
i = i(0+)e−t/τ = 7.5e−t/τ
1τ
=R
L=
40 + 120‖6080 × 10−3 = 1000
i = 7.5e−1000t
i(200µs) = 7.5e−103(200×10−6) = 7.5e−0.2 = 6.14 A
7–62 CHAPTER 7. Response of First-Order RL and RC Circuits
[c] i(1ms) = 7.5e−1 = 2.7591 A
1 ms ≤ t < ∞
1τ
=R
L=
4080 × 10−3 = 500
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 = Ldi
dt= (80 × 10−3)(−1000)(7.5e−1000t) = −600e−1000t V
v(1−ms) = −600e−1 = −220.73 V
[e] 1 ms ≤ t ≤ ∞:
i = 2.7591e−500(t−0.001)
v = Ldi
dt= (80 × 10−3)(−500)(2.591e−500(t−0.001))
= −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 µFcapacitor. Thus we will find vc first.t < 0
vc(0) =315
(−75) = −15 V
Problems 7–63
0 ≤ t ≤ 800 µs:
τ = ReC, Re =(6000)(3000)
9000= 2 kΩ
τ = (2 × 103)(0.5 × 10−6) = 1 ms,1τ
= 1000
vc = −15e−1000t V, t ≥ 0
vc(800 µs) = −15e−0.8 = −6.74 V
800 µs ≤ t ≤ 1.1 ms:
τ = (6 × 103)(0.5 × 10−6) = 3 ms,1τ
= 333.33
vc = −6.74e−333.33(t−800×10−6) V
1.1 ms ≤ t < ∞:
τ = 1 ms,1τ
= 1000
vc(1.1ms) = −6.74e−333.33(1100−800)10−6= −6.74e−0.1 = −6.1 V
7–64 CHAPTER 7. Response of First-Order RL and RC Circuits
vc = −6.1e−1000(t−1.1×10−3) V
vc(1.5ms) = −6.1e−1000(1.5−1.1)10−3= −6.1e−0.4 = −4.09 V
vo = (4/6)(−4.09) = −2.73 V
P 7.72 w(0) =12(0.5 × 10−6)(−15)2 = 56.25 µJ
0 ≤ t ≤ 800 µs:
vc = −15e−1000t; v2c = 225e−2000t
p3k = 75e−2000t mW
w3k =∫ 800×10−6
075 × 10−3e−2000t dt
= 75 × 10−3 e−2000t
−2000
∣∣∣∣800×10−6
0
= −37.5 × 10−6(e−1.6 − 1) = 29.93 µJ
1.1 ms ≤ t ≤ ∞:
vc = −6.1e−1000(t−1.1×10−3) V; v2c = 37.19e−2000(t−1.1×10−3)
p3k = 12.4e−2000(t−1.1×10−3) mW
w3k =∫ ∞
1.1×10−312.4 × 10−3e−2000(t−1.1×10−3) dt
= 12.4 × 10−3 e−2000(t−1.1×10−3)
−2000
∣∣∣∣∞1.1×10−3
= −6.2 × 10−6(0 − 1) = 6.2 µJ
w3k = 29.93 + 6.2 = 36.13 µJ
% =36.1356.25
(100) = 64.23%
Problems 7–65
P 7.73 For t < 0:
i(0) =1015
(15) = 10 A
0 ≤ t ≤ 10 ms:
i = 10e−100t A
i(10ms) = 10e−1 = 3.68 A
10 ms ≤ t ≤ 20 ms:
Req =(5)(20)
25= 4 Ω
1τ
=R
L=
450 × 10−3 = 80
i = 3.68e−80(t−0.01) A
20 ms ≤ t ≤ ∞:
i(20ms) = 3.68e−80(0.02−0.01) = 1.65 A
7–66 CHAPTER 7. Response of First-Order RL and RC Circuits
i = 1.65e−100(t−0.02) A
vo = Ldi
dt; L = 50 mH
di
dt= 1.65(−100)e−100(t−0.02) = −165e−100(t−0.02)
vo = (50 × 10−3)(−165)e−100(t−0.02)
= −8.26e−100(t−0.02) V, 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
w(0) =12(50 mH)(10 A)2 = 2.5 J
0.04w(0) = 0.1 J
·. .12(50 × 10−3)i2L = 0.1 so 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) =3.682
so t − 0.01 = 0.0076 ·. . t = 17.6 ms
P 7.75 0 ≤ t ≤ 10 µs:
τ = RC = (4 × 103)(20 × 10−9) = 80 µs; 1/τ = 12,500
Problems 7–67
vo(0) = 0 V; vo(∞) = −20 V
vo = −20 + 20e−12,500t V 0 ≤ t ≤ 10 µs
10 µs ≤ t ≤ ∞:
t = ∞:
i =−50 V20 kΩ
= −2.5 mA
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 kΩ‖16 kΩ = 3.2 kΩ
τ = (3200)(20 × 10−9) = 64 µs; 1/τ = 15,625
vo = −10 + 7.65e−15,625(t − 10×10−6) 10 µs ≤ t ≤ ∞
P 7.76 0 ≤ t ≤ 200 µs;
Re = 150‖100 = 60 kΩ; τ =(10
3× 10−9
)(60,000) = 200µs
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 = 30‖60 + 120‖40 = 20 + 30 = 50 kΩ
τ =(10
3× 10−9
)(50,000) = 166.67 µs;
1τ
= 6000
vc = 110.36e−6000(t − 200 µs) V
vc(300 µs) = 110.36e−6000(100 µs) = 60.57 V
io(300 µs) =60.5750,000
= 1.21 mA
i1 =6090
io =23io; i2 =
40160
io =14io
isw = i1 − i2 =23io − 1
4io =
512
io =512
(1.21 × 10−3) = 0.50 mA
P 7.77 t < 0:
vc(0−) = (20 × 10−3)(500) = 10 V = vc(0+)
Problems 7–69
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; vo = 10e−1000(t − 0.05) V
Summary:
vo = 10 V, 0 ≤ t ≤ 50 ms
vo = 10e−1000(t − 0.05) V, 50 ms ≤ t ≤ ∞
P 7.78 t < 0:
iL(0−) = 10 V/5 Ω = 2 A = iL(0+)
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 ≤ 5 s
5 ≤ t ≤ ∞:
τ =51
= 5 s; 1/τ = 0.2
iL(t) = 2e−0.2(t −5) A, t ≥ 5 s
P 7.79 [a] 0 ≤ t ≤ 2.5 ms
vo(0+) = 80 V; vo(∞) = 0
τ =L
R= 2 ms; 1/τ = 500
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)
20= 2.85 A
vo(2.5+ ms) = −20(2.85) = −57.08 V
vo(∞) = 0; τ = 2 ms; 1/τ = 500
vo = −57.08e−500(t − 0.0025) V 2.5+ ms ≤ t ≤ ∞
Problems 7–71
[b]
[c] vo(5 ms) = −16.35 V
io =+16.35
20= 817.68 mA
P 7.80 [a] io(0) = 0; io(∞) = 25 mA
1τ
=R
L=
2000250
× 103 = 8000
io = (25 − 25e−8000t) mA, 0 ≤ t ≤ 75 µs
vo = 0.25diodt
= 50e−8000t V, 0+ ≤ t ≤ 75− µs
75+ µs ≤ t ≤ ∞:
io(75µs) = 25 − 25e−0.6 = 11.28 mA; io(∞) = 0
io = 11.28e−8000(t−75×10−6) mA
vo = (0.25)diodt
= −22.56e−8000(t−75µs)
·. . 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
7–72 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.81 [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 = 00 ≤ t ≤ 4 ms:
τ = (200 × 103)(0.025 × 10−6) = 5 ms; 1/τ = 200
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
Problems 7–73
[b]
[c] t ≤ 0 : vo = 00 ≤ t ≤ 4 ms:
τ = (50 × 103)(0.025 × 10−6) = 1.25 ms 1/τ = 800
vo = 100 − 100e−800t V, 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 ≤ ∞
7–74 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.83 [a] τ = RC = (20,000)(0.2 × 10−6) = 4 ms; 1/τ = 250
io = vo = 0 t < 0
io(0+) = 20(16
20
)= 16 mA, io(∞) = 0
·. . io = 16e−250t mA 0+ ≤ t ≤ 2− ms
i16kΩ = 20 − 16e−250t mA
·. . vo = 320 − 256e−250t V 0+ ≤ t ≤ 2− ms
vc = vo − 4 × 103io = 320 − 320e−250t V 0 ≤ t ≤ 2 ms
vc(2 ms) = 320 − 320e−0.5 = 125.91 V
·. . io(2+ ms) = 16e−0.5 = 9.7 mA
io(∞) = 0
vc = 125.91e−250(t−0.002), 2+ ms ≤ t ≤ ∞
io = Cdvc
dt= (0.2 × 10−6)(−250)(125.91)e−250(t−0.002)
= −6.3e−250(t−0.002) mA, 2+ ms ≤ t ≤ ∞vo = 4000io + vc = 100.73e−250(t −0.002) V 2+ ms ≤ t ≤ ∞Summary part (a)
io = 0 t < 0
io = 16e−250t mA (0+ ≤ t ≤ 2− ms)
io = −6.3e−250(t −0.002) mA 2+ ms ≤ t ≤ ∞vo = 0 t < 0
vo = 320 − 256e−250t V, 0 ≤ t ≤ 2− ms
vo = 100.73e−250(t −0.002) V, 2+ ms ≤ t ≤ ∞[b] io(0−) = 0
io(0+) = 16 mA
io(2− ms) = 16e−0.5 = 9.7 mA
io(2+ ms) = −6.3 mA
Problems 7–75
[c] vo(0−) = 0
vo(0+) = 64 V
vo(2− ms) = 320 − 256e−0.5 = 164.73 V
vo(2+ ms) = 100.73
[d]
[e]
P 7.84 [a]
Using Ohm’s law,
vT = 5000iσ
Using current division,
iσ =20,000
20,000 + 5000(iT + βiσ) = 0.8iT + 0.8βiσ
7–76 CHAPTER 7. Response of First-Order RL and RC Circuits
Solve for iσ:
iσ(1 − 0.8β) = 0.8iT
iσ =0.8iT
1 − 0.8β; vT = 5000iσ =
4000iT(1 − 0.8β)
Find β such that RTh = −5 kΩ:
RTh =vT
iT=
40001 − 0.8β
= −5000
1 − 0.8β = −0.8 ·. . β = 2.25
[b] Find VTh;
Write a KCL equation at the top node:
VTh − 405000
+VTh
20,000− 2.25iσ = 0
The constraint equation is:
iσ =(VTh − 40)
5000= 0
Solving,
VTh = 50 V
Write a KVL equation around the loop:
50 = −5000i + 0.2di
dt
Rearranging:
di
dt= 250 + 25,000i = 25,000(i + 0.01)
Problems 7–77
Separate the variables and integrate to find i;
di
i + 0.01= 25,000 dt
∫ i
0
dx
x + 0.01=∫ t
025,000 dx
·. . i = −10 + 10e25,000t mA
di
dt= (10 × 10−3)(25,000)e25,000t = 250e25,000t
Solve for the arc time:
v = 0.2di
dt= 50e25,000t = 45,000; e25,000t = 900
·. . t =ln 90025,000
= 272.1 µs
P 7.85 Find the Thévenin equivalent with respect to the terminals of the capacitor.RTh calculation:
iT =vT
2000+
vT
5000− 4
vT
5000
·. .iTvT
=5 + 2 − 810,000
= − 110,000
vT
iT= −10,000
1= −10 kΩ
Open circuit voltage calculation:
7–78 CHAPTER 7. Response of First-Order RL and RC Circuits
The node voltage equations:
voc
2000+
voc − v1
1000− 4i∆ = 0
v1 − voc
1000+
v1
4000− 5 × 10−3 = 0
The constraint equation:
i∆ =v1
4000
Solving, voc = −80 V, v1 = −60 V
vc(0) = 0; vc(∞) = −80 V
τ = RC = (−10,000)(1.6 × 10−6) = −16 ms;1τ
= −62.5
vc = vc(∞) + [vc(0+) − vc(∞)]e−t/τ = −80 + 80e62.5t = 14,400
Solve for the time of the maximum voltage rating:
e62.5t = 181; 62.5t = ln 181; t = 83.18 ms
P 7.86
vT = 2000iT + 4000(iT − 2 × 10−3vφ) = 6000iT − 8vφ
= 6000iT − 8(2000iT )
Problems 7–79
vT
iT= −10,000
τ =10
−10,000= −1 ms; 1/τ = −1000
i = 25e1000t mA
·. . 25e1000t × 10−3 = 5; t =ln 2001000
= 5.3 ms
P 7.87 t > 0:
vT = 12 × 104i∆ + 16 × 103iT
i∆ = − 20100
iT = −0.2iT
·. . vT = −24 × 103iT + 16 × 103iT
RTh =vT
iT= −8 kΩ
τ = RC = (−8 × 103)(2.5 × 10−6) = −0.02 1/τ = −50
vc = 20e50t V; 20e50t = 20,000
50t = ln 1000 ·. . t = 138.16 ms
7–80 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.88 [a]
τ = (25)(2) × 10−3 = 50 ms; 1/τ = 20
vc(0+) = 80 V; vc(∞) = 0
vc = 80e−20t V
·. . 80e−20t = 5; e20t = 16; t =ln 1620
= 138.63 ms
[b] 0+ < t < 138.63 ms:
i = (2 × 10−6)(−1600e−20t) = −3.2e−20t mA
138.63+ ms < t ≤ ∞:
τ = (2)(4) × 10−3 = 8 ms; 1/τ = 125
vc(138.63+ ms) = 5 V; vc(∞) = 80 V
vc = 80 − 75e−125(t−0.13863) V, 138.63+ ms ≤ t ≤ ∞
i = 2 × 10−6(9375)e−125(t−0.13863)
= 18.75e−125(t−0.13863) mA, 138.63+ ms ≤ t ≤ ∞
[c] 80 − 75e−125∆t = 0.85(80) = 68
80 − 68 = 75e−125∆t = 12
e125∆t = 6.25; ∆t =ln 6.2512.5
∼= 14.66 ms
P 7.89 Use voltage division to find the voltage at the non-inverting terminal:
vp =80100
(−45) = −36 V = vn
Problems 7–81
Write a KCL equation at the inverting terminal:
−36 − 1480,000
+ 2.5 × 10−6 d
dt(−36 − vo) = 0
·. . 2.5 × 10−6dvo
dt=
−5080,000
Separate the variables and integrate:
dvo
dt= −250 ·. . dvo = −250dt
∫ vo(t)
vo(0)dx = −250
∫ t
0dy ·. . vo(t) − vo(0) = −250t
vo(0) = −36 + 56 = 20 V
vo(t) = −250t + 20
Find the time when the voltage reaches 0:
0 = −250t + 20 ·. . t =20250
= 80 ms
P 7.90 The equation for an integrating amplifier:
vo =1
RC
∫ t
0(vb − va) dy + vo(0)
Find the values and substitute them into the equation:
RC = (100 × 103)(0.05 × 10−6) = 5 ms
1RC
= 200; vb − va = −15 − (−7) = −8 V
vo(0) = −4 + 12 = 8 V
vo = 200∫ t
0−8 dx + 8 = (−1600t + 8) V, 0 ≤ t ≤ tsat
RC circuit analysis for v2:
v2(0+) = −4 V; v2(∞) = −15 V; τ = RC = (100 k)(0.05 µ) = 5 ms
7–82 CHAPTER 7. Response of First-Order RL and RC Circuits
v2 = v2(∞) + [v2(0+) − v2(∞)]e−t/τ
= −15 + (−4 + 15)e−200t = −15 + 11e−200t V, 0 ≤ t ≤ tsat
vf + v2 = vo·. . vf = vo − v2 = 23 − 1600t − 11e−200t V, 0 ≤ t ≤ tsat
Note that
−1600tsat + 8 = −20 ·. . tsat =−28
−1600= 17.5 ms
so the op amp operates in its linear region until it saturates at 17.5 ms.
P 7.91 vo = − 1R(0.5 × 10−6)
∫ t
04 dx + 0 =
−4tR(0.5 × 10−6)
−4(15 × 10−3)R(0.5 × 10−6)
= −10
·. . R =−4(15 × 10−3)
−10(0.5 × 10−6)= 12 kΩ
P 7.92 vo =−4t
R(0.5 × 10−6)+ 6 =
−4(40 × 10−3)R(0.5 × 10−6)
+ 6 = −10
·. . R =−4(40 × 10−3)
−16(0.5 × 10−6)= 20 kΩ
P 7.93 [a]Cdvp
dt+
vp − vb
R= 0; therefore
dvp
dt+
1RC
vp =vb
RC
vn − va
R+ C
d(vn − vo)dt
= 0;
thereforedvo
dt=
dvn
dt+
vn
RC− va
RC
But vn = vp
Thereforedvn
dt+
vn
RC=
dvp
dt+
vp
RC=
vb
RC
Thereforedvo
dt=
1RC
(vb − va); vo =1
RC
∫ t
0(vb − va) dy
[b] The output is the integral of the difference between vb and va and then scaled bya factor of 1/RC.
Problems 7–83
[c] vo =1
RC
∫ t
0(vb − va) dx
RC = (50 × 103)(10 × 10−9) = 0.5 ms
vb − va = −25 mV
vo =1
0.0005
∫ t
0−25 × 10−3dx = −50t
−50tsat = −6; tsat = 120 ms
P 7.94 [a] RC = (25 × 103)(0.4 × 10−6) = 10 ms;1
RC= 100
vo = 0, t < 0
[b] 0 ≤ t ≤ 250 ms :
vo = −100∫ t
0−0.20 dx = 20t V
[c] 250 ms ≤ t ≤ 500 ms;
vo(0.25) = 20(0.25) = 5 V
vo(t) = −100∫ t
0.250.20 dx + 5 = −20(t − 0.25) + 5 = −20t + 10 V
[d] 500 ms ≤ t ≤ ∞ :
vo(0.5) = −10 + 10 = 0 V
vo(t) = 0 V
7–84 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.95 [a] vo = 0, t < 0
RC = (25 × 103)(0.4 × 10−6) = 10 ms1
RC= 100
[b] RfCf = (5 × 106)(0.4 × 10−6) = 2;1
RfCf
= 0.5
vo =−5 × 106
25 × 103 (−0.2)[1 − e−0.5t] = 40(1 − e−0.5t) V, 0 ≤ t ≤ 250 ms
[c] vo(0.25) = 40(1 − e−0.125) ∼= 4.70 V
vo =−VmRf
Rs
+VmRf
Rs
(2 − e−0.125)e−0.5(t−0.25)
= −40 + 40(2 − e−0.125)e−0.5(t−0.25)
= −40 + 44.70e−0.5(t−0.25) V, 250 ms ≤ t ≤ 500 ms
[d] vo(0.5) = −40 + 44.70e−0.125 ∼= −0.55 V
vo = −0.55e−0.5(t−0.5) V, 500 ms ≤ t ≤ ∞
P 7.96 [a] RC = (1000)(800 × 10−12) = 800 × 10−9;1
RC= 1,250,000
0 ≤ t ≤ 1 µs:
vg = 2 × 106t
vo = −1.25 × 106∫ t
02 × 106x dx + 0
= −2.5 × 1012x2
2
∣∣∣∣t0= −125 × 1010t2 V, 0 ≤ t ≤ 1 µs
Problems 7–85
vo(1 µs) = −125 × 1010(1 × 10−6)2 = −1.25 V
1 µs ≤ t ≤ 3 µs:
vg = 4 − 2 × 106t
vo = −125 × 104∫ t
1×10−6(4 − 2 × 106x) dx − 1.25
= −125 × 104
[4x
∣∣∣∣t1×10−6
−2 × 106x2
2
∣∣∣∣t1×10−6
]− 1.25
= −5 × 106t + 5 + 125 × 1010t2 − 1.25 − 1.25= 125 × 1010t2 − 5 × 106t + 2.5 V, 1 µs ≤ t ≤ 3 µs
vo(3 µs) = 125 × 1010(3 × 10−6)2 − 5 × 106(3 × 10−6) + 2.5
= −1.25
3 µs ≤ t ≤ 4 µs:
vg = −8 + 2 × 106t
vo = −125 × 104∫ t
3×10−6(−8 + 2 × 106x) dx − 1.25
= −125 × 104
[−8x
∣∣∣∣t3×10−6
+2 × 106x2
2
∣∣∣∣t3×10−6
]− 1.25
= 107t − 30 − 125 × 1010t2 + 11.25 − 1.25= −125 × 1010t2 + 107t − 20 V, 3 µs ≤ t ≤ 4 µs
vo(4 µs) = −125 × 1010(4 × 10−6)2 + 107(4 × 10−6) − 20 = 0
[b]
[c] The output voltage will also repeat. This follows from the observation that att = 4 µs the output voltage is zero, hence there is no energy stored in thecapacitor. This means the circuit is in the same state at t = 4 µs as it was att = 0, thus as vg repeats itself, so will vo.
7–86 CHAPTER 7. Response of First-Order RL and RC Circuits
P 7.97 [a] While T2 has been ON, C2 is charged to VCC , positive on the left terminal. Atthe instant T1 turns ON the capacitor C2 is connected across b2 − e2, thusvbe2 = −VCC . This negative voltage snaps T2 OFF. Now the polarity of thevoltage on C2 starts to reverse, that is, the right-hand terminal of C2 starts tocharge toward +VCC . At the same time, C1 is charging toward VCC , positiveon the right. At the instant the charge on C2 reaches zero, vbe2 is zero, T2 turnsON. This makes vbe1 = −VCC and T1 snaps OFF. Now the capacitors C1 andC2 start to charge with the polarities to turn T1 ON and T2 OFF. This switchingaction repeats itself over and over as long as the circuit is energized. At theinstant T1 turns ON, the voltage controlling the state of T2 is governed by thefollowing circuit:
It follows that vbe2 = VCC − 2VCCe−t/R2C2 .
[b] While T2 is OFF and T1 is ON, the output voltage vce2 is the same as the voltageacross C1, thus
It follows that vce2 = VCC − VCCe−t/RLC1 .
[c] T2 will be OFF until vbe2 reaches zero. As soon as vbe2 is zero, ib2 will becomepositive and turn T2 ON. vbe2 = 0 when VCC − 2VCCe−t/R2C2 = 0, or whent = R2C2 ln 2.
[d] When t = R2C2 ln 2, we have
vce2 = VCC − VCCe−[(R2C2 ln 2)/(RLC1)] = VCC − VCCe−10 ln 2 ∼= VCC
[e] Before T1 turns ON, ib1 is zero. At the instant T1 turns ON, we have
Problems 7–87
ib1 =VCC
R1+
VCC
RLe−t/RLC1
[f] At the instant T2 turns back ON, t = R2C2 ln 2; therefore
ib1 =VCC
R1+
VCC
RLe−10 ln 2 ∼= VCC
R1
When T2 turns OFF, ib1 drops to zero instantaneously.
[g]
[h]
P 7.98 [a] tOFF2 = R2C2 ln 2 = 14.43 × 103(1 × 10−9) ln 2 ∼= 10 µs
[b] tON2 = R1C1 ln 2 ∼= 10 µs
[c] tOFF1 = R1C1 ln 2 ∼= 10 µs
[d] tON1 = R2C2 ln 2 ∼= 10 µs
[e] ib1 =10
1000+
1014,430
∼= 10.69 mA
[f] ib1 =10
14,430+
101000
e−10 ∼= 0.693 mA
7–88 CHAPTER 7. Response of First-Order RL and RC Circuits
[g] vce2 = 10 − 10e−10 ∼= 10 V
P 7.99 [a] tOFF2 = R2C2 ln 2 = (14.43 × 103)(0.8 × 10−9) ln 2 ∼= 8 µs
[b] tON2 = R1C1 ln 2 ∼= 10 µs
[c] tOFF1 = R1C1 ln 2 ∼= 10 µs
[d] tON1 = R2C2 ln 2 = 8 µs
[e] ib1 = 10.69 mA
[f] ib1 =10
14,430+
101000
e−8 ∼= 0.693 mA
[g] vce2 = 10 − 10e−8 ∼= 10 V
Note in this circuit T2 is OFF 8 µs and ON 10 µs of every cycle, whereas T1 isON 8 µs and OFF 10 µs every cycle.
P 7.100 If R1 = R2 = 50RL = 100 kΩ, then
C1 =48 × 10−6
100 × 103 ln 2= 692.49 pF; C2 =
36 × 10−6
100 × 103 ln 2= 519.37 pF
If R1 = R2 = 6RL = 12 kΩ, then
C1 =48 × 10−6
12 × 103 ln 2= 5.77 nF; C2 =
36 × 10−6
12 × 103 ln 2= 4.33 nF
Therefore 692.49 pF ≤ C1 ≤ 5.77 nF and 519.37 pF ≤ C2 ≤ 4.33 nF
P 7.101 [a] T2 is normally ON since its base current ib2 is greater than zero, i.e.,ib2 = VCC/R when T2 is ON. When T2 is ON, vce2 = 0, therefore ib1 = 0.When ib1 = 0, T1 is OFF. When T1 is OFF and T2 is ON, the capacitor C ischarged to VCC , positive at the left terminal. This is a stable state; there isnothing to disturb this condition if the circuit is left to itself.
[b] When S is closed momentarily, vbe2 is changed to −VCC and T2 snaps OFF. Theinstant T2 turns OFF, vce2 jumps to VCCR1/(R1 + RL) and ib1 jumps toVCC/(R1 + RL), which turns T1 ON.
[c] As soon as T1 turns ON, the charge on C starts to reverse polarity. Since vbe2 isthe same as the voltage across C, it starts to increase from −VCC toward+VCC . However, T2 turns ON as soon as vbe2 = 0. The equation for vbe2 isvbe2 = VCC − 2VCCe−t/RC . vbe2 = 0 when t = RC ln 2, therefore T2 staysOFF for RC ln 2 seconds.
P 7.102 [a] For t < 0, vce2 = 0. When the switch is momentarily closed, vce2 jumps to
vce2 =(
VCC
R1 + RL
)R1 =
6(5)25
= 1.2 V
Problems 7–89
T2 remains open for (23,083)(250) × 10−12 ln 2 ∼= 4 µs.
[b] ib2 =VCC
R= 259.93 µA, −5 ≤ t ≤ 0 µs
ib2 = 0, 0 < t < RC ln 2
ib2 =VCC
R+
VCC
RLe−(t−RC ln 2)/RLC
= 259.93 + 300e−0.2×106(t−4×10−6) µA, RC ln 2 < t
P 7.103 [a] We want the lamp to be in its nonconducting state for no more than 10 s, thevalue of to:
10 = R(10 × 10−6) ln1 − 64 − 6
and R = 1.091 MΩ
[b] When the lamp is conducting
VTh =20 × 103
20 × 103 + 1.091 × 106 (6) = 0.108 V
RTh = 20 k||1.091 M = 19,640 Ω
7–90 CHAPTER 7. Response of First-Order RL and RC Circuits
So,
(tc − to) = (19,640)(10 × 10−6) ln4 − 0.1081 − 0.108
= 0.289 s
The flash lasts for 0.289 s.
P 7.104 [a] At t = 0 we have
τ = (800)(25) × 10−3 = 20 sec; 1/τ = 0.05
vc(∞) = 40 V; vc(0) = 5 V
vc = 40 − 35e−0.05t V, 0 ≤ t ≤ to
40 − 35e−0.05to = 15; ·. . e0.05to = 1.4
to = 20 ln 1.4 s = 6.73 s
At t = to we have
The Thévenin equivalent with respect to the capacitor is
τ =(800
81
)(25) × 10−3 =
2081
s;1τ
=8120
= 4.05
vc(to) = 15 V; vc(∞) =4081
V
vc(t) =4081
+(15 − 40
81
)e−4.05(t−to) V =
4081
+117581
e−4.05(t−to)
·. .4081
+117581
e−4.05(t−to) = 5
Problems 7–91
117581
e−4.05(t−to) =36581
e4.05(t−to) =1175365
= 3.22
t − to =1
4.05ln 3.22 ∼= 0.29 s
One cycle = 7.02 seconds.
N = 60/7.02 = 8.55 flashes per minute
[b] At t = 0 we have
τ = 25R × 10−3; 1/τ = 40/R
vc = 40 − 35e−(40/R)t
40 − 35e−(40/R)to = 15
·. . to =R
40ln 1.4, R in kΩ
At t = to:
vTh =10
R + 10(40) =
400R + 10
; RTh =10R
R + 10kΩ
τ =(25)(10R) × 10−3
R + 10=
0.25RR + 10
;1τ
=4(R + 10)
R
vc =400
R + 10+(15 − 400
R + 10
)e− 4(R+10)
R(t−to)
·. .400
R + 10+[15R − 250
R + 10
]e− 4(R+10)
R(t−to) = 5
or(15R − 250
R + 10
)e− 4(R+10)
R(t−to) =
5R − 350(R + 10)
7–92 CHAPTER 7. Response of First-Order RL and RC Circuits
·. . e4(R+10)
R(t−to) =
3R − 50R − 70
·. . t − to =R
4(R + 10)ln(3R − 50
R − 70
)
At 12 flashes per minute to + (t − to) = 5 s
·. .R
40ln 1.4︸ ︷︷ ︸+
R
4(R + 10)ln(3R − 50
R − 70
)= 5
dominantterm
Start the trial-and-error procedure by setting (R/40) ln 1.4 = 5, thenR = 200/(ln 1.4) or 594.40 kΩ. If R = 594.40 kΩ then t − to ∼= 0.29 s.Second trial set (R/40) ln 1.4 = 4.7 s or R = 558.74 kΩ.
With R = 558.74 kΩ, t − to ∼= 0.30 s
The procedure converges to R = 559.3 kΩ
P 7.105 [a] to = RC ln(
Vmin − Vs
Vmax − Vs
)= (3700)(250 × 10−6) ln
(−700−100
)
= 1.80 s
tc − to =RCRL
R + RLln(
Vmax − VTh
Vmin − VTh
)
RL
R + RL=
1.31.3 + 3.7
= 0.26 RC = (3700)(250 × 10−6) = 0.925 s
VTh =1000(1.3)1.3 + 3.7
= 260 V RTh = 3.7 k||1.3 k = 962 Ω
·. . tc − to = (0.925)(0.26) ln(640/40) = 0.67 s
·. . tc = 1.8 + 0.67 = 2.47 s
flashes/min =60
2.47= 24.32
[b] 0 ≤ t ≤ to:
vL = 1000 − 700e−t/τ1
τ1 = RC = 0.925 s
to ≤ t ≤ tc:
vL = 260 + 640e−(t−to)/τ2
τ2 = RThC = 962(250) × 10−6 = 0.2405 s
Problems 7–93
0 ≤ t ≤ to : i =1000 − vL
3700=
737
e−t/0.925 A
to ≤ t ≤ tc : i =1000 − vL
3700=
74370
− 64370
e−(t−to)/0.2405
Graphically, i versus t is
The average value of i will equal the areas (A1 + A2) divided by tc.
·. . iavg =A1 + A2
tc
A1 =737
∫ to
0e−t/0.925 dt
=6.47537
(1 − e− ln 7) = 0.15 A–s
A2 =∫ tc
to
74 − 64e−(t−to)/0.2405
370dt
=74370
(tc − to) +15.392370
(e− ln 16 − 1)
=17.797370
ln 16 − 15.392370
(1 − e− ln 16)
= 0.09436 A–s
iavg =(0.15 + 0.09436)
0.925 ln 7 + 0.2405 ln 16(1000) = 99.06 mA
[c] Pavg = (1000)(99.06 × 10−3) = 99.06 W
No. of kw hrs/yr =(99.06)(24)(365)
1000= 867.77
Cost/year = (867.77)(0.05) = 43.39 dollars/year
P 7.106 [a] Replace the circuit attached to the capacitor with its Thévenin equivalent, wherethe equivalent resistance is the parallel combination of the two resistors, andthe open-circuit voltage is obtained by voltage division across the lampresistance. The resulting circuit is
7–94 CHAPTER 7. Response of First-Order RL and RC Circuits
RTh = R‖RL =RRL
R + RL; VTh =
RL
R + RLVs
From this circuit,
vC(∞) = VTh; vC(0) = Vmax; τ = RThC
Thus,
vC(t) = VTh + (Vmax − VTh)e−(t−to)/τ
where
τ =RRLC
R + RL
[b] Now, set vC(tc) = Vmin and solve for (tc − to):
VTh + (Vmax − VTh)e−(tc−to)/τ = Vmin
e−(tc−to)/τ =Vmin − VTh
Vmax − VTh
−(tc − to)τ
= lnVmin − VTh
Vmax − VTh
(tc − to) = − RRLC
R + RLln
Vmin − VTh
Vmax − VTh
(tc − to) =RRLC
R + RLln
Vmax − VTh
Vmin − VTh
P 7.107 [a] 0 ≤ t ≤ 0.5:
i =2160
+(30
60− 21
60
)e−t/τ where τ = L/R.
i = 0.35 + 0.15e−60t/L
i(0.5) = 0.35 + 0.15e−30/L = 0.40
·. . e30/L = 3; L =30ln 3
= 27.31 H
Problems 7–95
[b] 0 ≤ t ≤ tr, where tr is the time the relay releases:
i = 0 +(30
60− 0
)e−60t/L = 0.5e−60t/L
·. . 0.4 = 0.5e−60tr/L; e60tr/L = 1.25
tr =27.31 ln 1.25
60∼= 0.10 s
8Natural and Step Responses of RLC
Circuits
Assessment Problems
AP 8.1 [a]1
(2RC)2 =1
LC, therefore C = 500 nF
[b] α = 5000 =1
2RC, therefore C = 1 µF
s1,2 = −5000 ±√
25 × 106 − (103)(106)20
= (−5000 ± j5000) rad/s
[c]1√LC
= 20,000, therefore C = 125 nF
s1,2 =[−40 ±
√(40)2 − 202
]103,
s1 = −5.36 krad/s, s2 = −74.64 krad/s
AP 8.2 iL =1
50 × 10−3
∫ t
0[−14e−5000x + 26e−20,000x] dx + 30 × 10−3
= 20−14e−5000x
−5000
∣∣∣∣t0+
26e−20,000t
−20,000
∣∣∣∣t0
+ 30 × 10−3
= 56 × 10−3(e−5000t − 1) − 26 × 10−3(e−20,000t − 1) + 30 × 10−3
= [56e−5000t − 56 − 26e−20,000t + 26 + 30] mA
= 56e−5000t − 26e−20,000t mA, t ≥ 0
AP 8.3 From the given values of R, L, and C, s1 = −10 krad/s and s2 = −40 krad/s.
[a] v(0−) = v(0+) = 0, therefore iR(0+) = 0
8–1
8–2 CHAPTER 8. Natural and Step Responses of RLC Circuits
[b] iC(0+) = −(iL(0+) + iR(0+)) = −(−4 + 0) = 4 A
[c] Cdvc(0+)
dt= ic(0+) = 4, therefore
dvc(0+)dt
=4C
= 4 × 108 V/s
[d] v = [A1e−10,000t + A2e
−40,000t] V, t ≥ 0+
v(0+) = A1 + A2,dv(0+)
dt= −10,000A1 − 40,000A2
Therefore A1 + A2 = 0, −A1 − 4A2 = 40,000; A1 = 40,000/3 V
[e] A2 = −40,000/3 V
[f] v = [40,000/3][e−10,000t − e−40,000t] V, t ≥ 0
AP 8.4 [a]1
2RC= 8000, therefore R = 62.5 Ω
[b] iR(0+) =10 V
62.5 Ω= 160 mA
iC(0+) = −(iL(0+) + iR(0+)) = −80 − 160 = −240 mA = Cdv(0+)
dt
Thereforedv(0+)
dt=
−240 mC
= −240 kV/s
[c] B1 = v(0+) = 10 V,dvc(0+)
dt= ωdB2 − αB1
Therefore 6000B2 − 8000B1 = −240,000, B2 = (−80/3) V
[d] iL = −(iR + iC); iR = v/R; iC = Cdv
dt
v = e−8000t[10 cos 6000t − 803
sin 6000t] V
Therefore iR = e−8000t[160 cos 6000t − 12803
sin 6000t] mA
iC = e−8000t[−240 cos 6000t +4603
sin 6000t] mA
iL = 10e−8000t[8 cos 6000t +823
sin 6000t] mA, t ≥ 0
AP 8.5 [a]( 1
2RC
)2
=1
LC=
106
4, therefore
12RC
= 500, R = 100 Ω
[b] 0.5CV 20 = 12.5 × 10−3, therefore V0 = 50 V
[c] 0.5LI20 = 12.5 × 10−3, I0 = 250 mA
Problems 8–3
[d] D2 = v(0+) = 50,dv(0+)
dt= D1 − αD2
iR(0+) =50100
= 500 mA
Therefore iC(0+) = −(500 + 250) = −750 mA
Thereforedv(0+)
dt= −750 × 10−3
C= −75,000 V/s
Therefore D1 − αD2 = −75,000; α =1
2RC= 500, D1 = −50,000 V/s
[e] v = [50e−500t − 50,000te−500t] V
iR =v
R= [0.5e−500t − 500te−500t] A, t ≥ 0+
AP 8.6 [a] iR(0+) =V0
R=
40500
= 0.08 A
[b] iC(0+) = I − iR(0+) − iL(0+) = −1 − 0.08 − 0.5 = −1.58 A
[c]diL(0+)
dt=
Vo
L=
400.64
= 62.5 A/s
[d] α =1
2RC= 1000;
1LC
= 1,562,500; s1,2 = −1000 ± j750 rad/s
[e] iL = if + B′1e
−αt cos ωdt + B′2e
−αt sin ωdt, if = I = −1 A
iL(0+) = 0.5 = if + B′1, therefore B′
1 = 1.5 A
diL(0+)dt
= 62.5 = −αB′1 + ωdB
′2, therefore B′
2 = (25/12) A
Therefore iL(t) = −1 + e−1000t[1.5 cos 750t + (25/12) sin 750t] A, t ≥ 0
[f] v(t) =ŁdiLdt
= 40e−1000t[cos 750t − (154/3) sin 750t]V t ≥ 0
AP 8.7 [a] i(0+) = 0, since there is no source connected to L for t < 0.
[b] vc(0+) = vC(0−) =(
15 k15 k + 9 k
)(80) = 50 V
[c] 50 + 80i(0+) + Ldi(0+)
dt= 100,
di(0+)dt
= 10,000 A/s
[d] α = 8000;1
LC= 100 × 106; s1,2 = −8000 ± j6000 rad/s
[e] i = if + e−αt[B′1 cos ωdt + B′
2 sin ωdt]; if = 0, i(0+) = 0
Therefore B′1 = 0;
di(0+)dt
= 10,000 = −αB′1 + ωdB
′2
Therefore B′2 = 1.67 A; i = 1.67e−8000t sin 6000t A, t ≥ 0
8–4 CHAPTER 8. Natural and Step Responses of RLC Circuits
AP 8.8 vc(t) = vf + e−αt[B′1 cos ωdt + B′
2 sin ωdt], vf = 100 V
vc(0+) = 50 V;dvc(0+)
dt= 0; therefore 50 = 100 + B′
1
B′1 = −50 V; 0 = −αB′
1 + ωdB′2
Therefore B′2 =
α
ωd
B′1 =
(80006000
)(−50) = −66.67 V
Therefore vc(t) = 100 − e−8000t[50 cos 6000t + 66.67 sin 6000t] V, t ≥ 0
Problems
P 8.1 [a] α =1
2RC=
12(1000)(2 × 10−6)
= 250
ω2o =
1LC
=1
(12.5)(2 × 10−6)= 40,000
s1,2 = −250 ±√
2502 − 40,000 = −250 ± 150
s1 = −100 rad/s
s2 = −400 rad/s
[b] overdamped
[c] Note — we want ωd = 120 rad/s:
ωd =√
ω2o − α2
·. . α2 = ω2o − ω2
d = 40,000 − (120)2 = 25,600
α = 160
12RC
= 160; ·. . R =1
2(160)(2 × 10−6)= 1562.5 Ω
[d] s1, s2 = −160 ±√
1602 − 40,000 = −160 ± j120 rad/s
[e] α =√
40,000 =1
2RC; ·. . R =
12(200)(2 × 10−6)
= 1250 Ω
Problems 8–5
P 8.2 [a] −α +√
α2 − ω2o = −250
−α −√
α2 − ω2o = −1000
Adding the above equations, − 2α = −1250
α = 625 rad/s
12RC
=1
2R(0.1 × 10−6)= 625
R = 8 kΩ
2√
α2 − ω2o = 750
4(α2 − ω2o) = 562,500
·. . ωo = 500 rad/s
ω2o = 25 × 104 =
1LC
·. . L =1
(25 × 104)(0.1 × 10−6)= 40 H
[b] iR =v(t)R
= −1e−250t + 4e−1000t mA, t ≥ 0+
iC = Cdv(t)dt
= 0.2e−250t − 3.2e−1000t mA, t ≥ 0+
iL = −(iR + iC) = 0.8e−250t − 0.8e−1000t mA, t ≥ 0
P 8.3 [a] iR(0) =15200
= 75mA
iL(0) = −45mA
iC(0) = −iL(0) − iR(0) = 45 − 75 = −30 mA
[b] α =1
2RC=
12(200)(0.2 × 10−6)
= 12,500
ω2o =
1LC
=1
(50 × 10−3)(0.2 × 10−6)= 108
s1,2 = −12,500 ±√
1.5625 × 108 − 108 = −12,500 ± 7500
s1 = −5000 rad/s; s2 = −20,000 rad/s
v = A1e−5000t + A2e
−20,000t
v(0) = A1 + A2 = 15
8–6 CHAPTER 8. Natural and Step Responses of RLC Circuits
dv
dt(0) = −5000A1 − 20,000A2 =
−30 × 10−3
0.2 × 10−6 = −15 × 104V/s
Solving, A1 = 10; A2 = 5
v = 10e−5000t + 5e−20,000t V, t ≥ 0
[c] iC = Cdv
dt
= 0.2 × 10−6[−50,000e−5000t − 100,000e−20,000t]
= −10e−5000t − 20e−20,000t mA
iR = 50e−5000t + 25e−20,000t mA
iL = −iC − iR = −40e−5000t − 5e−20,000t mA, t ≥ 0
P 8.41
2RC=
12(312.5)(0.2 × 10−6)
= 8000
1LC
=1
(50 × 10−3)(0.2 × 10−6)= 108
s1,2 = −8000 ±√
80002 − 108 = −8000 ± j6000 rad/s
·. . response is underdamped
v(t) = B1e−8000t cos 6000t + B2e
−8000t sin 6000t
v(0+) = 15 V = B1; iR(0+) =15
312.5= 48 mA
iC(0+) = [−iL(0+) + iR(0+)] = −[−45 + 48] = −3 mA
dv(0+)dt
=−3 × 10−3
0.2 × 10−6 = −15,000 V/s
dv(0)dt
= −8000B1 + 6000B2 = −15,000
6000B2 = 8000(15) − 15,000; ·. . B2 = 17.5 V
v(t) = 15e−8000t cos 6000t + 17.5e−8000t sin 6000t V, t ≥ 0
Problems 8–7
P 8.5 α =1
2RC=
12(250)(0.2 × 10−6)
= 104
α2 = 108; ·. . α2 = ω2o
Critical damping:
v = D1te−αt + D2e
−αt
iR(0+) =15250
= 60 mA
iC(0+) = −[iL(0+) + iR(0+)] = −[−45 + 60] = −15 mA
v(0) = D2 = 15
dv
dt= D1[t(−αe−αt) + e−αt] − αD2e
−αt
dv
dt(0) = D1 − αD2 =
iC(0)C
=−15 × 10−3
0.2 × 10−6 = −75,000
D1 = αD2 − 75,000 = (104)(15) − 75,000 = 75,000
v = (75,000t + 15)e−10,000t V, t ≥ 0
P 8.6 α = 1000/2 = 500
R =1
2αC=
12(500)(2.5 × 10−6)
= 400 Ω
v(0+) = 3(1 + 1) = 6 V
iR(0+) =6
400= 15 mA
dv
dt= −300e−100t − 2700e−900t
dv(0+)dt
= −300 − 2700 = −3000 V/s
iC(0+) = 2.5 × 10−6(−3000) = −7.5 mA
iL(0+) = −[iR(0+) + iC(0+)] = −[15 − 7.5] = −7.5 mA
8–8 CHAPTER 8. Natural and Step Responses of RLC Circuits
P 8.7 [a] α = 20,000; ωd = 15,000
ωd =√
ω2o − α2
·. . ω2o = ω2
d + α2 = 225 × 106 + 400 × 106 = 625 × 106
1LC
= 625 × 106
L =1
(625 × 106)(40 × 10−9)= 40 mH
[b] α =1
2RC
·. . R =1
2αC=
12(20,000)(40 × 10−9)
= 625 Ω
[c] Vo = v(0) = 100 V
[d] Io = iL(0) = −iR(0) − iC(0)
iR(0) =Vo
R=
100625
= 160 mA
iC(0) = Cdv
dt(0)
dv
dt= 100e−20,000t[−15,000 sin 15,000t − 30,000 cos 15,000t]−
20,000e−20,000t[cos 15,000t − 2 sin 15,000t]
dv
dt(0) = 1001(−30,000) − 20,000 = −500 × 104
Cdv
dt(0) = −500 × 104(40 × 10−9) = −200 mA
·. . Io = 200 − 160 = 40 mA
[e]dv
dt= 100e−20,000t[25,000 sin 15,000t − 50,000 cos 15,000t]
= 25 × 105e−20,000t[sin 15,000t − 2 cos 15,000t]
Cdv
dt= 0.1e−20,000t(sin 15,000t − 2 cos 15,000t)
iC(t) = 0.1e−20,000t(sin 15,000t − 2 cos 15,000t) A
iR(t) = 0.16e−20,000t(cos 15,000t − 2 sin 15,000t) A
iL(t) = −iR(t) − iC(t)
= e−20,000t(40 cos 15,000t + 220 sin 15,000t) mA, t ≥ 0
Problems 8–9
P 8.8 [a] 2α = 1000; α = 500 rad/s
2√
α2 − ω2o = 600; ωo = 400 rad/s
C =1
2αR=
12(500)(250)
= 4 µF
L =1
ω2oC
=1
(400)2(4 × 10−6)= 1.5625 H
iC(0+) = A1 + A2 = 45 mA
diCdt
+diLdt
+diRdt
= 0
diC(0)dt
= −diL(0)dt
− diR(0)dt
diL(0)dt
=15
1.5625= 9.6 A/s
diR(0)dt
=1R
dv(0)dt
=1R
iC(0)C
=45 × 10−3
(250)(4 × 10−6)= 45 A/s
·. .diC(0)
dt= −9.6 − 45 = −54.6 A/s
·. . 200A1 + 800A2 = 54.6 A1 + A2 = 0.045
Solving, A1 = −31 mA; A2 = 76 mA
·. . iC = −31e−200t + 76e−800t mA, t ≥ 0+
[b] By hypothesis
v = A3e−200t + A4e
−800t, t ≥ 0
v(0) = A3 + A4 = 15
dv(0)dt
=45 × 10−3
4 × 10−6 = 11,250 V/s
−200A3 − 800A4 = 11,250; ·. . A3 = 38.75 V; A4 = −23.75 V
v = 38.75e−200t − 23.75e−800t V, t ≥ 0
[c] iR(t) =v
250= 155e−200t − 95e−800t mA, t ≥ 0+
[d] iL = −iR − iC
iL = −124e−200t + 19e−800t mA, t ≥ 0
8–10 CHAPTER 8. Natural and Step Responses of RLC Circuits
P 8.9 [a]( 1
2RC
)2
=1
LC= (500)2
·. . C =1
(500)2(4)= 1 µF
12RC
= 500
·. . R =1
2(500)(10−6)= 1 kΩ
v(0) = D2 = 8 V
iR(0) =8
1000= 8mA
iC(0) = −8 + 10 = 2 mA
dv
dt(0) = D1 − 500D2 =
2 × 10−3
10−6 = 2000 V/s
·. . D1 = 2000 + 500(8) = 6000 V/s
[b] v = 6000te−500t + 8e−500t V, t ≥ 0
dv
dt= [−3 × 106t + 2000]e−500t
iC = Cdv
dt= (−3000t + 2)e−500t mA, t ≥ 0+
P 8.10 [a] α =1
2RC= 0.5 rad/s
ω2o =
1LC
= 25.25
ωd =√
25.25 − (0.5)2 = 5 rad/s
·. . v = B1e−t/2 cos 5t + B2e
−t/2 sin 5t
v(0) = B1 = 0; v = B2e−t/2 sin 5t
iR(0+) = 0 A; iC(0+) = 4 A;dv
dt(0+) =
40.08
= 50 V/s
50 = −αB1 + ωdB2 = −0.5(0) + 5B2
·. . B2 = 10
·. . v = 10e−t/2 sin 5t V, t ≥ 0
Problems 8–11
[b]dv
dt= −5e−t/2 sin 5t + 10e−t/2(5 cos 5t)
dv
dt= 0 when 10 cos 5t = sin 5t or tan 5t = 10
·. . 5t1 = 1.47, t1 = 294.23 ms
5t2 = 1.47 + π, t2 = 922.54 ms
5t3 = 1.47 + 2π, t3 = 1550.86 ms
[c] t3 − t1 = 1256.6 ms; Td =2πωd
=2π5
= 1256.6 ms
[d] t2 − t1 = 628.3 ms;Td
2=
1256.62
= 628.3 ms
[e] v(t1) = 10e−(0.147115) sin 5(0.29423) = 8.59 V
v(t2) = 10e−(0.46127) sin 5(0.92254) = −6.27 V
v(t3) = 10e−(0.77543) sin 5(1.55086) = 4.58 V
[f]
P 8.11 [a] α = 0; ωd = ωo =√
25.25 = 5.02 rad/s
v = B1 cos ωot + B2 sin ωot; v(0) = B1 = 0; v = B2 sin ωot
Cdv
dt(0) = −iL(0) = 4
50 = −αB1 + ωdB2 = −0 +√
25.25B2
·. . B2 = 50/√
25.25 = 9.95 V
v = 9.95 sin 5.02t V, t ≥ 0
[b] 2πf = 5.02; f =5.022π
∼= 0.80 Hz
8–12 CHAPTER 8. Natural and Step Responses of RLC Circuits
[c] 9.95 V
P 8.12 [a] ω2o =
1LC
=1
(12.5)(3.2 × 10−9)= 25 × 106
ωo = 5000 rad/s
12RC
= 5000; R =1
2(5000)(3.2 × 10−9)= 31.25 kΩ
[b] v(t) = D1te−5000t + D2e
−5000t
v(0) = 100 V = D2
dv
dt= (D1t + 100)(−5000e−5000t) + D1e
−5000t
dv
dt(0) = −500 × 103 + D1 =
iC(0)C
iC(0) = −iR(0) − iL(0)
iR(0) =100
31,500= 3.2 mA
·. . iC(0) = −(3.2 + 6.4) = −9.6 mA
·. .dv
dt(0) = −9.6 × 10−3
3.2 × 10−9 = −3 × 106
·. . −500 × 103 + D1 = −3 × 106
D1 = −25 × 105V/s
·. . v(t) = (−25 × 105t + 100)e−5000t V, t ≥ 0
[c] iC(t) = 0 whendv
dt(t) = 0
dv
dt= (−25 × 105t + 100)(−5000)e−5000t + e−5000t(−25 × 105)
= (125 × 108t − 30 × 105)e−5000t
dv
dt= 0 when 125 × 108t1 = 3 × 106; ·. . t1 = 240 µs
v(240µs) = e−1.2[(−25 × 105)(240 × 10−6) + 100] = −150.6 V
Problems 8–13
[d] iL(240µs) = −iR(240µs) =−150.631,250
= −4.82 mA
ωC(240µs) =12(3.2 × 10−9)(−150.6)2 = 36.29 µJ
ωL(240µs) =12(12.5)(−4.82 × 10−3)2 = 145.2 µJ
ω(240µs) = ωC + ωL = 181.49 µJ
ω(0) =12(12.5)(6.4 × 10−3)2 +
12(3.2 × 10−9)(100)2 = 272 µJ
% remaining =181.49272
(100) = 66.72%
P 8.13 [a] α =1
2RC= 1250, ωo = 103, therefore overdamped
s1 = −500, s2 = −2000
therefore v = A1e−500t + A2e
−2000t
v(0+) = 0 = A1 + A2;[dv(0+)
dt
]=
iC(0+)C
= 98,000 V/s
Therefore − 500A1 − 2000A2 = 98,000
A1 =+98015
, A2 =−98015
v(t) =[980
15
][e−500t − e−2000t] V, t ≥ 0
[b]
Example 8.4: vmax∼= 74.1 V at 1.4 ms
8–14 CHAPTER 8. Natural and Step Responses of RLC Circuits
Example 8.5: vmax∼= 36.1 V at 1.0 ms
Problem 8.13: vmax∼= 30.9 at 0.92 ms
P 8.14 From the form of the solution we have
v(0) = A1 + A2
dv(0+)dt
= −α(A1 + A2) + jωd(A1 − A2)
We know both v(0) and dv(0+)/dt will be real numbers. To facilitate the algebra welet these numbers be K1 and K2, respectively. Then our two simultaneous equationsare
K1 = A1 + A2
K2 = (−α + jωd)A1 + (−α − jωd)A2
The characteristic determinate is
∆ =
∣∣∣∣∣∣∣1 1
(−α + jωd) (−α − jωd)
∣∣∣∣∣∣∣ = −j2ωd
The numerator determinates are
N1 =
∣∣∣∣∣∣∣K1 1
K2 (−α − jωd)
∣∣∣∣∣∣∣ = −(α + jωd)K1 − K2
and N2 =
∣∣∣∣∣∣∣1 K1
(−α + jωd) K2
∣∣∣∣∣∣∣ = K2 + (α − jωd)K1
It follows that A1 =N1
∆=
ωdK1 − j(αK1 + K2)2ωd
and A2 =N2
∆=
ωdK1 + j(αK1 + K2)2ωd
We see from these expressions that A1 = A∗2
Problems 8–15
P 8.15 By definition, B1 = A1 + A2. From the solution to Problem 8.14 we have
A1 + A2 =2ωdK1
2ωd
= K1
But K1 is v(0), therefore, B1 = v(0), which is identical to Eq. (8.30).By definition, B2 = j(A1 − A2). From Problem 8.14 we have
B2 = j(A1 − A2) =j[−2j(αK1 + K2)]
2ωd
=αK1 + K2
ωd
It follows that
K2 = −αK1 + ωdB2, but K2 =dv(0+)
dtand K1 = B1
Thus we have
dv
dt(0+) = −αB1 + ωdB2,
which is identical to Eq. (8.31).
P 8.16 t < 0 : Vo = 15 V, Io = −60 mA
t > 0:
iR(0) =15100
= 150 mA; iL(0) = −60 mA
iC(0) = −150 − (−60) = −90 mA
α =1
2RC=
12(100)(10−6)
= 5000 rad/s
8–16 CHAPTER 8. Natural and Step Responses of RLC Circuits
ω2o =
1LC
=1
(62.5 × 10−3)(10−6)= 16 × 106
s1,2 = −5000 ±√
25 × 106 − 16 × 106 = −5000 ± 3000
s1 = −2000 rad/s; s2 = −8000 rad/s
·. . vo = A1e−2000t + A2e
−8000t
A1 + A2 = vo(0) = 15
dvo
dt(0) = −2000A1 − 8000A2 =
−90 × 10−3
10−6 = −90,000
Solving, A1 = 5 V, A2 = 10 V
·. . vo = 5e−2000t + 10e−8000t V, t ≥ 0
P 8.17 ω2o =
1LC
=1
(62.5 × 10−3)(10−6)= 16 × 106
α =1
2RC=
12(250)(10−6)
= 2500
s1,2 = −2500 ±√
25002 − 16 × 106 = −2500 ± j3122.5rad/s
vo(t) = B1e−2500t cos 3122.5t + B2e
−2500t sin 3122.5t
vo(0) = B1 = 15 V
iR(0) =15200
= 75 mA
iL(0) = −60 mA
iC(0) = −iR(0) − iL(0) = −15 mA ·. .iC(0)
C= −15,000 V/s
dvo
dt(0) = −2500B1 + 3122.5B2 = −15,000 V/s
·. . 3122.5B2 = 2500(15) − 15,000 ·. . B2 = 7.21 V
vo(t) = 15e−2500t cos 3122.5t + 7.21e−2500t sin 3122.5t V, t ≥ 0
Problems 8–17
P 8.18 ω2o =
1LC
=1
(62.5 × 10−3)(10−6)= 16 × 106
α =1
2RC=
12(125)(10−6)
= 4000
·. . α2 = ω2o (critical damping)
vo(t) = D1te−4000t + D2e
−4000t
vo(0) = D2 = 15 V
iR(0) =15125
= 120 mA
iL(0) = −60 mA
iC(0) = −60 mA
dvo
dt(0) = −4000D2 + D1
iC(0)C
=−60 × 10−3
10−6 = −60,000
D1 − 4000D2 = −60,000; D1 = 0
vo(t) = 15e−4000t V, t ≥ 0
P 8.19
vT = −2 × 104iφ + 16 × 103iT ; iφ =20100
(−iT )
= 4000it + 16,000iT = 20,000iT
vT
iT= 20 kΩ
8–18 CHAPTER 8. Natural and Step Responses of RLC Circuits
Vo =30005000
(50) = 30 V; Io = 0
iC(0) = −iR(0) − iL(0) = − 3020,000
= −1.5 mA
iC(0)C
=−1.5 × 10−3
0.25 × 10−6 = −6000
ω2o =
1LC
=1
(40)(0.25 × 10−6)= 105
α =1
2RC=
12(20 × 103)(0.25 × 10−6)
= 100 rad/s
ωd =√
105 − 1002 = 300 rad/s
vo = B1e−100t cos 300t + B2e
−100t sin 300t
vo(0) = B1 = 30 V
dvo
dt(0) = 300B2 − 100B1 = −6000
·. . 300B2 = 100(30) − 6000; ·. . B2 = −10 V
vo = 30e−100t cos 300t − 10e−100t sin 300t V, t ≥ 0
P 8.20 [a] v = L
(diLdt
)= 16[e−20,000t − e−80,000t] V, t ≥ 0
[b] iR =v
R= 40[e−20,000t − e−80,000t] mA, t ≥ 0+
[c] iC = I − iL − iR = [−8e−20,000t + 32e−80,000t] mA, t ≥ 0+
P 8.21 [a] v = L
(diLdt
)= 40e−32,000t sin 24,000t V, t ≥ 0
[b] iC(t) = I − iR − iL = 24 × 10−3 − v
625− iL
= [24e−32,000t cos 24,000t − 32e−32,000t sin 24,000t] mA, t ≥ 0+
P 8.22 v = L
(diLdt
)= 960,000te−40,000t V, t ≥ 0
Problems 8–19
P 8.23 t < 0 : iL = 9/3000 = 3 mAt > 0:
6 k‖3 k = 2 kΩ
iL(0) = 3 mA, iL(∞) = 9 mA
ω2o =
1LC
=1
(62.5)(2.5 × 10−6)= 6400; ωo = 80 rad/s
α =1
2RC=
12(2000)(2.5 × 10−6)
= 100; α2 = 104
α2 − ω2o = 104 − 6400 = 3600
s1,2 = −100 ± 60 rad/s
s1 = −40 rad/s; s2 = −160 rad/s
iL = If + A′1e
−40t + A′2e
−160t
iL(∞) = If = 9mA
iL(0) = A′1 + A′
2 + If = 3 mA
·. . A′1 + A′
2 + 9 m = 3 m so A′1 + A′
2 = −6 mA
diLdt
(0) = 0 = −40A′1 − 160A′
2
Solving, A′1 = −8 mA, A′
2 = 2 mA
iL = 9 − 8e−40t + 2e−160t mA, t ≥ 0
8–20 CHAPTER 8. Natural and Step Responses of RLC Circuits
P 8.24 ω2o =
1LC
=1
(50 × 10−3)(0.2 × 10−6)= 108; ωo = 104 rad/s
α =1
2RC=
12(200)(0.2 × 10−6)
= 12,500 rad/s ·. . overdamped
s1,2 = −12,500 ±√
(12,500)2 − 108 = −12,500 ± 7500 rad/s
s1 = −5000 rad/s; s2 = −20,000 rad/s
If = 60 mA
iL = 60 × 10−3 + A′1e
−5000t + A′2e
−20,000t
·. . −45 × 10−3 = 60 × 10−3 + A′1 + A′
2; A′1 + A′
2 = −105 × 10−3
diLdt
= −5000A′1 − 20,000A′
2 =15
0.05= 300
Solving, A′1 = −120 mA; A′
2 = 15 mA
iL = 60 − 120e−5000t + 15e−20,000t mA, t ≥ 0
P 8.25 α =1
2RC=
12(312.5)(0.2 × 10−6)
= 8000; α2 = 64 × 106
ωo = 104 underdamped
s1,2 = −8000 ± j√
80002 − 108 = −8000 ± j6000 rad/s
iL = 60 × 10−3 + B′1e
−8000t cos 6000t + B′2e
−8000t sin 6000t
−45 × 10−3 = 60 × 10−3 + B′1
·. . B′1 = −105 mA
diLdt
(0) = −8000B′1 + 6000B′
2 = 300
·. . B′2 = −90 mA
iL = 60 − 105e−8000t cos 6000t − 90e−8000t sin 6000t mA, t ≥ 0
Problems 8–21
P 8.26 α =1
2RC=
12(250)(0.2 × 10−6)
= 104
α2 = 108 = ω2o critical damping
iL = If + D′1te
−104t + D′2e
−104t = 60 × 10−3 + D′1te
−104t + D′2e
−104t
iL(0) = −45 × 10−3 = 60 × 10−3 + D′2; ·. . D′
2 = −105 mA
diLdt
(0) = −104D′2 + D′
1 = 300 A/s
·. . D′1 = 300 + 104(−105 × 10−3) = −750 A/s
iL = 60 − 750,000te−104t − 105e−104t mA, t ≥ 0
P 8.27 For t > 0
α =1
2RC= 1000;
1LC
= 64 × 104
s1,2 = −1000 ± 600 rad/s
s1 = −400 rad/s; s2 = −1600 rad/s
vo = Vf + A′1e
−400t + A′2e
−1600t
Vf = 0; vo(0+) = 0; iC(0+) = 30 mA
·. . A′1 + A′
2 = 0
dvo(0+)dt
=iC(0+)
1.25 × 10−6 = 24,000 V/s
dvo(0+)dt
= −400A′1 − 1600A′
2 = 24,000
Solving,
A′1 = 20 V; A′
2 = −20 V
vo = 20e−400t − 20e−1600t V, t ≥ 0
8–22 CHAPTER 8. Natural and Step Responses of RLC Circuits
P 8.28 [a] From the solution to Prob. 8.27 s1 = −400 rad/s and s2 = −1600 rad/s,therefore
io = If + A′1e
−400t + A′2e
−1600t
If = 30 mA; io(0+) = 0;dio(0+)
dt= 0
·. . 0 = 30 × 10−3 + A′1 + A′
2; −400A′1 − 1600A′
2 = 0
Solving
A′1 = −40 mA; A′
2 = 10 mA
·. . io = 30 − 40e−400t + 10e−1600t mA, t ≥ 0
[b]diodt
= 16e−400t − 16e−1600t
vo = Ldiodt
= 20e−400t − 20e−1600t V, t ≥ 0
This agrees with the solution to Problem 8.27
P 8.29 α =1
2RC=
12(400)(1.25 × 10−6)
= 1000
ω2o =
1LC
=1
(1.25 × 10−6)(1.25)= 64 × 104
s1,2 = −1000 ±√
10002 − 64 × 104 = −1000 ± 600 rad/s
s1 = −400 rad/s; s2 = −1600 rad/s
vo(∞) = 0 = Vf
·. . vo = A′1e
−400t + A′2e
−1600t
vo(0) = 12 = A′1 + A′
2
Note: iC(0+) = 0
·. .dvo
dt(0) = 0 = −400A′
1 − 1600A′2
Solving, A′1 = 16 V, A′
2 = −4 V
vo(t) = 16e−400t − 4e−1600t V, t > 0
Problems 8–23
P 8.30 [a] io = If + A′1e
−400t + A′2e
−1600t
If =12400
= 30mA; io(0) = 0
0 = 30 × 10−3 + A′1 + A′
2,·. . A′
1 + A′2 = −30 × 10−3
diodt
(0) =12
1.25= −400A′
1 − 1600A′2
Solving, A′1 = −32 mA; A′
2 = 2 mA
io = 30 − 32e−400t + 2e−1600t mA, t ≥ 0
[b]diodt
= [12.8e−400t − 3.2e−1600t]
vo = Ldiodt
= 16e−400t − 4e−1600t V, t ≥ 0
This agrees with the solution to Problem 8.29.
P 8.31 iL(0−) = iL(0+) = 37.5 mA
For t > 0
iL(0−) = iL(0+) = 37.5 mA
α =1
2RC= 100 rad/s; ω2
o =1
LC= 6400
s1 = −40 rad/s s2 = −160 rad/s
vo(∞) = 0 = Vf
vo = A′1e
−40t + A′2e
−160t
iC(0+) = −37.5 + 37.5 + 0 = 0
·. .dvo
dt= 0
8–24 CHAPTER 8. Natural and Step Responses of RLC Circuits
dvo
dt(0) = −40A′
1 − 160A′2
·. . A′1 + 4A′
2 = 0; A′1 + A′
2 = 0
·. . A′1 = 0; A′
2 = 0
·. . vo = 0 for t ≥ 0
Note: vo(0) = 0; vo(∞) = 0;dvo(0)
dt= 0
Hence the 37.5 mA current circulates between the current source and the idealinductor in the equivalent circuit. In the original circuit the 7.5 V source sustains acurrent of 37.5 mA in the inductor. This is an example of a circuit going directly intosteady state when the switch is closed. There is no transient period, or interval.
P 8.32 t < 0:
vo(0−) = vo(0+) =625
781.25(25) = 20 V
iL(0−) = iL(0+) = 0
t > 0
−160 × 10−3 +20125
+ iC(0+) + 0 = 0; ·. . iC(0+) = 0
12RC
=1
2(125)(5 × 10−6)= 800 rad/s
ω2o =
1LC
=1
(312.5 × 10−3)(5 × 10−6)= 64 × 104
·. . α2 = ω2o critically damped
Problems 8–25
[a] vo = Vf + D′1te
−800t + D′2e
−800t
Vf = 0
dvo(0)dt
= −800D′2 + D′
1 = 0
vo(0+) = 20 = D′2
D′1 = 800D′
2 = 16,000 V/s
·. . vo = 16,000te−800t + 20e−800t V, t ≥ 0+
[b] iL = If + D′3te
−800t + D′4e
−800t
iL(0+) = 0; If = 160 mA;diL(0+)
dt=
20312.5 × 10−3 = 64 A/s
·. . 0 = 160 + D′4; D′
4 = −160 mA;
−800D′4 + D′
3 = 64; D′3 = −64 A/s
·. . iL = 160 − 64,000te−800t − 160e−800t mA t ≥ 0
P 8.33 [a] wL =∫ ∞
0pdt =
∫ ∞
0voiL 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
−1,024,000t2e−1600t − 3.2e−1600t W
wL = 3.2∫ ∞
0e−800t dt + 2560
∫ ∞
0te−800t dt − 3480
∫ ∞
0te−1600t dt
−1,024,000∫ ∞
0t2e−1600t dt − 3.2
∫ ∞
0e−1600t dt
= 3.2e−800t
−800
∣∣∣∣∣∞
0
+2560
(800)2 e−800t(−800t − 1)∣∣∣∣∣∞
0
− 3840(1600)2 e−1600t(−1600t − 1)
∣∣∣∣∣∞
0
− 1,024,000(−1600)3 e−1600t(16002t2 + 3200t + 2)
∣∣∣∣∣∞
0
− 3.2e−1600t
(−1600)
∣∣∣∣∣∞
0
8–26 CHAPTER 8. Natural and Step Responses of RLC Circuits
All the upper limits evaluate to zero hence
wL =3.2800
+25608002 − 3840
16002 − (1,024,000)(2)16003 − 3.2
1600= 4 mJ
Note this value corresponds to the final energy stored in the inductor, i.e.
wL(∞) =12(312.5 × 10−3)(0.16)2 = 4 mJ.
[b] v = 16,000te−800t + 20e−800t V
iR =v
125= 128te−800t + 0.16e−800t A
pR = viR = 2,048,000t2e−1600t + 5120te−1600t + 3.2e−1600t
wR =∫ ∞
0pR dt
= 2,048,000∫ ∞
0t2e−1600t dt + 5120
∫ ∞
0te−1600t dt + 3.2
∫ ∞
0e−1600t dt
=2,048,000e−1600t
−16003 [16002t2 + 3200t + 2]∣∣∣∣∞0
+
5120e−1600t
16002 (−1600t − 1)∣∣∣∣∞0
+3.2e−1600t
(−1600)
∣∣∣∣∞0
Since all the upper limits evaluate to zero we have
wR =2,048,000(2)
16003 +512016002 +
3.21600
= 5 mJ
[c] 160 = iR + iC + iL (mA)
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]
= −1,024,000t2e−1600t − 1280e−1600t
wC = −1,024,000∫ ∞
0t2e−1600t dt − 1280
∫ ∞
0te−1600t dt
wC =−1,024,000e−1600t
−16003 [16002t2 + 3200t + 2]∣∣∣∣∞0
−1280e−1600t
16002 (−1600t − 1)∣∣∣∣∞0
Since all upper limits evaluate to zero we have
wC =−1,024,000(2)
16003 − 1280(1)16002 = −1 mJ
Problems 8–27
Note this 1 mJ corresponds to the initial energy stored in the capacitor, i.e.,
wC(0) =12(5 × 10−6)(20)2 = 1 mJ.
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∫ ∞
0e−800t dt +
∫ ∞
02560te−800t dt
=3.2e−800t
−800
∣∣∣∣∞0
+2560e−800t
8002 (−800t − 1)∣∣∣∣∞0
=3.2800
+2560800
= 8 mJ
[e] wL = 4 mJ (absorbed)
wR = 5 mJ (absorbed)
wC = 1 mJ (delivered)
wS = 8 mJ (delivered)∑
wdel = wabs = 9 mJ.
P 8.34 vC(0+) =3.75 × 103
11.25 × 103 (150) = 50 V
iL(0+) = 100 mA; iL(∞) =1507500
= 20 mA
α =1
2RC=
12(2500)(0.25 × 10−6)
= 800
ω2o =
1LC
=1
(4)(0.25 × 10−6)= 106
α2 = 64 × 104; α2 < ω2o ; ·. . underdamped
s1,2 = −800 ± j√
8002 − 106 = −800 ± j600 rad/s
8–28 CHAPTER 8. Natural and Step Responses of RLC Circuits
iL = If + B′1e
−αt cos ωdt + B′2e
−αt sin ωdt
= 20 + B′1e
−800t cos 600t + B′2e
−800t sin 600t
iL(0) = 20 × 10−3 + B′1; B′
1 = 100 m − 20 m = 80 mA
diLdt
(0) = 600B′2 − 800B′
1 =504
= 12.5
·. . 600B2 = 800(80 × 10−3) + 12.5; B′2 = 127.5 mA
·. . iL = 20 + 80e−800t cos 600t + 127.5e−800t sin 600t mA, t ≥ 0
P 8.35 [a] 2α = 5000; α = 2500 rad/s√α2 − ω2
o = 1500; ω2o = 4 × 106; ωo = 2000 rad/s
α =R
2L= 2500; R = 5000L
ω2o =
1LC
= 4 × 106; L =109
4 × 106(50)= 5H
R = 25,000 Ω
[b] i(0) = 0
Ldi(0)dt
= vc(0);12(50) × 10−9v2
c (0) = 90 × 10−6
·. . v2c (0) = 3600; vc(0) = 60 V
di(0)dt
=605
= 12 A/s
[c] i(t) = A1e−1000t + A2e
−4000t
i(0) = A1 + A2 = 0
di(0)dt
= −1000A1 − 4000A2 = 12
Solving,
·. . A1 = 4 mA; A2 = −4 mA
i(t) = 4e−1000t − 4e−4000t mA t ≥ 0
Problems 8–29
[d]di(t)dt
= −4e−1000t + 16e−4000t
di
dt= 0 when 16e−4000t = 4e−1000t
or e3000t = 4
·. . t =ln 43000
µs = 462.10 µs
[e] imax = 4e−0.4621 − 4e−1.8484 = 1.89 mA
[f] vL(t) = 5di
dt= [−20e−1000t + 80e−4000t] V, t ≥ 0+
P 8.36 α = 2000 rad/s; ωd = 1500 rad/s
ω2o − α2 = 225 × 104; ω2
o = 625 × 104; wo = 2500 rad/s
α =R
2L= 2000; R = 4000L
1LC
= 625 × 104; L =1
(625 × 104)(80 × 10−9)= 2 H
·. . R = 8 kΩ
i(0+) = B1 = 7.5 mA; at t = 0+
60 + vL(0+) − 30 = 0; ·. . vL(0+) = −30 V
di(0+)dt
=−302
= −15 A/s
·. .di(0+)
dt= 1500B2 − 2000B1 = −15
·. . 1500B2 = 2000(7.5 × 10−3) − 15; ·. . B2 = 0 A
·. . i = 7.5e−2000t sin 1500t mA, t ≥ 0
8–30 CHAPTER 8. Natural and Step Responses of RLC Circuits
P 8.37 From Prob. 8.36 we know vc will be of the form
vc = B3e−2000t cos 1500t + B4e
−2000t sin 1500t
From Prob. 8.36 we have
vc(0) = −30 V = B3
and
dvc(0)dt
=iC(0)
C=
7.5 × 10−3
80 × 10−9 = 93.75 × 103
dvc(0)dt
= 1500B4 − 2000B3 = 93,750
·. . 1500B4 = 2000(−30) + 93,750; B4 = 22.5 V
vc(t) = −30e−2000t cos 1500t + 22.5e−2000t sin 1500t V t ≥ 0
P 8.38 [a] ω2o =
1LC
=1
(80 × 10−3)(0.5 × 10−6)= 25 × 106
α =R
2L= ωo = 5000 rad/s
·. . R = (5000)(2)L = 800 Ω
[b] i(0) = iL(0) = 30 mA
vc(0) = 800i(0) + 80 × 10−3di(0)dt
20 − 800(30 × 10−3)80 × 10−3 =
di(0)dt
·. .di(0)dt
= −50 A/s
[c] vC = D1te−5000t + D2e
−5000t
vC(0) = D2 = 20 V
dvC
dt(0) = D1 − 5000D2 =
iC(0)C
=−iL(0)
C
D1 − 100,000 = − 30 × 10−3
0.5 × 10−6 = −60,000 ·. . D1 = 40,000 V/s
vC = 40,000te−5000t + 20e−5000t V, t ≥ 0
Problems 8–31
P 8.39 [a] For t > 0:
Since i(0−) = i(0+) = 0
va(0+) = 72 V
[b] va = 5000i +1
0.1 × 10−6
∫ t
0i dx + 72
dva
dt= 5000
di
dt+ 10 × 106i
dva(0+)dt
= 5000di(0+)
dt+ 10 × 106i(0+) = 5000
di(0+)dt
−Ldi(0+)
dt= 72
di(0+)dt
= − 722.5
= −28.8 A/s
·. .dva(0+)
dt= −144,000 V/s
[c] α =R
2L=
12,5002(2.5)
= 2500 rad/s
ω2o =
1LC
=1
(2.5)(0.1 × 10−6)= 4 × 106
s1,2 = −2500 ±√
25002 − 4 × 106 = −2500 ± 1500 rad/s
Overdamped:
va = A1e−1000t + A2e
−4000t
va(0) = 72 = A1 + A2
dva(0)dt
= −144,000 = −1000A1 − 4000A2
Solving, A1 = 48; A2 = 24
va = 48e−1000t + 24e−4000t V, t ≥ 0+
8–32 CHAPTER 8. Natural and Step Responses of RLC Circuits
P 8.40 ω2o =
1LC
=1
(10)(4 × 10−3)= 25
α =R
2L=
802(10)
= 4; α2 = 16
α2 < ω2o
·. . underdamped
s1,2 = −4 ± j√
9 = −4 ± j3 rad/s
i = B1e−4t cos 3t + B2e
−4t sin 3t
i(0) = B1 = −240/100 = −2.4 A
di
dt(0) = 3B2 − 4B1 = 0
·. . B2 = −3.2 A
i = −2.4e−4t cos 3t − 3.2 sin 3t A, t ≥ 0
P 8.41 t < 0:
i(0) =240
8 + 30‖70 + 11=
24040
= 6 A
vo(0) = 240 − 8(6) − 70100
(6)(20) = 108 V
Problems 8–33
t > 0:
α =R
2L=
202(1)
= 10, α2 = 100
ω2o =
1LC
=1
(1)(5 × 10−3)= 200
ω2o > α2 underdamped
s1,2 = −100 ± √100 − 200 = −10 ± j10 rad/s
vo = B1e−10t cos 10t + B2e
−10t sin 10t
vo(0) = B1 = 108 V
Cdvo
dt(0) = −6,
dvo
dt=
−65 × 10−3 = −1200 V/s
dvo
dt(0) = −10B1 + 10B2 = −1200
10B2 = −1200 + 10B1 = −1200 + 1080; B2 = −120/10 = −12 V
·. . vo = 108e−10t cos 10t − 12e−10t sin 10t V, t ≥ 0
P 8.42 [a] t < 0:
io =80800
= 100 mA; vo = 500io = (500)(0.01) = 50 V
t > 0:
α =R
2L=
5002(2.5 × 10−3)
= 105 rad/s
ω2o =
1LC
=1
(2.5 × 10−3)(40 × 10−9)= 100 × 108
α2 = ω2o
·. . critically damped
8–34 CHAPTER 8. Natural and Step Responses of RLC Circuits
·. . io(t) = D1te−105t + D2e
−105t
io(0) = D2 = 100 mA
diodt
(0) = −αD2 + D1 = 0
·. . D1 = 105(100 × 10−3) = 10,000
io(t) = 10,000te−105t + 0.1e−105t A, t ≥ 0
[b] vo(t) = D3te−105t + D4e
−105t
vo(0) = D4 = 50
Cdvo
dt(0) = −0.1
dvo
dt(0) =
−0.140 × 10−9 = −25 × 105 V/s = −αD4 + D3
·. . D3 = 105(50) − 25 × 105 = 25 × 105
vo(t) = 25 × 105te−105t + 50e−105t V, t ≥ 0
P 8.43 α =R
2L=
80002(1)
= 4000 rad/s
ω2o =
1LC
=1
(1)(50 × 10−9)= 20 × 106
s1,2 = −4000 ±√
40002 − 20 × 106 = −4000 ± j2000 rad/s
vo = Vf + B′1e
−4000t cos 2000t + B′2e
−4000t sin 2000t
vo(0) = 0 = Vf + B′1
vo(∞) = 80 V; ·. . B′1 = −80 V
dvo(0)dt
= 0 = 2000B′2 − 4000B′
1
·. . 2000B′2 = 4000(−80) ·. . B′
2 = −160 V
vo = 80 − 80e−4000t cos 2000t − 160e−4000t sin 2000t V, t ≥ 0
Problems 8–35
P 8.44 t < 0:
iL(0) =−15030
= −5 A
vC(0) = 18iL(0) = −90 V
t > 0:
α =R
2L=
102(0.1)
= 50 rad/s
ω2o =
1LC
=1
(0.1)(2 × 10−3)= 5000
ωo > α2 ·. . underdamped
s1,2 = −50 ±√
502 − 5000 = −50 ± j50
vc = 60 + B′1e
−50t cos 50t + B′2e
−50t sin 50t
vc(0) = −90 = 60 + B′1
·. . B′1 = −150
Cdvc
dt(0) = −5;
dvc
dt(0) =
−52 × 10−3 = −2500
dvc
dt(0) = −50B′
1 + 50B2 = −2500 ·. . B′2 = −200
vc = 60 − 150e−50t cos 50t − 200e−50t sin 50t V, t ≥ 0
8–36 CHAPTER 8. Natural and Step Responses of RLC Circuits
P 8.45 iC(0) = 0; vo(0) = 50 V
α =R
2L=
80002(160 × 10−3)
= 25,000 rad/s
ω2o =
1LC
=1
(160 × 10−3)(10 × 10−9)= 625 × 106
·. . α2 = ω2o ; critical damping
vo(t) = Vf + D′1te
−25,000t + D′2e
−25,000t
Vf = 250 V
vo(0) = 250 + D′2 = 50; D′
2 = −200 V
dvo
dt(0) = −25,000D′
2 + D′1 = 0
D′1 = 25,000D′
2 − 5 × 106 V/s
vo = 250 − 5 × 106te−25,000t − 200e−25,000t V, t ≥ 0
P 8.46 [a] t < 0:
io(0−) =120
30,000= 4 mA
vC(0−) = 80 − (10,000)(0.004) = 40 V
t = 0+:
5 kΩ‖20 kΩ = 4 kΩ
·. . vo(0+) = −(0.004)(4000) + 40 = 40 − 16 = 24 V
Problems 8–37
[b] vo(t) = vc − 4000io
dvo
dt(0+) =
dvc
dt(0+) − 4000
diodt
(0+)
dvc
dt(0+) =
−4 × 10−3
(125/64) × 10−6 = −2048 V/s
−vL(0+) + vo(0+) + 40 = 0 vL = 64 V
diodt
(0+) =645
= 12.8 A/s
dvo
dt(0+) = −2048 − 4000(12.8) = −53,248 V/s
[c] ω2o =
1LC
=1
(5)[(125/64) × 10−6]= 10.24 × 104
α =R
2L=
40002(5)
= 400 rad/s; α2 = 16 × 104
α2 > ω2o overdamped
s1,2 = −400 ± 240 rad/s
vo(t) = Vf + A′1e
−160t + A′2e
−640t
Vf = vo(∞) = −40 V
−40 + A′1 + A′
2 = 24
−160A′1 − 640A′
2 = −53,248
Solving, A′1 = −25.6; A′
2 = 89.6
·. . vo(t) = −40 − 25.6e−160t + 89.6e−640t V, t ≥ 0+
P 8.47 [a] vc = Vf + [B′1 cos ωdt + B′
2 sin ωdt] e−αt
dvc
dt= [(ωdB
′2 − αB′
1) cos ωdt − (αB′2 + ωdB
′1) sin ωdt]e−αt
Since the initial stored energy is zero,
vc(0+) = 0 anddvc(0+)
dt= 0
It follows that B′1 = −Vf and B′
2 =αB′
1
ωd
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=(
α2
ωd
+ ωd
)Vfe
−αt sin ωdt
But Vf = V andα2
ωd
+ ωd =α2 + ω2
d
ωd
=ω2
o
ωd
Thereforedvc
dt=(
ω2o
ωd
)V e−αt sin ωdt
[b]dvc
dt= 0 when sin ωdt = 0, or ωdt = nπ
where n = 0, 1, 2, 3, . . .
Therefore t =nπ
ωd
[c] When tn =nπ
ωd
, cos ωdtn = cos nπ = (−1)n
and sin ωdt = sin nπ = 0
Therefore vc(tn) = V [1 − (−1)ne−αnπ/ωd ]
[d] It follows from [c] that
v(t1) = V + V e−(απ/ωd) and vc(t3) = V + V e−(3απ/ωd)
Thereforevc(t1) − V
vc(t3) − V=
e−(απ/ωd)
e−(3απ/ωd) = e(2απ/ωd)
But2πωd
= t3 − t1 = Td, thus α =1Td
ln[vc(t1) − V ][vc(t3) − V ]
P 8.48 α =1Td
ln
vc(t1) − V
vc(t3) − V
; Td = t3 − t1 =
3π7
− π
7=
2π7
ms
α =70002π
ln[63.8426.02
]≈ 1000; ωd =
2πTd
= 7000 rad/s
ω2o = ω2
d + α2 = 49 × 106 + 106 = 50 × 106
L =1
(50 × 106)(0.1 × 10−6)= 200 mH; R = 2αL = 400 Ω
Problems 8–39
P 8.49 [a] Let i be the current in the direction of the voltage drop vo(t). Then by hypothesis
i = if + B′1e
−αt cos ωdt + B′2e
−αt sin ωdt
if = i(∞) = 0, i(0) =Vg
R= B′
1
Therefore i = B′1e
−αt cos ωdt + B′2e
−αt sin ωdt
Ldi(0)dt
= 0, thereforedi(0)dt
= 0
di
dt= [(ωdB
′2 − αB′
1) cos ωdt − (αB′2 + ωdB
′1) sin ωdt] e−αt
Therefore ωdB′2 − αB′
1 = 0; B′2 =
α
ωd
B′1 =
α
ωd
Vg
R
Therefore
vo = Ldi
dt= −
L
(α2Vg
ωdR+
ωdVg
R
)sin ωdt
e−αt
= −
LVg
R
(α2
ωd
+ ωd
)sin ωdt
e−αt
= −VgL
R
(α2 + ω2
d
ωd
)e−αt sin ωdt
vo = − Vg
RCωd
e−αt sin ωdt V, t ≥ 0+
[b]dvo
dt= − Vg
ωdRCωd cos ωdt − α sin ωdte−αt
dvo
dt= 0 when tan ωdt =
ωd
α
Therefore ωdt = tan−1(ωd/α) (smallest t)
t =1ωd
tan−1(
ωd
α
)
P 8.50 [a] From Problem 8.49 we have
vo =−Vg
RCωd
e−αt sin ωdt
α =R
2L=
48002(64 × 10−3)
= 37,500 rad/s
ω2o =
1LC
=1
(64 × 10−3)(4 × 10−9)= 3906.25 × 106
8–40 CHAPTER 8. Natural and Step Responses of RLC Circuits
ωd =√
ω2o − α2 = 50 krad/s
−Vg
RCωd
=−(−72)
(4800)(4 × 10−9)(50 × 103)= 75
·. . vo = 75e−37,500t sin 50,000t V, t ≥ 0
[b] From Problem 8.49
td =1ωd
tan−1(
ωd
α
)=
150,000
tan−1
(50,00037,500
)
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]d2vo
dt2=
1R1C1R2C2
vg
1R1C1R2C2
=10−6
(100)(400)(0.5)(0.2) × 10−6 × 10−6 = 250
·. .d2vo
dt2= 250vg
0 ≤ t ≤ 0.5−:
vg = 80 mV
d2vo
dt2= 20
Let g(t) =dvo
dt, then
dg
dt= 20 or dg = 20 dt
∫ g(t)
g(0)dx = 20
∫ t
0dy
g(t) − g(0) = 20t, g(0) =dvo
dt(0) = 0
g(t) =dvo
dt= 20t
Problems 8–41
dvo = 20t dt
∫ vo(t)
vo(0)dx = 20
∫ t
0x dx; vo(t) − vo(0) = 10t2, vo(0) = 0
vo(t) = 10t2 V, 0 ≤ t ≤ 0.5−
dvo1
dt= − 1
R1C1vg = −20vg = −1.6
dvo1 = −1.6 dt
∫ vo1(t)
vo1(0)dx = −1.6
∫ t
0dy
vo1(t) − vo1(0) = −1.6t, vo1(0) = 0
vo1(t) = −1.6t V, 0 ≤ t ≤ 0.5−
0.5+ ≤ t ≤ tsat:
d2vo
dt2= −10, let g(t) =
dvo
dt
dg(t)dt
= −10; dg(t) = −10 dt
∫ g(t)
g(0.5+)dx = −10
∫ t
0.5dy
g(t) − g(0.5+) = −10(t − 0.5) = −10t + 5
g(0.5+) =dvo(0.5+)
dt
Cdvo
dt(0.5+) =
0 − vo1(0.5+)400 × 103
vo1(0.5+) = vo(0.5−) = −1.6(0.5) = −0.80 V
·. . Cdvo(0.5+)
dt=
0.800.4 × 106 = 2 µA
dvo
dt(0.5+) =
2 × 10−6
0.2 × 10−6 = 10 V/s
·. . g(t) = −10t + 5 + 10 = −10t + 15 =dvo
dt
·. . dvo = −10t dt + 15 dt
∫ vo(t)
vo(0.5+)dx =
∫ t
0.5+−10y dy +
∫ t
0.5+15 dy
8–42 CHAPTER 8. Natural and Step Responses of RLC Circuits
vo(t) − vo(0.5+) = −5y2∣∣∣∣t0.5
+ 15y∣∣∣∣t0.5
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, 0.5+ ≤ t ≤ tsat
dvo1
dt= −20(−0.04) = 0.8, 0.5+ ≤ t ≤ tsat
dvo1 = 0.8 dt;∫ vo1(t)
vo1(0.5+)dx = 0.8
∫ t
0.5+dy
vo1(t) − vo1(0.5+) = 0.8t − 0.4; vo1(0.5+) = vo1(0.5−) = −0.8 V
·. . vo1(t) = 0.8t − 1.2 V, 0.5+ ≤ t ≤ tsat
Summary:
0 ≤ t ≤ 0.5−s : vo1 = −1.6t V, vo = 10t2 V
0.5+s ≤ t ≤ tsat : 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
1τ1
= 2; τ2 = (5 × 106)(0.2 × 10−6) = 1 s; ·. .1τ2
= 1
·. .d2vo
dt2+ 3
dvo
dt+ 2vo = 20
s2 + 3s + 2 = 0
(s + 1)(s + 2) = 0; s1 = −1, s2 = −2
vo = Vf + A′1e
−t + A′2e
−2t; Vf =202
= 10 V
vo = 10 + A′1e
−t + A′2e
−2t
Problems 8–43
vo(0) = 0 = 10 + A′1 + A′
2;dvo
dt(0) = 0 = −A′
1 − 2A′2
·. . A′1 = −20, A′
2 = 10 V
vo(t) = 10 − 20e−t + 10e−2t V, 0 ≤ t ≤ 0.5 s
dvo1
dt+ 2vo1 = −1.6; ·. . vo1 = −0.8 + 0.8e−2t V, 0 ≤ t ≤ 0.5 s
vo(0.5) = 10 − 20e−0.5 + 10e−1 = 1.55 V
vo1(0.5) = −0.8 + 0.8e−1 = −0.51 V
At t = 0.5 s
iC =0 + 0.51400 × 103 − 1.55 − 0
5 × 106 = 0.954 µA
Cdvo
dt= 0.954 µA;
dvo
dt=
0.9540.2
= 4.773 V/s
t ≥ 0.5 s
d2vo
dt2+ 3
dvo
dt+ 2vo = −10
vo(∞) = −5
·. . vo = −5 + A′1e
−(t−0.5) + A′2e
−2(t−0.5)
1.55 = −5 + A′1 + A′
2
8–44 CHAPTER 8. Natural and Step Responses of RLC Circuits
dvo
dt(0.5) = 4.773 = −A′
1 − 2A′2
·. . A′1 + A′
2 = 6.55; −A′1 − 2A′
2 = 4.773
Solving,
A′1 = 17.87 V; A′
2 = −11.32 V
·. . vo = −5 + 17.87e−(t−0.5) − 11.32e−2(t−0.5) V, t ≥ 0.5 s
dvo1
dt+ 2vo1 = 0.8
·. . 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 theoperational amplifiers are ideal, the current in the 500 kΩ is zero. Therefore therecannot be an instantaneous change in the current in the 1 µF capacitor. Since thecapacitor current equals C(dvo/dt), the derivative must be zero.
P 8.54 [a] From Example 8.13d2vo
dt2= 2
thereforedg(t)dt
= 2, g(t) =dvo
dt
g(t) − g(0) = 2t; g(t) = 2t + g(0); g(0) =dvo(0)
dt
iR =5
500× 10−3 = 1 µA = iC = −C
dvo(0)dt
dvo(0)dt
=−1 × 10−6
1 × 10−6 = −1 = g(0)
dvo
dt= 2t − 1
dvo = 2t dt − dt
vo − vo(0) = t2 − t; vo(0) = 8 V
vo = t2 − t + 8, 0 ≤ t ≤ tsat
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; C1 = 0.1 µF; Rb = 25 kΩ; C2 = 1 µF
d2vo
dt2=( 1
RaC1
)( 1RbC2
)vg;
1RaC1
= 1001
RbC2= 40
vg = 250 × 10−3; therefored2vo
dt2= 1000
[b] Since vo(0) = 0 =dvo(0)
dt, our solution is vo = 500t2 V
The second op-amp will saturate when
vo = 6 V, or tsat =√
6/500 ∼= 0.1095 s
[c]dvo1
dt= − 1
RaC1vg = −25
[d] Since vo1(0) = 0, vo1 = −25t V
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-ampsaturates, 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 isthe same as Example 8.14.
[b] vo = 5 + A′1e
−10t + A′2e
−20t (from Example 8.14)
vo(0) = 4 = 5 + A′1 + A′
2
8–46 CHAPTER 8. Natural and Step Responses of RLC Circuits
4100
+ iC(0+) − 225
= 0
iC(0+) =4
100mA = C
dvo(0+)dt
dvo(0+)dt
=0.04 × 10−3
10−6 = 40 V/s
dvo
dt= −10A′
1e−10t − 20A′
2e−20t
dvo
dt(0+) = −10A′
1 − 20A′2 = 40
Therefore −A′1 − 2A′
2 = 4 and A′1 + A′
2 = −1Thus, A′
1 = 2 and A′2 = −3
vo = 5 + 2e−10t − 3e−20t V, t ≥ 0
[c] Same as Example 8.14:
dvo1
dt+ 20vo1 = −25
[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, t ≥ 0
P 8.56 [a]
2Cdva
dt+
va − vg
R+
va
R= 0
(1) Thereforedva
dt+
va
RC=
vg
2RC;
0 − va
R+ C
d(0 − vb)dt
= 0
(2) Thereforedvb
dt+
va
RC= 0, va = −RC
dvb
dt
Problems 8–47
2vb
R+ C
dvb
dt+ C
d(vb − vo)dt
= 0
(3) Thereforedvb
dt+
vb
RC=
12
dvo
dt
From (2) we havedva
dt= −RC
d2vb
dt2and va = −RC
dvb
dt
When these are substituted into (1) we get
(4) − RCd2vb
dt2− dvb
dt=
vg
2RC
Now differentiate (3) to get
(5)d2vb
dt2+
1RC
dvb
dt=
12
d2vo
dt2
But from (4) we have
(6)d2vb
dt2+
1RC
dvb
dt= − vg
2R2C2
Now substitute (6) into (5)
d2vo
dt2= − vg
R2C2
[b] When R1C1 = R2C2 = RC :d2vo
dt2=
vg
R2C2
The two equations are the same except for a reversal in algebraic sign.
[c] Two integrations of the input signal with one operational amplifier.
P 8.57 [a] f(t) = inertial force + frictional force + spring force
= M [d2x/dt2] + D[dx/dt] + Kx
[b]d2x
dt2=
f
M−(
D
M
)(dx
dt
)−(
K
M
)x
Given vA =d2x
dt2, then
vB = − 1R1C1
∫ t
0
(d2x
dy2
)dy = − 1
R1C1
dx
dt
vC = − 1R2C2
∫ t
0vB dy =
1R1R2C1C2
x
vD = −R3
R4· vB =
R3
R4R1C1
dx
dt
8–48 CHAPTER 8. Natural and Step Responses of RLC Circuits
vE =[R5 + R6
R6
]vC =
[R5 + R6
R6
]· 1R1R2C1C2
· x
vF =[−R8
R7
]f(t), vA = −(vD + vE + vF )
Therefored2x
dt2=[R8
R7
]f(t) −
[R3
R4R1C1
]dx
dt−[
R5 + R6
R6R1R2C1C2
]x
Therefore M =R7
R8, D =
R3R7
R8R4R1C1and K =
R7(R5 + R6)R8R6R1R2C1C2
Box Number Function
1 inverting and scaling
2 inverting and scaling
3 integrating and scaling
4 integrating and scaling
5 inverting and scaling
6 noninverting and scaling
P 8.58 [a] Given that the current response is underdamped we know i will be of the form
i = If + [B′1 cos ωdt + B′
2 sin ωdt]e−αt
where α =R
2L
and ωd =√
ω2o − α2 =
√1
LC− α2
The capacitor will force the final value of i to be zero, therefore If = 0.By hypothesis i(0+) = Vdc/R therefore B′
1 = 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
dt= [(ωdB
′2 − αB′
1) cos ωdt − (ωdB′1 + αB′
2) sin ωdt]e−αt
Hencedi(0+)
dt= ωdB
′2 − αB′
1 = 0
Thus
B′2 =
α
ωd
B′1 =
αVdc
ωdR
It follows directly that
i =Vdc
R
[cos ωdt +
α
ωd
sin ωdt]e−αt
Problems 8–49
[b] Since ωdB′1 − αB′
1 = 0 it follows that
di
dt= −(ωdB
′1 + αB′
2)e−αt sin ωdt
But αB′2 =
α2Vdc
ωdRand ωdB
′1 =
ωdVdc
R
Therefore
ωdB′1 + αB′
2 =ωdVdc
R+
α2Vdc
ωdR=
Vdc
R
[ω2
d + α2
ωd
]
But ω2d + α2 = ω2
o =1
LC
Hence
ωdB′1 + αB′
2 =Vdc
ωdRLC
Now since v1 = Ldi
dtwe get
v1 = −LVdc
ωdRLCe−αt sin ωdt = − Vdc
ωdRCe−αt sin ωdt
[c] vc = Vdc − iR − Ldi
dt
iR = Vdc
(cos ωdt +
α
ωd
sin ωdt)
e−αt
vc = Vdc − Vdc
(cos ωdt +
α
ωd
sin ωdt)
e−αt +Vdc
ωdRCe−αt sin ωdt
= Vdc − Vdce−αt cos ωdt +
(Vdc
ωdRC− αVdc
ωd
)e−αt sin ωdt
= Vdc
[1 − e−αt cos ωdt +
1ωd
( 1RC
− α)
e−αt sin ωdt]
= Vdc [1 − e−αt cos ωdt + Ke−αt sin ωdt]
P 8.59 vsp = Vdc
[1 − a
ωdRCe−αt sin ωdt
]
dvsp
dt=
−aVdc
ωdRC
d
dt[e−αt sin ωdt]
=−aVdc
ωdRC[−αe−αt sin ωdt + ωd cos ωdte
−αt]
=aVdce
−αt
ωdRC[α sin ωdt − ωd cos ωdt]
8–50 CHAPTER 8. Natural and Step Responses of RLC Circuits
dvsp
dt= 0 when α sin ωdt = ωd cos ωdt
or tan ωdt =ωd
α; ωdt = tan−1
(ωd
α
)
·. . tmax =1ωd
tan−1(
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 bemaximum when t has its smallest positive value. Hence
tmax =tan−1(ωd/α)
ωd
.
P 8.60 [a] vc = Vdc[1 − e−αt cos ωdt + Ke−αt sin ωdt]
dvc
dt= Vdc
d
dt[1 + e−αt(K sin ωdt − cos ωdt)]
= Vdc(−αe−αt)(K sin ωdt − cos ωdt)+
e−αt[ωdK cos ωdt + ωd sin ωdt]= Vdce
−αt[(ωd − αK) sin ωdt + (α + ωdK) cos ωdt]
dvc
dt= 0 when (ωd − αK) sin ωdt = −(α + ωdK) cos ωdt
or tan ωdt =[α + ωdK
αK − ωd
]
·. . ωdt ± nπ = tan−1[α + ωdK
αK − ωd
]
tc =1ωd
tan−1
(α + ωdK
αK − ωd
)± nπ
α =R
2L=
4 × 103
6= 666.67 rad/s
ωd =
√109
1.2− (666.67)2 = 28,859.81 rad/s
Problems 8–51
K =1ωd
( 1RC
− α)
= 21.63
tc =1ωd
tan−1(−43.29) + nπ
=
1ωd
−1.55 + nπ
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 ωdtc max + Ke−αtc max sin ωdtc max]
= 262.42 V
[c] From the text example the voltage across the spark plug reaches its maximumvalue in 53.63 µs. If the spark plug does not fire the capacitor voltage peaks in55.23 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 [a] w =12L[i(0+)]2 =
12(5)(16) × 10−3 = 40 mJ
[b] α =R
2L=
3 × 103
10= 300 rad/s
ωd =
√109
1.25− (300)2 = 28,282.68 rad/s
1RC
=106
0.75=
4 × 106
3
tmax =1ωd
tan−1(
ωd
α
)= 55.16 µs
vsp (tmax) = 12 − 12(50)(4 × 106)3(28,282.68)
e−αtmax sin ωdtmax = −27,808.04 V
[c] vc (tmax) = 12[1 − e−αtmax cos ωdtmax + Ke−αtmax sin ωdtmax]
K =1ωd
[ 1RC
− α]
= 47.13
vc (tmax) = 568.15 V
9Sinusoidal 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
9–2 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
AP 9.4 [a] XC =−1ωC
=−1
4000(5 × 10−6)= −50 Ω
[b] ZC = jXC = −j50 Ω
[c] I =VZC
=30/25
50/−90 = 0.6/115 A
[d] i = 0.6 cos(4000t + 115) A
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, therefore i4 = 0 A
AP 9.6 [a] I =125/−60
|Z|/θz
=125|Z| /(−60 − θZ)
But −60 − θZ = −105 ·. . θZ = 45
Z = 90 + j160 + jXC
·. . XC = −70 Ω; XC = − 1ωC
= −70
·. . C =1
(70)(5000)= 2.86 µF
[b] I =Vs
Z=
125/−60
(90 + j90)= 0.982/−105A; ·. . |I| = 0.982 A
AP 9.7 [a]
ω = 2000 rad/s
ωL = 10 Ω,−1ωC
= −20 Ω
Zxy = 20‖j10 + 5 + j20 =20(j10)
(20 + j10)+ 5 − j20
= 4 + j8 + 5 − j20 = (9 − j12) Ω
Problems 9–3
[b] ωL = 40 Ω,−1ωC
= −5 Ω
Zxy = 5 − j5 + 20‖j40 = 5 − j5 +[(20)(j40)20 + j40
]
= 5 − j5 + 16 + j8 = (21 + j3) Ω
[c] Zxy =[
20(jωL)20 + jωL
]+(
5 − j106
25ω
)
=20ω2L2
400 + ω2L2 +j400ωL
400 + ω2L2 + 5 − j106
25ωThe impedance will be purely resistive when the j terms cancel, i.e.,
400ωL
400 + ω2L2 =106
25ωSolving for ω yields ω = 4000 rad/s.
[d] Zxy =20ω2L2
400 + ω2L2 + 5 = 10 + 5 = 15 Ω
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
Zxy=
150/0
15= 10/0 A
Using current division,
IL =20
20 + j20(10) = 5 − j5 = 7.07/−45 A
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 itsequivalent 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
14 + 12 − j16 + 4= 4/28.07 A
AP 9.10 V1 = 240/53.13 = 144 + j192 V
V2 = 96/−90 = −j96 V
jωL = j(4000)(15 × 10−3) = j60 Ω
1jωC
= −j6 × 106
(4000)(25)= −j60 Ω
Perform source transformations:
V1
j60=
144 + j192j60
= 3.2 − j2.4 A
V2
20= −j
9620
= −j4.8 A
Combine the parallel impedances:
Y =1
j60+
130
+1
−j60+
120
=j5j60
=112
Z =1Y
= 12 Ω
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 Ωresistor and VTh = node voltage across the capacitor. Writing the node voltageequations gives us
V1
20− 2/45 +
V1 − 10Ix
j10= 0 and VTh =
−j1010 − j10
(10Ix)
We also have
Ix =V1
20
Solving these equations for VTh gives VTh = 10/45V. To find the Théveninimpedance, we remove the independent current source and apply a test voltagesource at the terminals a, b. Thus
It follows from the circuit that
10Ix = (20 + j10)Ix
Therefore
Ix = 0 and IT =VT
−j10+
VT
10
ZTh =VT
IT
, 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
−10 +V5
+V
−j(20/9)+
Vj5
+V − 100/−90
20= 0
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.
AP 9.14 [a] M = 0.4√
0.0625 = 0.1 H, ωM = 80 Ω
Z22 = 40 + j800(0.125) + 360 + j800(0.25) = (400 + j300) Ω
Therefore |Z22| = 500 Ω, Z∗22 = (400 − j300) Ω
Zr =( 80
500
)2
(400 − j300) = (10.24 − j7.68) Ω
[b] I1 =245.20
184 + 100 + j400 + Zr
= 0.50/− 53.13 A
i1 = 0.5 cos(800t − 53.13) A
[c] I2 =(
jωM
Z22
)I1 =
j80500/36.87 (0.5/− 53.13) = 0.08/0 A
i2 = 80 cos 800t mA
Problems 9–7
AP 9.15 I1 =Vs
Z1 + Z2/a2 =25 × 103/0
1500 + j6000 + (25)2(4 − j14.4)
= 4 + j3 = 5/36.87 A
V1 = Vs − Z1I1 = 25,000/0 − (4 + j3)(1500 + j6000)
= 37,000 − j28,500
V2 = − 125
V1 = −1480 + j1140 = 1868.15/142.39 V
I2 =V2
Z2=
1868.15/142.39
4 − j14.4= 125/− 143.13 A
Also, I2 = −25I1
9–8 CHAPTER 9. Sinusoidal Steady State Analysis
Problems
P 9.1 [a] ω = 2πf = 3769.91 rad/s, f =ω
2π= 600 Hz
[b] T = 1/f = 1.67 ms
[c] Vm = 10 V
[d] v(0) = 10 cos(−53.13) = 6 V
[e] φ = −53.13; φ =−53.13(2π)
360 = −0.9273 rad
[f] V = 0 when 3769.91t − 53.13 = 90. Now resolve the units:
(3769.91 rad/s)t =143.13
(180/π)= 2.498 rad, t = 662.64 µs
[g] (dv/dt) = (−10)3769.91 sin(3769.91t − 53.13)
(dv/dt) = 0 when 3769.91t − 53.13 = 0
or 3769.91t =53.13
57.3/rad= 0.9273 rad
Therefore t = 245.97 µs
P 9.2 Vrms =
√1T
∫ T/2
0V 2
m sin2 2πT
t dt
∫ T/2
0V 2
m sin2(2π
T
)t dt =
V 2m
2
∫ T/2
0
(1 − cos
4πT
t)
dt =V 2
mT
4
Therefore Vrms =
√1T
V 2mT
4=
Vm
2
P 9.3 [a] 40 V
[b] 2πf = 100π; f = 50Hz
[c] ω = 100π = 314.159 rad/s
[d] θ(rad) =2π
360 (60) =π
3= 1.05 rad
[e] θ = 60
[f] T =1f
=150
= 20 ms
[g] v = −40 when
100πt +π
3= π; ·. . t = 6.67 ms
Problems 9–9
[h] v = 40 cos[100π
(t − 0.01
3
)+
π
3
]
= 40 cos[100πt − (π/3) + (π/3)]
= 40 cos 100πt V
[i] 100π(t − to) + (π/3) = 100πt − (π/2)
·. . 100πto =5π6
; to = 8.33 ms
[j] 100π(t + to) + (π/3) = 100πt + 2π
·. . 100πto =5π3
; 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
dt= −80ω sin(ωt + θ)
·. . 80ω = 80,000; ω = 1000 rad/s
[b] f =ω
2π= 159.155 Hz; T =
1f
= 6.28 ms
−2π/36.28
= −0.3333, ·. . θ = −90 − (−0.3333)(360) = 30
·. . v = 80 cos(1000t + 30) V
9–10 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.6 [a]T
2= 8 + 2 = 10 ms; T = 20 ms
f =1T
=1
20 × 10−3 = 50Hz
[b] v = Vm sin(ωt + θ)
ω = 2πf = 100π rad/s
100π(−2 × 10−3) + θ = 0; ·. . θ =π
5rad = 36
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
toV 2
m cos2(ωt + φ) dt
= V 2m
∫ to+T
to
12
+12
cos(2ωt + 2φ) dt
=V 2
m
2
∫ to+Tto
dt +∫ to+T
tocos(2ωt + 2φ) dt
=V 2
m
2
T +
12ω
[sin(2ωt + 2φ) |to+T
to
]
=V 2
m
2
T +
12ω
[sin(2ωto + 4π + 2φ) − sin(2ωto + 2φ)]
= V 2m
(T
2
)+
12ω
(0) = V 2m
(T
2
)
P 9.8 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, ωL = 60√
R2 + ω2L2 = 100
φ = 25, θ = tan−1 60/80, θ = 36.87
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 mASteady-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 9–11
[e] The current lags the voltage by 36.87.
P 9.10 [a] From Eq. 9.9 we have
Ldi
dt=
VmR cos(φ − θ)√R2 + ω2L2
e−(R/L)t − ωLVm sin(ωt + φ − θ)√R2 + ω2L2
Ri =−VmR cos(φ − θ)e−(R/L)t
√R2 + ω2L2
+VmR cos(ωt + φ − θ)√
R2 + ω2L2
Ldi
dt+ Ri = Vm
[R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)√
R2 + ω2L2
]
ButR√
R2 + ω2L2= cos θ and
ωL√R2 + ω2L2
= sin θ
Therefore the right-hand side reduces to
Vm cos(ωt + φ)
At t = 0, Eq. 9.9 reduces to
i(0) =−Vm cos(φ − θ)√
R2 − ω2L2+
Vm cos(φ − θ)√R2 + ω2L2
= 0
[b] iss =Vm√
R2 + ω2L2cos(ωt + φ − θ)
Therefore
Ldissdt
=−ωLVm√R2 + ω2L2
sin(ωt + φ − θ)
and
Riss =VmR√
R2 + ω2L2cos(ωt + φ − θ)
Ldissdt
+ Riss = Vm
[R cos(ωt + φ − θ) − ωL sin(ωt + φ − θ)√
R2 + ω2L2
]
= 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)
9–12 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
[c] I =200/0
jωL=
200ωL
/− 90 = 25/− 90; θi = −90
[d]200ωL
= 25; ωL =20025
= 8 Ω
[e] L =8
2π(1000)= 1.27 mH
[f] ZL = jωL = j8 Ω
P 9.13 [a] ω = 2πf = 314,159.27 rad/s
[b] I =VZC
=10 × 10−3/0
1/jωC= jωC(10 × 10−3)/0 = 10 × 10−3ωC/90
·. . θi = 90
[c] 628.32 × 10−6 = 10 × 10−3 ωC
1ωC
=10 × 10−3
628.32 × 10−6 = 15.92 Ω, ·. . XC = −15.92 Ω
[d] C =1
15.92(ω)=
1(15.92)(100π × 103)
C = 0.2 µF
[e] Zc = j(−1
ωC
)= −j15.92 Ω
P 9.14 [a] jωL = j(2 × 104)(300 × 10−6) = j6 Ω
1jωC
= −j1
(2 × 104)(5 × 10−6)= −j10 Ω; Ig = 922/30 A
Problems 9–13
[b] Vo = 922/30Ze
Ze =1Ye
; Ye =110
+ j110
+1
8 + j6
Ye = 0.18 + j0.04 S
Ze =1
0.18 + j0.04= 5.42/− 12.53 Ω
Vo = (922/30 )(5.42/− 12.53 ) = 5000.25/17.47 V
[c] vo = 5000.25 cos(2 × 104t + 17.47) V
P 9.15 [a] ZL = j(8000)(5 × 10−3) = j40 Ω
ZC =−j
(8000)(1.25 × 10−6)= −j100 Ω
[b] I =600/20
40 + j40 − j100= 8.32/76.31 A
[c] i = 8.32 cos(8000t + 76.31) A
P 9.16 Z = 4 + j(50)(0.24) − j1
(50)(0.0025)= 4 + j4 = 5.66/45 Ω
Io =VZ
=0.1/− 90
5.66/45 = 17.68/− 135 mA
io(t) = 17.68 cos(50t − 135) mA
P 9.17 [a] Y =1
3 + j4+
116 − j12
+1
−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
9–14 CHAPTER 9. Sinusoidal Steady State Analysis
[d] I = 8/0 A, V =IY
=8
0.2/36.87 = 40/−36.87 V
IC =VZC
=40/−36.87
4/−90 = 10/53.13 A
iC = 10 cos(ωt + 53.13) A, Im = 10 A
P 9.18 ZL = j(2000)(60 × 10−3) = j120 Ω
ZC =−j
(2000)(12.5 × 10−6)= −j40 Ω
Construct the phasor domain equivalent circuit:
Using current division:
I =(120 − j40)
120 − j40 + 40 + j120(0.5) = 0.25 − j0.25 A
Vo = j120I = 30 + j30 = 42.43/45 V
vo = 42.43 cos(2000t + 45) V
P 9.19 [a] Vg = 300/78; Ig = 6/33
·. . Z =Vg
Ig
=300/78
6/33 = 50/45 Ω
[b] ig lags vg by 45:
2πf = 5000π; f = 2500 Hz; T = 1/f = 400 µs
·. . ig lags vg by45
360 (400 µs) = 50 µs
Problems 9–15
P 9.201
jωC=
1(1 × 10−6)(50 × 103)
= −j20 Ω
jωL = j50 × 103(1.2 × 10−3) = j60 Ω
Vg = 40/0 V
Ze = −j20 + 30‖j60 = 24 − j8 Ω
Ig =40/0
24 − j8= 1.5 + j0.5 mA
Vo = (30‖j60)Ig =30(j60)30 + j60
(1.5 + j0.5) = 30 + j30 = 42.43/45 V
vo = 42.43 cos(50,000t + 45) V
P 9.21 [a] Z1 = R1 − j1
ωC1
Z2 =R2/jωC2
R2 + (1/jωC2)=
R2
1 + jωR2C2=
R2 − jωR22C2
1 + ω2R22C
22
Z1 = Z2 when R1 =R2
1 + ω2R22C
22
and
1ωC1
=ωR2
2C2
1 + ω2R22C
22
or C1 =1 + ω2R2
2C22
ω2R22C2
[b] R1 =1000
1 + (40 × 103)2(1000)2(50 × 10−9)2 = 200 Ω
C1 =1 + (40 × 103)2(1000)2(50 × 10−9)2
(40 × 103)2(1000)2(50 × 10−9)= 62.5 nF
9–16 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.22 [a] Y2 =1R2
+ jωC2
Y1 =1
R1 + (1/jωC1)=
jωC1
1 + jωR1C1=
ω2R1C21 + jωC1
1 + ω2R21C
21
Therefore Y1 = Y2 when
R2 =1 + ω2R2
1C21
ω2R1C21
and C2 =C1
1 + ω2R21C
21
[b] R2 =1 + (50 × 103)2(1000)2(40 × 10−9)2
(50 × 103)2(1000)(40 × 10−9)2 = 1250 Ω
C2 =40 × 10−9
1 + (50 × 103)2(1000)2(40 × 10−9)2 = 8 nF
P 9.23 [a] Z1 = R1 + jωL1
Z2 =R2(jωL2)R2 + jωL2
=ω2L2
2R2 + jωL2R22
R2 + ω2L22
Z1 = Z2 when R1 =ω2L2
2R2
R22 + ω2L2
2and L1 =
R22L2
R22 + ω2L2
2
[b] R1 =(4000)2(1.25)2(5000)50002 + 40002(1.25)2 = 2500 Ω
L1 =(5000)2(1.25)
50002 + 40002(1.25)2 = 625 mH
P 9.24 [a] Y2 =1R2
− j
ωL2
Y1 =1
R1 + jωL1=
R1 − jωL1
R21 + ω2L2
1
Therefore Y2 = Y1 when
R2 =R2
1 + ω2L21
R1and L2 =
R21 + ω2L2
1
ω2L1
[b] R2 =80002 + 10002(4)2
8000= 10 kΩ
L2 =80002 + 10002(4)2
10002(4)= 20 H
Problems 9–17
P 9.25 Vg = 500/30 V; Ig = 0.1/83.13 mA
Z =Vg
Ig
= 5000/− 53.13 Ω = 3000 − j4000 Ω
z = 3000 + j
(ω − 32 × 103
ω
)
ω − 32 × 103
ω= −4000
ω2 + 4000ω − 32 × 103 = 0
ω = 7.984 rad/s
P 9.26 [a] Zeq =50,000
3+
−j20 × 106
ω‖(1200 + j0.2ω)
=50,000
3+
−j20 × 106
ω
(1200 + j0.2ω)1200 + j[0.2ω − 20×106
ω]
=50,000
3+
−j20×106
ω(1200 + j0.2ω)
[1200 − j
(0.2ω − 20×106
ω
)]12002 +
(0.2ω − 20×106
ω
)2
Im(Zeq) = −20 × 106
ω(1200)2 − 20 × 106
ω
[0.2ω
(0.2ω − 20 × 106
ω
)]= 0
−20 × 106(1200)2 − 20 × 106
[0.2ω
(0.2ω − 20 × 106
ω
)]= 0
−(1200)2 = 0.2ω(
0.2ω − 20 × 106
ω
)
0.22ω2 − 0.2(20 × 106) + 12002 = 0
ω2 = 64 × 106 ·. . ω = 8000 rad/s
·. . f = 1273.24 Hz
[b] Zeq =50,000
3+ −j2500‖(1200 + j1600)
=50,000
3+
(−j2500)(1200 + j1600)1200 − j900
= 20,000 Ω
Ig =30/0
20,000= 1.5/0 mA
ig(t) = 1.5 cos 8000t mA
9–18 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.27 [a] Find the equivalent impedance seen by the source, as a function of L, and set theimaginary part of the equivalent impedance to 0, solving for L:
ZC =−j
(500)(2 × 10−6)= −j1000 Ω
Zeq = −j1000 + j500L‖2000 = −j1000 +2000(j500L)2000 + j500L
= −j1000 +2000(j500L)(2000 − j500L)
20002 + (500L)2
Im(Zeq) = −1000 +20002(500L)
20002 + (500L)2 = 0
·. .20002(500L)
20002 + (500L)2 = 1000
·. . 5002L2 − 1220002L + 20002 = 0
Solving the quadratic equation, L = 4 H
[b] Ig =100/0
−j1000 + j2000‖2000=
100/0
1000= 0.1/0 A
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)
ω2C2R2 + 1
Im(Zab) = ωL − ωCR2
ω2C2R2 + 1= 0
·. . L =CR2
ω2C2R2 + 1
·. . ω2C2R2 + 1 =CR2
L
·. . ω2 =(CR2/L) − 1
C2R2 =(25×10−9)(100)2
160×10−6 − 1(25 × 10−9)2(100)2 = 900 × 108
ω = 300 krad/s
Problems 9–19
[b] Zab(300 × 103) = j48 +(100)(−j133.33)100 − j133.33
= 64 Ω
P 9.29 jωL = j100 × 103(0.6 × 10−3) = j60 Ω
1jωC
=−j
(100 × 103)(0.4 × 10−6)= −j25 Ω
VT = −j25IT + 5I∆ − 30I∆
I∆ =−j60
30 + j60IT
VT = −j25IT + 25j60
30 + j60IT
VT
IT
= Zab = 20 − j15 = 25/− 36.87 Ω
P 9.30 [a] Z1 = 400 − j106
500(2.5)= 400 − j800 Ω
Z2 = 2000‖j500L =j106L
2000 + j500L
ZT = Z1 + Z2 = 400 − j800 +j106L
2000 + j500L
= 400 +500 × 106L2
20002 + 5002L2 − j800 + j2 × 109L
20002 + 5002L2
ZT is resistive when
2 × 109L
20002 + 5002L2 = 800 or 5002L2 − 25 × 105L + 20002 = 0
Solving, L1 = 8 H and L2 = 2 H.
9–20 CHAPTER 9. Sinusoidal Steady State Analysis
[b] When L = 8 H:
ZT = 400 +500 × 106(8)2
20002 + 5002(8)2 = 2000 Ω
Ig =200/0
2000= 100/0 mA
ig = 100 cos 500t mA
When L = 2 H:
ZT = 400 +500 × 106(2)2
20002 + 500(2)2 = 800 Ω
Ig =200/0
800= 250/0 mA
ig = 250 cos 500t mA
P 9.31 [a] Y1 =11
2500 × 103 = 4.4 × 10−6 S
Y2 =1
14,000 + j5ω
=14,000
196 × 106 + 25ω2 − j5ω
196 × 106 + 25ω2
Y3 = jω2 × 10−9
YT = Y1 + 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 + 25ω2
or
25ω2 + 196 × 106 =5
2 × 10−9
·. . 25ω2 = 2304 × 106 ·. . ω = 9600 rad/s
[b] YT = 4.4 × 10−6 +14,000
196 × 106 + 25(9600)2 = 10 × 10−6 S
·. . ZT = 100 kΩ
Vo = (0.25 × 10−3/0)(100 × 103) = 25/0 V
vo = 25 cos 9600t V
Problems 9–21
P 9.32 [a] Zg = 500 − j106
ω+
103(j0.5ω)103 + j0.5ω
= 500 − j106
ω+
500jω(1000 − j0.5ω)106 + 0.25ω2
= 500 − j106
ω+
250ω2
106 + 0.25ω2 + j5 × 105ω
106 + 0.25ω2
·. . If Zg is purely real,106
ω=
5 × 105ω
106 + 0.25ω2
2(106 + 0.25ω2) = ω2 ·. . 4 × 106 = ω2
·. . ω = 2000 rad/s
[b] When ω = 2000 rad/s
Zg = 500 − j500 + (j1000‖1000) = 1000 Ω
·. . Ig =20/0
1000= 20/0 mA
Vo = Vg − IgZ1
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) + (40‖j20)
= 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
2 − j6+
112 + j4
+12
+1
j0.5= 0.625 − j1.875 S
Zp =1Yp
=1
0.625 − j1.875= 0.16 + j0.48 Ω
Zab = −j4.48 + 0.16 + j0.48 + 2.84 = 3 − j4 Ω
Yab =1
Zab=
13 − j4
= 120 + j160 mS
= 200/53.13 mS
9–22 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.35 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)
1 + j1 − j1= j1 Ω
Z2 =(−j1)(1)
1 + j1 − j1= −j1 Ω
Z3 =(j1)(−j1)1 + j1 − j1
= 1 Ω
Convert the lower right delta to a wye:
Z4 =(−j1)(1)
1 + j1 − j1= −j1 Ω
Z5 =(−j1)(j1)1 + j1 − j1
= 1 Ω
Z6 =(j1)(1)
1 + j1 − j1= j1 Ω
The resulting circuit is shown below:
Simplify the middle portion of the circuit by making series and parallelcombinations:
(1 + j1 − j1)‖(1 + 1) = 1‖2 = 2/3 Ω
Zab = −j1 + 2/3 + j1 = 2/3 Ω
Problems 9–23
P 9.36 Vo = VgZo
ZT
=500 − j1000
300 + j1600 + 500 − j1000(100/0) = 111.8/− 100.3 V
vo = 111.8 cos(8000t − 100.3) V
P 9.371
jωC= −j400 Ω
jωL = j1200 Ω
Let Z1 = 200 − j400 Ω; Z2 = 600 + j1200 Ω
Ig = 400/0 mA
Io =Z2
Z1 + Z2Ig =
600 + j1200800 + j800
(0.4/0)
= 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
4 − j3= 2.4 + j1.8 A
I2 = I1 − j5 = (2.4 + j1.8) − j5 = 2.4 − j3.2 A
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 =VZ
IZ
=−1 − j123.8 − j3.4
= 1.42 − j1.88 Ω
P 9.39 Is = 3/0 mA
1jωC
= −j0.4 Ω
jωL = j0.4 Ω
After source transformation we have
Vo =−j0.4‖j0.4‖5
28 + −j0.4‖j0.4‖5(66 × 10−3) = 10 mV
vo = 10 cos 200t mV
P 9.40 [a]
Va = (50 + j150)(2/0) = 100 + j300 V
Ib =100 + j300120 − j40
= j2.5 A = 2.5/90 A
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 9–25
[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
1jωC
= −j106
(1000)(10)= −j100 Ω
Using voltage division,
Vab =(100 + j100)‖(−j100)
j100 + (100 + j100)‖(−j100)(247.49/45) = 350/0
VTh = Vab = 350/0 V
[b] Remove the voltage source and combine impedances in parallel to findZTh = Zab:
Yab =1
j100+
1100 + j100
+1
−j100= 5 − j5 mS
ZTh = Zab =1
Yab= 100 + j100 Ω
[c]
9–26 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.42 Using voltage division:
VTh =36
36 + j60 − j48(240) = 216 − j72 = 227.68/− 18.43 V
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
10+ 88Iφ +
V2 − 15V2
−j50= 0
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
10+ 88Iφ +
0.8Vt
−j50
Iφ =−VT /5
200
IT = VT
(110
− 881/5200
+0.8
−j50
)
Problems 9–27
·. .VT
IT
= 30 − j40 = ZTh
IN =VTh
ZTh=
−66 + j8830 − j40
= −2.2 + j0 A = 2.2/180 A
The Norton equivalent circuit:
P 9.44 Short circuit current
Iβ =−6Iβ
2
2Iβ = −6Iβ; ·. . Iβ = 0
I1 = 0; ·. . Isc = 10/−45 A = IN
The Norton impedance is the same as the Thévenin impedance. Find it using a testsource
VT = 6Iβ + 2Iβ = 8Iβ, Iβ =j1
2 + j1IT
9–28 CHAPTER 9. Sinusoidal Steady State Analysis
ZTh =VT
IT
=8Iβ
[(2 + j1)/j1]Iβ
=j8
2 + j1= 1.6 + j3.2 Ω
P 9.45 Using current division:
IN = Isc =50
80 + j60(4) = 1.6 − j1.2 = 2/− 36.87 A
ZN = −j100‖(80 + j60) = 100 − j50 Ω
The Norton equivalent circuit:
P 9.46 ω = 2π(200/π) = 400 rad/s
Zc =−j
400(10−6)= −j2500 Ω
VT = (10,000 − j2500)IT + 100(200)IT
ZTh =VT
IT
= 30 − j2.5 kΩ
P 9.47
IN =5 − j15
ZN+ (1 − j3) mA, ZN in kΩ
Problems 9–29
IN =−18 − j13.5
ZN
+ 4.5 − j6 mA, ZN in kΩ
5 − j15ZN
+ 1 − j3 =−18 − j13.5
ZN
+ (4.5 − j6)
23 − j1.5ZN
= 3.5 − j3 ·. . ZN = 4 + j3 kΩ
IN =5 − j154 + j3
+ 1 − j3 = −j6 mA = 6/− 90 mA
P 9.48 Open circuit voltage:
V1 − 25020 + j10
− 0.03Vo +V1
50 − j100= 0
·. . Vo =−j100
50 − j100V1
9–30 CHAPTER 9. Sinusoidal Steady State Analysis
V1
20 + j10+
j3V1
50 − j100+
V1
50 − j100=
25020 + j10
V1 = 500 − j250 V; Vo = 300 − j400 V = VTh = 500/− 53.13 V
Short circuit current:
Isc =250/0
70 + j10= 3.5 − j0.5 A
ZTh =VTh
Isc=
300 − j4003.5 − j0.5
= 100 − j100 Ω
The Thévenin equivalent circuit:
P 9.49 Open circuit voltage:
(9 + j4)Ia − Ib = −60/0
Problems 9–31
−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
Isc=
10/0
2.07/0 = 4.83 Ω
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 − j49 − j4
Z3 =16 − j169 − j4
Za = 4 + j4 +4
9 − j4=
56 + j209 − j4
Zb = 4 +4 − j49 − j4
=40 − j209 − j4
Za‖Zb =2640 − j320864 − j384
Z3 + Za‖Zb =16 − j169 − j4
+2640 − j320864 − j384
=4176 − j1856864 − j384
= 4.83 Ω
Problems 9–33
P 9.50 [a]
IT =VT
1000+
VT − αVT
−j1000
IT
VT
=1
1000− (1 − α)
j1000=
j − 1 + α
j1000
·. . ZTh =VT
IT
=j1000
α − 1 + j
ZTh is real when α = 1.
[b] ZTh = 1000 Ω
[c] ZTh = 500 − j500 =j1000
α − 1 + j
=1000
(α − 1)2 + 1+ j
1000(α − 1)(α − 1)2 + 1
Equate the real parts:
1000(α − 1)2 + 1
= 500 ·. . (α − 1)2 + 1 = 2
·. . (α − 1)2 = 1 so α = 0
Check the imaginary parts:
(α − 1)1000(α − 1)2 + 1
∣∣∣∣α=1
= −500
Thus, α = 0.
[d] ZTh =1000
(α − 1)2 + 1+ j
1000(α − 1)(α − 1)2 + 1
For Im(ZTh) > 0, α must be greater than 1. So ZTh is inductive for1 < α ≤ 10.
9–34 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.51
V1 − 240j10
+V1
50+
V1
30 + j10= 0
Solving for V1 yields
V1 = 198.63/− 24.44 V
Vo =30
30 + j10(V1) = 188.43/− 42.88 V
P 9.52 jωL = j(2000)(1 × 10−3) = j2 Ω
1jωC
= −j106
(2000)(100)= −j5 Ω
Vg1 = 20/− 36.87 = 16 − j12 V
Vg2 = 50/−106.26 = −14 − j48 V
Vo − (16 − j12)j2
+Vo
10+
Vo − (−14 − j48)−j5
= 0
Solving,
Vo = 36/0 V
vo(t) = 36 cos 2000t V
Problems 9–35
P 9.53 From the solution to Problem 9.52 the phasor-domain circuit is
Making two source transformations yields
Ig1 =16 − j12
j2= −6 − j8 A
Ig2 =−14 − j48
−j5= 9.6 − j2.8 A
Y =1j2
+110
+1
−j5= (0.1 − j0.3) S
Z =1Y
= 1 + j3 Ω
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
9–36 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.54 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-handsource removed is
V′o =
10‖ − j5j2 + 10‖ − j5
(16 − j12) = 18 − j26 V
Problems 9–37
With the left hand source removed
V′′o =
10‖j2−j5 + 10‖j2
(−14 − j48) = 18 + j26 V
Vo = V′o + V′′
o = 18 − j26 + 18 + j26 = 36 V
vo(t) = 36 cos 2000t V
P 9.56 Write a KCL equation at the top node:
Vo
−j8+
Vo − 2.4I∆
j4+
Vo
5− (10 + j10) = 0
The constraint equation is:
I∆ =Vo
−j8
Solving,
Vo = j80 = 80/90 V
P 9.57
Write node voltage equations:
Left Node:
V1
40+
V1 − Vo/8j20
= 0.025/0
Right Node:
Vo
50+
Vo
j25+ 16Io = 0
9–38 CHAPTER 9. Sinusoidal Steady State Analysis
The constraint equation is
Io =V1 − Vo/8
j20
Solution:
Vo = (4 + j4) = 5.66/45 VV1 = (0.8 + j0.6) = 1.0/36.87 VIo = (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 9–39
−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:
At V1: − 5/0 +V1 − V2
−j8+
V1 − 20/90
−j4= 0
At V2:V2 − V1
−j8+
V2
j4+
V2 − 20/90
12= 0
In standard form:
V1
(1
−j8+
1−j4
)+ V2
(− 1
−j8
)= 5/0 +
20/90
−j4
V1
(− 1
−j8
)+ V2
(1
−j8+
1j4
+112
)=
20/90
12
Solving on a calculator:
V1 = −83
+ j43
V V2 = −8 + j4 V
Thus
V0 = V1 − 20/90 = −83
− j563
= 18.86/− 98.13 V
9–40 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.61 jωL = j5000(60 × 10−3) = j300 Ω
1jωC
=−j
(5000)(2 × 10−6)= −j100 Ω
−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 9–41
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)20
+Va
j5+
Va − (140 + j30)12 + j16
= 0
Solving,
Va = 40 + j30 V
IZ + (30 + j20) − 140 + j30−j10
+(40 + j30) − (140 + j30)
12 + j16= 0
Solving,
IZ = −30 − j10 A
Z =(100 − j50) − (140 + j30)
−30 − j10= 2 + j2 Ω
9–42 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.64 [a]1
jωC= −j50 Ω
jωL = j120 Ω
Ze = 100‖ − j50 = 20 − j40 Ω
Ig = 2/0
Vg = IgZe = 2(20 − j40) = 40 − j80 V
Vo =j120
80 + j80(40 − j80) = 90 − j30 = 94.87/− 18.43 V
vo = 94.87 cos(16 × 105t − 18.43) V
[b] ω = 2πf = 16 × 105; f =8 × 105
π
T =1f
=π
8 × 105 = 1.25π µs
·. .18.43360
(1.25π µs) = 201.09 ns
·. . vo lags ig by 201.09 ns
P 9.65 jωL = j106(10 × 10−6) = j10 Ω
1jωC
=−j
(106)(0.1 × 10−6)= −j10 Ω
Va = 50/− 90 = −j50 V
Vb = 25/90 = j25 V
(10 − j10)I1 + j10I2 − 10I3 = −j50
Problems 9–43
j10I1 + 10I2 − 10I3 = 0
−10I1 − 10I2 + 20I3 = j25
Solving,
I1 = 0.5 − j1.5 A; I3 = −1 + j0.5 A I2 = −2.5 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(106t + 108.43) A
ib = 1.12 cos(106t − 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 =M√L1L2
=2√16
= 0.5
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
w =12L1i
21 +
12L2i
22 + Mi1i2 =
12(2 × 10−3)(9) + 0 + 0 = 9 mJ
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
w =12(2 × 10−3)(16) +
12(8 × 10−3)(1) + 2 × 10−3(−4)(1) = 12 mJ
P 9.67 Remove the voltage source to find the equivalent impedance:
ZTh = 45 + j125 +(
20|5 + j5|
)2
(5 − j5) = 85 + j85 Ω
Using voltage division:
VTh = Vcd = j20I1 = j20(
4255 + j5
)= 850 + j850 V = 1202.1/45 V
P 9.68 [a] jωL1 = j(200 × 103)(10−3) = j200 Ω
jωL2 = j(200 × 103)(4 × 10−3) = j800 Ω
1jωC
=−j
(200 × 103)(12.5 × 10−9)= −j400 Ω
·. . Z22 = 100 + 200 + j800 − j400 = 300 + j400 Ω
·. . Z∗22 = 300 − j400 Ω
M = k√
L1L2 = 2k × 10−3
Problems 9–45
ωM = (200 × 103)(2k × 10−3) = 400k
Zr =[400k500
]2
(300 − j400) = k2(192 − j256) Ω
Zin = 200 + j200 + 192k2 − j256k2
|Zin| = [(200 + 192k2)2 + (200 − 256k2)2]12
d|Zin|dk
=12[(200 + 192k2)2 + (200 − 256k2)2]−
12 ×
[2(200 + 192k2)384k + 2(200 − 256k2)(−512k)]
d|Zin|dk
= 0 when
768k(200 + 192k2) − 1024k(200 − 256k2) = 0
·. . k2 = 0.125; ·. . k =√
0.125 = 0.3536
[b] Zin (min) = 200 + 192(0.125) + j[200 − 0.125(256)]
= 224 + j168 = 280/36.87 Ω
I1 (max) =560/0
224 + j168= 2/− 36.87 A
·. . i1 (peak) = 2 A
Note — You can test that the k value obtained from setting d|Zin|/dk = 0leads 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
9–46 CHAPTER 9. Sinusoidal Steady State Analysis
1jωC
= −j20 Ω
jωM = j(4 × 103)k√
(12.5)(8) × 10−3 = j40k Ω
Z22 = 5 + j32 − j20 = 5 + j12 Ω
Z∗22 = 5 − j12 Ω
Zr =[
40k|5 + j12|
]2
(5 − j12) = 47.337k2 − j113.609k2
Zab = 20 + j50 + 47.337k2 − j113.609k2 = (20 + 47.337k2) + j(50 − 113.609k2)
Zab is resistive when
50 − 113.609k2 = 0 or k2 = 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
Z∗22 = 800 − j600 Ω
ωM = 270 Ω
Zr =( 270
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
I3= 80/60 Ω
Problems 9–47
V2
10=
V3
1; 10I2 = 1I3
V1
8= −V2
1; 8I1 = −1I2
Zab =V1
I1
Substituting,
Zab =V1
I1=
−8V2
−I2/8=
82V2
I2
=82(10V3)
I3/10=
(8)2(10)2V3
I3= (8)2(10)2ZL = 512, 000/60 Ω
P 9.72 In Eq. 9.69 replace ω2M2 with k2ω2L1L2 and then write Xab as
Xab = ωL1 − k2ω2L1L2(ωL2 + ωLL)R2
22 + (ωL2 + ωLL)2
= ωL1
1 − k2ωL2(ωL2 + ωLL)
R222 + (ωL2 + ωLL)2
For Xab to be negative requires
R222 + (ωL2 + ωLL)2 < k2ωL2(ωL2 + ωLL)
or
R222 + (ωL2 + ωLL)2 − k2ωL2(ωL2 + ωLL) < 0
which reduces to
R222 + ω2L2
2(1 − k2) + ωL2ωLL(2 − k2) + ω2L2L < 0
But k ≤ 1 hence it is impossible to satisfy the inequality. Therefore Xab can neverbe negative if XL is an inductive reactance.
9–48 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.73 [a]
Zab =Vab
I1 + I2=
V2
I1 + I2=
V2
(1 + N1/N2)I1
N1I1 = N2I2, I2 =N1
N2I1
V1
V2=
N1
N2, V1 =
N1
N2V2
V1 + V2 = ZLI1 =(
N1
N2+ 1
)V2
Zab =I1ZL
(N1/N2 + 1)(1 + N1/N2)I1
·. . Zab =ZL
[1 + (N1/N2)]2Q.E.D.
[b] Assume dot on the N2 coil is moved to the lower terminal. Then
V1 = −N1
N2V2 and I2 = −N1
N2I1
As before
Zab =V2
I1 + I2and V1 + V2 = ZLI1
·. . Zab =V2
(1 − N1/N2)I1=
ZLI1
[1 − (N1/N2)]2I1
Zab =ZL
[1 − (N1/N2)]2Q.E.D.
Problems 9–49
P 9.74 [a]
Zab =Vab
I1=
V1 + V2
I1
V1
N1=
V2
N2, V2 =
N2
N1V1
N1I1 = N2I2, I2 =N1
N2I1
V2 = (I1 + I2)ZL = I1
(1 +
N1
N2
)ZL
V1 + V2 =(
N1
N2+ 1
)V2 =
(1 +
N1
N2
)2
ZLI1
·. . Zab =(1 + N1/N2)2ZLI1
I1
Zab =(1 +
N1
N2
)2
ZL Q.E.D.
[b] Assume dot on N2 is moved to the lower terminal, then
V1
N1=
−V2
N2, V1 =
−N1
N2V2
N1I1 = −N2I2, I2 =−N1
N2I1
As in part [a]
V2 = (I2 + I1)ZL and Zab =V1 + V2
I1
Zab =(1 − N1/N2)V2
I1=
(1 − N1/N2)(1 − N1/N2)ZLI1
I1
Zab = [1 − (N1/N2)]2 ZL Q.E.D.
9–50 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.75 [a] I =24024
+240j32
= (10 − j7.5) A
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 =240j7.5
= −j32 Ω
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
2− IRx; I =
Vm/0
Rx − jXC
As Rx varies from 0 to ∞, the amplitude of vo remains constant and its phase angledecreases from 0 to −180, as shown in the following phasor diagram:
Problems 9–51
P 9.77 [a]
I =1207.5
+120j12
= 16 − j10 A
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 =1202.5
+120j4
= 48 − j30 A
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 thandoubled) to maintain the load voltage at 120 V.
9–52 CHAPTER 9. Sinusoidal Steady State Analysis
[d] I =1202.5
+120j4
+120−j2
= 48 + j30 A
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 maintainthe load voltage at 120 V.
P 9.78 Vg = 4/0 V;1
jωC= −j20 kΩ
Let Va = voltage across the capacitor, positive at upper terminalThen:
Va − 4/0
20,000+
Va
−j20,000+
Va
20,000= 0; ·. . Va = (1.6 − j0.8) V
0 − Va
20,000+
0 − Vo
10,000= 0; Vo = −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
20,000+ jωCoVa +
Va
20,000= 0
Va =4
2 + j20,000ωCo
Vo = −Va
2(see solution to Prob. 9.78)
Problems 9–53
Vo =−2
2 + j4 × 106Co
=2/180
2 + j4 × 106Co
·. . denominator angle = 45
so 4 × 106Co = 2 ·. . Co = 0.5 µF
[b] Vo =2/180
2 + j2= 0.707/135 V
vo = 0.707 cos(200t + 135) V
P 9.801
jωC1= −j10 kΩ
1jωC2
= −j100 kΩ
Va − 25000
+Va
−j10,000+
Va
20,000+
Va − Vo
100,000= 0
20Va − 40 + j10Va + 5Va + Va − Vo = 0
·. . (26 + j10)Va − Vo = 40
0 − Va
20,000+
0 − Vo
−j100,000= 0
j5Va − Vo = 0
Solving,
Vo = 1.43 + j7.42 = 7.55/79.11 V
vo(t) = 7.55 cos(106t + 79.11) V
9–54 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.81 [a] Vg = 25/0 V
Vp =20100
Vg = 5/0; Vn = Vp = 5/0 V
580,000
+5 − Vo
Zp= 0
Zp = −j80,000‖40,000 = 32,000 − j16,000 Ω
Vo =5Zp
80,000+ 5 = 7 − j1 = 7.07/− 8.13 V
vo = 7.07 cos(50,000t − 8.13) V
[b] Vp = 0.2Vm/0; Vn = Vp = 0.2Vm/0
0.2Vm
80,000+
0.2Vm − Vo
32,000 − j16,000= 0
·. . Vo = 0.2Vm +32,000 − j16,000
80,000Vm(0.2) = 0.2Vm(1.4 − j0.2)
·. . |0.2Vm(1.4 − j0.2)| ≤ 10
·. . Vm ≤ 35.36 V
P 9.82 [a]1
jωC= −j20 Ω
Vn
20+
Vn − Vo
−j20= 0
Vo
−j20=
Vn
20+
Vn
−j20
Vo = −j1Vn + Vn = (1 − j1)Vn
Vp =Vg(1/jωCo)5 + (1/jωCo)
=Vg
1 + j(5)(105)Co
Vg = 6/0 V
Vp =6/0
1 + j5 × 105Co
= Vn
·. . Vo =(1 − j1)6/0
1 + j5 × 105Co
|Vo| =√
2(6)√1 + 25 × 1010C2
o
= 6
Solving,
Co = 2 µF
Problems 9–55
[b] Vo =6(1 − j1)1 + j1
= −j6 V
vo = 6 cos(105t − 90) V
P 9.83 [a]
Because the op-amps are ideal Iin = Io, thus
Zab =Vab
Iin=
Vab
Io
; Io =Vab − Vo
Z
Vo1 = Vab; Vo2 = −(
R2
R1
)Vo1 = −KVo1 = −KVab
Vo = Vo2 = −KVab
·. . Io =Vab − (−KVab)
Z=
(1 + K)Vab
Z
·. . Zab =Vab
(1 + K)VabZ =
Z
(1 + K)
[b] Z =1
jωC; Zab =
1jωC(1 + K)
; ·. . Cab = C(1 + K)
P 9.84 [a] I1 =12024
+240
8.4 + j6.3= 23.29 − j13.71 = 27.02/−30.5 A
I2 =12012
− 12024
= 5/0 A
I3 =12012
+240
8.4 + j6= 28.29 − j13.71 = 31.44/−25.87 A
I4 =12024
= 5/0 A; I5 =12012
= 10/0 A
I6 =240
8.4 + j6.3= 18.29 − j13.71 = 22.86/−36.87 A
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 of22.86 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 Ic = 19.40/0 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
Problems 9–57
[b] Let N1 be the number of turns on the primary winding; because the secondarywinding is center-tapped, let 2N2 be the total turns on the secondary. FromFig. 9.58,
13,200N1
=2402N2
orN2
N1=
1110
The ampere turn balance requires
N1Ip = N2I1 + N2I3
Therefore,
Ip =N2
N1(I1 + I3) =
1110
(24 + 21.96) = 0.42/0 A
Check voltages —
V4 = 10I4 = 194/0 VV5 = 20I5 = 92/0 VV6 = 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; Ic = 19.2/0 A
·. . I2 = Ia − Ib = 0 A
[b] Ip =N2
N1(I1 + I3) =
N2
N1(Ia + Ib
=1
110(24 + 24) = 0.436/0 A
9–58 CHAPTER 9. Sinusoidal Steady State Analysis
[c] Check voltages —
V4 = 10I4 = 10Ic = 192/0 VV5 = 20I5 = 20(Ia − Ic) = 96/0 VV6 = 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 lowfor 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 the125 V loads.
9–60 CHAPTER 9. Sinusoidal Steady State Analysis
P 9.88 [a] Let N1 = primary winding turns and 2N2 = secondary winding turns. Then
14,000N1
=2502N2
; ·. .N2
N1=
1112
= a
In part c),
Ip = 2aIa
·. . Ip =2N2Ia
N1=
156
Ia
=156
(25.28 − j0.10)
Ip = 451.4 − j1.8 mA = 451.4/− 0.23 mA
In part d),
IpN1 = I1N2 + I2N2
·. . Ip =N2
N1(I1 + I2)
=1
112(26.97 − j0.11 + 25.10 − j0.10)
=1
112(52.07 − j0.22)
Ip = 464.9 − j1.9 mA = 464.9/− 0.24 mA
[b] Yes, because the neutral conductor carries non-zero current whenever the load isnot balanced.
10Sinusoidal Steady State Power
Calculations
Assessment Problems
AP 10.1 [a] V = 100/− 45 V, I = 20/15 A
Therefore
P =12(100)(20) cos[−45 − (15)] = 500 W, A → B
Q = 1000 sin −60 = −866.03 VAR, B → A
[b] V = 100/− 45, I = 20/165
P = 1000 cos(−210) = −866.03 W, B → A
Q = 1000 sin(−210) = 500 VAR, A → B
[c] V = 100/− 45, I = 20/− 105
P = 1000 cos(60) = 500 W, A → B
Q = 1000 sin(60) = 866.03 VAR, A → B
[d] V = 100/0, 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
AP 10.3 From Ex. 9.4 Ieff =Iρ√3
=0.18√
3A
P = I2effR =
(0.03243
)(5000) = 54 W
AP 10.4 [a] Z = (39 + j26)‖(−j52) = 48 − j20 = 52/− 22.62 Ω
Therefore I =250/0
48 − j20 + 1 + j4= 4.85/18.08 A(rms)
VL = ZI = (52/− 22.62)(4.85/18.08) = 252.20/− 4.54 V(rms)
IL =VL
39 + j26= 5.38/− 38.23 A(rms)
[b] SL = VLI∗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|21 = (4.85)2 · 1 = 23.52 W; Q = |I|24 = 94.09 VAR
[d] Sg(delivering) = 250I∗ = (1152.62 − j376.36) VA
Therefore the source is delivering 1152.62 W and absorbing 376.36magnetizing VAR.
[e] Qcap =|VL|2−52
=(252.20)2
−52= −1223.18 VAR
Therefore the capacitor is delivering 1223.18 magnetizing VAR.
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 =VI
=250
200/36.87 = 1.25/− 36.87 = (1 − j0.75) Ω
Therefore R = 1 Ω, XC = −0.75 Ω
Problems 10–3
Parallel circuit derivation:
P =(250)2
R; therefore R =
(250)2
40,000= 1.5625 Ω
Q =(250)2
XC; therefore XC =
(250)2
−30,000= −2.083 Ω
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 shownbelow:Phasor domain equivalent circuit:
Thévenin equivalent:
VTh = 3−j800
20 − j40= 48 − j24 = 53.67/− 26.57 V
ZTh = 4 + j18 +−j800
20 − j40= 20 + j10 = 22.36/26.57 Ω
For maximum power transfer, ZL = (20 − j10) Ω
10–4 CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] I =53.67/− 26.57
40= 1.34/− 26.57 A
Therefore P =(
1.34√2
)2
20 = 18 W
[c] RL = |ZTh| = 22.36 Ω
[d] I =53.67/− 26.57
42.36 + j10= 1.23/− 39.85 A
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; ·. . P =12(3.5)2(100) = 612.50 W
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
·. . P =12(0.25)(400) = 50 W
Problems 10–5
[b] I1 − I2 = 0.4 − j0.32 A
P375 =12|I1 − I2|2(375) = 49.20 W
[c] Pg =12(248)(0.8) = 99.20 W
∑Pabs = 50 + 49.2 = 99.20 W (checks)
AP 10.10 [a] VTh = 210/0 V; V2 = 14V1; I1 = 1
4I2
Short circuit equations:
840 = 80I1 − 20I2 + V1
0 = 20(I2 − I1) − V2
·. . I2 = 14 A; RTh =21014
= 15 Ω
[b] Pmax =(210
30
)2
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; RTh =−584−0.4
= 1460 Ω
[b] P =(−584
2920
)2
1460 = 58.40 W
10–6 CHAPTER 10. Sinusoidal Steady State Power Calculations
Problems
P 10.1 [a] P =12(100)(10) cos(50 − 15) = 500 cos 35 = 409.58 W (abs)
Q = 500 sin 35 = 286.79 VAR (abs)
[b] P =12(40)(20) cos(−15 − 60) = 400 cos(−75) = 103.53 W (abs)
Q = 400 sin(−75) = −386.37 VAR (del)
[c] P =12(400)(10) cos(30 − 150) = 2000 cos(−120) = −1000 W (del)
Q = 2000 sin(−120) = −1732.05 VAR (del)
[d] P =12(200)(5) cos(160 − 40) = 500 cos(120) = −250 W (del)
Q = 500 sin(120) = 433.01 VAR (abs)
P 10.2 p = P + P cos 2ωt − Q sin 2ωt;dp
dt= −2ωP sin 2ωt − 2ωQ cos 2ωt
dp
dt= 0 when − 2ωP sin 2ωt = 2ωQ cos 2ωt or tan 2ωt = −Q
P
cos 2ωt =P√
P 2 + Q2; sin 2ωt = − Q√
P 2 + Q2
Let θ = tan−1(−Q/P ), then p is maximum when 2ωt = θ and p is minimum when2ωt = (θ + π).
Therefore pmax = P + P · P√P 2 + Q2
− Q(−Q)√P 2 + Q2
= P +√
P 2 + Q2
and pmin = P − P · P√P 2 + Q2
− Q · Q√P 2 + Q2
= P −√
P 2 + Q2
Problems 10–7
P 10.3 [a] hair dryer = 600 W vacuum = 630 W
sun lamp = 279 W air conditioner = 860 W
television = 240 W∑
P = 2609 W
Therefore Ieff =2609120
= 21.74 A
Yes, the breaker will trip.
[b]∑
P = 2609 − 909 = 1700 W; Ieff =1700120
= 14.17 A
Yes, the breaker will not trip if the current is reduced to 14.17 A.
P 10.4 [a] Ieff = 40/115 ∼= 0.35 A; [b] Ieff = 130/115 ∼= 1.13 A
P 10.5 Wdc =V 2
dc
RT ; Ws =
∫ to+T
to
v2s
Rdt
·. .V 2
dc
RT =
∫ to+T
to
v2s
Rdt
V 2dc =
1T
∫ to+T
tov2
s dt
Vdc =
√1T
∫ to+T
tov2
s dt = Vrms = Veff
P 10.6 [a] Area under one cycle of v2g :
A = (52)(2)(30 × 10−6) + 22(2)(37.5 × 10−6)
= 1800 × 10−6
Mean value of v2g :
M.V. =A
200 × 10−6 =1800 × 10−6
200 × 10−6 = 9
·. . Vrms =√
9 = 3 V(rms)
[b] P =V 2
rms
R=
32
2.25= 4 W
P 10.7 i(t) = 200t 0 ≤ t ≤ 75 ms
i(t) = 60 − 600t 75 ms ≤ t ≤ 100 ms
Irms =
√1
0.1
∫ 0.075
0(200)2t2 dt +
∫ 0.1
0.075(60 − 600t)2 dt
=√
10(5.625) + 10(1.875) =√
75 = 8.66 A(rms)
10–8 CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.8 P = I2rmsR
·. . R =3 × 103
75= 40 Ω
P 10.9 Ig = 40/0 mA
jωL = j10,000 Ω;1
jωC= −j10,000 Ω
Io =j10,0005000
(40/0) = 80/90 mA
P =12|Io|2(5000) =
12(0.08)2(5000) = 16 W
Q =12|Io|2(−10,000) = −32 VAR
S = P + jQ = 16 − j32 VA
|S| = 35.78 VA
P 10.10 Ig = 4/0 mA;1
jωC= −j1250 Ω; jωL = j500 Ω
Zeq = 500 + [−j1250‖(1000 + j500)] = 1500 − j500 Ω
Pg = −12|I|2ReZeq = −1
2(0.004)2(1500) = −12 mW
The source delivers 12 mW of power to the circuit.
Problems 10–9
P 10.11 jωL = j105(0.5 × 10−3) = j50 Ω;1
jωC=
1j105[(1/3) × 10−6]
= −j30 Ω
−4 +Vo
j50+
Vo − 50I∆
40 − j30= 0
I∆ =Vo
j50
Place the equations in standard form:
Vo
(1
j50+
140 − j30
)+ I∆
( −5040 − j30
)= 4
Vo
(1
j50
)+ I∆(−1) = 0
Solving,
Vo = 200 − j400 V; I∆ = −8 − j4 A
Io = 4 − (−8 − j4) = 12 + j4 A
P40Ω =12|Io|2(40) =
12(160)(40) = 3200 W
P 10.12 [a] line loss = 7500 − 2500 = 5 kW
line loss = |Ig|220 ·. . |Ig|2 = 250
|Ig| =√
250 A
10–10 CHAPTER 10. Sinusoidal Steady State Power Calculations
|Ig|2RL = 2500 ·. . RL = 10 Ω
|Ig|2XL = −5000 ·. . XL = −20 Ω
Thus,
|Z| =√
(30)2 + (X − 20)2 |Ig| =500√
900 + (X − 20)2
·. . 900 + (X − 20)2 =25 × 104
250= 1000
Solving, (X − 20) = ±10.
Thus, X = 10 Ω or X = 30 Ω
[b] If X = 30 Ω:
Ig =500
30 + j10= 15 − j5 A
Sg = −500I∗g = −7500 − j2500 VA
Thus, the voltage source is delivering 7500 W and 2500 magnetizing vars.
Qj30 = |Ig|2X = 250(30) = 7500 VAR
Therefore the line reactance is absorbing 7500 magnetizing vars.
Q−j20 = |Ig|2XL = 250(−20) = −5000 VAR
Therefore the load reactance is generating 5000 magnetizing vars.∑Qgen = 7500 VAR =
∑Qabs
If X = 10 Ω:
Ig =500
30 − j10= 15 + j5 A
Sg = −500I∗g = −7500 + j2500 VA
Thus, the voltage source is delivering 7500 W and absorbing 2500 magnetizingvars.
Qj10 = |Ig|2(10) = 250(10) = 2500 VAR
Therefore the line reactance is absorbing 2500 magnetizing vars. The loadcontinues to generate 5000 magnetizing vars.∑
Qgen = 5000 VAR =∑
Qabs
Problems 10–11
P 10.13 Zf = −j10,000‖20,000 = 4000 − j8000 Ω
Zi = 2000 − j2000 Ω
·. .Zf
Zi=
4000 − j80002000 − j2000
= 3 − j1
Vo = −Zf
ZiVg; Vg = 1/0 V
Vo = (3 − j1)(1) = 3 − j1 = 3.16/− 18.43 V
P =12
V 2m
R=
12
(10)1000
= 5 × 10−3 = 5 mW
P 10.14 [a] P =12
(240)2
480= 60 W
− 1ωC
=−9 × 106
(5000)(5)= −360 Ω
Q =12
(240)2
(−360)= −80 VAR
pmax = P +√
P 2 + Q2 = 60 +√
(60)2 + (80)2 = 160 W(del)
[b] pmin = 60 −√
602 + 802 = −40 W(abs)
[c] P = 60 W from (a)
[d] Q = −80 VAR from (a)
[e] generate, because Q < 0
[f] pf = cos(θv − θi)
I =240480
+240
−j360= 0.5 + j0.67 = 0.83/53.13 A
·. . pf = cos(0 − 53.13) = 0.6 leading
[g] rf = sin(−53.13) = −0.8
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 = −VgI∗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
∑Pdiss = 170 + 510 = 680 W
·. .∑
Pdel =∑
Pdiss = 680 W
[f]∑
Qabs = 170 + 680 = 850 VAR∑
Qdev = 850 VAR
·. .∑
mag VAR dev =∑
mag VAR abs = 850
Problems 10–13
P 10.16 [a]1
jωC= −j40 Ω; jωL = j80 Ω
Zeq = 40‖ − j40 + j80 + 60 = 80 + j60 Ω
Ig =40/0
80 + j60= 0.32 − j0.24 A
Sg = −12
VgI∗g = −1
240(0.32 + j0.24) = −6.4 − j4.8 VA
P = 6.4 W(del); Q = 4.8 VAR(del)
|S| = |Sg| = 8 VA
[b] I1 =−j40
40 − j40Ig = 0.04 − j0.28 A
P40Ω =12|I1|2(40) = 1.6 W
P60Ω =12|Ig|2(60) = 4.8 W
∑Pdiss = 1.6 + 4.8 = 6.4 W =
∑Pdev
[c] I−j40Ω = Ig − I1 = 0.28 + j0.04 A
Q−j40Ω =12|I−j40Ω|2(−40) = −1.6 VAR(del)
Qj80Ω =12|Ig|2(80) = 6.4 VAR(abs)
∑Qabs = 6.4 − 1.6 = 4.8 VAR =
∑Qdev
P 10.17 [a] Z1 = 240 + j70 = 250/16.26 Ω
pf = cos(16.26) = 0.96 lagging
rf = sin(16.26) = 0.28
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 =1Y
= 37.44/− 42.03 Ω
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; ·. . I = 120 − j40 A
Z =250
120 − j40= 1.875 + j0.625 Ω = 1.98/18.43 Ω
[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 Q = (16,000)(0.08) = 1280 VAR
[c] Ps = 30,000 + 160 = 30.16 kW Qs = 10,000 + 1280 = 11.28 kVAR
[d] η =30
30.16(100) = 99.47%
Problems 10–15
P 10.20 ST = 4500 − j45000.96
(0.28) = 4500 − j1312.5 VA
S1 =27000.8
(0.8 + j0.6) = 2700 + j2025 VA
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; ·. . I2 = 8.33 + j4.167 A(rms)
I3 =2400/0
144= 16.67 + j0 A; I4 =
2400/0
j96= 0 − j25 A
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|2Z∗
2=
(2500)2
24 − j7= 240,000 + j70,000 VA
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)
10–16 CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] T =1f
=160
= 16.67 ms
2.735
360 =t
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
Sg = −12
VoI∗g = −1
2(j80)(10 − j10) = −400 − j400 VA
Therefore, the independent current source is delivering 400 W and 400magnetizing vars.
I1 =Vo
5= j16 A
P5Ω =12(16)2(5) = 640 W
Therefore, the 8 Ω resistor is absorbing 640 W.
I∆ =Vo
−j8= −10 A
Qcap =12(10)2(−8) = −400 VAR
Therefore, the −j8 Ω capacitor is developing 400 magnetizing vars.
2.4I∆ = −24 V
I2 =Vo − 2.4I∆
j4=
j80 + 24j4
= 20 − j6 A = 20.88/− 16.7 A
Problems 10–17
Qj4 =12|I2|2(4) = 872 VAR
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]∑
Pgen = 400 + 240 = 640 W =∑
Pabs
[c]∑
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
S100V = −12(100)I∗
a = −50(j10) = −j500 VA
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 magnetizingvars.
P10Ω =12|Ia|2(10) = 500 W
Thus the 10 Ω resistor is absorbing 500 W.
Q−j10Ω =12|Ib|2(−10) = −2500 VAR
Thus the −j10 Ω capacitor is developing 2500 magnetizing vars.
Qj5Ω =12|Io|2(5) = 2000 VAR
Thus the j5 Ω inductor is absorbing 2000 magnetizing vars.
[b]∑
Pdev = 500 W =∑
Pabs
10–18 CHAPTER 10. Sinusoidal Steady State Power Calculations
[c]∑
Qdev = 500 + 2500 = 3000 VAR∑
Qabs = 1000 + 2000 = 3000 VAR =∑
Qdev
P 10.25 [a] I =465/0
124 + j93= 2.4 − j1.8 = 3/− 36.87 A(rms)
P = (3)2(4) = 36 W
[b] YL =1
120 + j90= 5.33 − j4 mS
·. . XC =1
−4 × 10−3 = −250 Ω
[c] ZL =1
5.33 × 10−3 = 187.5 Ω
[d] I =465/0
191.5 + j3= 2.43/− 0.9 A
P = (2.43)2(4) = 23.58 W
[e] % =23.5836
(100) = 65.5%
Thus the power loss after the capacitor is added is 65.6% of the power lossbefore the capacitor is added.
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 =50012.5
+500j50
= 40 − j10 A(rms)
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 magnetizingvars.
[b]∑
Pgen = 17.5 + 12.8 = 30.3 kW
∑Pabs = 7500 + 2800 +
(500)2
12.5= 30.3kW =
∑Pgen
∑Qdel = 9600 + 5000 = 14.6 kVAR
∑Qabs = 2500 + 7100 +
(500)2
50= 14.6 kVAR =
∑Qdel
P 10.27 S1 = 1200 + 1196 = 2396 + j0 VA
·. . I1 =2396120
= 19.97 A
S2 = 860 + 600 + 240 = 1700 + j0 VA
·. . I2 =1700120
= 14.167 A
S3 = 4474 + 12,200 = 16,674 + j0 VA
·. . I3 =16,674240
= 69.48 A
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
125= 32 − j8 A (rms)
10–20 CHAPTER 10. Sinusoidal Steady State Power Calculations
I2 =5000 − j2000
125= 40 − j16 A (rms)
I3 =10,000 + j0
250= 40 + j0 A (rms)
·. . 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 thecircuit.
[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 =
∑Qabs
P 10.29 [a] Let VL = Vm/0:
SL = 600(0.8 + j0.6) = 480 + j360 VA
I∗ =
480Vm
+ j360Vm
; I =480Vm
− j360Vm
Problems 10–21
120/θ = Vm +(480
Vm
− j360Vm
)(1 + j2)
120Vm/θ = V 2m + (480 − j360)(1 + j2) = V 2
m + 1200 + j600
120Vm cos θ = V 2m + 1200; 120Vm sin θ = 600
(120)2V 2m = (V 2
m + 1200)2 + 6002
14,400V 2m = V 4
m + 2400V 2m + 18 × 105
or
V 4m − 12,000V 2
m + 18 × 105 = 0
Solving,
Vm = 108.85 V and Vm = 12.326 V
If Vm = 108.85 V:
sin θ =600
(108.85)(120)= 0.045935; ·. . θ = 2.63
If Vm = 12.326 V:
sin θ =600
(12.326)(120)= 0.405647; ·. . 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
10–22 CHAPTER 10. Sinusoidal Steady State Power Calculations
[c](125)2
XC= −10,535.65; XC = −1.483 Ω
− 1ωC
= −1.48; C =1
(1.48)(120π)= 1788.59 µF
[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
4800= 32 − j24 A(rms)
IC =4800−jXC
= j4800XC
= jIC
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
·. . 104I2C − 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 = − 480031.57
= −152.04
·. . C =1
(152.04)(120π)= 17.45 µF
Problems 10–23
P 10.32 ZL = |ZL|/θ = |ZL| cos θ + j|ZL| sin θ
Thus |I| =|VTh|√
(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
dP
d|ZL| =(0.5|VTh|2 cos θ)(D · 1 − |ZL|dD/d|ZL|)
D2
dD
d|ZL| = 2(RTh + |ZL| cos θ) cos θ + 2(XTh + |ZL| sin θ) sin θ
dP
d|ZL| = 0 when D = |ZL|(
dD
d|ZL|)
Substituting the expressions for D and (dD/d|ZL|) into this equation gives us therelationship R2
Th + X2Th = |ZL|2 or |ZTh| = |ZL|.
P 10.33 [a] ZTh = j40‖40 − j40 = 20 − j20
·. . ZL = Z∗Th = 20 + j20 Ω
[b] VTh =40
40 + j40(120) = 60 − j60 V
I =60 − j60
40= 1.5 − j1.5 A
Pload =12|I|2(20) = 45 W
P 10.34 [a]115.2 + j33.6 − 240
ZTh+
115.2 + j33.680 − j60
= 0
·. . ZTh = 40 − j100 Ω
·. . ZL = 40 + j100 Ω
10–24 CHAPTER 10. Sinusoidal Steady State Power Calculations
[b] I =24080
= 3 A(rms)
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 =−j8
3 − j4(112.5) = 144 − j108 V(rms)
I =144 − j10835.21 − j15
= 4.45 − j1.14
P = |I|2(21.21) = 447.35 W
P 10.36 [a] Open circuit voltage:
V1 = 5Iφ = 5100 − 5Iφ
25 + j10
(25 + j10)Iφ = 100 − 5Iφ
Iφ =100
30 + j10= 3 − j1 A
VTh =j3
1 + j3(5Iφ) = 15/0 V
Problems 10–25
Short circuit current:
V2 = 5Iφ =100 − 5Iφ
25 + j10
Iφ = 3 − j1 A
Isc =5Iφ
1= 15 − j5 A
ZTh =15
15 − j5= 0.9 + j0.3 Ω
ZL = Z∗Th = 0.9 − j0.3 Ω
IL =151.8
= 8.33 A(rms)
P = |IL|2(0.9) = 62.5 W
[b] VL = (0.9 − j0.3)(8.33) = 7.5 − j2.5 V(rms)
I1 =VL
j3= −0.833 − j2.5 A(rms)
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 =62.5375
(100) = 16.67%
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 =
∑Pdev
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 − 1005
+Vφ
j5− 0.1Vφ = 0
·. . Vφ = 40 + j80 V(rms)
Problems 10–27
VTh = Vφ + 0.1Vφ(−j5) = Vφ(1 − j0.5) = 80 + j60 V(rms)
Short circuit current:
Isc = 0.1Vφ +Vφ
−j5= (0.1 + j0.2)Vφ
Vφ − 1005
+Vφ
j5+
Vφ
−j5= 0
·. . Vφ = 100 V(rms)
Isc = (0.1 + j0.2)(100) = 10 + j20 A(rms)
ZTh =VTh
Isc=
80 + j6010 + j20
= 4 − j2 Ω
·. . Ro = |ZTh| = 4.47 Ω
[b]
I =80 + j60
4 +√
20 − j2= 7.36 + j8.82 A(rms)
P = (11.49)2(√
20) = 590.17 W
10–28 CHAPTER 10. Sinusoidal Steady State Power Calculations
[c]
I =80 + j60
8= 10 + j7.5 A(rms)
P = (102 + 7.52)(4) = 625 W
[d]
Vφ − 1005
+Vφ
j5+
Vφ − (25 + j50)−j5
= 0
Vφ = 50 + j25 V(rms)
0.1Vφ = 5 + j2.5
5 + j2.5 + IC = 10 + j7.5
IC = 5 + j5 A(rms)
IL =Vφ
j5= 5 − j10 A(rms)
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 =6251500
(100) = 41.67%
Problems 10–29
P 10.38 [a] First find the Thévenin equivalent:
jωL = j3000 Ω
ZTh = 6000‖12,000 + j3000 = 4000 + j3000 Ω
VTh =12,000
6000 + 12,000(180) = 120/0 V
−j
ωC= −j1000 Ω
I =120
6000 + j2000= 18 − j6 mA
P =12|I|2(2000) = 360 mW
[b] Set Co = 0.1 µF so −j/ωC = −j2000 Ω j3000 − j2000 = j1000 ΩSet Ro as close as possible to
Ro =√
40002 + 10002 = 4123.1 Ω
·. . Ro = 4000 Ω
[c] I =120
8000 + j1000= 14.77 − j1.85 mA
P =12|I|2(4000) = 443.1 mW
Yes; 443.1 mW > 360 mW
[d] I =1208000
= 15 mA
P =12(0.015)2(4000) = 450 mW
[e] Ro = 4000 Ω; Co = 66.67 nF
[f] Yes; 450 mW > 443.1 mW
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
8123.1 + j1000= 14.55 − j1.79 mA
P =12|I|2(4123.1) = 443.18 mW
[b] Yes; 443.18 mW > 360 mW
[c] Yes; 443.18 mW < 450 mW
P 10.40 [a]1
ωC= 100 Ω; C =
1(100)(120π)
= 26.53 µF
[b] Iwo =13,800300
+13,800j100
= 46 − j138 A(rms)
Vswo = 13,800 + (46 − j138)(1.5 + j12) = 15,525 + j345
= 15,528.83/1.27 V(rms)
Iw =13,800300
= 46 A(rms)
Vsw = 13,800 + 46(1.5 + j12) = 13,869 + j552 = 13,879.98/2.28 V(rms)
% increase =(
15,528.8213,879.98
− 1)
(100) = 11.88%
[c] Pwo = |46 − j138|21.5 = 31.74 kW
Pw = 462(1.5) = 3174 W
% increase =(31,740
3174− 1
)(100) = 900%
P 10.41 [a] So = original load = 1600 + j16000.8
(0.6) = 1600 + j1200 kVA
Sf = final load = 1920 + j19200.96
(0.28) = 1920 + j560 kVA
·. . 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 10–31
[d] I∗L =
(1600 + j1200) × 103
2400= 666.67 + j500 A
IL = 666.67 − j500 = 833.33/− 36.87 A(rms)
|IL| = 833.33 A(rms)
[e] I∗L =
(1920 + j560) × 103
2400= 800 + j233.33
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 ·. . Pg = −3240 W
% delivered =12963240
(100) = 40%
10–32 CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.44 [a] Open circuit voltage:
180 = 3I1 + j4I1 − j3I1 + j9I1 − j3I1
·. . I1 =180
3 + j7= 9.31 − j21.72 A(rms)
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 I1 = 30 − j20 A
ZTh =VTh
Isc=
130.34 + j55.8620 − j20
= 1.86 + j4.66 Ω
IL =130.34 + j55.86
3.72= 35 + j15 = 38.08/23.20 A
Problems 10–33
PL = (38.12)2(1.86) = 2700 W
[b] I1 =Zo + j9
j6I2 =
1.86 − j4.66 + j9j6
(35 + j15) = 30/0 A(rms)
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 =63648
(100) = 9.72%
P 10.46 [a]
Open circuit:
VTh = −j3I1 + j2I1 = −jI1
I1 =54
1 + j2= 10.8 − j21.6 A
VTh = −21.6 − j10.8 V
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
Isc=
−21.6 − j10.8−3.32 + j5.82
= 0.2 + 3.6j = 3.6/86.82 Ω
·. . RL = |ZTh| = 3.606 Ω
[b]
I =−21.6 − j10.83.806 + j3.6
= 4.610/163.2 A
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 − I2j(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
1 + j2= 10.8 − j21.6 A(rms)
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, formaximum power transfer, ZL = Z∗
Th = 85 − j85 Ω:
I2 =850 + j850
170= 5 + j5 A
425/0 = (5 + j5)I1 − j20(5 + j5)
·. . I1 =325 + j100
5 + j5= 42.5 − j22.5 A
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
18,062.5(100) = 23.53%
10–36 CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.49 [a] From Problem 9.70,
Zab = 100 + j136.26 so
I1 =50
100 + j13.74 + 100 + 136.26=
50200 + j150
= 160 − j120 mA
I2 =jωM
Z22I1 =
j270800 + j600
(0.16 − j0.12) = 51.84 + j15.12 mA
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
4(100) = 21.87%
P 10.50 [a]
Open circuit:
VTh =120
16 + j12(j10) = 36 + j48 V
Short circuit:
(16 + j12)I1 − j10Isc = 120
−j10I1 + (11 + j23)Isc = 0
Solving,
Isc = 2.4/0 A
ZTh =36 + j48
2.4= 15 + j20 Ω
·. . ZL = Z∗Th = 15 − j20 Ω
IL =VTh
ZTh + ZL
=36 + j48
30= 1.2 + j1.6 A(rms) = 2.0/53.13 A(rms)
PL = |IL|2(15) = 60 W
Problems 10–37
[b] I1 =Z22I2
jωM=
26 + j3j10
(1.2 + j1.6) = 5.23/− 30.29 A)rms)
Ptransformer = (120)(5.23) cos(−30.29) − (5.23)2(4) = 432.8 W
% delivered =60
432.8(100) = 13.86%
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
[b] k =M√L1L2
=0.5√
1= 0.5
[c] When t = 50π µs:
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
w =12L1i
21 +
12L2i
22 + Mi1i2 =
12(10−3)(15)2 + 0 + 0 = 112.5 mJ
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
w =12(10−3)(10)2 +
12(10−3)(5)2 + 0.5 × 10−3(−10)(5) = 37.5 mJ
10–38 CHAPTER 10. Sinusoidal Steady State Power Calculations
[d] From (a), Im = 5 A,
·. . P =12(5)2(15) = 187.5 W
[e] Open circuit:
VTh =200
5 + j10(−j5) = −80 − j40 V
Short circuit:
200 = (5 + j10)I1 + j5Isc
0 = j10Isc + j5I1
Solving,
Isc = −11.094/123.69 A; I1 = 22.188/− 56.31 A
ZTh =VTh
Isc=
−80 − j4011.094/123.69 = 1 + j8 Ω
·. . RL = 8.962 Ω
[f]
I =−80 − j409.062 + j8
= 7.399/165.13 A
P =12(7.399)2(8.062) = 220.70 W
[g] ZL = Z∗Th = 1 − j8 Ω
[h] I =−80 − j40
2= 44.72/− 153.43 A
P =12(44.72)2(1) = 1000 W
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
[b]
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
Sa = −10I∗a = −62.5 + j75 VA
Sb = VbI∗b = 6.25 + j1.5625 VA
10–40 CHAPTER 10. Sinusoidal Steady State Power Calculations
Sc = VcI∗c = 31.25 + j0 VA
Sd = VdI∗d = −6.25 + j25 VA
Se = VeI∗e = 0 − j101.5625 VA
Sf = VfI∗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 VARThe 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
1 + j2= 2 − j4 A
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 10–41
0 = −j3.2I1 + j5.4Isc
Solving,
Isc = 5.89/− 5.92 A
ZTh =14.31/26.57
5.89/− 5.92 = 2.43/32.49 = 2.048 + j1.304 Ω
·. . I2 =14.31/26.57
4.096= 3.49/26.57 A
10/0 = (1 + j2)I1 − j3.2I2
·. . I1 =10 + j3.2I2
1 + j2=
10 + j3.2(3.49/26.57)1 + j2
= 5/0 A
Zg =10/0
5/0 = 2 + j0 = 2/0 Ω
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
10–42 CHAPTER 10. Sinusoidal Steady State Power Calculations
Solving,
Ig = 20 − j4 A(rms); I2 = 24/0 A(rms)
P8Ω = (24)2(8) = 4608 W
[b] Pg(developed) = (272)(20) = 5440 W
[c] Zab =Vg
Ig
− 2 =272
20 − j4− 2 = 11.08 + j2.62 = 11.38/13.28 Ω
[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
V2 =14
V1; 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
30= Ia; Vo = 10Io therefore
Va
Ia= 9 kΩ
Vb
1=
−Va
20; Ib = −20Ia; therefore
Vb
Ib=
9000400
= 22.5 Ω
Therefore Ib = [50/(2.5 + 22.5)] = 2 A (rms); since the ideal transformers arelossless, P10Ω = P22.5Ω, and the power delivered to the 22.5 Ω resistor is 22(22.5) or90 W.
P 10.57 [a]Vb
Ib=
a210400
= 2.5 Ω; therefore a2 = 100, a = 10
[b] Ib =505
= 10 A; P = (100)(2.5) = 250 W
P 10.58 [a] ZTh = 720 + j1500 +(200
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 or N2 = N1 = 1000 turns
[b] VTh =255/0
40 + j30(j200) = 1020/53.13 V
IL =1020/53.13
3060 + j1020= 0.316/34.7 A(rms)
Since the transformer is ideal, P6800 = P1700.
P = |IL|2(1700) = 170 W
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 =8831053
(100) = 83.85%
P 10.59 [a]
For maximum power transfer, Zab = 90 kΩ
Zab =(1 +
N1
N2
)2
ZL
·. .(1 +
N1
N2
)2
=90,000400
= 225
1 +N1
N2= ±15;
N1
N2= 15 − 1 = 14
[b] P = |Ii|2(90,000) =(
180180,000
)2
(90,000) = 90 mW
[c] V1 = RiIi = (90,000)(
180180,000
)= 90 V
[d]
Vg = (2.25 × 10−3)(100,000‖80,000) = 100 V
Problems 10–45
Pg(del) = (2.25 × 10−3)(100) = 225 mW
% delivered =90225
(100) = 40%
P 10.60 [a] Zab =(1 +
N1
N2
)2
(1 − j2) = 25 − j50 Ω
·. . I1 =100/0
15 + j50 + 25 − j50= 2.5/0 A
I2 =N1
N2I1 = 10/0 A
·. . 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 =∑
Pdev
P 10.61 [a] 25a21 + 4a2
2 = 500
I25 = a1I; P25 = a21I2(25)
I4 = a2I; P4 = a22I2(4)
P4 = 4P25; a22I24 = 100a2
1I2
·. . 100a21 = 4a2
2
25a21 + 100a2
1 = 500; a1 = 2
25(4) + 4a22 = 500; a2 = 10
[b] I =2000/0
500 + 500= 2/0 A(rms)
I25 = a1I = 4 A
P25Ω = (16)(25) = 400 W
[c] I4 = a2I = 10(2) = 20 A(rms)
V4 = (20)(4) = 80/0 V(rms)
10–46 CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.62 [a] Open circuit voltage:
500 = 100I1 + V1
V2 = 400I2
V1
1=
V2
2·. . V2 = 2V1
I1 = 2I2
Substitute and solve:
2V1 = 400I1/2 = 200I1·. . V1 = 100I1
500 = 100I1 + 100I1·. . I1 = 500/200 = 2.5 A
·. . I2 =12
I1 = 1.25 A
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 = 40I1
2+ 360(Isc + 0.5I1)
500 = 80(I1 + Isc) + 20I1 + V1
Problems 10–47
Solving,
Isc = −1.47 A; I1 = 4.41 A; V1 = 176.47 V
RTh =VTh
Isc=
−150−1.47
= 102 Ω
P =752
102= 55.15 W
[b]
500 = 80[I1 − (75/102)] − 75 + 360[I2 − (75/102)]
575 +6000102
+27,000102
= 80I1 + 180I1
·. . I1 = 3.456 A
Psource = (500)[3.456 − (75/102)] = 1360.29 W
% delivered =55.15
1360.29(100) = 4.05%
[c] P80Ω = 80(
I1 − 75102
)2
= 592.13 W
P20Ω = 20I21 = 238.86 W
P40Ω = 40I22 = 119.43 W
P102Ω =752
102= 55.15 W
P360Ω = 360(
I2 − 75102
)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
4= −I3; 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; I3 = 1 A; Isc = I1/4 + I3 = 2.5 A
RTh =VTh
Isc=
402.5
= 16 Ω
Problems 10–49
I =40/0
32= 1.25/0 A(rms)
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; I3 = −0.25 A; I1 + I3 = 5.75/0 A; V1 = 11/0 V
P40V (developed) = 40(5.75) = 230 W
·. . % delivered =25230
(100) = 10.87%
[c] PRL= 25 W; 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)(20‖j10) = 60 + j120 V
ZTh = 2 + 20‖j10 = 6 + j8 Ω
10–50 CHAPTER 10. Sinusoidal Steady State Power Calculations
Transfer the secondary impedance to the primary side:
Zp =125
(100 + jXC) = 4 + jXC
25Ω
Now maximize I by setting (XC/25) = −8 Ω:
·. . C =1
200(20 × 103)= 0.25 µF
[b] I =60 + j120
10= 6 + j12 A
P = |I|2(4) = 720 W
[c]Ro
25= 6 Ω; ·. . Ro = 150 Ω
[d] I =60 + j120
12= 5 + j10 A
P = |I|2(6) = 750 W
P 10.65 [a] Zab = 50 − j400 =(1 − N1
N2
)2
ZL =(1 − 2800
700
)2
ZL = 9ZL
·. . ZL =19(50 − j400) = 5.556 − j44.444 Ω
[b]
I1 =24100
= 240/0 mA
Problems 10–51
N1I1 = −N2I2
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 theresistor R2. Then,
Pmed = 500 W =V 2
R2=
1202
R2
Thus,
R2 =1202
500= 28.8 Ω
[b] Now move to the LOW setting, as shown in Fig. 10.30, which involves theresistors R1 and R2 connected in series:
Plow =V 2
R1 + R2=
V 2
R1 + 28.8= 250 W
Thus,
R1 =1202
250− 28.8 = 28.8 Ω
[c] Note that the HIGH setting has R1 and R2 in parallel:
Phigh =V 2
R1‖R2=
1202
28.8‖28.8= 1000 W
If the HIGH setting has required power other than 1000 W, this problem souldnot have been solved. In other words, the HIGH power setting was chosen insuch a way that it would be satisfied once the two resistor values werecalculated to satisfy the LOW and MEDIUM power settings.
10–52 CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.67 [a] PL =V 2
R1 + R2; R1 + R2 =
V 2
PL
PM =V 2
R2; R2 =
V 2
PM
PH =V 2(R1 + R2)
R1R2
R1 + R2 =V 2
PL; R1 =
V 2
PL− V 2
PM
PH =V 2V 2/PL(
V 2
PL− V 2
PM
) (V 2
PM
) =PMPLPM
PL(PM − PL)
PH =P 2
M
PM − PL
[b] PH =(750)2
(750 − 250)= 1125 W
P 10.68 First solve the expression derived in P10.67 for PM as a function of PL and PH. Thus
PM − PL =P 2
M
PHor
P 2M
PH− PM + PL = 0
P 2M − PMPH + PLPH = 0
·. . PM =PH
2±√(
PH
2
)2
− PLPH
=PH
2± PH
√14
−(
PL
PH
)
For the specified values of PL and PH
PM = 500 ± 1000√
0.25 − 0.24 = 500 ± 100
·. . PM1 = 600 W; PM2 = 400 W
Note in this case we design for two medium power ratingsIf PM1 = 600 W
R2 =(120)2
600= 24 Ω
Problems 10–53
R1 + R2 =(120)2
240= 60 Ω
R1 = 60 − 24 = 36 Ω
CHECK: PH =(120)2(60)(36)(24)
= 1000 W
If PM2 = 400 W
R2 =(120)2
400= 36 Ω
R1 + R2 = 60 Ω (as before)
R1 = 24 Ω
CHECK: PH = 1000 W
P 10.69 R1 + R2 + R3 =(120)2
600= 24 Ω
R2 + R3 =(120)2
900= 16 Ω
·. . R1 = 24 − 16 = 8 Ω
R3 + R1‖R2 =(120)2
1200= 12 Ω
·. . 16 − R2 +8R2
8 + R2= 12
R2 − 8R2
8 + R2= 4
8R2 + R22 − 8R2 = 32 + 4R2
R22 − 4R2 − 32 = 0
R2 = 2 ± √4 + 32 = 2 ± 6
·. . R2 = 8 Ω; ·. . R3 = 8 Ω
10–54 CHAPTER 10. Sinusoidal Steady State Power Calculations
P 10.70 R2 =(220)2
500= 96.8 Ω
R1 + R2 =(220)2
250= 193.6 Ω
·. . R1 = 96.8 Ω
CHECK: R1‖R2 = 48.4 Ω
PH =(220)2
48.4= 1000 W
11Balanced 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/− 30V, and thephase 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, wecan 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 thephase 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, wecan 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 isthe 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 wewere not given any phase angles in the problem statement, we can assume thatthe phase voltage given, VAN, has a phase angle of 0.
2400/0 = IaA(16 + j12)
so
IaA =2400/0
16 + j12= 96 − j72 = 120/− 36.87 A
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 relationshipbetween 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
Vc′n = 2482.05/− 118.07 V
AP 11.4 IcC = (√
3/− 30)ICA = (√
3/− 30) · 8/− 15 = 13.86/− 45 A
AP 11.5 IaA = 12/(65 − 120) = 12/− 55
IAB =[(
1√3
)/− 30
]IaA =
(/− 30
√3
)· 12/− 55
= 6.93/− 85 A
AP 11.6 [a] IAB =[(
1√3
)/30
][69.28/− 10] = 40/20 A
Therefore Zφ =4160/0
40/20 = 104/− 20 Ω
[b] IAB =[(
1√3
)/− 30
][69.28/− 10] = 40/− 40 A
Therefore Zφ = 104/40 Ω
AP 11.7 Iφ =110
3.667+
110j2.75
= 30 − j40 = 50/− 53.13 A
Therefore |IaA| =√
3Iφ =√
3(50) = 86.60 A
Problems 11–5
AP 11.8 [a] |S| =√
3(208)(73.8) = 26,587.67 VA
Q =√
(26,587.67)2 − (22,659)2 = 13,909.50 VAR
[b] pf =22,659
26,587.67= 0.8522 lagging
AP 11.9 [a] VAN =(
2450√3
)/0 V; VANI∗
aA = Sφ = 144 + j192 kVA
Therefore
I∗aA =
(144 + j192)10002450/
√3
= (101.8 + j135.7) A
IaA = 101.8 − j135.7 = 169.67/− 53.13 A
|IaA| = 169.67 A
[b] P =(2450)2
R; therefore R =
(2450)2
144,000= 41.68 Ω
Q =(2450)2
X; therefore X =
(2450)2
192,000= 31.26 Ω
[c] Zφ =VAN
IaA=
2450/√
3169.67/− 53.13 = 8.34/53.13 = (5 + j6.67) Ω
·. . R = 5 Ω, X = 6.67 Ω
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:
/V′a = 27 − 27 = 0
/V′b = 147 − 27 = 120
/V′c = −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:
/V′a = −18 + 18 = 0
/V′b = −138 + 18 = −120
/V′c = 102 + 18 = 120
Compare the result to Eqs. 11.1 and 11.2:
Therefore abc
P 11.2 [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 positivesequence 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 isstill zero.
P 11.4 I =Va + Vb + Vc
3(RW + jXW)= 0
P 11.5 [a] IaA =20025
= 8/0 A
IbB =200/− 120
30 − j40= 4/− 66.87 A
IcC =200/120
80 + j60= 2/83.13 A
The magnitudes are unequal and the phase angles are not 120 apart.
b] Io = IaA + IbB + IcC = 9.96/− 9.79 A
11–8 CHAPTER 11. Balanced Three-Phase Circuits
P 11.6 [a] IaA =277/0
80 + j60= 2.77/− 36.87 A
IbB =277/− 120
80 + j60= 2.77/− 156.87 A
IcC =277/120
80 + j60= 2.77/83.13 A
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 − 24060 + j80
+VN − 240/120
40 + j30+
VN − 240/− 120
20 + j15+
VN
10= 0
Solving for VN yields
VN = 42.94/− 156.32 V
Io =VN
10= 4.29/− 156.32 A
P 11.8 Make a sketch of the load in the frequency domain. Note that we convert the timedomain 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 threeline 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’slaw for the a-phase load impedance. Since we are only concerned with findingvoltage and current magnitudes, the phase sequence doesn’t matter and wearbitrarily assume a positive phase sequence. Since we are not given any phaseangles in the problem statement, we can assume the angle of VAB is 0. UseFig. 11.9(a) to find VAN from VAB.
VAN =660√
3/(0 − 30) = 381.05/− 30 V
Now find IaA using Ohm’s law:
IaA =VAN
ZL=
381.05/− 30
30.48 + j22.86= 3.993 − j9.20 = 10/− 66.87 V
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 thethree-phase circuit,
Vab = VaA + VAB + VBb
= ZIaA + VAB + ZIBb
= ZIaA + VAB − ZIbB
= (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
0.1 + j0.8 + 19.9 + j14.2=
125/0
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 theline 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 usingIaA and the load impedance:
VAN = IaAZL = (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 acbsequence:
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 fromthe 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
60 + j45= 6.4/− 36.87 A
IBC = 6.4/− 156.87 A
ICA = 6.4/83.13 A
[b] IaA =√
3/− 30IAB = 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:
ZY =Z∆
3=
60 + j453
= 20 + j15 Ω
The single-phase equivalent circuit is:
Van = 277.13/− 30 + (0.8 + j0.6)(11.09/− 66.87)
= 288.21/− 30 V
Vab =√
3/30Van = 499.20/0 V
Vbc = 499.20/− 120 V
Vca = 499.20/120 V
P 11.12 [a]
IaA =7650
72 + j21+
765050
= 252.54/− 6.49 A
|IaA| = 252.54 A
[b] IAB =7650
√3/30
150= 88.33/30 A
|IAB| = 88.33 A
Problems 11–13
[c] IAN =7650/0
72 + j21= 102/− 16.26 A
|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
ZY =13Z∆ = 0.9 + j4.5 Ω/φ
[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
11–14 CHAPTER 11. Balanced Three-Phase Circuits
[c]
Iba =4156
2.7 + j13.5= 301.87/− 78.69 A
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 wehave:
IbB = 522.86/71.31 A
IcC = 522.86/− 168.69 A
Problems 11–15
[d]
IaA =2399.47/30
0.9 + j4.5= 522.86/− 48.69 A
Since we have a balanced three-phase circuit and a negative phase sequence wehave:
IbB = 522.86/71.31 A
IcC = 522.86/− 168.69 A
P 11.14 [a]
[b] IaA =2399.47/30
1920 − j560= 1.2/46.26 A
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
[c] |Iab| =1.2√
3= 0.69 A
[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√
3(2.375 + j1.349)= 2917/− 29.6 A
|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
[e] |IAB| =|IaA|√
3= 1684.13 A
[f] |Iab| = |IAB| = 1684.13 A
P 11.16 [a] IAB =4160/0
160 + j120= 20.8/− 36.87 A
IBC = 20.8/83.13 A
ICA = 20.8/− 156.87 A
[b] IaA =√
3/30IAB = 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 11–17
P 11.17 [a] IAB =480/0
2.4 − j0.7= 192/16.26 A
IBC =480/120
8 + j6= 48/83.13 A
ICA =480/− 120
20= 24/− 120 A
[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]
I1 =24,000
√3/0
400 + j300= 66.5 − j49.9 A
I2 =24,000
√3/0
800 − j600= 33.3 + j24.9 A
I∗3 =
57,600 + j734,40024,000
√3
= 1.4 + j17.7
I3 = 1.4 − j17.7 A
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 = VanI∗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
% delivered =(4205.1
4229.2
)(100) = 99.4%
P 11.20 [a]
IaA =1365/0
30 + j40= 27.3/− 53.13 A
ICA =IaA√
3/150 = 15.76/96.87 A
[b] Sg/φ = −1365I∗aA = −22,358.7 − j29,811.6 VA
·. . Pdeveloped/phase = 22.359 kW
Pabsorbed/phase = |IaA|228.5 = 21.241 kW
% delivered =21.24122.359
(100) = 95%
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 = vaniaA = [Vm cos ωt][Im cos(ωt − θφ)]
pb = vbnibB = [Vm cos(ωt − 120)][Im cos(ωt − θφ − 120)]
pc = vcnicC = [Vm cos(ωt + 120)][Im cos(ωt − θφ + 120)]
The total instantaneous power is pT = pa + pb + pc, so
pT = VmIm[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 coefficientsof cos(ωt − θφ) and sin(ωt − θφ). We get
pT = VmIm[cos ωt(1 + 2 cos2 120) cos(ωt − θφ)
+2 sin ωt sin2 120 sin(ωt − θφ)]
= 1.5VmIm[cos ωt cos(ωt − θφ) + sin ωt sin(ωt − θφ)]
= 1.5VmIm 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
·. . I∗aA =
120,000 + j30,0002400
= 50 + j12.5
·. . IaA = 50 − j12.5 A
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
122,656.25(100) = 97.83%
P 11.23 [a] S1 = (4.864 + j3.775) kVA
S2 = 17.636(0.96) + j17.636(0.28) = (16.931 + j4.938) kVA√
3VLIL sin θ3 = 13,853; sin θ3 =13,853√
3(208)(73.8)= 0.521
Therefore cos θ3 = 0.854
Therefore
P3 =13,8530.521
× 0.854 = 22,693.58 W
S3 = 22.694 + j13.853 kVA
ST = S1 + S2 + S3 = 44.49 + j22.57 kVA
ST/φ =13ST = 14.83 + j7.52 kVA
208√3
I∗aA = (14.83 + j7.52)103; I∗
aA = 123.49 + j62.64 A
IaA = 123.49 − j62.64 = 138.46/− 26.90 A (rms)
[b] pf = cos(−26.90) = 0.892 lagging
P 11.24
4000I∗1 = (210 + j280)103
Problems 11–21
I∗1 =
2104
+ j2804
= 52.5 + j70 A
I1 = 52.5 − j70 A
I2 =4000/0
15.36 − j4.48= 240 + j70 A
·. . 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√
3VLIL cos θ = 76,907.22
IL =76,907.22√3(208)(0.88)
= 242.58 A
[b] Q =√
3VLIL sin φ =√
3(208)(242.58)(0.475) = 41,510.12 VAR
P 11.26 [a] I∗aA =
(160 + j46.67)103
1200= 133.3 + j38.9
IaA = 133.3 − j38.9 A
Van = 1200 + (133.3 − j38.9)(0.18 + j1.44) = 1280 + j185 V
IC =1280 + j185
−j60= −3.1 + j21.3 A
Ina = (IaA + IC) = −130.3 − j17.6 = 131.4/7.7 A
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]
Ss/φ =13(60)(0.96 − j0.28) × 103 = 19.2 − j5.6 kVA
S1/φ = 15 kVA
S2/φ = Ss/φ − S1/φ = 4.2 − j5.6 kVA
·. . I∗2 =
4200 − j5600630/
√3
= 11.547 − j15.396 A
I2 = 11.547 + j15.396 A
Zy =630/0 /
√3
I2= 11.34 − j15.12 Ω
Z∆ = 3Zy = 34.02 − j45.36 Ω
Problems 11–23
[b] R =(630/
√3)2
4200= 31.5 Ω; R∆ = 3R = 94.5 Ω
XL =(630/
√3)2
−5600= −23.625 Ω; X∆ = 3XL = −70.875 Ω
P 11.28 Assume a ∆-connect load (series):
Sφ =13(96 × 103)(0.8 + j0.6) = 25,600 + j19,200 VA
Z∗∆φ =
|480|225,600 + j19,200
= 5.76 − j4.32 Ω
Z∆φ = 5.76 + 4.32 Ω
Now assume a Y-connected load (series):
ZY φ =13Z∆φ = 1.92 + j1.44 Ω
11–24 CHAPTER 11. Balanced Three-Phase Circuits
Now assume a ∆-connected load (parallel):
Pφ =|480|2R∆
R∆φ =|480|225,600
= 9 Ω
Qφ =|480|2X∆
X∆φ =|480|219,200
= 12 Ω
Now assume a Y-connected load (parallel):
RY φ =13R∆φ = 3 Ω
XY φ =13X∆φ = 4 Ω
Problems 11–25
P 11.29 Sg/φ =13(41.6)(0.707 + j0.707) × 103 = 9803.73 + j9806.69 VA
I∗aA =
9803.73 + j9803.73240/
√3
= 70.75 + j70.77 A
IaA = 70.75 − j70.77 A
VAN =240√
3− (0.04 + j0.03)(70.75 − j70.77)
= 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)
11–26 CHAPTER 11. Balanced Three-Phase Circuits
P 11.30 [a]
SL/φ =13
[720 + j
7200.8
(0.6)]103 = 240,000 + j180,000 VA
I∗aA =
240,000 + j180,0002400
= 100 + j75 A
IaA = 100 − j75 A
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])
S2 = 0 − j13(576) × 103 = −j192,000 VAR
I∗2 =
−j192,0002400
= −j80 A
·. . 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 11–27
[c] |IaA| = 125 A
Ploss/φ = (125)2(0.8) = 12,500 W
Pg/φ = 240,000 + 12,500 = 252.5 kW
% η =240
252.5(100) = 95.05%
[d] |IaA| = 100.125 A
P/φ = (100.125)2(0.8) = 8020 W
% η =240,000248,200
(100) = 96.77%
[e] Zcap/Y = −j24002
−192,000= −j30 Ω
Zcap/∆ = 3Zcap/Y = −j90 Ω
·. .1
ωC= 90; C =
1(90)(120π)
= 29.47 µF
P 11.31 [a] From Assessment Problem 11.9, IaA = (101.8 − j135.7) A
Therefore Icap = j135.7 A
Therefore ZCY =2450/
√3
j135.7= −j10.42 Ω
Therefore CY =1
(10.42)(2π)(60)= 254.5 µF
ZC∆ = (−j10.42)(3) = −j31.26 Ω
Therefore C∆ =254.5
3= 84.84 µF
[b] CY = 254.5 µF
[c] |IaA| = 101.8 A
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 11–29
P 11.34 [a] W2 − W1 = VLIL[cos(θ − 30) − cos(θ + 30)]
= VLIL[cos θ cos 30 + sin θ sin 30
− cos θ cos 30 + sin θ sin 30]
= 2VLIL sin θ sin 30 = VLIL sin θ,
therefore√
3(W2 − W1) =√
3VLIL 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
11–30 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
VBC = VBN − VCN = 680√
3/− 90 V
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
160 − j120= 69,222.4 + j51,916.8 VA
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: PT = 207.7 kW = Wm1 + Wm2.
P 11.38 [a] I∗aA =
144(0.96 − j0.28)103
7200= 20/− 16.26 A
VBN = 7200/− 120 V; VCN = 7200/120 V
VBC = VBN − VCN = 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 11–31
P 11.39 [a] W1 = |VBA||IbB| cos θ1
Negative phase sequence:
VBA = 240√
3/150 V
IaA =240/0
13.33/− 30 = 18/30 A
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
IAB =240
√3/− 30
20/30 = 20.78/− 60 A
IBC =240
√3/90
60/0 = 6.93/90 A
ICA =240
√3/− 150
40/− 30 = 10.39/− 120 A
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
11–32 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
P 11.41 tan φ =√
3(W2 − W1)W1 + W2
= 0.7498
·. . φ = 36.86
·. . 2400|IL| cos 66.87 = 40,823.09
|IL| = 43.3 A
|Zφ| =2400/
√3
43.3= 32 Ω ·. . Zφ = 32/36.87 Ω
P 11.42 [a] Z =13Z∆ = 4.48 + j15.36 = 16/73.74 Ω
IaA =600/0
16/73.74 = 37.5/− 73.74 A
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 11–33
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 (round-off differences)
P 11.44 [a] For one phase,
[b]
[c]
11–34 CHAPTER 11. Balanced Three-Phase Circuits
[d]
P 11.45 [a] Q =|V|2XC
·. . |XC| =(13,800)2
1.2 × 106 = 158.70 Ω
·. .1
ωC= 158.70; C =
12π(60)(158.70)
= 16.71 µF
[b] |XC| =(13,800/
√3)2
1.2 × 106 =13(158.70)
·. . C = 3(16.71) = 50.14 µF
P 11.46 If the capacitors remain connected when the substation drops its load, the expressionfor the line current becomes
13,800√3
I∗aA = −j1.2 × 106
or I∗aA = −j150.61 A
Hence IaA = j150.61 A
Now,
Van =13,800√
3/0 + (0.6 + j4.8)(j150.61) = 7244.49 + j90.37 = 7245.05/0.71 V
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 amountof power loss in the lines, which in this example is cut in half.
P 11.48 [a]13,800√
3I∗aA = 80 × 103 + j200 × 103 − j1200 × 103
I∗aA =
80√
3 − j1000√
313.8
= 10.04 − j125.51 A
·. . IaA = 10.04 + j125.51 A
Van =13,800√
3/0 + (0.6 + j4.8)(10.04 + j125.51)
= 7371.01 + j123.50 = 7372.04/0.96 V
·. . |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 theallowable minimum of 13 kV.
P 11.49 [a]13,800√
3I∗aA = (80 + j200) × 103
I∗aA =
80√
3 + j200√
313.8
= 10.04 + j25.1 A
·. . IaA = 10.04 − j25.1 A
Van =13,800√
3/0 + (0.6 + j4.8)(10.04 − j25.1)
= 8093.95 + j33.13 = 8094.02/0.23 V
·. . |Vab| =√
3(8094.02) = 14,019.25 V
[b] Yes: 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 lossis greatly reduced.
12Introduction to the Laplace Transform
Assessment Problems
AP 12.1 [a] cosh βt =eβt + e−βt
2Therefore,
Lcosh βt =12
∫ ∞
0−[e−(s−β)t + e−(s+β)t]dt
=12
[e−(s−β)t
−(s − β)
∣∣∣∣∞0−
+e−(s+β)t
−(s + β)
∣∣∣∣∞0−
]
=12
(1
s − β+
1s + β
)=
s
s2 − β2
[b] sinh βt =eβt − e−βt
2Therefore,
Lsinh βt =12
∫ ∞
0−
[e−(s−β)t − e−(s+β)t
]dt
=12
[e−(s−β)t
−(s − β)
]∞
0−− 1
2
[e−(s+β)t
−(s + β)
]∞
0−
=12
(1
s − β− 1
s + β
)=
β
(s2 − β2)
AP 12.2 [a] Let f(t) = te−at:
F (s) = Lte−at =1
(s + a)2
Now, Ltf(t) = −dF (s)ds
12–1
12–2 CHAPTER 12. Introduction to the Laplace Transform
So, Lt · te−at = − d
ds
[1
(s + a)2
]=
2(s + a)2
[b] Let f(t) = e−at sinh βt, then
Lf(t) = F (s) =β
(s + a)2 − β2
L
df(t)dt
= sF (s) − f(0−) =
s(β)(s + a)2 − β2 − 0 =
βs
(s + a)2 − β2
[c] Let f(t) = cos ωt. Then
F (s) =s
(s2 + ω2)and
dF (s)ds
=−(s2 − ω2)(s2 + ω2)2
Therefore Lt cos ωt = −dF (s)ds
=s2 − ω2
(s2 + ω2)2
AP 12.3 F (s) =6s2 + 26s + 26
(s + 1)(s + 2)(s + 3)=
K1
s + 1+
K2
s + 2+
K3
s + 3
K1 =6 − 26 + 26
(1)(2)= 3; K2 =
24 − 52 + 26(−1)(1)
= 2
K3 =54 − 78 + 26
(−2)(−1)= 1
Therefore f(t) = [3e−t + 2e−2t + e−3t] u(t)
AP 12.4 F (s) =7s2 + 63s + 134
(s + 3)(s + 4)(s + 5)=
K1
s + 3+
K2
s + 4+
K3
s + 5
K1 =63 − 189 + 134
1(2)= 4; K2 =
112 − 252 + 134(−1)(1)
= 6
K3 =175 − 315 + 134
(−2)(−1)= −3
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
Problems 12–3
F (s) =K1
s + 5+
K2
s + 5 − j12+
K∗2
s + 5 + j12
K1 =10(25 + 119)25 − 50 + 169
= 10
K2 =10[(−5 + j12)2 + 119]
(j12)(j24)= j4.167 = 4.167/90
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) =4s2 + 7s + 1
s(s + 1)2 =K0
s+
K1
(s + 1)2 +K2
s + 1
K0 =1
(1)2 = 1; K1 =4 − 7 + 1
−1= 2
K2 =d
ds
[4s2 + 7s + 1
s
]s=−1
=s(8s + 7) − (4s2 + 7s + 1)
s2
∣∣∣∣∣s=−1
=1 + 2
1= 3
Therefore f(t) = [1 + 2te−t + 3e−t] u(t)
AP 12.7 F (s) =40
(s2 + 4s + 5)2 =40
(s + 2 − j1)2(s + 2 + j1)2
=K1
(s + 2 − j1)2 +K2
(s + 2 − j1)+
K∗1
(s + 2 + j1)2
+K∗
2
(s + 2 + j1)
K1 =40
(j2)2 = −10 = 10/180 and K∗1 = −10
K2 =d
ds
[40
(s + 2 + j1)2
]s=−2+j1
=−80(j2)3 = −j10 = 10/− 90
K∗2 = j10
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 + 2)(s + 4)
=5s2 + 29s + 32
s2 + 6s + 8= 5 − s + 8
(s + 2)(s + 4)
s + 8(s + 2)(s + 4)
=K1
s + 2+
K2
s + 4
K1 =−2 + 8
2= 3; K2 =
−4 + 8−2
= −2
Therefore,
F (s) = 5 − 3s + 2
+2
s + 4
f(t) = 5δ(t) + [−3e−2t + 2e−4t]u(t)
AP 12.9 F (s) =2s3 + 8s2 + 2s − 4
s2 + 5s + 4= 2s − 2 +
4(s + 1)(s + 1)(s + 4)
= 2s − 2 +4
s + 4
f(t) = 2dδ(t)dt
− 2δ(t) + 4e−4t u(t)
AP 12.10
lims→∞ sF (s) = lim
s→∞
[7s3[1 + (9/s) + (134/7s2)]
s3[1 + (3/s)][1 + (4/s)][1 + (5/s)]
]= 7
·. . f(0+) = 7
lims→0
sF (s) = lims→0
[7s3 + 63s2 + 134s
(s + 3)(s + 4)(s + 5)
]= 0
·. . f(∞) = 0
lims→∞ sF (s) = lim
s→∞
[s3[4 + (7/s) + (1/s2)]
s3[1 + (1/s)]2
]= 4
·. . f(0+) = 4
Problems 12–5
lims→0
sF (s) = lims→0
[4s2 + 7s + 1
(s + 1)2
]= 1
·. . f(∞) = 1
lims→∞ sF (s) = lim
s→∞
[40s
s4[1 + (4/s) + (5/s2)]2
]= 0
·. . f(0+) = 0
lims→0
sF (s) = lims→0
[40s
(s2 + 4s + 5)2
]= 0
·. . f(∞) = 0
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 [a] A =(1
2
)bh =
(12
)(2ε)
(1ε
)= 1
[b] 0; [c] ∞
P 12.6 [a] I =∫ 3
−1(t3 + 2)δ(t) dt +
∫ 3
−18(t3 + 2)δ(t − 1) dt
= (03 + 2) + 8(13 + 2) = 2 + 8(3) = 26
[b] I =∫ 2
−2t2δ(t) dt +
∫ 2
−2t2δ(t + 1.5) dt +
∫ 2
−2t2δ(t − 3) dt
= 02 + (−1.5)2 + 0 = 2.25
P 12.7 f(t) =12π
∫ ∞
−∞(4 + jω)(9 + jω)
· πδ(ω) · ejtω dω =( 1
2π
)(4 + j09 + j0
πejt0
)=
29
P 12.8 As ε → 0 the amplitude → ∞; the duration → 0; and the area is independent of ε,i.e.,
A =∫ ∞
−∞ε
π
1ε2 + t2
dt = 1
P 12.9 F (s) =∫ ε
−ε
12ε
e−st dt =esε − e−sε
2εs
F (s) =12s
limε→0
[sesε + se−sε
1
]=
12s
· 2s1
= 1
P 12.10 [a] Let dv = δ′(t − a) dt, v = δ(t − a)
u = f(t), du = f ′(t) dt
Therefore∫ ∞
−∞f(t)δ′(t − a) dt = f(t)δ(t − a)
∣∣∣∣∞−∞−∫ ∞
−∞δ(t − a)f ′(t) dt
= 0 − f ′(a)
[b] Lδ′(t) =∫ ∞
0−δ′(t)e−st dt = −
[d(e−st)
dt
]t=0
= −[−se−st
]t=0
= s
12–8 CHAPTER 12. Introduction to the Laplace Transform
P 12.11
F (s) =∫ −ε/2
−ε
4ε3 e−st dt +
∫ ε/2
−ε/2
(−4ε3
)e−st dt +
∫ ε
ε/2
4ε3 e−st dt
Therefore F (s) =4
sε3 [esε − 2esε/2 + 2e−sε/2 − e−sε]
Lδ′′(t) = limε→0
F (s)
After applying L’Hopital’s rule three times, we have
limε→0
2s3
[sesε − s
4esε/2 − s
4e−sε/2 + se−sε
]=
2s3
(3s2
)
Therefore Lδ′′(t) = s2
P 12.12 L
dnf(t)dtn
= snF (s) − sn−1f(0−) − sn−2f ′(0−) − · · · ,
Therefore
Lδ(n)(t) = sn(1) − sn−1δ(0−) − sn−2δ′(0−) − · · · = sn
P 12.13 [a] Lt =1s2 ; therefore Lte−at =
1(s + a)2
[b] sin ωt =ejωt − e−jωt
j2Therefore
Lsin ωt =(
1j2
)(1
s − jω− 1
s + jω
)=(
1j2
)( 2jωs2 + ω2
)
=ω
s2 + ω2
Problems 12–9
[c] sin(ωt + θ) = (sin ωt cos θ + cos ωt sin θ)Therefore
Lsin(ωt + θ) = cos θLsin ωt + sin θLcos ωt=
ω cos θ + s sin θ
s2 + ω2
[d] Lt =∫ ∞
0te−st dt =
e−st
s2 (−st − 1)∣∣∣∣∞0
= 0 − 1s2 (0 − 1) =
1s2
[e] f(t) = cosh t cosh θ + sinh t sinh θ
From Assessment Problem 12.1(a)
Lcosh t =s
s2 − 1From Assessment Problem 12.1(b)
Lsinh t =1
s2 − 1
·. . Lcosh(t + θ) = cosh θ[
s
s2 − 1
]+ sinh θ
[ 1s2 − 1
]
=sinh θ + s[cosh θ]
s2 − 1
P 12.14 [a] Lte−at =∫ ∞
0−te−(s+a)t dt
=e−(s+a)t
(s + a)2
[− (s + a)t − 1
]∞0−
= 0 +1
(s + a)2
·. . Lte−at =1
(s + a)2
[b]
L
d
dt(te−at)
=
s
(s + a)2 − 0
=s
(s + a)2
[c]d
dt(te−at) = −ate−at + e−at
L−ate−at + e−at =−a
(s + a)2 +1
(s + a)=
−a
(s + a)2 +s + a
(s + a)2
·. . L
d
dt(te−at)
=
s
(s + a)2 CHECKS
12–10 CHAPTER 12. Introduction to the Laplace Transform
P 12.15 [a] Lf ′(t) =∫ ε
−ε
e−st
2εdt +
∫ ∞
ε− ae−a(t−ε)e−st dt
=1
2sε(esε − e−sε) −
(a
s + a
)e−sε = F (s)
limε→0
F (s) = 1 − a
s + a=
s
s + a
[b] Le−at =1
s + a
Therefore Lf ′(t) = sF (s) − f(0−) =s
s + a− 0 =
s
s + a
P 12.16 Le−atf(t) =∫ ∞
0−[e−atf(t)]e−st dt =
∫ ∞
0−f(t)e−(s+a)t dt = F (s + a)
P 12.17 [a] L∫ t
0−e−ax dx
=
F (s)s
=1
s(s + a)
[b] L∫ t
0−y dy
=
1s
( 1s2
)=
1s3
[c]∫ t
0−e−ax dx =
1a
− e−at
a
L
1a
− e−at
a
=
1a
[1s
− 1s + a
]=
1s(s + a)
∫ t
0−y dy =
t2
2; L
t2
2
=
12
· 2s3 =
1s3
P 12.18 [a] L
d sin ωt
dt
=
sω
s2 + ω2 − 0
[b] L
d cos ωt
dt
=
s2
s2 + ω2 − 0
[c] L
d3(t2)dt3
= s3
( 2s3
)− s2(0) − s(0) − 2(0) = 2
[d]d sin ωt
dt= (cos ωt) · ω, Lω cos ωt =
ωs
s2 + ω2
d cos ωt
dt= −ω sin ωt + δ(t)
L−ω sin ωt + δ(t) = − ω2
s2 + ω2 + 1 =s2
s2 + ω2
d2(t2)dt2
= 2u(t);d3(t2)dt3
= 2δ(t); L2δ(t) = 2
Problems 12–11
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)
·. . F (s) =5[1 − 2e−2s + 2e−6s − e−8s]
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)
Lf ′(t) =5[1 − 2e−2s + 2e−6s − e−8s]
s
[c]
f ′′(t) = 5δ(t) − 10δ(t − 2) + 10δ(t − 6) − 5δ(t − 8)
Lf ′′(t) = 5[1 − 2e−2s + 2e−6s − e−8s]
12–12 CHAPTER 12. Introduction to the Laplace Transform
P 12.20 [a]∫ t
0−x dx =
t2
2
L
t2
2
=
12
∫ ∞
0−t2e−st dt
=12
[e−st
−s3 (s2t2 + 2st + 2)∣∣∣∣∣∞
0−
]
=1
2s3 (2) =1s3
·. . L∫ t
0−x dx
=
1s3
[b] L∫ t
0−x dx
=
Lts
=1/s2
s=
1s3
·. . L∫ t
0−x dx
=
1s3 CHECKS
P 12.21 [a] L40e−8(t−3)u(t − 3) =40e−3s
(s + 8)[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) =5[e−2s − 2e−4s + 2e−8s − e−10s]
s2
P 12.22 Lf(at) =∫ ∞
0−f(at)e−st dt
Let u = at, du = a dt, u = 0− when t = 0−
and u = ∞ when t = ∞
Therefore Lf(at) =∫ ∞
0−f(u)e−(u/a)s du
a=
1aF (s/a)
P 12.23 [a] f1(t) = e−at sin ωt; F1(s) =ω
(s + a)2 + ω2
F (s) = sF1(s) − f1(0−) =sω
(s + a)2 + ω2 − 0
Problems 12–13
[b] f1(t) = e−at cos ωt; F1(s) =s + a
(s + a)2 + ω2
F (s) =F1(s)
s=
s + a
s[(s + a)2 + ω2]
[c]d
dt[e−at sin ωt] = ωe−at cos ωt − ae−at sin ωt
Therefore F (s) =ω(s + a) − ωa
(s + a)2 + ω2 =ωs
(s + a)2 + ω2
∫ t
0−e−ax cos ωx dx =
−ae−at cos ωt + ωe−at sin ωt + a
a2 + ω2
Therefore
F (s) =1
a2 + ω2
[ −a(s + a)(s + a)2 + ω2 +
ω2
(s + a)2 + ω2 +a
s
]
=s + a
s[(s + a)2 + ω2]
P 12.24 [a]dF (s)
ds=
d
ds
[∫ ∞
0−f(t)e−st dt
]= −
∫ ∞
0−tf(t)e−st dt
Therefore Ltf(t) = −dF (s)ds
[b]d2F (s)
ds2 =∫ ∞
0−t2f(t)e−st dt;
d3F (s)ds3 =
∫ ∞
0−−t3f(t)e−st dt
ThereforednF (s)
dsn= (−1)n
∫ ∞
0−tnf(t)e−st dt = (−1)nLtnf(t)
[c] Lt5 = Lt4t = (−1)4 d4
ds4
( 1s2
)=
120s6
Lt sin βt = (−1)1 d
ds
(β
s2 + β2
)=
2βs
(s2 + β2)2
Lte−t cosh t:
From Assessment Problem 12.1(a),
F (s) = Lcosh t =s
s2 − 1
dF
ds=
(s2 − 1)1 − s(2s)(s2 − 1)2 = − s2 + 1
(s2 − 1)2
Therefore − dF
ds=
s2 + 1(s2 − 1)2
12–14 CHAPTER 12. Introduction to the Laplace Transform
Thus
Lt cosh t =s2 + 1
(s2 − 1)2
Le−tt cosh t =(s + 1)2 + 1
[(s + 1)2 − 1]2=
s2 + 2s + 2s2(s + 2)2
P 12.25 [a]∫ ∞
sF (u)du =
∫ ∞
s
[∫ ∞
0−f(t)e−ut dt
]du =
∫ ∞
0−
[∫ ∞
sf(t)e−ut du
]dt
=∫ ∞
0−f(t)
∫ ∞
se−ut du dt =
∫ ∞
0−f(t)
[e−tu
−t
∣∣∣∣∞s
]dt
=∫ ∞
0−f(t)
[−e−st
−t
]dt = L
f(t)
t
[b] Lt sin βt =2βs
(s2 + β2)2
therefore L
t sin βt
t
=∫ ∞
s
[2βu
(u2 + β2)2
]du
Let ω = u2 + β2, then ω = s2 + β2 when u = s, and ω = ∞ when u = ∞;also dω = 2u du, thus
L
t sin βt
t
= β
∫ ∞
s2+β2
[dω
ω2
]= β
(−1ω
) ∣∣∣∣∞s2+β2
=β
s2 + β2
P 12.26 Ig(s) =1.2s
s2 + 1;
1RC
= 1.6;1
LC= 1;
1C
= 1.6
V (s)R
+1L
V (s)s
+ C[sV (s) − v(0−)] = Ig(s)
V (s)[ 1R
+1Ls
+ sC]
= Ig(s)
V (s) =Ig(s)
1R
+ 1Ls
+ sC=
LsIg(s)LRs + 1 + s2LC
=1CsIg(s)
s2 + 1RC
s + 1LC
=(1.6)(1.2)s2
(s2 + 1.6s + 1)(s2 + 1)=
1.92s2
(s2 + 1.6s + 1)(s2 + 1)
P 12.27 [a]vo − Vdc
R+
1L
∫ t
0vo dx + C
dvo
dt= 0
·. . vo +R
L
∫ t
0vo dx + RC
dvo
dt= Vdc
Problems 12–15
[b] Vo +R
L
Vo
s+ RCsVo =
Vdc
s
·. . sLVo + RVo + RCLs2Vo = LVdc
·. . Vo(s) =(1/RC)Vdc
s2 + (1/RC)s + (1/LC)
[c] io =1L
∫ t
0vo dx
Io(s) =Vo
sL=
(1/RCL)Vdc
s[s2 + (1/RC)s + (1/LC)]
P 12.28 [a]1
LC=
1(200 × 10−3)(100 × 10−9)
= 50 × 106
1RC
=1
(5000)(100 × 10−9)= 2000
Vo(s) =70,000
s2 + 2000s + 50 × 106
s1,2 = −1000 ± j7000 rad/s
Vo(s) =70,000
(s + 1000 − j7000)(s + 1000 + j7000)
=K1
s + 1000 − j7000+
K∗1
s + 1000 + j7000
K1 =70,000j14,000
= 5/− 90
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)
=K1
s+
K2
s + 1000 − j7000+
K∗2
s + 1000 + j7000
K1 =35(10,000)50 × 106 = 7 mA
K2 =35(10,000)
(−1000 + j7000)(j14,000)= 3.54/171.87 mA
io(t) = [7 + 7.07e−1000t cos(7000t + 171.87)]u(t) mA
12–16 CHAPTER 12. Introduction to the Laplace Transform
P 12.29 [a] Idc =1L
∫ t
0vo dx +
vo
R+ C
dvo
dt
[b]Idc
s=
Vo(s)sL
+Vo(s)
R+ sCVo(s)
·. . Vo(s) =Idc/C
s2 + (1/RC)s + (1/LC)
[c] io = Cdvo
dt
·. . Io(s) = sCVo(s) =sIdc
s2 + (1/RC)s + (1/LC)
P 12.30 [a]1
RC=
1(1 × 103)(2 × 10−6)
= 500
1LC
=1
(12.5)(2 × 10−6)= 40,000
Vo(s) =500,000Idc
s + 500s + 40,000
=500,000Idc
(s + 100)(s + 400)
=15,000
(s + 100)(s + 400)
=K1
s + 100+
K2
s + 400
K1 =15,000300
= 50; K2 =15,000−300
= −50
Vo(s) =50
s + 100− 50
s + 400
vo(t) = [50e−100t − 50e−400t]u(t) V
[b] Io(s) =0.03s
(s + 100)(s + 400)
=K1
s + 100+
K2
s + 400
K1 =0.03(−100)
300= −0.01
K2 =0.03(−400)
−300= 0.04
Problems 12–17
Io(s) =−0.01s + 100
+0.04
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 resistorcurrent is zero because the initial capacitor voltage is zero by hypothesis. Thusat t = 0 the source current appears in the capacitor.
P 12.31 [a] Cdv1
dt+
v1 − v2
R= ig
1L
∫ t
0v2 dτ +
v2 − v1
R= 0
or
Cdv1
dt+
v1
R− v2
R= ig
−v1
R+
v2
R+
1L
∫ t
0v2 dτ = 0
[b] CsV1(s) +V1(s)
R− V2(s)
R= Ig(s)
−V1(s)R
+V2(s)
R+
V2(s)sL
= 0
or
(RCs + 1)V1(s) − V2(s) = RIg(s)
−sLV1(s) + (R + sL)V2(s) = 0
Solving,
V2(s) =sIg(s)
C[s2 + (R/L)s + (1/LC)]
P 12.321C
= 5 × 106;1
LC= 25 × 106;
R
L= 8000
V2(s) =(6 × 10−3)(5 × 106)
s2 + 8000s + 25 × 106
s1,2 = −4000 ± j3000
V2(s) =30,000
(s + 4000 − j3000)(s + 4000 + j3000)
=K1
s + 4000 − j3000+
K∗1
s + 4000 + j3000
12–18 CHAPTER 12. Introduction to the Laplace Transform
K1 =30,000j6000
= −j5 = 5/− 90
v2(t) = 10e−4000t cos(3000t − 90)
= [10e−4000t sin 3000t]u(t) V
P 12.33 [a] For t ≥ 0+:
vo
R+ C
dvo
dt+ io = 0
vo = Ldiodt
;dvo
dt= L
d2iodt2
·. .L
R
diodt
+ LCd2iodt2
ord2iodt2
+1
RC
diodt
+1
LCio = 0
[b] s2Io(s) − sIdc − 0 +1
RC[sIo(s) − Idc] +
1LC
Io(s) = 0
Io(s)[s2 +
1RC
s +1
LC
]= Idc(s + 1/RC)
Io(s) =Idc[s + (1/RC)]
[s2 + (1/RC)s + (1/LC)]
P 12.341
RC= 8000;
1LC
= 16 × 106
Io(s) =0.005(s + 8000)
s2 + 8000s + 16 × 106
s1,2 = −4000
Io(s) =0.005(s + 8000)
(s + 4000)2 =K1
(s + 4000)2 +K2
s + 4000
K1 = 0.005(s + 8000)∣∣∣∣s=−4000
= 20
K2 =d
ds[0.005(s + 8000)]s=−4000 = 0.005
Io(s) =20
(s + 4000)2 +0.005
s + 4000
io(t) = [20te−4000t + 0.005e−4000t]u(t) V
Problems 12–19
P 12.35 [a] 300 = 60i1 + 25di1dt
+ 10d
dt(i2 − i1) + 5
d
dt(i1 − i2) − 10
di1dt
0 = 5d
dt(i2 − i1) + 10
di1dt
+ 40i2
Simplifying the above equations gives:
300 = 60i1 + 10di1dt
+ 5di2dt
0 = 40i2 + 5di1dt
+ 5di2dt
[b]300s
= (10s + 60)I1(s) + 5sI2(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) =K1
s+
K2
s + 4+
K3
s + 24
K1 =(60)(8)(4)(24)
= 5; K2 =(60)(4)
(−4)(20)= −3
K3 =(60)(−16)
(−24)(−20)= −2
I1(s) =(5
s− 3
s + 4− 2
s + 24
)
i1(t) = (5 − 3e−4t − 2e−24t)u(t) A
I2(s) =K1
s + 4+
K2
s + 24
K1 =−6020
= −3; K2 =−60−20
= 3
I2(s) =( −3
s + 4+
3s + 24
)
i2(t) = (3e−24t − 3e−4t)u(t) A
[e] i1(∞) = 5 A; i2(∞) = 0 A
12–20 CHAPTER 12. Introduction to the Laplace Transform
[f] Yes, at t = ∞
i1 =30060
= 5 A
Since i1 is a dc current at t = ∞ there is no voltage induced in the 10 Hinductor; hence, i2 = 0. Also note that i1(0) = 0 and i2(0) = 0. Thus oursolutions satisfy the condition of no initial energy stored in the circuit.
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
s + 0.8 − j0.6+
K∗1
s + 0.8 + j0.6+
K2
s − j1+
K∗2
s + j1
K1 =1.92s2
(s + 0.8 + j0.6)(s2 + 1)
∣∣∣∣s=−0.8+j0.6
= 1/− 126.87
K2 =1.92s2
(s + j1)(s2 + 1.6s + 1)
∣∣∣∣s=j1
= 0.6/0
Therefore
v(t) = [2e−0.8t cos(0.6t − 126.87) + 1.2 cos(t)]u(t) V
P 12.37 [a] F (s) =K1
s + 1+
K2
s + 2+
K3
s + 4
K1 =8s2 + 37s + 32(s + 2)(s + 4)
∣∣∣∣s=−1
= 1
K2 =8s2 + 37s + 32(s + 1)(s + 4)
∣∣∣∣s=−2
= 5
K3 =8s2 + 37s + 32(s + 1)(s + 2)
∣∣∣∣s=−4
= 2
f(t) = [e−t + 5e−2t + 2e−4t]u(t)
Problems 12–21
[b] F (s) =K1
s+
K2
s + 2+
K3
s + 3+
K4
s + 5
K1 =8s3 + 89s2 + 311s + 300
(s + 2)(s + 3)(s + 5)
∣∣∣∣s=0
= 10
K2 =8s3 + 89s2 + 311s + 300
s(s + 3)(s + 5)
∣∣∣∣s=−2
= 5
K3 =8s3 + 89s2 + 311s + 300
s(s + 2)(s + 5)
∣∣∣∣s=−3
= −8
K4 =8s3 + 89s2 + 311s + 300
s(s + 2)(s + 3)
∣∣∣∣s=−5
= 1
f(t) = [10 + 5e−2t − 8e−3t + e−5t]u(t)
[c] F (s) =K1
s + 1+
K2
s + 2 − j+
K∗2
s + 2 + j
K1 =22s2 + 60s + 58
s2 + 4s + 5
∣∣∣∣s=−1
= 10
K2 =22s2 + 60s + 58
(s + 1)(s + 2 + j)
∣∣∣∣s=−2+j
= 6 + j8 = 10/53.13
f(t) = [10e−t + 20e−2t cos(t + 53.13)]u(t)
[d] F (s) =K1
s+
K2
s + 7 − j+
K∗2
s + 7 + j
K1 =250(s + 7)(s + 14)
s2 + 14s + 50
∣∣∣∣s=0
= 490
K2 =250(s + 7)(s + 14)
s(s + 7 + j)
∣∣∣∣s=−7+j
= 125/− 163.74
f(t) = [490 + 250e−7t cos(t − 163.74)]u(t)
P 12.38 [a] F (s) =K1
s2 +K2
s+
K3
s + 5
K1 =100
s + 5
∣∣∣∣s=0
= 20
K2 =d
ds
[ 100s + 5
]=
−100(s + 5)2
∣∣∣∣s=0
= −4
K3 =100s2
∣∣∣∣s=−5
= 4
f(t) = [20t − 4 + 4e−5t]u(t)
12–22 CHAPTER 12. Introduction to the Laplace Transform
[b] F (s) =K1
s+
K2
(s + 1)2 +K3
s + 1
K1 =50(s + 5)(s + 1)2
∣∣∣∣s=0
= 250
K2 =50(s + 5)
s
∣∣∣∣s=−1
= −200
K3 =d
ds
[50(s + 5)
s
]=[50s
− 50(s + 5)s2
]s=−1
= −250
f(t) = [250 − 200te−t − 250e−t]u(t)
[c] F (s) =K1
s2 +K2
s+
K3
s + 3 − j+
K∗3
s + 3 + j
K1 =100(s + 3)
s2 + 6s + 10
∣∣∣∣s=0
= 30
K2 =d
ds
[100(s + 3)
s2 + 6s + 10
]
=[
100s2 + 6s + 10
− 100(s + 3)(2s + 6)(s2 + 6s + 10)2
]s=0
= 10 − 18 = −8
K3 =100(s + 3)
s2(s + 3 + j)
∣∣∣∣s=−3+j
= 4 + j3 = 5/36.87
f(t) = [30t − 8 + 10e−3t cos(t + 36.87)]u(t)
[d] F (s) =K1
s+
K2
(s + 1)3 +K3
(s + 1)2 +K4
s + 1
K1 =5(s + 2)2
(s + 1)3
∣∣∣∣s=0
= 20
K2 =5(s + 2)2
s
∣∣∣∣s=−1
= −5
K3 =d
ds
[5(s + 2)2
s
]=[10(s + 2)
s− 5(s + 2)2
s2
]s=−1
= −10 − 5 = −15
K4 =12
d
ds
[10(s + 2)
s− 5(s + 2)2
s2
]
=12
[10s
− 10(s + 2)s2 − 10(s + 2)
s2 +10(s + 2)2
s3
]s=−1
Problems 12–23
=12(−10 − 10 − 10 − 10) = −20
f(t) = [20 − 2.5t2e−t − 15te−t − 20e−t]u(t)
[e] F (s) =K1
s+
K2
(s + 2 − j)2 +K∗
2
(s + 2 + j)2 +K3
s + 2 − j+
K∗3
s + 2 − j
K1 =400
(s2 + 4s + 5)2
∣∣∣∣s=0
= 16
K2 =400
s(s + 2 + j)2
∣∣∣∣s=−2+j
= 44.72/26.57
K3 =d
ds
[400
s(s + 2 + j)2
]=[ −400s2(s + 2 + j)2 +
−800s(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
s2 + 6s + 8= 5 +
K1
s + 2+
K2
s + 4
K1 =8s + 40s + 4
∣∣∣∣s=−2
= 12
K2 =8s + 40s + 2
∣∣∣∣s=−4
= −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 +32s + 936
s2 + 48s + 625= 10 +
K1
s + 24 − j7+
K∗2
s + 24 + j7
K1 =32s + 936
s + 24 + j7
∣∣∣∣s=−24+j7
= 16 − j12 = 20/− 36.87
f(t) = 10δ(t) + [40e−24t cos(7t − 36.87)]u(t)
12–24 CHAPTER 12. Introduction to the Laplace Transform
[c] s − 10
F (s) = s2 + 15s + 50 s3 + 5s2 − 50s − 100
s3 + 15s2 + 50s
−10s2 − 100s − 100
−10s2 − 150s − 500
50s + 400
F (s) = s − 10 +K1
s + 5+
K2
s + 10
K1 =50s + 400
s + 10
∣∣∣∣s=−5
= 30
K2 =50s + 400
s + 5
∣∣∣∣s=−10
= 20
f(t) = δ′(t) − 10δ(t) + [30e−5t + 20e−10t]u(t)
P 12.40 [a] F (s) =K1
s2 +K2
s+
K3
s + 1 − j2+
K∗3
s + 1 + j2
K1 =100(s + 1)s2 + 2s + 5
∣∣∣∣s=0
= 20
K2 =d
ds
[100(s + 1)s2 + 2s + 5
]=[
100s2 + 2s + 5
− 100(s + 1)(2s + 2)(s2 + 2s + 5)2
]s=0
= 20 − 8 = 12
K3 =100(s + 1)
s2(s + 1 + j2)
∣∣∣∣s=−1+j2
= −6 + j8 = 10/126.87
f(t) = [20t + 12 + 20e−t cos(2t + 126.87)]u(t)
[b] F (s) =K1
s+
K2
(s + 5)3 +K3
(s + 5)2 +K4
s + 5
K1 =500
(s + 5)3
∣∣∣∣s=0
= 4
K2 =500s
∣∣∣∣s=−5
= −100
K3 =d
ds
[500s
]=
−500s2
∣∣∣∣s=−5
= −20
K4 =12
d
ds
[−500s2
]=
12
1000(s3)
∣∣∣∣s=−5
= −4
f(t) = [4 − 50t2e−5t − 20te−5t − 4e−5t]u(t)
Problems 12–25
[c] F (s) =K1
s+
K2
(s + 1)3 +K3
(s + 1)2 +K4
s + 1
K1 =40(s + 2)(s + 1)3
∣∣∣∣s=0
= 80
K2 =40(s + 2)
s
∣∣∣∣s=−1
= −40
K3 =d
ds
[40(s + 2)
s
]=[40s
− 40(s + 2)s2
]s=−1
= −40 − 40 = −80
K4 =12
d
ds
[40s
− 40(s + 2)s2
]
=12
[−40s2 − 40
s2 +80(s + 2)
s3
]s=−1
=12(−40 − 40 − 80) = −80
f(t) = [80 − 20t2e−t − 80te−t − 80e−t]u(t)
[d] F (s) =K1
s+
K2
(s + 1)4 +K3
(s + 1)3 +K4
(s + 1)2 +K5
s + 1
K1 =(s + 5)2
(s + 1)4
∣∣∣∣s=0
= 25
K2 =(s + 5)2
s
∣∣∣∣s=−1
= −16
K3 =d
ds
[(s + 5)2
s
]=[2(s + 5)
s− (s + 5)2
s2
]s=−1
= −8 − 16 = −24
K4 =12
d
ds
[2(s + 5)
s− (s + 5)2
s2
]
=12
[2s
− 2(s + 5)s2 − 2(s + 5)
s2 +2(s + 5)2
s3
]s=−1
=12(−2 − 8 − 8 − 32) = −25
K5 =16
d
ds
[2s
− 2(s + 5)s2 − 2(s + 5)
s2 +2(s + 5)2
s3
]
=16
[−2s2 − 2
s2 +4(s + 5)
s3 − 2s2 +
4(s + 5)s3 +
4(s + 5)s3 − 6(s + 5)2
s4
]s=−1
=16(−2 − 2 − 16 − 2 − 16 − 16 − 96) = −25
f(t) = [25 − (8/3)t3e−t − 12t2e−t − 25te−t − 25e−t]u(t)
12–26 CHAPTER 12. Introduction to the Laplace Transform
P 12.41 f(t) = L−1
K
s + α − jβ+
K∗
s + α + jβ
= Ke−αtejβt + K∗e−αte−jβt
= |K|e−αt[ejθejβt + e−jθe−jβt]
= |K|e−αt[ej(βt+θ) + e−j(βt+θ)]
= 2|K|e−αt cos(βt + θ)
P 12.42 [a] Ltnf(t) = (−1)n
[dnF (s)
dsn
]
Let f(t) = 1, then F (s) =1s, thus
dnF (s)dsn
=(−1)nn!s(n+1)
Therefore Ltn = (−1)n
[(−1)nn!s(n+1)
]=
n!s(n+1)
It follows that Lt(r−1) =(r − 1)!
sr
and Lt(r−1)e−at =(r − 1)!(s + a)r
ThereforeK
(r − 1)!Ltr−1e−at =
K
(s + a)r= L
Ktr−1e−at
(r − 1)!
[b] f(t) = L−1
K
(s + α − jβ)r+
K∗
(s + α + jβ)r
Therefore
f(t) =Ktr−1
(r − 1)!e−(α−jβ)t +
K∗tr−1
(r − 1)!e−(α+jβ)t
=|K|tr−1e−αt
(r − 1)!
[ejθejβt + e−jθe−jβt
]
=[2|K|tr−1e−αt
(r − 1)!
]cos(βt + θ)
P 12.43 [a] lims→∞ sV (s) = lim
s→∞
[1.92s3
s4[1 + (1.6/s) + (1/s2)][1 + (1/s2)]
]= 0
Therefore v(0+) = 0
[b] No, V has a pair of poles on the imaginary axis.
Problems 12–27
P 12.44 [a] sF (s) =8s3 + 37s2 + 32s
(s + 1)(s + 2)(s + 4)
lims→0
sF (s) = 0, ·. . f(∞) = 0
lims→∞ sF (s) = 8, ·. . f(0+) = 8
[b] sF (s) =8s3 + 89s2 + 311s + 300
(s + 2)(s2 + 8s + 15)
lims→0
sF (s) = 10; ·. . f(∞) = 10
lims→∞ sF (s) = 8, ·. . f(0+) = 8
[c] sF (s) =22s3 + 60s2 + 58s
(s + 1)(s2 + 4s + 5)
lims→0
sF (s) = 0, ·. . f(∞) = 0
lims→∞ sF (s) = 22, ·. . f(0+) = 22
[d] sF (s) =250(s + 7)(s + 14)
(s2 + 14s + 50)
lims→0
sF (s) =250(7)(14)
50= 490, ·. . f(∞) = 490
lims→∞ sF (s) = 250, ·. . f(0+) = 250
P 12.45 [a] sF (s) =100
s(s + 5)F (s) has a second-order pole at the origin so we cannot use the final valuetheorem.
lims→∞ sF (s) = 0, ·. . f(0+) = 0
[b] sF (s) =50(s + 5)(s + 1)2
lims→0
sF (s) = 250, ·. . f(∞) = 250
lims→∞ sF (s) = 0, ·. . f(0+) = 0
[c] sF (s) =100(s + 3)
s(s2 + 6s + 10)F (s) has a second-order pole at the origin so we cannot use the final valuetheorem.
lims→∞ sF (s) = 0, ·. . f(0+) = 0
12–28 CHAPTER 12. Introduction to the Laplace Transform
[d] sF (s) =5(s + 2)2
(s + 1)3
lims→0
sF (s) = 20, ·. . f(∞) = 20
lims→∞ sF (s) = 0, ·. . f(0+) = 0
[e] sF (s) =400
(s2 + 4s + 5)2
lims→0
sF (s) = 16, ·. . f(∞) = 16
lims→∞ sF (s) = 0, ·. . f(0+) = 0
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 theinitial 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)
lims→0
sVo(s) = 0, ·. . vo(∞) = 0
lims→∞ sVo(s) = 0, ·. . vo(0+) = 0
sIo(s) =Vdc/RCL)
s2 + (1/RC)s + (1/LC)
lims→0
sIo(s) =Vdc/RLC
1/LC=
Vdc
R, ·. . io(∞) =
Vdc
R
lims→∞ sIo(s) = 0, ·. . io(0+) = 0
P 12.48 sVo(s) =(Idc/C)s
s2 + (1/RC)s + (1/LC)
lims→0
sVo(s) = 0, ·. . vo(∞) = 0
lims→∞ sVo(s) = 0, ·. . vo(0+) = 0
sIo(s) =s2Idc
s2 + (1/RC)s + (1/LC)
lims→0
sIo(s) = 0, ·. . io(∞) = 0
lims→∞ sIo(s) = Idc, ·. . vo(0+) = Idc
Problems 12–29
P 12.49 [a] sF (s) =100(s + 1)
s(s2 + 2s + 5)F (s) has a second-order pole at the origin, so we cannot use the final valuetheorem here.
lims→∞ sF (s) = 0, ·. . f(0+) = 0
[b] sF (s) =500
(s + 5)3
lims→0
sF (s) = 4, ·. . f(∞) = 4
lims→∞ sF (s) = 0, ·. . f(0+) = 0
[c] sF (s) =40(s + 2)(s + 1)3
lims→0
sF (s) = 80, ·. . f(∞) = 80
lims→∞ sF (s) = 0, ·. . f(0+) = 0
[d] sF (s) =(s + 5)2
(s + 1)4
lims→0
sF (s) = 25, ·. . f(∞) = 25
lims→∞ sF (s) = 0, ·. . f(0+) = 0
P 12.50 sIo(s) =Idcs[s + (1/RC)]
s2 + (1/RC)s + (1/LC)
lims→0
sIo(s) = 0, ·. . io(∞) = 0
lims→∞ sIo(s) = Idc, ·. . io(0+) = Idc
13The Laplace Transform in Circuit
Analysis
Assessment Problems
AP 13.1 [a] Y =1R
+1sL
+ sC =C[s2 + (1/RC)s + (1/LC)
s
1RC
=106
(500)(0.025)= 80,000;
1LC
= 25 × 108
Therefore Y =25 × 10−9(s2 + 80,000s + 25 × 108)
s
[b] −z1,2 = −40,000 ±√
16 × 108 − 25 × 108 = −40,000 ± j30,000 rad/s
−z1 = −40,000 − j30,000 rad/s
−z2 = −40,000 + j30,000 rad/s
−p1 = 0 rad/s
AP 13.2 [a] Z = 2000 +1Y
= 2000 +4 × 107s
s2 + 80,000s + 25 × 108
=2000(s2 + 105s + 25 × 108)
s2 + 80,000s + 25 × 108 =2000(s + 50,000)2
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
13–2 CHAPTER 13. The Laplace Transform in Circuit Analysis
AP 13.3 [a] At t = 0−, 0.2v1 = 0.8v2; v1 = 4v2; v1 + v2 = 100 V
Therefore v1(0−) = 80 V = v1(0+); v2(0−) = 20 V = v2(0+)
I =(80/s) + (20/s)
5000 + [(5 × 106)/s] + (1.25 × 106/s)=
20 × 10−3
s + 1250
V1 =80s
− 5 × 106
s
(20 × 10−3
s + 1250
)=
80s + 1250
V2 =20s
− 1.25 × 106
s
(20 × 10−3
s + 1250
)=
20s + 1250
[b] i = 20e−1250tu(t) mA; v1 = 80e−1250tu(t) V
v2 = 20e−1250tu(t) V
AP 13.4 [a]
I =Vdc/s
R + sL + (1/sC)=
Vdc/L
s2 + (R/L)s + (1/LC)
Vdc
L= 40;
R
L= 1.2;
1LC
= 1.0
I =40
s2 + 1.2s + 1
Problems 13–3
[b] I =40
(s + 0.6 − j0.8)(s + 0.6 + j0.8)=
K1
s + 0.6 − j0.8+
K∗1
s + 0.6 + j0.8
K1 =40
j1.6= −j25 = 25/− 90; K∗
1 = 25/90
i = 50e−0.6t cos(0.8t − 90) = [50e−0.6t sin 0.8t]u(t) A
[c] V = sLI =160s
s2 + 1.2s + 1=
160s(s + 0.6 − j0.8)(s + 0.6 + j0.8)
=K1
s + 0.6 − j0.8+
K∗1
s + 0.6 + j0.8
K1 =160(−0.6 + j0.8)
j1.6= 100/36.87
[d] v(t) = [200e−0.6t cos(0.8t + 36.87)]u(t) V
AP 13.5 [a]
The two node voltage equations are
V1 − V2
s+ V1s =
5s
andV2
3+
V2 − V1
s+
V2 − (15/s)15
= 0
Solving for V1 and V2 yields
V1 =5(s + 3)
s(s2 + 2.5s + 1), V2 =
2.5(s2 + 6)s(s2 + 2.5s + 1)
[b] The partial fraction expansions of V1 and V2 are
V1 =15s
− 50/3s + 0.5
+5/3
s + 2and V2 =
15s
− 125/6s + 0.5
+25/3s + 2
It follows that
v1(t) =[15 − 50
3e−0.5t +
53e−2t
]u(t) V and
v2(t) =[15 − 125
6e−0.5t +
253
e−2t]u(t) V
[c] v1(0+) = 15 − 503
+53
= 0
v2(0+) = 15 − 1256
+253
= 2.5 V
13–4 CHAPTER 13. The Laplace Transform in Circuit Analysis
[d] v1(∞) = 15 V; v2(∞) = 15 V
AP 13.6 [a]
With no load across terminals a–b, Vx = 20/s:
12
[20s
− VTh
]s +
[1.2
(20s
)− VTh
]= 0
therefore VTh =20(s + 2.4)s(s + 2)
Vx = 5IT and ZTh =VT
IT
Solving for IT gives
IT =(VT − 5IT )s
2+ VT − 6IT
Therefore
14IT = VT s − 5sIT + 2VT ; therefore ZTh =5(s + 2.8)
s + 2[b]
I =VTh
ZTh + 2 + s=
20(s + 2.4)s(s + 3)(s + 6)
Problems 13–5
AP 13.7 [a] i2 = 1.25e−t − 1.25e−3t; thereforedi2dt
= −1.25e−t + 3.75e−3t
Thereforedi2dt
= 0 when
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 N1 = 60(s + 2)
Therefore I1 =N1
∆=
5(s + 2)(s + 1)(s + 3)
A partial fraction expansion leads to the expression
I1 =2.5
s + 1+
2.5s + 3
Therefore we get
i1 = 2.5[e−t + e−3t]u(t) A
[c]di1dt
= −2.5[e−t + 3e−3t];di1(0.54931)
dt= −2.89 A/s
[d] When i2 is at its peak value,
di2dt
= 0
Therefore L2
(di2dt
)= 0 and i2 = −
(M
12
)(di1dt
)
[e] i2(max) =−2(−2.89)
12= 481.13 mA (Checks)
AP 13.8 [a] The s-domain circuit with the voltage source acting alone is
V ′ − (20/s)2
+V ′
1.25s+
V ′s20
= 0
13–6 CHAPTER 13. The Laplace Transform in Circuit Analysis
V ′ =200
(s + 2)(s + 8)=
100/3s + 2
− 100/3s + 8
v′ =1003
[e−2t − e−8t]u(t) V
[b] With the current source acting alone,
V ′′
2+
V ′′
1.25s+
V ′′s20
=5s
V ′′ =100
(s + 2)(s + 8)=
50/3s + 2
− 50/3s + 8
v′′ =503
[e−2t − e−8t]u(t) V
[c] v = v′ + v′′ = [50e−2t − 50e−8t]u(t) V
AP 13.9 [a]Vo
s + 2+
Vos
10= Ig; therefore
Vo
Ig
= H(s) =10(s + 2)
s2 + 2s + 10[b] −z1 = −2 rad/s; −p1 = −1 + j3 rad/s; −p2 = −1 − j3 rad/s
AP 13.10 [a] Vo =10(s + 2)
s2 + 2s + 10· 1s
=K0
s+
K1
s + 1 − j3+
K∗1
s + 1 + j3
K0 = 2; K1 = (5/3)/− 126.87; K∗1 = (5/3)/126.87
vo = [2 + (10/3)e−t cos(3t − 126.87)]u(t) V
[b] Vo =10(s + 2)
s2 + 2s + 10· 1 =
K2
s + 1 − j3+
K∗2
s + 1 + j3
K2 = 5.27/− 18.43; K∗2 = 5.27/18.43
vo = [10.54e−t cos(3t − 18.43)]u(t) V
AP 13.11 [a] H(s) = Lh(t) = Lvo(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 13–7
Therefore H(s) =9600(s + 70)
(s + 70)2 + (240)2 − 2800(240)(s + 70)2 + (240)2
=9600s
s2 + 140s + 62,500
[b] Vo(s) = H(s) · 1s
=9600
s2 + 140s + 62,500
=K1
s + 70 − j240+
K∗1
s + 70 + j240
K1 =9600j480
= −j20 = 20/− 90
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)
s2 + 2s + 10
Therefore H(j4) =10(2 + j4)
10 − 16 + j8= 4.47/− 63.43
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, R2C = 2 × 10−5
then V1 = V2 =VgR2
R2 + (1/sC)
AlsoV1 − Vg
R1+
V1 − Vo
R1= 0
therefore Vo = 2V1 − Vg
Now solving for Vo/Vg, we get H(s) =R2Cs − 1R2Cs + 1
It follows that H(j50,000) =j − 1j + 1
= j1 = 1/90
Therefore vo = 10 cos(50,000t + 90) V
13–8 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Replacing R2 by Rx gives us H(s) =RxCs − 1RxCs + 1
Therefore
H(j50,000) =j20 × 10−6Rx − 1j20 × 10−6Rx + 1
=Rx + j50,000Rx − j50,000
Thus,
50,000Rx
= tan 60 = 1.7321, Rx = 28,867.51 Ω
Problems 13–9
Problems
P 13.1 Iscab = IN =−LI0
sL=
−I0
s; ZN = sL
Therefore, the Norton equivalent is the same as the circuit in Fig. 13.4.
P 13.2 i =1L
∫ t
0−vdτ + I0; therefore I =
( 1L
)(V
s
)+
I0
s=
V
sL+
I0
s
P 13.3 VTh = Vab = CVo
( 1sC
)=
Vo
s; ZTh =
1sC
P 13.4 [a] Z = R + sL +1
sC=
L[s2 + (R/L)s + (1/LC)]s
=0.0025[s2 + 16 × 107s + 1010]
s
[b] Zeros at −62.5 rad/s and −1.6 × 108 rad/sPole at 0.
P 13.5 [a] Y =1R
+1sL
+ sC =C[s2 + (1/RC)s + (1/LC)]
s
Z =1Y
=s/C
s2 + (1/RC)s + (1/LC)=
4 × 106s
s2 + 2000s + 64 × 104
[b] zero at −z1 = 0poles at −p1 = −400 rad/s and −p2 = −1600 rad/s
P 13.6 [a]
Z =(R + sL)(1/sC)R + sL + (1/sC)
=(1/C)(s + R/L)
s2 + (R/L)s + (1/LC)
R
L=
2500.08
= 3125;1
LC=
1(0.08)(0.5 × 10−6)
= 25 × 106
Z =2 × 106(s + 3125)
s2 + 3125s + 25 × 106
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 equivalentdelta-connection:
where
Za =(1/s)(1) + (1)(1/s) + (1)(1)
1/s= s + 2
Zb = Zc =(1/s)(1) + (1)(1/s) + (1)(1)
1=
s + 2s
Then
Zab = Za‖[(s‖Zc) + (s‖Zb)] = Za‖2(s‖Zb)
s‖Zb =s + 2
s + (s + 2)/s=
s(s + 2)s2 + s + 2
Zab = (s + 2)‖ 2s(s + 2)s2 + s + 2
=2s(s + 2)2
(s + 2)(s2 + s + 2) + 2s(s + 2)
=2s(s + 2)
s2 + 3s + 2=
2ss + 1
One zero at the origin (0 rad/s); one pole at −1 rad/s.
P 13.8 Z1 =16s
+ s‖4 =16s
+4s
s + 4=
4(s2 + 4s + 16)s(s + 4)
Zab = 4‖4(s2 + 4s + 16)s(s + 4)
=16(s2 + 4s + 16)8s2 + 32s + 64
Problems 13–11
=2(s2 + 4s + 16)
s2 + 4s + 8=
2(s + 2 + j3.46)(s + 2 − j3.46)(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 − j2rad/s.
P 13.9 [a] For t > 0:
[b] Vo =2.5s
(16 × 105)/s + 5000 + 2.5s
(−150s
)
=−150s
s2 + 2000s + 64 × 104
=−150s
(s + 400)(s + 1600)
[c] Vo =K1
s + 400+
K2
s + 1600
K1 =−150s
s + 1600
∣∣∣∣s=−400
= 50
K2 =−150ss + 400
∣∣∣∣s=−1600
= −200
Vo =50
s + 400− 200
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:
1Re
=18
+180
+120
= 0.1875; Re = 5.33 Ω
v1 = (9)(5.33) = 48 V
iL(0−) =4820
= 2.4 A
vC(0−) = −v1 = −48 V
For t = 0+:
s-domain circuit:
where
R = 20 Ω; C = 6.25 µF; γ = −48 V;
L = 6.4 mH; and ρ = −2.4 A
[b]Vo
R+ VosC − γC +
Vo
sL− ρ
s= 0
·. . Vo =γ[s + (ρ/γC)]
s2 + (1/RC)s + (1/LC)
ρ
γC=
−2.4(−48)(6.25 × 10−6)
= 8000
Problems 13–13
1RC
=1
(20)(6.25 × 10−6)= 8000
1LC
=1
(6.4 × 10−3)(6.25 × 10−6)= 25 × 106
Vo =−48(s + 8000)
s2 + 8000s + 25 × 106
[c] IL =Vo
sL− ρ
s=
Vo
0.0064s+
2.4s
=−7500(s + 8000)
s(s2 + 8000s + 25 × 106)+
2.4s
=2.4(s + 4875)
(s2 + 8000s + 25 × 106)
[d] Vo =−48(s + 8000)
s2 + 8000s + 25 × 106
=K1
s + 4000 − j3000+
K∗1
s + 4000 + j3000
K1 =−48(s + 8000)
s + 4000 + j3000
∣∣∣∣s=−4000+j3000
= 40/126.87
vo(t) = [80e−4000t cos(3000t + 126.87)]u(t) V
[e] IL =2.4(s + 4875)
s2 + 8000s + 25 × 106
=K1
s + 4000 − j3000+
K∗1
s + 4000 + j3000
K1 =2.4(s + 4875)
s + 4000 + j3000
∣∣∣∣s=−4000+j3000
= 1.25/− 16.26
iL(t) = [2.5e−4000t cos(3000t − 16.26)]u(t) A
P 13.11 For t < 0:
vo(0−) − 5005
+vo(0−)
25+
vo(0−)100
= 0
13–14 CHAPTER 13. The Laplace Transform in Circuit Analysis
25vo(0−) = 10,000 ·. . vo(0−) = 400 V
iL(0−) =vo(0−)
25=
40025
= 16 A
For t > 0 :
Vo + 40025 + 25s
+Vo
100+
Vo − (400/s)100/s
= 0
Vo
( 125 + 25s
+1
100+
s
100
)= 4 − 400
25 + 25s
·. . Vo =400(s − 3)s2 + 2s + 5
Io =Vo − (400/s)
100/s=
−20s − 20s2 + 2s + 5
=K1
s + 1 − j2+
K∗1
s + 1 + j2
K1 =−20(s + 1)s + 1 + j2
∣∣∣∣s=−1+j2
= −10
io(t) = [−20e−t cos 2t]u(t) A
Problems 13–15
P 13.12 [a] For t < 0:
V2 =10
10 + 40(450) = 90 V
For t > 0:
[b] V1 =25(450/s)
(125,000/s) + 25 + 1.25 × 10−3s
=9 × 106
s2 + 20, 000s + 108 =9 × 106
(s + 10,000)2
v1(t) = (9 × 106te−10,000t)u(t) V
[c] V2 =90s
− (25,000/s)(450/s)(125,000/s) + 1.25 × 10−3s + 25
=90(s + 20,000)
s2 + 20,000s + 108
=900,000
(s + 10,000)2 +90
s + 10,000
v2(t) = [9 × 105te−10,000t + 90e−10,000t]u(t) V
13–16 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.13 [a] For t < 0:
iL(0−) =−100
4 + 10‖40 + 8=
−10020
= −5 A
i1 =1050
(5) = 1 A
vC(0−) = 10(1) + 4(5) − 100 = −70 V
For t > 0:
[b] (20 + 2s + 100/s)I = 10 +70s
·. . I =5(s + 7)
s2 + 10s + 50
Vo =100s
I − 70s
=−70s2 − 200s
s(s2 + 10s + 50)=
−70(s + 20/7)s2 + 10s + 50
=K1
s + 5 − j5+
K∗1
s + 5 + j5
K1 =−70(s + 20/7)
s + 5 + j5
∣∣∣∣s=−5+j5
= 38.1/− 156.8
[c] vo(t) = 76.2e−5t cos(5t − 156.8)u(t) V
Problems 13–17
P 13.14 [a] iL(0−) = iL(0+) =243
= 8 A directed upward
VT = 25Iφ +[
20(10/s)20 + (10/s)
]IT =
25IT (10/s)20 + (10/s)
+( 200
10 + 20s
)IT
VT
IT
= Z =250 + 20020s + 10
=45
2s + 1
Vo
5+
Vo(2s + 1)45
+Vo
5.625s=
8s
[9s + (2s + 1)s + 8]Vo
45s=
8s
Vo[2s2 + 10s + 8] = 360
Vo =360
2s2 + 10s + 8=
180s2 + 5s + 4
[b] Vo =180
(s + 1)(s + 4)=
K1
s + 1+
K2
s + 4
K1 =1803
= 60; K2 =180−3
= −60
Vo =60
s + 1− 60
s + 4
vo(t) = [60e−t − 60e−4t]u(t) V
13–18 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.15 [a]
Vo − 35/s2
+ 0.4V∆ +Vo − 8Iφ
s + (250/s)= 0
V∆ =[
Vo − 8Iφ
s + (250/s)
]s; Iφ =
(35/s) − Vo
2
Solving for Vo yields:
Vo =29.4s2 + 56s + 1750
s(s2 + 2s + 50)=
29.4s2 + 56s + 1750s(s + 1 − j7)(s + 1 + j7)
Vo =K1
s+
K2
s + 1 − j7+
K∗2
s + 1 + j7
K1 =29.4s2 + 56s + 1750
s2 + 2s + 50
∣∣∣∣s=0
= 35
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 − 352
+ 0.4v∆ = 0; vo − 35 + 0.8v∆ = 0
vo = v∆ + 8iφ = v∆ + 8(0.4v∆) = 4.2v∆
vo + (0.8)vo
4.2= 35; ·. . vo(0+) = 29.4 V(Checks)
Problems 13–19
At t = ∞, the circuit is
v∆ = 0, iφ = 0 ·. . vo = 35 V(Checks)
P 13.16 [a] For t < 0:
Vc − 50400
+Vc
1200+
Vc − 137.5500
= 0
Vc
( 1400
+1
1200+
1500
)=
50400
+137.5500
Vc = 75 V
iL(0−) =75 − 137.5
500= −0.125 A
For t > 0:
[b] Vo =5 × 105
sI +
75s
0 = −137.5s
+ 100I +5 × 105
sI +
75s
− 1.25 × 10−3 + 0.01sI
13–20 CHAPTER 13. The Laplace Transform in Circuit Analysis
I
(100 +
5 × 105
s+ 0.01s
)=
62.5s
+ 1.25 × 10−3
·. . I =6250 + 0.125s
s2 + 104s + 5 × 107
Vo =5 × 105
s
( 6250 + 0.125ss2 + 104s + 5 × 107
)+
75s
=75s2 + 812,500s + 6875 × 106
s(s2 + 104s + 5 × 107)
[c] Vo =K1
s+
K2
s + 5000 − j5000+
K∗2
s + 5000 + j5000
K1 =75s2 + 812,500s + 6875 × 106
s2 + 104s + 5 × 107
∣∣∣∣s=0
= 137.5
K2 =75s2 + 812,500s + 6875 × 106
s(s + 5000 + j5000)
∣∣∣∣s=−5000+j5000
= 40.02/141.34
vo(t) = [137.5 + 80.04e−5000t cos(5000t + 141.34)]u(t) V
P 13.17
5 × 10−3
s=
Vo
200 + 4 × 106/s+ 3.75 × 10−3Vφ +
Vo
0.04s
Vφ =4 × 106/s
200 + 4 × 106/sVo =
4 × 106Vo
200s + 4 × 106
·. .5 × 10−3
s=
Vos
200s + 4 × 106 +15,000Vo
200s + 4 × 106 +25Vo
s
·. . Vo =s + 20,000
s2 + 20,000s + 108 =K1
(s + 10,000)2 +K2
s + 10,000
K1 = 10,000; K2 = 1
Vo =10,000
(s + 10,000)2 +1
s + 10,000
vo(t) = [10,000te−10,000t + e−10,000t]u(t) V
Problems 13–21
P 13.18 vo(0−) = vo(0+) = 0
−0.05s
+Vo
1000+
Vo
25− 21
Vo
1000+
Vo
107/s= 0
Vo
( 201000
+s
107
)=
0.05s
·. . Vo =500,000
s(s + 200,000)=
2.5s
− 2.5s + 200,000
vo(t) = [2.5 − 2.5e−200,000t]u(t) V
P 13.19 [a] io(0−) =20
4000= 5 mA
Io =20/s + Lρ
R + sL + 1/sC
=20/L + sρ
s2 + sR/L + 1/LC=
40 + s(0.005)s2 + 8000s + 16 × 106
Vo = −Lρ + sLIo = −0.0025 +0.0025s(s + 8000)
s2 + 8000s + 16 × 106
=−40,000
(s + 4000)2
vo(t) = −40,000te−4000tu(t) V
13–22 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Io =0.005(s + 8000)
s2 + 8000s + 16 × 106
=K1
(s + 4000)2 +K2
s + 4000
K1 = 20 K2 = 0.005
io(t) = [20te−4000t + 0.005e−4000t]u(t) A
P 13.20
Vo =5 × 106/s
1120 + 0.8s + 5 × 106/s
(240s
)
=12 × 108
s(0.8s2 + 1120s + 5 × 106)
=15 × 108
s(s2 + 1400s + 625 × 104)
=K1
s+
K2
s + 700 − j2400+
K∗2
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/s1120 + 0.8s
+Vos
5 × 106 + 14.4 × 10−6 = 0
Problems 13–23
Vo
( 11120 + 0.8s
+s
5 × 106
)=
240/s0.8s + 1120
− 14.4 × 10−6
Vo =−72s2 − 100,800s + 15 × 108
s(s2 + 1400s + 625 × 104)
=240s
+162.5/163.74
s + 700 − j2400+
162.5/− 163.74
s + 700 + j2400
·. . vo(t) = [240 + 325e−700t cos(2400t + 163.74)]u(t) V
P 13.22 [a]
Vo =(1/sC)(LIg)
R + sL + (1/sC)=
Ig/C
s2 + (R/L)s + (1/LC)
Ig
C=
150.1
= 150
R
L= 7;
1LC
= 10
Vo =150
s2 + 7s + 10
[b] sVo =150s
s2 + 7s + 10
lims→0
sVo = 0; ·. . vo(∞) = 0
lims→∞ sVo = 0; ·. . vo(0+) = 0
[c] Vo =150
(s + 2)(s + 5)=
50s + 2
+−50s + 5
vo = [50e−2t − 50e−5t]u(t) V
13–24 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.23 IL =Ig
s− Vo
1/sC=
Ig
s− sCVo
IL =15s
− 15s(s + 2)(s + 5)
=15s
−[ −10s + 2
+25
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
2500 + (125,000/s)=
0.01s + 50
V1 =(100,000/s)(25/s)2500 + (125,000/s)
=1000
s(s + 50)
V2 =(25,000/s)(25/s)
2500 + (125,000/s)=
250s(s + 50)
[c] i1(t) = 10e−50tu(t) mA
V1 =20s
− 20s + 50
·. . v1(t) = (20 − 20e−50t)u(t) V
V2 =5s
− 5s + 50
·. . v2(t) = (5 − 5e−50t)u(t) V
[d] i1(0+) = 10 mA
i1(0+) =25
2.5 × 10−3 = 10 mA(Checks)
v1(0+) = 0; v2(0+) = 0(Checks)
v1(∞) = 20 V; v2(∞) = 5 V(Checks)
Problems 13–25
v1(∞) + v2(∞) = 25 V(Checks)
(10 × 10−6)v1(∞) = 200 µC
(40 × 10−6)v2(∞) = 200 µC(Checks)
P 13.25 [a]
100‖5s =500s
5s + 100=
100ss + 20
Vo =100s
s + 20
[50
(s + 25)2
]=
5000s(s + 20)(s + 25)2
Io =Vo
100=
50s(s + 20)(s + 25)2
IL =Vo
5s=
1000(s + 20)(s + 25)2
[b] Vo =K1
s + 20+
K2
(s + 25)2 +K3
s + 25
K1 =5000s
(s + 25)2
∣∣∣∣s=−20
= −4000
K2 =5000s
(s + 20)
∣∣∣∣s=−25
= 25,000
K3 =d
ds
[ 5000ss + 20
]s=−25
=[
5000s + 20
− 5000s(s + 20)2
]s=−25
= 4000
vo(t) = [−4000e−20t + 25,000te−25t + 4000e−25t]u(t) V
Io =K1
s + 20+
K2
(s + 25)2 +K3
s + 25
K1 =50s
(s + 25)2
∣∣∣∣s=−20
= −40
K2 =50s
(s + 20)
∣∣∣∣s=−25
= 250
K3 =d
ds
[ 50ss + 20
]s=−25
=[
50s + 20
− 50s(s + 20)2
]s=−25
= 40
13–26 CHAPTER 13. The Laplace Transform in Circuit Analysis
io(t) = [−40e−20t + 250te−25t + 40e−25t]u(t) V
IL =K1
s + 20+
K2
(s + 25)2 +K3
s + 25
K1 =1000
(s + 25)2
∣∣∣∣s=−20
= 40
K2 =1000
(s + 20)
∣∣∣∣s=−25
= −200
K3 =d
ds
[ 1000s + 20
]s=−25
=[− 1000
(s + 20)2
]s=−25
= −40
iL(t) = [40e−20t − 200te−25t − 40e−25t]u(t) V
P 13.26
10s
= (s + 1)I1 − sI2
0 = −sI1 +(s + 1 +
1s
)I2
In standard form,
s(s + 1)I1 − s2I2 = 10
−s2I1 + (s2 + s + 1)I2 = 0
∆ =
∣∣∣∣∣∣∣s(s + 1) −s2
−s2 (s2 + s + 1)
∣∣∣∣∣∣∣ = 2s(s2 + s + 0.5)
N1 =
∣∣∣∣∣∣∣10 −s2
0 (s2 + s + 1)
∣∣∣∣∣∣∣ = 10(s2 + s + 1)
N2 =
∣∣∣∣∣∣∣s(s + 1) 10
−s2 0
∣∣∣∣∣∣∣ = 10s2
Problems 13–27
I1 =N1
∆; I2 =
N2
∆; I0 = I1 − I2
·. . Io =N1 − N2
∆=
5(s + 1)s(s2 + s + 0.5)
=K1
s+
K2
s + 0.5 − j0.5+
K∗2
s + 0.5 + j0.5
K1 =5
0.5= 10
K2 =5(−0.5 + j0.5 + 1)(−0.5 + j0.5)(j1)
= 5/− 180
io(t) = [10 − 10e−t/2 cos 0.5t]u(t) A
P 13.27 [a]
0 = 2.5s(I1 − 6/s) +5s(I1 − I2) + 10I1
−75s
=5s(I2 − I1) + 5(I2 − 6/s)
or
(s2 + 4s + 2)I1 − 2I2 = 6s
−I1 + (s + 1)I2 = −9
∆ =
∣∣∣∣∣∣∣(s2 + 4s + 2) −2
−1 (s + 1)
∣∣∣∣∣∣∣ = s(s + 2)(s + 3)
13–28 CHAPTER 13. The Laplace Transform in Circuit Analysis
N1 =
∣∣∣∣∣∣∣6s −2
−9 (s + 1)
∣∣∣∣∣∣∣ = 6(s2 + s − 3)
I1 =N1
∆=
6(s2 + s − 3)s(s + 2)(s + 3)
N2 =
∣∣∣∣∣∣∣(s2 + 4s + 2) 6s
−1 −9
∣∣∣∣∣∣∣ = −9s2 − 30s − 18
I2 =N2
∆=
−9s2 − 30s − 18s(s + 2)(s + 3)
[b] sI1 =6(s2 + s − 3)(s + 2)(s + 3)
lims→∞ sI1 = i1(0+) = 6 A; lim
s→0sI1 = i1(∞) = −3 A
sI2 =−9s2 − 30s − 18(s + 2)(s + 3)
lims→∞ sI2 = i2(0+) = −9 A; lim
s→0sI2 = i2(∞) = −3 A
[c] I1 =6(s2 + s − 3)
s(s + 2)(s + 3)=
K1
s+
K2
s + 2+
K3
s + 3
K1 =6(−3)
6= −3; K2 =
6(4 − 2 − 3)(−2)(1)
= 3
K3 =6(9 − 3 − 3)(−3)(−1)
= 6
i1(t) = [−3 + 3e−2t + 6e−3t]u(t) A
I2 =−9s2 − 30s − 18s(s + 2)(s + 3)
=K1
s+
K2
s + 2+
K3
s + 3
K1 =−186
= −3; K2 =−36 + 60 − 18
(−2)(1)= −3
K3 =−81 + 90 − 18
(−3)(−1)= −3
i2(t) = [−3 − 3e−2t − 3e−3t]u(t) A
Problems 13–29
P 13.28 [a]
54s
= 2I1 − I2 − I3
0 = −I1 +(2 +
45s
)I2 − I3
0 = −I1 − I2 + (2 + 0.36s)I3
∆ =
∣∣∣∣∣∣∣∣∣∣∣
2 −1 −1
−1 (2s + 45)/s −1
−1 −1 (0.36s + 2)
∣∣∣∣∣∣∣∣∣∣∣=
1.08(s + 5)(s + 25)s
N2 =
∣∣∣∣∣∣∣∣∣∣∣
2 (54/s) −1
−1 0 −1
−1 0 (0.36s + 2)
∣∣∣∣∣∣∣∣∣∣∣=
162s
(0.12s + 1)
N3 =
∣∣∣∣∣∣∣∣∣∣∣
2 −1 (54/s)
−1 (2s + 45)/s 0
−1 −1 0
∣∣∣∣∣∣∣∣∣∣∣=
162s2 (s + 15)
I2 =N2
∆=
150(0.12s + 1)(s + 5)(s + 25)
Vo =45s
I2 =6750(0.12s + 1)s(s + 5)(s + 25)
I3 =N3
∆=
150(s + 15)s(s + 5)(s + 25)
= Io
[b] Vo =K1
s+
K2
s + 5+
K3
s + 25
K1 =6750125
= 54; K2 =6750(−0.6 + 1)
(−5)(20)= −27
13–30 CHAPTER 13. The Laplace Transform in Circuit Analysis
K3 =6750(−3 + 1)(−25)(−20)
= −27
·. . vo(t) = [54 − 27e−5t − 27e−25t]u(t) V
Io =K1
s+
K2
s + 5+
K3
s + 25
K1 =150(15)(5)(25)
= 18; K2 =150(10)(−5)(20)
= −15
K3 =150(−10)
(−25)(−20)= −3
·. . 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 is54 V. Our solution predicts
vL(0+) = 0.36dio(0+)
dt= 0.36(75 + 75) = 54 V
At t = 0+ the current in the capacitive branch is (1/2)(54/1.5) = 18 A. Fromour solution we have
sI2 =150(0.12 + 1/s)
(1 + 5/s)(1 + 25/s)and lim
s→∞ sI2 = i2(0+) = 150(0.12) = 18 A
Problems 13–31
P 13.29 [a]
120s
= 50(I1 − 0.05Vφ) +250s
(I1 − I2)
120s
= 50I1 − 2.5(250
s
)(I2 − I1) +
250s
I1 − 250s
I2;
0 =250s
(I2 − I1) + 20s(I2 − 0.05Vφ) + 700I2
0 =250s
(I2 − I1) + 20s[I2 − 0.05
(250s
)(I2 − I1)
]Vφ) + 700I2
Simplifying,
(50s + 875)I1 − 875I2 = 120
250(s − 1)I1 + (20s2 + 450s + 250)I2 = 0
∆ =
∣∣∣∣∣∣∣(50s + 875) −875
250(s − 1) (20s2 + 450s + 250)
∣∣∣∣∣∣∣ = 1000s(s2 + 40s + 625)
N1 =
∣∣∣∣∣∣∣120 −875
0 (20s2 + 450s + 250)
∣∣∣∣∣∣∣ = 1200(2s2 + 45s + 25)
N2 =
∣∣∣∣∣∣∣(50s + 875) 120
250(s − 1) 0
∣∣∣∣∣∣∣ = −30,000(s − 1)
I1 =N1
∆=
1.2(2s2 + 45s + 25)s(s2 + 40s + 625)
I2 =N2
∆=
−30(s − 1)s(s2 + 40s + 625)
Io = I2 − 0.05Vφ = I2 − 0.05[250
s(I2 − I1)
]
I2 − I1 =−2.4s(s + 35)
s(s2 + 40s + 625)
13–32 CHAPTER 13. The Laplace Transform in Circuit Analysis
250s
(I2 − I1) =−600(s + 35)
s(s2 + 40s + 625)
·. . Io =−30(s − 1)
s(s2 + 40s + 625)+
30(s + 35)s(s2 + 40s + 625)
=1080
s(s2 + 40s + 625)
[b] sIo =1080
(s2 + 40s + 625)
io(0+) = lims→∞ sIo = 0
io(∞) = lims→0
sIo =1080625
= 1728 mA
[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
ia =1202500
= 48 mA
vφ = −700ia = −33.60 V
io(∞) = 48 × 10−3 − 0.05(−33.60) = 48 × 10−3 + 1.68 = 1728 mA (Checks)
Problems 13–33
[d] Io =1080
s(s2 + 40s + 625)=
K1
s+
K2
s + 20 − j15+
K∗2
s + 20 + j15
K1 =1080625
= 1.728
K2 =1080
(−20 + j15)(j30)= 1.44/126.87
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
10+
V1 − 50/s25/s
+V1 − Vo
4s= 0
−5s
+Vo − V1
4s+
Vo − 50/s30
= 0
Simplifying,
(4s2 + 10s + 25)V1 − 25Vo = 200s
−15V1 + (2s + 15)Vo = 400
∆ =
∣∣∣∣∣∣∣(4s2 + 10s + 25) −25
−15 (2s + 15)
∣∣∣∣∣∣∣ = 8s(s + 5)2
No =
∣∣∣∣∣∣∣(4s2 + 10s + 25) 200s
−15 400
∣∣∣∣∣∣∣ = 200(8s2 + 35s + 50)
Vo =No
∆=
200(8s2 + 35s + 50)8s(s + 5)2 =
K1
s+
K2
(s + 5)2 +K3
s + 5
K1 =(25)(50)
25= 50; K2 =
25(200 − 175 + 50)−5
= −375
K3 = 25d
ds
[8s2 + 35s + 50
s
]s=−5
= 25[s(16s + 35) − (8s2 + 35s + 50)
s2
]s=−5
13–34 CHAPTER 13. The Laplace Transform in Circuit Analysis
= −5(−45) − 75 = 150
·. . Vo =50s
− 375(s + 5)2 +
150s + 5
[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(∞)10
− 5 +vo(∞) − 50
30= 0
·. . 3vo(∞) − 150 + vo(∞) − 50 = 0; ·. . 4vo(∞) = 200
·. . vo(∞) = 50 V(Checks)
Problems 13–35
P 13.31 [a]
10s
I1 +10s
(I1 − I2) + 10(I1 − 9/s) = 0
10s
(I2 − 9/s) +10s
(I2 − I1) + 10I2 = 0
Simplifying,
(s + 2)I1 − I2 = 9
−I1 + (s + 2)I2 =9s
∆ =
∣∣∣∣∣∣∣(s + 2) −1
−1 (s + 2)
∣∣∣∣∣∣∣ = s2 + 4s + 3 = (s + 1)(s + 3)
N1 =
∣∣∣∣∣∣∣9 −1
9/s (s + 2)
∣∣∣∣∣∣∣ =9s2 + 18s + 9
s=
9s(s + 1)2
I1 =N1
∆=
9s
[(s + 1)2
(s + 1)(s + 3)
]=
9(s + 1)s(s + 3)
N2 =
∣∣∣∣∣∣∣(s + 2) 9
−1 9/s
∣∣∣∣∣∣∣ =18s
(s + 1)
I2 =N2
∆=
18(s + 1)s(s + 1)(s + 3)
=18
s(s + 3)
Ia =9s
− I1 =9s
− 9(s + 1)s(s + 3)
=6s
− 6s + 3
Ib = I1 =9(s + 1)s(s + 3)
=3s
+6
s + 3
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 =10s
Ib =10s
(3s
+6
s + 3
)
=30s2 +
60s(s + 3)
=30s2 +
20s
− 20s + 3
Vb =10s
(I2 − I1) =10s
[(6s
− 6s + 3
)−(3
s+
6s + 3
)]
=10s
[3s
− 12s + 3
]=
30s2 − 40
s+
40s + 3
Vc =10s
(9/s − I2) =10s
(9s
− 6s
+6
s + 3
)
=30s2 +
20s
− 20s + 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 timebecomes is negligible when compared to the other terms. Thus,
30t + 20 = 1000 or 30t − 40 = 1000
Thus,
t =98030
= 32.67 s or t =104030
= 34.67 s
Therefore, the breakdown will occur at t = 32.67 s.
Problems 13–37
P 13.32 [a]
20Iφ + 25s(Io − Iφ) + 25(Io − I1) = 0
50s
Iφ + 5I1 + 25(I1 − Io) + 25s(Iφ − Io) = 0
Iφ − I1 =100s
·. . I1 = Iφ − 100s
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)
5s(5s2 + 6s + 10) −25(s + 1)
∣∣∣∣∣∣∣ = −625(s + 1)(1 + 2/s)
N2 =
∣∣∣∣∣∣∣−5(5s + 1) −2500/s
5s(5s2 + 6s + 10) 3000/s
∣∣∣∣∣∣∣ = −12,500s2 − 4.8s − 10
s2
Io =N2
∆=
20(s2 − 4.8s − 10)s(s + 1)(s + 2)
[b] io(0+) = lims→∞ sIo = 20 A
io(∞) = lims→0
sIo =−200
2= −100 A
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 =20(s2 − 4.8s − 10)
s(s + 1)(s + 2)=
K1
s+
K2
s + 1+
K3
s + 2
K1 =−200(1)(2)
= −100; K2 =20(1 + 4.8 − 10)
(−1)(1)= 84
K3 =20(4 + 9.6 − 10)
(−2)(−1)= 36
Io =−100
s+
84s + 1
+36
s + 2
Problems 13–39
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 × 105te−5000t V, C = 5 µF; therefore
iC = C
(dvC
dt
)= 6e−5000t(1 − 5000t) A
iC > 0 when 1 > 5000t or iC < 0 when 0 < t < 200 µs
and iC < 0 when t > 200 µs
iC = 0 when 1 − 5000t = 0, or t = 200 µs
dvC
dt= 12 × 105e−5000t[1 − 5000t]
·. . iC = 0 whendvC
dt= 0
P 13.34 [a] The s-domain equivalent circuit is
I =Vg
R + sL=
Vg/L
s + (R/L), Vg =
Vm(ω cos φ + s sin φ)s2 + ω2
I =K0
s + R/L+
K1
s − jω+
K∗1
s + jω
K0 =Vm(ωL cos φ − R sin φ)
R2 + ω2L2 , K1 =Vm/φ − 90 − θ(ω)
2√
R2 + ω2L2
where tan θ(ω) = ωL/R. Therefore, we have
i(t) =Vm(ωL cos φ − R sin φ)
R2 + ω2L2 e−(R/L)t +Vm sin[ωt + φ − θ(ω)]√
R2 + ω2L2
[b] iss(t) =Vm√
R2 + ω2L2sin[ωt + φ − θ(ω)]
13–40 CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] itr =Vm(ωL cos φ − R sin φ)
R2 + ω2L2 e−(R/L)t
[d] I =Vg
R + jωL, Vg = Vm/φ − 90
Therefore I =Vm/φ − 90
√R2 + ω2L2/θ(ω)
=Vm√
R2 + ω2L2/φ − 90 − θ(ω)
Therefore iss =Vm√
R2 + ω2L2sin[ωt + φ − θ(ω)]
[e] The transient component vanishes when
ωL cos φ = R sin φ or tan φ =ωL
Ror φ = θ(ω)
P 13.35
VTh =10s
10s + 1000· 40
s=
40010s + 1000
=40
s + 100
ZTh = 1000 + 1000‖10s = 1000 +10,000s
10s + 1000=
2000(s + 50)s + 100
I =40/(s + 100)
(5 × 105)/s + 2000(s + 50)/(s + 100)=
40s2000s2 + 600,000s + 5 × 107
=0.02s
s2 + 300s + 25,000=
K1
s + 150 − j50+
K∗1
s + 150 + j50
K1 =0.02s
s + 150 + j50
∣∣∣∣s=−150+j50
= 31.62 × 10−3/71.57
i(t) = 63.25e−150t cos(50t + 71.57)u(t) mA
Problems 13–41
P 13.36 [a]
180s
= (100 + 15s)I1 + 10sI2
0 = 10sI1 + (20s + 200)I2
∆ =
∣∣∣∣∣∣∣15s + 100 10s
10s 20s + 200
∣∣∣∣∣∣∣ = 200(s + 5)(s + 20)
N2 =
∣∣∣∣∣∣∣15s + 100 180/s
10s 0
∣∣∣∣∣∣∣ = −1800
I2 =N2
∆=
−9(s + 5)(s + 20)
Vo = 160I2 =−1440
(s + 5)(s + 20)
[b] sVo =−1440s
(s + 5)(s + 20)
lims→0
sVo = vo(∞) = 0 V
lims→∞ sVo = vo(0+) = 0 V
[c] Vo =−96s + 5
+96
s + 20
vo(t) = [−96e−5t + 96e−20t]u(t) V
P 13.37
180s
= (100 + 15s)I1 − 10sI2
0 = −10sI1 + (20s + 200)I2
13–42 CHAPTER 13. The Laplace Transform in Circuit Analysis
∆ =
∣∣∣∣∣∣∣15s + 100 −10s
−10s 20s + 200
∣∣∣∣∣∣∣ = 200(s + 5)(s + 20)
N2 =
∣∣∣∣∣∣∣15s + 100 180/s
−10s 0
∣∣∣∣∣∣∣ = 1800
I2 =N2
∆=
9(s + 5)(s + 20)
Vo = 160I2 =1440
(s + 5)(s + 20)=
96s + 5
− 96s + 20
vo(t) = [96e−5t − 96e−20t]u(t) V
P 13.38 [a] W =12L1i
21 +
12L2i
22 + Mi1i2
W = 4(15)2 + 9(100) + 150(6) = 2700 J
[b] 120i1 + 8di1dt
− 6di2dt
= 0
270i2 + 18di2dt
− 6di1dt
= 0
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
∆ =
∣∣∣∣∣∣∣8s + 120 −6s
−6s 18s + 270
∣∣∣∣∣∣∣ = 108(s + 10)(s + 30)
N1 =
∣∣∣∣∣∣∣180 −6s
−270 18s + 270
∣∣∣∣∣∣∣ = 1620(s + 30)
N2 =
∣∣∣∣∣∣∣8s + 120 180
−6s −270
∣∣∣∣∣∣∣ = −1080(s + 30)
Problems 13–43
I1 =N1
∆=
1620(s + 30)108(s + 10)(s + 30)
=15
s + 10
I2 =N2
∆=
−1080(s + 30)108(s + 10)(s + 30)
=−10
s + 10
[c] i1(t) = 15e−10tu(t) A; i2(t) = −10e−10tu(t) A
[d] W120Ω =∫ ∞
0(225e−20t)(120) dt = 27,000
e−20t
−20
∣∣∣∣∞0
= 1350 J
W270Ω =∫ ∞
0(100e−20t)(270) dt = 27,000
e−20t
−20
∣∣∣∣∞0
= 1350 J
W120Ω + W270Ω = 2700 J (Checks)
[e] W =12L1i
21 +
12L2i
22 + Mi1i2 = 900 + 900 − 900 = 900 J
With the dot reversed the s-domain equations are
(8s + 120)I1 + 6sI2 = 60
6sI1 + (18s + 270)I2 = −90
As before, ∆ = 108(s + 10)(s + 30). Now,
N1 =
∣∣∣∣∣∣∣60 6s
−90 18s + 270
∣∣∣∣∣∣∣ = 1620(s + 10)
N2 =
∣∣∣∣∣∣∣8s + 120 60
6s −90
∣∣∣∣∣∣∣ = −1080(s + 10)
I1 =N1
∆=
15s + 30
; I2 =N2
∆=
−10s + 30
i1(t) = 15e−30tu(t) A; i2(t) = −10e−30tu(t) A
W270Ω =∫ ∞
0(100e−60t)(270) dt = 450 J
W120Ω =∫ ∞
0(225e−60t)(120) dt = 450 J
W120Ω + W270Ω = 900 J (Checks)
13–44 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.39 [a] s-domain equivalent circuit is
Note: i2(0+) = −2010
= −2 A
[b]24s
= (120 + 3s)I1 + 3sI2 + 6
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
s 5s + 120
∣∣∣∣∣∣∣ = 4(s + 20)(s + 60)
N1 =
∣∣∣∣∣∣∣(8/s) − 2 s
−10 5s + 120
∣∣∣∣∣∣∣ =−200(s − 4.8)
s
I1 =N1
∆=
−50(s − 4.8)s(s + 20)(s + 60)
[c] sI1 =−50(s − 4.8)
(s + 20)(s + 60)
lims→∞ sI1 = i1(0+) = 0 A
lims→0
sI1 = i1(∞) =(−50)(−4.8)
(20)(60)= 0.2 A
[d] I1 =K1
s+
K2
s + 20+
K3
s + 60
K1 =2401200
= 0.2; K2 =−50(−20) + 240
(−20)(40)= −1.55
Problems 13–45
K3 =−50(−60) + 240
(−60)(−40)= 1.35
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
13–46 CHAPTER 13. The Laplace Transform in Circuit Analysis
(s − 10)I1 + (2s + 10)I2 = 54
·. . ∆ =
∣∣∣∣∣∣∣(3s + 30) (s − 10)
(s − 10) (2s + 10)
∣∣∣∣∣∣∣ = 5(s + 2)(s + 20)
N1 =
∣∣∣∣∣∣∣72 (s − 10)
54 (2s + 10)
∣∣∣∣∣∣∣ = 90s + 1260
N2 =
∣∣∣∣∣∣∣(3s + 30) 72
(s − 10) 54
∣∣∣∣∣∣∣ = 90s + 2340
Io = I1 − I2 =N1
∆− N2
∆=
−10805(s + 2)(s + 20)
=−216
(s + 2)(s + 20)− 12
s + 2− 12
s + 20
io(t) = [12e−2t + 12e−20t]u(t) A
P 13.41 The s-domain equivalent circuit is
V1 − 48/s4 + (100/s)
+V1 + 9.6
0.8s+
V1
0.8s + 20= 0
V1 =−1200
s2 + 10s + 125
Vo =20
0.8s + 20V1 =
−30,000(s + 25)(s + 5 − j10)(s + 5 + j10)
=K1
s + 25+
K2
s + 5 − j10+
K∗2
s + 5 + j10
Problems 13–47
K1 =−30,000
s2 + 10s + 125
∣∣∣∣s=−25
= −60
K2 =−30,000
(s + 25)(s + 5 + j10)
∣∣∣∣s=−5+j10
= 67.08/63.43
vo(t) = [−60e−25t + 134.16e−5t cos(10t + 63.43)]u(t) V
P 13.42 [a] Voltage source acting alone:
Vo1 − 60/s10
+Vo1s
80+
Vo1
20 + 10s= 0
·. . Vo1 =480(s + 2)
s(s + 4)(s + 6)
Current source acting alone:
Vo2
10+
Vo2s
80+
Vo2 − 30/s10(s + 2)
= 0
·. . Vo2 =240
s(s + 4)(s + 6)
Vo = Vo1 + Vo2 =480(s + 2) + 240s(s + 4)(s + 6)
=480(s + 2.5)
s(s + 4)(s + 6)
13–48 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] Vo =K1
s+
K2
s + 4+
K3
s + 6
K1 =(480)(2.5)
(4)(6)= 50; K2 =
480(−1.5)(−4)(2)
= 90; K3 =480(−3.5)(−6)(−2)
= −140
vo(t) = [50 + 90e−4t − 140e−6t]u(t) V
P 13.43 ∆ =
∣∣∣∣∣∣∣Y11 Y12
Y12 Y22
∣∣∣∣∣∣∣ = Y11Y22 − Y 212
N2 =
∣∣∣∣∣∣∣Y11 [(Vg/R1) + γC − (ρ/s)]
Y12 (Ig − γC)
∣∣∣∣∣∣∣
V2 =N2
∆
Substitution and simplification lead directly to Eq. 13.90.
P 13.44
Va − 4.8/s0.8
+Va
1/s+
Va − Vo
1/s= 0
0 − Va
1/s+
0 − Vo
1.25= 0
Va =−Vo
1.25s
Va(2s + 1.25) − sVo = 6/s
Problems 13–49
−Vo
[(2s + 1.25)
1.25s+ s
]= 6/s
−Vo
[125s2 + 2s + 1.25
1.25s
]= 6/s
Vo =−7.5
1.25s2 + 2s + 1.25=
−6s2 + 1.6s + 1
=K1
s + 0.8 − j0.6+
K∗1
s + 0.8 + j0.6
K1 =−6
s + 0.8 + j0.6
∣∣∣∣s=−0.8+j0.6
= 5/90
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
Zi
Vg
Zf =107
s‖1000 =
1010/s
107/s + 1000=
1010
1000s + 107 =107
s + 104
Zi =2 × 106
s+ 400 =
400s + 2 × 106
s=
400s
(s + 5000)
Vg =20,000
s2
·. . Vo =−107/(s + 104)
(400/s)(s + 5000)· 20,000
s2 =−5 × 108
s(s + 5000)(s + 10,000)
[b] Vo =K1
s+
K2
s + 5000+
K3
s + 10,000
K1 =−5 × 108
(s + 5000)(s + 10,000)
∣∣∣∣s=0
= −10
K2 =−5 × 108
s(s + 10,000)
∣∣∣∣s=−5000
= 20
K3 =−5 × 108
s(s + 5000)
∣∣∣∣s=−10,000
= −10
·. . vo(t) = [−10 + 20e−5000t − 10e−10,000t]u(t) V
13–50 CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] −10 + 20e−5000ts − 10e−10,000ts = −5Let x = e−5000ts . Then
10x2 − 20x + 5 = 0
Solving,
x = 0.292893
e−5000ts = 0.292893 ·. . ts = 245.6 µs
[d] vg = m tu(t); Vg =m
s2
Vo =−107s
400(s + 5000)(s + 10,000)· m
s2
=−25,000m
s(s + 5000)(s + 10,000)
K1 =−25,000m
(5000)(10,000)= −5 × 10−4m
·. . −5 = −5 × 10−4m ·. . m = 10,000 V/s
P 13.46 [a]
Vp =50/s
5 + 50/sVg2 =
505s + 50
Vg2
Vp − 40/s20
+Vp − Vo
5+
Vp − Vo
100/s= 0
Vp
( 120
+15
+s
100
)− Vo
(15
+s
100
)=
2s
s + 25100
( 505s + 50
) 16s
− 2s
= Vo
(15
+s
100
)= Vo
(s + 20100
)
Problems 13–51
Vo =100
s + 20
[16(s + 25)
10(s + 10)(s)− 2
s
]=
−40s + 2000s(s + 10)(s + 20)
=K1
s+
K2
s + 10+
K3
s + 20
K1 = 10; K2 = −24; K3 = 14
·. . vo(t) = [10 − 24e−10t + 14e−20t]u(t) V
[b] 10 − 24e−10t + 14e−20t = 5Let 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 =−Zf1
ZA1Vg where Zf1 = 25 × 103 Ω
ZA1 = 25,000 +25,000(20 × 104/s)
25,000 + (20 × 104/s)
=25,000(s + 16)
(s + 8)Ω
·. . Vo1 =−(s + 8)(s + 16)
Vg
Also,
Vo =−Zf2
ZA2Vo1 where Zf2 =
2 × 108
sΩ and ZA2 = 25,000 Ω
·. . Vo =−8000
sVo1 =
−8000s
[−(s + 8)(s + 16)
]Vg
=8000(s + 8)s(s + 16)
Vg
vg(t) = 16u(t) mV; ·. . Vg =16 × 10−3
s
13–52 CHAPTER 13. The Laplace Transform in Circuit Analysis
Vo =128(s + 8)s2(s + 16)
=K1
s2 +K2
s+
K3
s + 16
K1 =128(8)
16= 64
K2 = 128d
ds
[s + 8s + 16
]s=0
= 4
K3 =128(−8)
256= −4
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 ort > 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
106
0.5s=
2 × 106
sand
106
0.25s=
4 × 106
s; Vg =
0.5s
sVa
2 × 106 +Va − (0.5/s)
200,000+
Va
200,000= 0
Problems 13–53
sVa + 10Va − 5s
+ 10Va = 0
Va =5
s(s + 20)
0 − Va
200,000+
(0 − Vb)s4 × 106 = 0
·. . Vb = −20s
Va =−100
s2(s + 20)
Vb
100,000+
(Vb − 0)s4 × 106 +
(Vb − Vo)s4 × 106 = 0
40Vb + sVb + sVb = sVo
·. . Vo =2(s + 20)Vb
s; Vo = 2
(−100s3
)=
−200s3
[b] vo(t) = −100t2u(t) V
[c] −100t2 = −4; t = 0.2 s = 200 ms
P 13.49 [a]Vo
Vi
=1/sC
R + 1/sC
H(s) =(1/RC)
s + (1/RC)=
200s + 200
; −p1 = −200 rad/s
[b]Vo
Vi
=R
R + 1/sC=
RCs
RCs + 1=
s
s + (1/RC)
=s
s + 200; z1 = 0, −p1 = −200 rad/s
[c]Vo
Vi
=sL
R + sL=
s
s + R/L=
s
s + 8000
z1 = 0; −p1 = −8000 rad/s
[d]Vo
Vi
=R
R + sL=
R/L
s + (R/L)=
8000s + 8000
−p1 = −8000 rad/s
[e]
Vos
4 × 106 +Vo
10,000+
Vo − Vi
40,000= 0
13–54 CHAPTER 13. The Laplace Transform in Circuit Analysis
sVo + 400Vo + 100Vo = 100Vi
H(s) =Vo
Vi
=100
s + 500
−p1 = −500 rad/s
P 13.50 [a] Let R1 = 250 kΩ; R2 = 125 kΩ; C2 = 1.6 nF; and Cf = 0.4 nF. Then
Zf =(R2 + 1/sC2)1/sCf(
R2 + 1sC2
+ 1sCf
) =(s + 1/R2C2)
Cfs(s + C2+Cf
C2Cf R2
)1
Cf
= 2.5 × 109
1R2C2
=62.5 × 107
125 × 103 = 5000 rad/s
C2 + Cf
C2CfR2=
2 × 10−9
(0.64 × 10−18)(125 × 103)= 25,000 rad/s
·. . Zf =2.5 × 109(s + 5000)
s(s + 25,000)Ω
Zi = R1 = 250 × 103 Ω
H(s) =Vo
Vg
=−Zf
Zi
=−104(s + 5000)s(s + 25,000)
[b] −z1 = −5000 rad/s
−p1 = 0; −p2 = −25,000 rad/s
P 13.51 [a]
Va − Vg
1000+
sVa
5 × 106 +(Va − Vo)s5 × 106 = 0
5000Va − 5000Vg + 2sVa − sVo = 0
Problems 13–55
(5000 + 2s)Va − sVo = 5000Vg
(0 − Va)s5 × 106 +
0 − Vo
5000= 0
−sVa − 1000Vo = 0; ·. . Va =−1000
sVo
(2s + 5000)(−1000
s
)Vo − sVo = 5000Vg
1000Vo(2s + 5000) + s2Vo = −5000sVg
Vo(s2 + 2000s + 5 × 106) = −5000sVg
Vo
Vg
=−5000s
s2 + 2000s + 5 × 106
s1,2 = −1000 ±√
106 − 5 × 106 = −1000 ± j2000
Vo
Vg
=−5000s
(s + 1000 − j2000)(s + 1000 + j2000)
[b] z1 = 0; −p1 = −1000 + j2000; −p2 = −1000 − j2000
P 13.52 [a] Zi = 1000 +5 × 106
s=
1000(s + 5000)s
Zf =40 × 106
s‖40,000 =
40 × 106
s + 1000
H(s) = −Zf
Zi
=−40 × 106/(s + 1000)
1000(s + 5000)/s=
−40,000s(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 =Vi
500,000 + [(20 × 106)/s](500,000) =
s
s + 40Vi
0.2Vi = Vo + Va
13–56 CHAPTER 13. The Laplace Transform in Circuit Analysis
·. . Vo = 0.2Vi − s
s + 40Vi
Vo
Vi
=0.2(s + 40) − s
s + 40=
−0.8s + 8s + 40
=−0.8(s − 10)
s + 40
[b] −z1 = 10 rad/s
−p1 = −40 rad/s
P 13.54
Vg = 25sI1 − 35sI2
0 = −35sI1 +(
50s + 10,000 +16 × 106
s
)I2
∆ =
∣∣∣∣∣∣∣25s −35s
−35s 50s + 10,000 + 16 × 106/s
∣∣∣∣∣∣∣ = 25(s + 2000)(s + 8000)
N2 =
∣∣∣∣∣∣∣25s Vg
−35s 0
∣∣∣∣∣∣∣ = 35sVg
I2 =N2
∆=
35sVg
25(s + 2000)(s + 8000)
Vo =16 × 106
sI2 =
22.4 × 106Vg
(s + 2000)(s + 8000)
H(s) =Vo
Vg
=22.4 × 106
(s + 2000)(s + 8000)
·. . −p1 = −2000 rad/s; −p2 = −8000 rad/s
Problems 13–57
P 13.55 [a]
Vo
5000+
Vo
0.2s+ Vo(10−7)s = Ig
·. . Vo =10 × 106s
s2 + 2000s + 50 × 106 · Ig
Ig =0.1s
s2 + 108 ; Io =Vos
10 × 106
·. . H(s) =s2
s2 + 2000s + 50 × 106
[b] Io =(s2)(0.1s)
(s + 1000 − j7000)(s + 1000 + j7000)(s2 + 108)
Io =0.1s3
(s + 1000 − j7000)(s + 1000 + j7000)(s + j104)(s − j104)
[c] Damped sinusoid of the form
Me−1000t cos(7000t + θ1)
[d] Steady-state sinusoid of the form
N cos(104t + θ2)
[e] Io =K1
s + 1000 − j7000+
K∗1
s + 1000 + j7000+
K2
s − j104 +K∗
2
s + j104
K1 =0.1(−1000 + j7000)3
(j14,000)(−1000 − j3000)(−1000 + j17,000)= 46.90 × 10−3/− 140.54
K2 =0.1(j104)3
(j20,000)(1000 + j3000)(1000 + j17,000)= 92.85 × 10−3/21.80
io(t) = [93.8e−1000t cos(7000t − 140.54) + 185.7 cos(104t + 21.80)] mA
Test:
io(0) = 93.8 cos(−140.54) + 185.7 cos(21.80) mA = 100 mA
Z =1Y
; Y =1
5000+
1j2000
+1
−j1000=
2 + j510,000
·. . Z =10,0002 + j5
= 1856.95/− 68.2 Ω
13–58 CHAPTER 13. The Laplace Transform in Circuit Analysis
Vo = IgZ = (0.1/0)(1856.95/− 68.2) = 185.695/− 68.2 V
Io =Vo
−j1000= 185.7/21.80 mA
ioss = 185.7 cos(104t + 21.80) mA(Checks)
P 13.56 [a]
2000(Io − Ig) + 8000Io + µ(Ig − Io)(2000) + 2sIo = 0
·. . Io =1000(1 − µ)
s + 1000(5 − µ)Ig
·. . H(s) =1000(1 − µ)
s + 1000(5 − µ)
[b] µ < 5
[c]µ H(s) Io
−3 4000/(s + 8000) 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:
Io =2.5s
− 2.5(s + 8000)
; io = [2.5 − 2.5e−8000t]u(t) A
µ = 0:
Io =1s
− 1s + 5000
; io = [1 − e−5000t]u(t) A
µ = 4:
Io =−15s
+15
s + 1000; io = [−15 + 15e−1000t]u(t) A
µ = 5:
Io =−20,000
s2 ; io = −20,000t u(t) A
Problems 13–59
µ = 6:
Io =25s
− 25s − 1000
; io = 25[1 − e1000t]u(t) A
P 13.57 H(s) =Vo
Vi
=1
s + 1; h(t) = e−t
For 0 ≤ t ≤ 1:
vo =∫ t
0e−λ dλ = (1 − e−t) V
For 1 ≤ t ≤ ∞:
vo =∫ t
t−1e−λ dλ = (e − 1)e−t V
P 13.58 H(s) =Vo
Vi
=s
s + 1= 1 − 1
s + 1; h(t) = δ(t) − e−t
h(λ) = δ(λ) − e−λ
For 0 ≤ t ≤ 1:
vo =∫ t
0[δ(λ) − e−λ] dλ = 1 + [e−λ] |t0= e−t V
For 1 ≤ t ≤ ∞:
vo =∫ t
t−1(−e−λ) dλ = e−λ
∣∣∣∣tt−1
= (1 − e)e−t V
13–60 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.59 [a]
t < 0 : y(t) = 0
0 ≤ t ≤ 10 : y(t) =∫ t
0625 dλ = 625t
10 ≤ t ≤ 20 : y(t) =∫ 10
t−10625 dλ = 625(10 − t + 10) = 625(20 − t)
20 ≤ t < ∞ : y(t) = 0
[b]
t < 0 : y(t) = 0
0 ≤ t ≤ 10 : y(t) =∫ t
0312.5 dλ = 312.5t
Problems 13–61
10 ≤ t ≤ 20 : y(t) =∫ t
t−10312.5 dλ = 3125
20 ≤ t ≤ 30 : y(t) =∫ 20
t−10312.5 dλ = 312.5(30 − t)
30 ≤ t < ∞ : y(t) = 0
[c]
t < 0 : y(t) = 0
0 ≤ t ≤ 1 : y(t) =∫ t
0625 dλ = 625t
1 ≤ t ≤ 10 : y(t) =∫ t
t−1625 dλ = 625
10 ≤ t ≤ 11 : y(t) =∫ 10
t−1625 dλ = 625(11 − t)
11 ≤ t < ∞ : y(t) = 0
13–62 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.60 [a] 0 ≤ t ≤ 40:
y(t) =∫ t
0(10)(1)(dλ) = 10λ
∣∣∣∣t0= 10t
40 ≤ t ≤ 80:
y(t) =∫ 40
t−40(10)(1)(dλ) = 10λ
∣∣∣∣40
t−40= 10(80 − t)
t ≥ 80 : y(t) = 0
Problems 13–63
[b] 0 ≤ t ≤ 10:
y(t) =∫ t
040 dλ = 40λ
∣∣∣∣t0= 40t
10 ≤ t ≤ 40:
y(t) =∫ t
t−1040 dλ = 40λ
∣∣∣∣tt−10
= 400
13–64 CHAPTER 13. The Laplace Transform in Circuit Analysis
40 ≤ t ≤ 50:
y(t) =∫ 40
t−1040 dλ = 40λ
∣∣∣∣40
t−10= 40(50 − t)
t ≥ 50 : y(t) = 0
[c] The expressions are
0 ≤ t ≤ 1 : y(t) =∫ t
0400 dλ = 400λ
∣∣∣∣t0= 400t
1 ≤ t ≤ 40 : y(t) =∫ t
t−1400 dλ = 400λ
∣∣∣∣tt−1
= 400
40 ≤ t ≤ 41 : y(t) =∫ 40
t−1400 dλ = 400λ
∣∣∣∣40
t−1= 400(41 − t)
41 ≤ t < ∞ : y(t) = 0
[d]
[e] Yes, note that h(t) is approaching 40δ(t), therefore y(t) must approach 40x(t),i.e.
y(t) =∫ t
0h(t − λ)x(λ) dλ →
∫ t
040δ(t − λ)x(λ) dλ
→ 40x(t)
This can be seen in the plot, e.g., in part (c), y(t) ∼= 40x(t).
Problems 13–65
P 13.61 [a] −1 ≤ t ≤ 4:
vo =∫ t+1
010λ dλ = 5λ2
∣∣∣∣t+1
0= 5t2 + 10t + 5 V
4 ≤ t ≤ 9:
vo =∫ t+1
t−410λ dλ = 5λ2
∣∣∣∣t+1
t−4= 50t − 75 V
9 ≤ t ≤ 14:
vo = 10∫ 10
t−4λ dλ + 10
∫ t+1
1010 dλ
= 5λ2∣∣∣∣10
t−4+100λ
∣∣∣∣t+1
10= −5t2 + 140t − 480 V
14 ≤ t ≤ 19:
vo = 100∫ t+1
t−4dλ = 500 V
19 ≤ t ≤ 24:
vo =∫ 20
t−4100 dλ +
∫ t+1
2010(30 − λ) dλ
= 100λ∣∣∣∣20
t−4+ 300λ
∣∣∣∣t+1
20− 5λ2
∣∣∣∣t+1
20
= −5t2 + 190t − 1305 V
24 ≤ t ≤ 29:
vo = 10∫ t+1
t−4(30 − λ) dλ = 300λ
∣∣∣∣t+1
t−4− 5λ2
∣∣∣∣t+1
t−4
= 1575 − 50t V
29 ≤ t ≤ 34:
vo = 10∫ 30
t−4(30 − λ) dλ = 300λ
∣∣∣∣30
t−4− 5λ2
∣∣∣∣30
t−4
= 5t2 − 340t + 5780 V
Summary:
vo = 0 − ∞ ≤ t ≤ −1
vo = 5t2 + 10t + 5 V − 1 ≤ t ≤ 4
vo = 50t − 75 V 4 ≤ t ≤ 9
vo = −5t2 + 140t − 480 V 9 ≤ t ≤ 14
13–66 CHAPTER 13. The Laplace Transform in Circuit Analysis
vo = 500 V 14 ≤ t ≤ 19
vo = −5t2 + 190t − 1305 V 19 ≤ t ≤ 24
vo = 1575 − 50t V 24 ≤ t ≤ 29
vo = 5t2 − 340t + 5780 V 29 ≤ t ≤ 34
vo = 0 V 34 ≤ t ≤ ∞[b]
P 13.62 [a] h(λ) =25λ 0 ≤ λ ≤ 5
h(λ) =(4 − 2
5λ)
5 ≤ λ ≤ 10
0 ≤ t ≤ 5:
vo = 10∫ t
0
25λ dλ = 2t2
Problems 13–67
5 ≤ t ≤ 10:
vo = 10∫ 5
0
25λ dλ + 10
∫ t
5
(4 − 2
5λ)
dλ
=4λ2
2
∣∣∣∣50
+ 40λ∣∣∣∣t5
− 4λ2
2
∣∣∣∣t5
= −100 + 40t − 2t2
10 ≤ t ≤ ∞:
vo = 10∫ 5
0
25λ dλ + 10
∫ 10
5
(4 − 2
5λ)
dλ
=4λ2
2
∣∣∣∣50
+ 40λ∣∣∣∣10
5− 4λ2
2
∣∣∣∣10
5
= 50 + 200 − 150 = 100
Summary:
vo = 2t2 V 0 ≤ t ≤ 5
vo = 40t − 100 − 2t2 V 5 ≤ t ≤ 10
vo = 100 V 10 ≤ t ≤ ∞[b]
[c] Area =12(10)(2) = 10 ·. .
12(4)h = 10 so h = 5
h(λ) =52λ 0 ≤ λ ≤ 2
h(λ) =(10 − 5
2λ)
2 ≤ λ ≤ 4
13–68 CHAPTER 13. The Laplace Transform in Circuit Analysis
0 ≤ t ≤ 2:
vo = 10∫ t
0
52λ dλ = 12.5t2
2 ≤ t ≤ 4:
vo = 10∫ 2
0
52λ dλ + 10
∫ t
2
(10 − 5
2λ)
dλ
=25λ2
2
∣∣∣∣20
+ 100λ∣∣∣∣t2
− 25λ2
2
∣∣∣∣t2
= −100 + 100t − 12.5t2
4 ≤ t ≤ ∞:
vo = 10∫ 2
0
52λ dλ + 10
∫ 4
2
(10 − 5
2λ)
dλ
=25λ2
2
∣∣∣∣20
+ 100λ∣∣∣∣42
− 25λ2
2
∣∣∣∣42
= 50 + 200 − 150 = 100
vo = 12.5t2 V 0 ≤ t ≤ 2
vo = 100t − 100 − 12.5t2 V 2 ≤ t ≤ 4
vo = 100 V 4 ≤ t ≤ ∞[d] The waveform in part (c) is closer to replicating the input waveform because in
part (c) h(λ) is closer to being an ideal impulse response. That is, the area waspreserved as the base was shortened.
Problems 13–69
P 13.63 [a]
vo =∫ t
010(10e−4λ) dλ
= 100e−4λ
−4
∣∣∣∣t0= −25[e−4t − 1]
= 25(1 − e−4t) V, 0 ≤ t ≤ ∞
[b]
0 ≤ t ≤ 0.5:
vo =∫ t
0100(1 − 2λ) dλ = 100(λ − λ2)
∣∣∣∣t0= 100t(1 − t)
0.5 ≤ t ≤ ∞:
vo =∫ 0.5
0100(1 − 2λ) dλ = 100(λ − λ2)
∣∣∣∣0.5
0= 25
13–70 CHAPTER 13. The Laplace Transform in Circuit Analysis
[c]
P 13.64 [a] From Problem 13.49(a)
H(s) =200
s + 200
h(λ) = 200e−200λ
0 ≤ t ≤ 5 ms:
vo =∫ t
020(200)e−200λ dλ = 20(1 − e−200t) V
5 ms ≤ t ≤ ∞:
vo =∫ t
t−5×10−320(200)e−200λ dλ = 20(e1 − 1)e−200t V
[b]
P 13.65 [a] H(s) =2000
s + 2000·. . h(λ) = 2000e−2000λ
Problems 13–71
0 ≤ t ≤ 5 ms:
vo =∫ t
020(2000)e−2000λ dλ = 20(1 − e−2000t) V
5 ms ≤ t ≤ ∞:
vo =∫ t
t−5×10−320(2000)e−2000λ dλ = 20(e10 − 1)e−2000t V
[b] decrease
[c] The circuit with R = 5 kΩ.
P 13.66 [a] Ig =Vo
105 +Vos
5 × 106 =Vo(s + 50)5 × 106
Vo
Ig
= H(s) =5 × 106
s + 50
h(λ) = 5 × 106e−50λu(λ)
13–72 CHAPTER 13. The Laplace Transform in Circuit Analysis
0 ≤ t ≤ 0.1 s:
vo =∫ t
0(50 × 10−6)(5 × 106)e−50λ dλ = 250
e−50λ
−50
∣∣∣∣t0
= 5(1 − e−50t) V
0.1 s ≤ t ≤ 0.2 s:
vo =∫ t−0.1
0(−50 × 10−6)(5 × 106e−50λ dλ)
+∫ t
t−0.1(50 × 10−6)(5 × 106e−50λ dλ)
= −250e−50λ
−50
∣∣∣∣t−0.1
0+ 250
e−50λ
−50
∣∣∣∣tt−0.1
= 5[e−50(t−0.1) − 1
]− 5
[e−50t − e−50(t−0.1)
]vo = [10e−50(t−0.1) − 5e−50t − 5] V
Problems 13–73
0.2 s ≤ t ≤ ∞:
vo =∫ t−0.1
t−0.2−250e−50λ dλ +
∫ t
t−0.1250e−50λ dλ
= 5e−50λ
∣∣∣∣t−0.1
t−0.2− 5e−50λ
∣∣∣∣tt−0.1
vo = [10e−50(t−0.1) − 5e−50(t−0.2) − 5e−50t] V
Summary:
vo = 5(1 − e−50t) V 0 ≤ t ≤ 0.1 s
vo = [10e−50(t−0.1) − 5e−50t − 5] V 0.1 s ≤ t ≤ 0.2 s
vo = [10e−50(t−0.1) − 5e−50(t−0.2) − 5e−50t] V 0.2 s ≤ t ≤ ∞
[b] Io =Vos
5 × 106 =s
5 × 106 · 5 × 106Ig
s + 50
Io
Ig
= H(s) =s
s + 50= 1 − 50
s + 50
h(λ) = δ(λ) − 50e−50λ
0 < t < 0.1 s:
io =∫ t
0(50 × 10−6)[δ(λ) − 50e−50λ] dλ
= 50 × 10−6 −[50 × 50 × 10−6 e−50λ
−50
] ∣∣∣∣t0
= 50 × 10−6 + 50 × 10−6[e−50t − 1] = 50e−50t µA
13–74 CHAPTER 13. The Laplace Transform in Circuit Analysis
0.1 s < t < 0.2 s:
io =∫ t−0.1
0(−50 × 10−6)[δ(λ) − 50e−50λ] dλ
+∫ t
t−0.1(50 × 10−6)(−50e−50λ) dλ
= −50 × 10−6 + 2500 × 10−6 e−50λ
−50
∣∣∣∣t−0.1
0− 2500 × 10−6 e−50λ
−50
∣∣∣∣tt−0.1
= −50 × 10−6 − 50 × 10−6[e−50(t−0.1) − 1] + 50 × 10−6[e−50t − e−50(t−0.1)]
= 50e−50t − 100e−50(t−0.1) µA
Problems 13–75
0.2 s < t < ∞:
io =∫ t−0.1
t−0.2(−50 × 10−6)(−50e−50λ) dλ
+∫ t
t−0.1(50 × 10−6)(−50e−50λ) dλ
= 50e−50t − 100e−50(t−0.1) + 50e−50(t−0.2) µA
Summary:
i0 = 50e−50t µA 0 ≤ t ≤ 0.1 s
i0 = 50e−50t − 100e−50(t−0.1) µA 0.1 s ≤ t ≤ 0.2 s
i0 = 50e−50t − 100e−50(t−0.1) + 50e−50(t−0.2) µA 0.2 s ≤ t ≤ ∞[c] At t = 0.1−:
vo = 5(1 − e−5) = 4.97 V; i100kΩ =4.970.1
= 49.66 µA; ig = 50 µA
·. . io = 50 − 49.66 = 0.34 µA
From the solution for io we have io(0.1−) = 50e−5 = 0.34 µA (Checks)At t = 0.1+:
vo(0.1+) = vo(0.1−) = 4.97 µV; i100kΩ = 49.66 µA; ig = −50 µA
·. . io(0.1+) = −(50 + 49.66) = −99.66 µA
From the solution for io we have
io(0.1+) = 50e−5 − 100 = −99.66 µA (Checks)
13–76 CHAPTER 13. The Laplace Transform in Circuit Analysis
At t = 0.2−:
vo = 10e−5 − 5e−10 − 5 = −4.93 µV
i100kΩ = −49.33 µA ig = −50 µA
io = ig − i100kΩ = −50 + 49.33 = −0.67 µA
From the solution for io, io(0.2−) = 50e−10 − 100e−5 = −0.67 µA (Checks)At t = 0.2+:
vo(0.2+) = io(0.2−) = −4.93 V; i100kΩ = −49.33 µA; ig = 0
io = ig − i100kΩ = 49.33 µA
From the solution for io,io(0.2+) = 50e−10 − 100e−5 + 50 = 49.33 µA(Checks)
P 13.67 H(s) =Vo
Vi
=5
5 + 2.5s=
2s + 2
h(λ) = 2e−2λ; h(t − λ) = 2e−2(t−λ) = 2e−2te2λ
T
2=
π
2; T = π s; f =
1π
Hz
vi(λ) = (20 sin 2λ)[u(λ) − u(λ − π/2)]
(π/2) s ≤ t ≤ ∞:
vo =∫ π/2
0(2e−2te2λ)(20 sin 2λ) dλ = 40e−2t
∫ π/2
0e2λ sin 2λ dλ
= 40e−2t
[e2λ
8(2 sin 2λ − 2 cos 2λ)
]π/2
0
= 10e−2t[eπ(sin π − cos π) − 1(0 − 1)]
= 10e−2t(eπ + 1) = 10(eπ + 1)e−2t V
vo(2.2) = 241.41e−4.4 = 2.96 V
Problems 13–77
P 13.68 [a] Vo =1620
Vg
·. . H(s) =Vo
Vg
=45
h(λ) = 0.8δ(λ)
[b]
0 < t < 0.5 s : vo =∫ t
075[0.8δ(λ)] dλ = 60 A
13–78 CHAPTER 13. The Laplace Transform in Circuit Analysis
0.5 s ≤ t ≤ 1.0 s:
vo =∫ t−0.5
0−75[0.8δ(λ)] dλ = −60 A
1 s < t < ∞ : vo = 0
[c]
Yes, because the circuit has no memory.
P 13.69 [a]
Vo − Vg
5+
Vos
4+
Vo
20= 0
(5s + 5)Vo = 4Vg
H(s) =Vo
Vg
=0.8
s + 1; h(λ) = 0.8e−λu(λ)
Problems 13–79
[b]
0 ≤ t ≤ 0.5 s;
vo =∫ t
075(0.8e−λ) dλ = 60
e−λ
−1
∣∣∣∣t0
vo = 60 − 60e−t V, 0 ≤ t ≤ 0.5 s
0.5 s ≤ t ≤ 1 s:
vo =∫ t−0.5
0(−75)(0.8e−λ) dλ +
∫ t
t−0.575(0.8e−λ) dλ
= −60e−λ
−1
∣∣∣∣t−0.5
0+ 60
e−λ
−1
∣∣∣∣tt−0.5
= 120e−(t−0.5) − 60e−t − 60 V, 0.5 s ≤ t ≤ 1 s
13–80 CHAPTER 13. The Laplace Transform in Circuit Analysis
1 s ≤ t ≤ ∞;
vo =∫ t−0.5
t−1(−75)(0.8e−λ) dλ +
∫ t
t−0.575(0.8e−λ) dλ
= −60e−λ
−1
∣∣∣∣t−0.5
t−1+ 60
e−λ
−1
∣∣∣∣tt−0.5
= 120e−(t−0.5) − 60e−(t−1) − 60e−t V, 1 s ≤ t ≤ ∞[c]
[d] No, the circuit has memory because of the capacitive storage element.
P 13.70
Vo =20 × 103
5000 + 25 × 105/s + 20 × 103 (5000Ig)
Vo
Ig
= H(s) =4000s
s + 100
H(s) = 4000[1 − 100
s + 100
]= 4000 − 4 × 105
s + 100
h(λ) = 4000δ(λ) − 400,000e−100λu(λ)
Problems 13–81
vo =∫ 10−3
0(−20 × 10−3)[4000δ(λ) − 400,000e−100λ] dλ
+∫ 5×10−3
10−3(10 × 10−3)[−400,000e−100λ] dλ
= −80 + 8000∫ 10−3
0e−100λ dλ −
∫ 5×10−3
10−34000e−100λ dλ
= −80 − 80(e−0.1 − 1) + 40(e−0.5 − e−0.1)
vo(5 × 10−3) = 40e−0.5 − 120e−0.1 = 24.26 − 108.58 = −84.32 V
Alternate solution (not using the convolution integral):
Ig =∫ 4×10−3
0(10 × 10−3)e−st dt +
∫ 6×10−3
4×10−3(−20 × 10−3)e−st dt
= 10−3 e−st
−s
∣∣∣∣4×10−3
0− 20 × 10−3 e−st
−s
∣∣∣∣6×10−3
4×10−3
= 10 × 10−3
[1s
− e−4×10−3s
s
]+ 20 × 10−3
[e−6×10−3s − e−4×10−3s
s
]
=10 × 10−3
s− 30 × 10−3
se−4×10−3s +
20 × 10−3
se−6×10−3s
Vo = IgH(s) =40
s + 100− 120e−4×10−3s
s + 100+
80e−6×10−3s
s + 100
Now use the operational transform L−1e−asF (s) = f(t − a)u(t − a):
vo = 40e−100t − 120e−100(t−4×10−3)u(t − 4 × 10−3)
+ 80e−100(t−6×10−3)u(t − 6 × 10−3) V
vo(5 × 10−3) = 40e−0.5 − 120e−0.1 + 80(0) = −84.32 V (Checks)
P 13.71 [a] H(s) =Vo
Vi
=1/LC
s2 + (R/L)s + (1/LC)
=100
s2 + 20s + 100=
100(s + 10)2
h(λ) = 100λe−10λu(λ)
13–82 CHAPTER 13. The Laplace Transform in Circuit Analysis
0 ≤ t ≤ 0.5:
vo = 500∫ t
0λe−10λ dλ
= 500
e−10λ
100(−10λ − 1)
∣∣∣∣t0
= 5[1 − e−10t(10t + 1)]
0.5 ≤ t ≤ ∞:
vo = 500∫ t
t−0.5λe−10λ dλ
= 500
e−10λ
100(−10λ − 1)
∣∣∣∣tt−0.5
= 5e−10t[e5(10t − 4) − 10t − 1]
[b]
Problems 13–83
P 13.72 H(s) =16s
40 + 4s + 16s=
0.8ss + 2
= 0.8(1 − 2
s + 2
)= 0.8 − 1.6
s + 2
h(λ) = 0.8δ(λ) − 1.6e−2λu(λ)
vo =∫ t
075[0.8δ(λ) − 1.6e−2λ] dλ =
∫ t
060δ(λ) dλ − 120
∫ t
0e−2λ dλ
= 60 − 120e−2λ
−2
∣∣∣∣t0= 60 + 60(e−2t − 1)
= 60e−2tu(t) V
P 13.73 [a] Y (s) =∫ ∞
0y(t)e−st dt
Y (s) =∫ ∞
0e−st
[∫ ∞
0h(λ)x(t − λ) dλ
]dt
=∫ ∞
0
∫ ∞
0e−sth(λ)x(t − λ) dλ dt
=∫ ∞
0h(λ)
∫ ∞
0e−stx(t − λ) dt dλ
But x(t − λ) = 0 when t < λ
Therefore Y (s) =∫ ∞
0h(λ)
∫ ∞
λe−stx(t − λ) dt dλ
Let u = t − λ; du = dt; u = 0, t = λ; u = ∞, t = ∞
Y (s) =∫ ∞
0h(λ)
∫ ∞
0e−s(u+λ)x(u) du dλ
=∫ ∞
0h(λ)e−sλ
∫ ∞
0e−sux(u) du dλ
=∫ ∞
0h(λ)e−sλX(s) dλ = H(s) X(s)
We are using one-sided Laplace transforms; therefore h(t) and X(t) areassumed zero for t < 0.
13–84 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] F (s) =a
s(s + a)2 =1s
· a
(s + a)2 = H(s)X(s)
·. . h(t) = u(t), x(t) = at e−atu(t)
·. . f(t) =∫ t
0(1)aλe−aλ dλ = a
[e−aλ
a2 (−aλ − 1)]∣∣∣∣∣
t
0
=1a[e−at(−at − 1) − 1(−1)] =
1a[1 − e−at − ate−at]
=[1a
− 1ae−at − te−at
]u(t)
Check:
F (s) =a
s(s + a)2 =K0
s+
K1
(s + a)2 +K2
s + a
K0 =1a; K1 = −1; K2 =
d
ds
(a
s
)s=−a
= −1a
f(t) =[1a
− te−at − 1ae−at
]u(t)
P 13.74 [a] The s-domain circuit is
The node-voltage equation isV
sL1+
V
R+
V
sL2=
ρ
s
Therefore V =ρR
s + (R/Le)where Le =
L1L2
L1 + L2
Therefore v = ρRe−(R/Le)tu(t) V
Problems 13–85
[b] I1 =V
R+
V
sL2=
ρ[s + (R/L2)]s[s + (R/Le)]
=K0
s+
K1
s + (R/Le)
K0 =ρL1
L1 + L2; K1 =
ρL2
L1 + L2
Thus we have i1 =ρ
L1 + L2[L1 + L2e
−(R/Le)t]u(t) A
[c] I2 =V
sL2=
(ρR/L2)s[s + (R/Le)]
=K2
s+
K3
s + (R/Le)
K2 =ρL1
L1 + L2; K3 =
−ρL1
L1 + L2
Therefore i2 =ρL1
L1 + L2[1 − e−(R/Le)t]u(t)
[d] λ(t) = L1i1 + L2i2 = ρL1
P 13.75 [a] As R → ∞, v(t) → ρLeδ(t) since the area under the impulse generatingfunction is ρLe.
i1(t) → ρL1
L1 + L2as R → ∞
i2(t) → ρL1
L1 + L2as R → ∞
13–86 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b] The s-domain circuit is
V
sL1+
V
sL2=
ρ
s; therefore V =
ρL1L2
L1 + L2= ρLe
Therefore v(t) = ρLeδ(t)
I1 = I2 =V
sL2=(
ρL1
L1 + L2
)(1s
)
Therefore i1 = i2 =ρL1
L1 + L2u(t) A
P 13.76 H(j3) =4(3 + j3)
−9 + j24 + 41= 0.42/8.13
·. . vo(t) = 16.97 cos(3t + 8.13) V
P 13.77 [a] H(s) =−Zf
Zi
Zf =(1/Cf )
s + (1/RfCf )=
108
s + 1000
Zi =Ri[s + (1/RiCi)]
s=
10,000(s + 400)s
H(s) =−104s
(s + 400)(s + 1000)
[b] H(j400) =−104(j400)
(400 + j400)(1000 + j400)= 6.565/− 156.8
vo(t) = 13.13 cos(400t − 156.8) V
Problems 13–87
P 13.78 [a]
Vp =0.01s
80 + 0.01sVg =
s
s + 8000Vg
Vn
5000+
Vn − Vo
25,000+ (Vn − Vo)8 × 10−9s = 0
5Vn + Vn − Vo + (Vn − Vo)2 × 10−4s = 0
6Vn + 2 × 10−4sVn = Vo + 2 × 10−4sVo
2 × 10−4Vn(s + 30,000) = 2 × 10−4Vo(s + 5000)
Vn = Vp
Vo =s + 30,000s + 5000
Vf =(
s + 30,000s + 5000
)(sVg
s + 8000
)
H(s) =Vo
Vg
=s(s + 30,000)
(s + 5000)(s + 8000)
[b] vg = 0.6u(t); Vg =0.6s
Vo =0.6(s + 30,000)
(s + 5000)(s + 8000)=
K1
s + 5000+
K2
s + 8000
K1 =0.6(25,000)
3000= 5; K2 =
0.6(22,000)−3000
= −4.4
·. . vo(t) = (5e−5000t − 4.4e−8000t)u(t) V
[c] Vg = 2 cos 10,000t V
H(jω) =j10,000(30,000 + j10,000)
(5000 + j10,000)(8000 + j10,000)= 2.21/− 6.34
·. . vo = 4.42 cos(10,000t − 6.34) V
13–88 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.79 Vo =50
s + 8000− 20
s + 5000=
30(s + 3000)(s + 5000)(s + 8000)
Vo = H(s)Vg = H(s)(30
s
)
·. . H(s) =s(s + 3000)
(s + 5000)(s + 8000)
H(j6000) =(j6000)(3000 + j6000)
(5000 + j6000)(8000 + j6000)= 0.52/66.37
·. . vo(t) = 61.84 cos(6000t + 66.37) V
P 13.80 Original charge on C1; q1 = V0C1
The charge transferred to C2; q2 = V0Ce =V0C1C2
C1 + C2
The charge remaining on C1; q′1 = q1 − q2 =
V0C21
C1 + C2
Therefore V2 =q2
C2=
V0C1
C1 + C2and V1 =
q′1
C1=
V0C1
C1 + C2
P 13.81 [a] Z1 =1/C1
s + 1/R1C1=
25 × 1010
s + 20 × 104 Ω
Z2 =1/C2
s + 1/R2C2=
6.25 × 1010
s + 12,500Ω
Vo
Z2+
Vo − 10/sZ1
= 0
Vo(s + 12,500)6.25 × 1010 +
Vo(s + 20 × 104)25 × 1010 =
10s
(s + 20 × 104)25 × 1010
Vo =2(s + 200,000)s(s + 50,000)
=K1
s+
K2
s + 50,000
K1 =2(200,000)
50,000= 8
K2 =2(150,000)−50,000
= −6
·. . vo = [8 − 6e−50,000t]u(t) V
Problems 13–89
[b] I0 =V0
Z2=
2(s + 200,000)(s + 12,500)s(s + 50,000)6.25 × 1010
= 32 × 10−12
[1 +
162,500s + 25 × 108
s(s + 50,000)
]
= 32 × 10−12
[1 +
K1
s+
K2
s + 50,000
]
K1 = 50,000; K2 = 112,500
io = 32δ(t) + [1.6 × 106 + 3.6 × 106e−50,000t]u(t) pA
[c] When C1 = 64 pF
Z1 =156.25 × 108
s + 12,500Ω
V0(s + 12,500)625 × 108 +
V0(s + 12,500)156.25 × 108 =
10s
(s + 12,500)156.25 × 108
·. . V0 + 4V0 =40s
V0 =8s
vo = 8u(t) V
I0 =V0
Z2=
8s
(s + 12,500)6.25 × 1010 = 128 × 10−12
[1 +
12,500s
]
io(t) = 128δ(t) + 1.6 × 10−6u(t) pA
P 13.82 Let a =1
R1C1=
1R2C2
Then Z1 =1
C1(s + a)and Z2 =
1C2(s + a)
Vo
Z2+
Vo
Z1=
10/sZ1
VoC2(s + a) + VoC1(s + a) = (10/s)C1(s + a)
Vo =10s
(C1
C1 + C2
)
Thus, vo is the input scaled by the factorC1
C1 + C2.
13–90 CHAPTER 13. The Laplace Transform in Circuit Analysis
P 13.83 [a] For t < 0, 0.5v1 = 2v2; therefore v1 = 4v2
v1 + v2 = 100; therefore v1(0−) = 80 V
[b] v2(0−) = 20 V
[c] v3(0−) = 0 V
[d] For t > 0:
I =100/s
3.125/s× 10−6 = 32 × 10−6
i(t) = 32δ(t) µA
[e] v1(0+) = −106
0.5
∫ 0+
0−32 × 10−6δ(t) dt + 80 = −64 + 80 = 16 V
[f] v2(0+) = −106
2
∫ 0+
0−32 × 10−6δ(t) dt + 20 = −16 + 20 = 4 V
[g] V3 =0.625 × 106
s· 32 × 10−6 =
20s
v3(t) = 20u(t) V; v3(0+) = 20 V
Check: v1(0+) + v2(0+) = v3(0+)
P 13.84 [a] For t < 0:
Req = 0.8 kΩ‖4 kΩ‖16 kΩ = 0.64 kΩ; v = 5(640) = 3200 V
i1(0−) =32004000
= 0.8 A; i2(0−) =3200
16,000= 0.2 A
Problems 13–91
[b] For t > 0:
i1 + i2 = 0
8(∆i1) = 2(∆i2)
i1(0−) + ∆i1 + i2(0−) + ∆i2 = 0; therefore ∆i1 = −0.2 A
∆i2 = −0.8 A; i1(0+) = 0.8 − 0.2 = 0.6 A
[c] i2(0−) = 0.2 A
[d] i2(0+) = 0.2 − 0.8 = −0.6 A
[e] The s-domain equivalent circuit for t > 0 is
I1 =0.006
0.01s + 20,000=
0.6s + 2 × 106
i1(t) = 0.6e−2×106tu(t) A
[f] i2(t) = −i1(t) = −0.6e−2×106tu(t) A
[g] V = −0.0064 + (0.008s + 4000)I1 =−0.0016(s + 6.5 × 106)
s + 2 × 106
= −1.6 × 10−3 − 7200s + 2 × 106
v(t) = [−1.6 × 10−3δ(t)] − [7200e−2×106tu(t)] V
P 13.85 [a]
Vo =0.5
50,000 + 5 × 106/s· 106
s
13–92 CHAPTER 13. The Laplace Transform in Circuit Analysis
500,00050,000s + 5 × 106 =
10s + 100
vo = 10e−100tu(t) V
[b] At t = 0 the current in the 1 µF capacitor is 10δ(t) µA
·. . vo(0+) = 106∫ 0+
0−10 × 10−6δ(t) dt = 10 V
After the impulsive current has charged the 1 µF capacitor to 10 V it dischargesthrough the 50 kΩ resistor.
Ce =C1C2
C1 + C2=
0.251.25
= 0.2 µF
τ = (50,000)(0.2 × 10−6) = 10−2
1τ
= 100 (Checks)
Note – after the impulsive current passes the circuit becomes
The solution for vo in this circuit is also
vo = 10e−100tu(t) V
P 13.86 [a] After making a source transformation, the circuit is as shown. The impulsecurrent will pass through the capacitive branch since it appears as a shortcircuit to the impulsive current,
Therefore vo(0+) = 106∫ 0+
0−
[δ(t)1000
]dt = 1000 V
Problems 13–93
Therefore wC = (0.5)Cv2 = 0.5 J
[b] iL(0+) = 0; therefore wL = 0 J
[c] Vo(10−6)s +Vo
250 + 0.05s+
Vo
1000= 10−3
Therefore
Vo =1000(s + 5000)
s2 + 6000s + 25 × 106
=K1
s + 3000 − j4000+
K∗1
s + 3000 + j4000
K1 = 559.02/− 26.57; K∗1 = 559.02/26.57
vo = [1118.03e−3000t cos(4000t − 26.57)]u(t) V
[d] The s-domain circuit is
Vos
106 +Vo
250 + 0.05s+
Vo
1000= 10−3
Note that this equation is identical to that derived in part [c], therefore thesolution for Vo will be the same.
P 13.87 [a]
20 = sI1 − 0.5sI2
0 = −0.5sI1 +(s +
3s
)I2
13–94 CHAPTER 13. The Laplace Transform in Circuit Analysis
∆ =
∣∣∣∣∣∣∣s −0.5s
−0.5s (s + 3/s)
∣∣∣∣∣∣∣ = s2 + 3 − 0.25s2 = 0.75(s2 + 4)
N1 =
∣∣∣∣∣∣∣20 −0.5s
0 (s + 3/s)
∣∣∣∣∣∣∣ = 20s +60s
=20s2 + 60
s=
20(s2 + 3)s
I1 =N1
∆=
20(s2 + 3)s(0.75)(s2 + 4)
=803
· s2 + 3s(s2 + 4)
=K0
s+
K1
s − j2+
K∗1
s + j2
K0 =803
(34
)= 20; K1 =
803
[ −4 + 3(j2)(j4)
]=
103
/0
·. . i1 =[20 +
203
cos 2t]u(t) A
[b] N2 =
∣∣∣∣∣∣∣s 20
−0.5s 0
∣∣∣∣∣∣∣ = 10s
I2 =N2
∆=
10s0.75(s2 + 4)
=403
(s
s2 + 4
)=
K1
s − j2+
K∗1
s + j2
K1 =403
(j2j4
)=
203
/0
i2 =403
(cos 2t)u(t) A
[c] V0 =3sI2 =
(3s
) 403
(s
s2 + 4
)=
40s2 + 4
=K1
s − j2=
K∗1
s + j2
K1 =40j4
= −j10 = 10/90
vo = 20 cos(2t − 90) = 20 sin 2t
vo = [20 sin 2t]u(t) V
[d] Let us begin by noting i1 jumps from 0 to (80/3) A between 0− and 0+ and inthis same interval i2 jumps from 0 to (40/3) A. Therefore in the derivatives ofi1 and i2 there will be impulses of (80/3)δ(t) and (40/3)δ(t), respectively.Thus
di1dt
=803
δ(t) − 403
sin 2t A/s
Problems 13–95
di2dt
=403
δ(t) − 803
sin 2t A/s
From the circuit diagram we have
20δ(t) = 1di1dt
− 0.5di2dt
=803
δ(t) − 403
sin 2t − 20δ(t)3
+403
sin 2t
= 20δ(t)
Thus our solutions for i1 and i2 are in agreement with known circuit behavior.Let us also note the impulsive voltage will impart energy into the circuit. Sincethere is no resistance in the circuit, the energy will not dissipate. Thus the factthat i1, i2, and vo exist for all time is consistent with known circuit behavior.Also note that although i1 has a dc component, i2 does not. This follows fromknown transformer behavior.Finally we note the flux linkage prior to the appearance of the impulsivevoltage is zero. Now since v = dλ/dt, the impulsive voltage source must bematched to an instantaneous change in flux linkage at t = 0+ of 20.For the given polarity dots and reference directions of i1 and i2 we have
λ(0+) = L1i1(0+) + Mi1(0+) − L2i2(0+) − Mi2(0+)
λ(0+) = 1(80
3
)+ 0.5
(803
)− 1
(403
)− 0.5
(403
)
=1203
− 603
= 20 (Checks)
P 13.88 [a]
V1
104 +V1
[(2 × 105)/s)] + [(5 × 104)/s]= 10−5
13–96 CHAPTER 13. The Laplace Transform in Circuit Analysis
V1
104 +sV1
25 × 104 = 10−5
25V1 + sV1 = 2.5
V1 =2.5
s + 25
Vo =(
sV1
25 × 104
)(5 × 104
s
)=
15V1
·. . Vo =0.5
s + 25; vo = 0.5e−25tu(t) V
[b] vo(0+) = 0.5 V
vo(0+) =106
20
∫ 0+
0−10 × 10−6δ(x) dx = 0.5 V (Checks)
Ce =(5)(20)
25= 4 µF
τ = RCe = (10 × 103)(4 × 10−6) = 4 × 10−2 s;1τ
=1004
= 25 (Checks)
Yes, the impulsive current establishes an instantaneous charge on eachcapacitor. After the impulsive current vanishes the capacitors dischargeexponentially to zero volts.
P 13.89 [a] The circuit parameters are
Ra =1202
1200= 12 Ω Rb =
1202
1800= 8 Ω Xa =
1202
350=
144035
Ω
The branch currents are
I1 =120/0
12= 10/0 A(rms) I2 =
120/0
j1440/35= −j
3512
=3512
/− 90 A(rms)
I3 =120/0
8= 15/0 A(rms)
·. . IL = I1 + I2 + I3 = 25 − j3512
= 25.17/− 6.65 A(rms)
Therefore,
i2 =(35
12
)√2 cos(ωt − 90) A and iL = 25.17
√2 cos(ωt − 6.65) A
Thus,
i2(0−) = i2(0+) = 0 A and iL(0−) = iL(0+) = 25√
2 A
Problems 13–97
[b] Begin by using the s-domain circuit in Fig. 13.60 to solve for V0 symbolically.Write a single node voltage equation:
V0 − (Vg + LI0)sL
+V0
Ra
+V0
sLa
= 0
·. . V0 =(Ra/L)Vg + I0Ra
s + [Ra(La + L)]/LaL
where L = 1/120π H, La = 12/35π H, Ra = 12 Ω, and I0Ra = 300√
2 V.Also,
Vg = V0 + IL(j) = 120 +(25 − j
3512
)j = 122.92 + 25j V(rms)
vg(t) = 122.92√
2 cos ωt − 25√
2 sin ωt V, with ω = 120π rad/s.
Thus,
V0 =1440π(122.92
√2s − 3000π
√2)
(s + 1475π)(s2 + 14,400π2)+
300√
2s + 1475π
=K1
s + 1475π+
K2
s − j120π+
K∗2
s + j120π+
300√
2s + 1475π
The coefficients are
K1 = −121.18√
2 V K2 = 61.03√
2/6.85 V K∗2 = 61.03
√2/− 6.85
Note that K1 + 300√
2 = 178.82√
2 V. Thus, the inverse transform of V0 is
v0 = 178.82√
2e−1475πt + 122.06√
2 cos(120πt + 6.85) V
Initially,
v0(0+) = 178.82√
2 + 122.06√
2 cos 6.85 = 300√
2 V
Note that at t = 0+ the initial value of iL, which is 25√
2 A, exists in the 12 Ωresistor Ra. Thus, the initial value of V0 is (25
√2)(12) = 300
√2 V.
[c] The phasor domain equivalent circuit has a j1 Ω inductive impedance in serieswith the parallel combination of a 12 Ω resistive impedance and a j1440/35 Ωinductive impedance (remember that ω = 120π rad/s). Note thatVg = 120/0 + (25.17/− 6.65)(j1) = 125.43/11.50 V(rms). The nodevoltage equation in the phasor domain circuit is
V0 − 125.43/11.50
j1+
V0
12+
35V0
j1440= 0
·. . V0 = 122.06/6.85 V(rms)
Therefore, v0 = 122.06√
2 cos(120πt + 6.85) V, agreeing with thesteady-state component of the result in part (b).
13–98 CHAPTER 13. The Laplace Transform in Circuit Analysis
[d] A plot of v0, generated in Excel, is shown below.
P 13.90 [a] At t = 0− the phasor domain equivalent circuit is
I1 =−j120
12= −j10 = 10/− 90A (rms)
I2 =−j120(35)
j1440= −35
12=
3512
/180A (rms)
I3 =−j120
8= −j15 = 15/− 90A (rms)
IL = I1 + I2 + I3 = −3512
− j25 = 25.17/− 96.65A (rms)
iL = 25.17√
2 cos(120πt − 96.65)A
iL(0−) = iL(0+) = −2.92√
2A
Problems 13–99
i2 =3512
√2 cos(120πt + 180)A
i2(0−) = i2(0+) = −3512
√2 = −2.92
√2A
Vg = Vo + j1IL
Vg = −j120 + 25 − j3512
= 25 − j122.92
vg = 25√
2 cos 120πt + 122.92√
2 sin 120πt
·. . Vg =25
√2s + 122.92
√2(120π)
s2 + (120π)2
Use a variation of the s-domain circuit in Fig.13.60, where
Ll =1
120πH; La =
1235π
H; Ra = 12 Ω
iL(0) = −2.92√
2A; i2(0) = −2.92√
2A
The node voltage equation is
0 =Vo − (Vg + iL(0)Ll)
sLl
+Vo
Ra
+Vo + i2(0)La
sLa
Solving for Vo yields
Vo =VgRa/Ll
[s + Ra(Ll + La)/LaLl]+
Ra[iL(0) − i2(0)][s + Ra(Ll + La)/LlLa]
Ra
Ll
= 1440π
Ra(Ll + La)LlLa
=12( 1
120π+ 12
35π)
( 1235π
)( 1120π
)= 1475π
iL(0) − i2(0) = −2.92√
2 + 2.92√
2 = 0
·. . Vo =1440π[25
√2s + 122.92
√2(120π)]
(s + 1475π)[s2 + (120π)2]
=K1
s + 1475π+
K2
s − j120π+
K∗2
s + j120π
K1 = −14.55√
2 K2 = 61.03√
2/− 83.15
·. . vo(t) = −14.55√
2e−1475πt + 122.06√
2 cos(120πt − 83.15)V
Check:
vo(0) = (−14.55 + 14.55)√
2 = 0
13–100 CHAPTER 13. The Laplace Transform in Circuit Analysis
[b]
Problems 13–101
13–102 CHAPTER 13. The Laplace Transform in Circuit Analysis
Problems 13–103
[c] In Prob. 13.89 the line-to-neutral voltage spikes at 300√
2 V. In part (a) theline-to-neutral voltage has no spike. Thus the amount of voltage disturbancedepends on what part of the cycle the sinusoidal steady-state voltage isswitched.
P 13.91 [a] First find Vg before Rb is disconnected. The phasor domain circuit is
IL =120/− θ
Ra
+120/− θ
Rb
+120/− θ
jXa
=120/− θ
RaRbXa
[(Ra + Rb)Xa − jRaRb]
Since Xl = 1 Ω we have
Vg = 120/− θ +120/− θ
RaRbXa
[RaRb + j(Ra + Rb)Xa]
Ra = 12 Ω; Rb = 8 Ω; Xa =144035
Ω
Vg =120/− θ
1440(1475 + j300)
=2512
/− θ(59 + j12) = 125.43/(−θ + 11.50)
vg = 125.43√
2 cos(120πt − θ + 11.50)V
Let β = −θ + 11.50. Then
vg = 125.43√
2(cos 120πt cos β − sin 120πt sin β)V
Therefore
Vg =125.43
√2(s cos β − 120π sin β)s2 + (120π)2
13–104 CHAPTER 13. The Laplace Transform in Circuit Analysis
The s-domain circuit becomes
where ρ1 = iL(0+) and ρ2 = i2(0+).The s-domain node voltage equation is
Vo − (Vg + ρ1Ll)sLl
+Vo
Ra
+Vo + ρ2La
sLa
= 0
Solving for Vo yields
Vo =VgRa/Ll + (ρ1 − ρ2)Ra
[s + (La+Ll)Ra
LaLl]
Substituting the numerical values
Ll =1
120πH; La =
1235π
H; Ra = 12 Ω; Rb = 8 Ω;
gives
Vo =1440πVg + 12(ρ1 − ρ2)
(s + 1475π)
Now determine the values of ρ1 and ρ2.
ρ1 = iL(0+) and ρ2 = i2(0+)
IL =120/− θ
RaRbXa
[(Ra + Rb)Xa − jRaRb]
=120/− θ
96(1440/35)
[(20)(1440)
35− j96
]
= 25.17/(−θ − 6.65)A(rms)
·. . iL = 25.17√
2 cos(120πt − θ − 6.65)A
iL(0+) = ρ1 = 25.17√
2 cos(−θ − 6.65)A
·. . ρ1 = 25√
2 cos θ − 2.92√
2 sin θA
I2 =120/− θ
j(1440/35)=
3512
/(−θ − 90)
Problems 13–105
i2 =3512
√2 cos(120πt − θ − 90)A
ρ2 = i2(0+) = −3512
√2 sin θ = −2.92
√2 sin θA
·. . ρ1 − ρ2 = 25√
2 cos θ
(ρ1 − ρ2)Ra = 300√
2 cos θ
·. . Vo =1440π
s + 1475π· Vg +
300√
2 cos θ
s + 1475π
=1440π
s + 1475π
[125.43
√2(s cos β − 120π sin β)s2 + 14,400π2
]+
300√
2 cos θ
s + 1475π
=K1 + 300
√2 cos θ
s + 1475π+
K2
s − j120π+
K∗2
s + j120π
Now
K1 =(1440π)(125.43
√2)[−1475π cos β − 120π sin β]
14752π2 + 14,400π2
=−1440(125.43
√2)[1475 cosβ + 120 sin β]
14752 + 14,400
Since β = −θ + 11.50, K1 reduces to
K1 = −121.18√
2 cos θ − 14.55√
2 sin θ
From the partial fraction expansion for Vo we see vo(t) will go directly intosteady state when K1 = −300
√2 cos θ. It follow that
−14.55√
2 sin θ = −178.82√
2 cos θ
or tan θ = 12.29
Therefore, θ = 85.35
[b] When θ = 85.35, β = −73.85
·. . K2 =1440π(125.43
√2)[−120π sin(−73.85) + j120π cos(−73.85)
(1475π + j120π)(j240π)
=720
√2(120.48 + j34.88)−120 + j1475
= 61.03√
2/− 78.50
·. . vo = 122.06√
2 cos(120πt − 78.50)V t > 0
= 172.61 cos(120πt − 78.50)V t > 0
13–106 CHAPTER 13. The Laplace Transform in Circuit Analysis
[c] vo1 = 169.71 cos(120πt − 85.35)V t < 0
vo2 = 172.61 cos(120πt − 78.50)V t > 0
Problems 13–107
13–108 CHAPTER 13. The Laplace Transform in Circuit Analysis
14Introduction to Frequency-Selective
Circuits
Assessment Problems
AP 14.1 fc = 8 kHz, ωc = 2πfc = 16π krad/s
ωc =1
RC; R = 10 kΩ;
·. . C =1
ωcR=
1(16π × 103)(104)
= 1.99 nF
AP 14.2 [a] ωc = 2πfc = 2π(2000) = 4π krad/s
L =R
ωc
=50004000π
= 0.40 H
[b] H(jω) =ωc
ωc + jω=
4000π4000π + jω
When ω = 2πf = 2π(50,000) = 100,000π rad/s
H(j100,000π) =4000π
4000π + j100,000π=
11 + j25
= 0.04/− 87.71
·. . |H(j100,000π)| = 0.04
[c] ·. . θ(100,000π) = −87.71
AP 14.3 ωc =R
L=
50003.5 × 10−3 = 1.43 Mrad/s
14–1
14–2 CHAPTER 14. Introduction to Frequency-Selective Circuits
AP 14.4 [a] ωc =1
RC=
106
R=
106
100= 10 krad/s
[b] ωc =106
5000= 200 rad/s
[c] ωc =106
3 × 104 = 33.33 rad/s
AP 14.5 Let Z represent the parallel combination of (1/sC) and RL. Then
Z =RL
(RLCs + 1)
Thus H(s) =Z
R + Z=
RL
R(RLCs + 1) + RL
=(1/RC)
s + R+RL
RL
(1
RC
) =(1/RC)
s + 1K
(1
RC
)
where K =RL
R + RL
AP 14.6 ω2o =
1LC
so L =1
ω2oC
=1
(24π × 103)2(0.1 × 10−6)= 1.76 mH
Q =ωo
β=
ωo
R/Lso R =
ωoL
Q=
(24π × 103)(1.76 × 10−3)6
= 22.10 Ω
AP 14.7 ωo = 2π(2000) = 4000π rad/s;
β = 2π(500) = 1000π rad/s; R = 250 Ω
β =1
RCso C =
1βR
=1
(1000π)(250)= 1.27 µF
ω2o =
1LC
so L =1
ω2oC
=106
(4000π)2(1.27)= 4.97 mH
AP 14.8 ω2o =
1LC
so L =1
ω2oC
=1
(104π)2(0.2 × 10−6)= 5.07 mH
β =1
RCso R =
1βC
=1
400π(0.2 × 10−6)= 3.98 kΩ
Problems 14–3
AP 14.9 ω2o =
1LC
so L =1
ω2oC
=1
(4000π)2(0.2 × 10−6)= 31.66 mH
Q =fo
β=
5 × 103
200= 25 = ωoRC
·. . R =Q
ωoC=
25(4000π)(0.2 × 10−6)
= 9.95 kΩ
AP 14.10ωo = 8000π rad/s
C = 500 nF
ω2o =
1LC
so L =1
ω2oC
= 3.17 mH
Q =ωo
β=
ωoL
R=
1ωoCR
·. . R =1
ωoCQ=
1(8000π)(500 × 10−9)(5)
= 15.92 Ω
AP 14.11ωo = 2πfo = 2π(20,000) = 40π krad/s; R = 100 Ω; Q = 5
Q =ωo
β=
ωoL
RCso L =
RQ
ωo
=100
40π × 103 = 3.98 mH
ω2o =
1LC
so C =1
ω2oL
=1
(40π × 103)2(3.98 × 10−3)= 15.92 nF
14–4 CHAPTER 14. Introduction to Frequency-Selective Circuits
Problems
P 14.1 [a] ωc =R
L=
12710 × 10−3 = 12.7 krad/s
·. . fc =ωc
2π=
12,7002π
= 2021.27 Hz
[b] H(s) =ωc
s + ωc
=12,700
s + 12,700
H(jω) =12,700
12,700 + jω
H(jωc) =12,700
12,700 + j12,700= 0.7071/− 45
H(j0.2ωc) =12,700
12,700 + j2540= 0.981/− 11.31
H(j5ωc) =12,700
12,700 + j63,500= 0.196/− 78.69
[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
L= 2000π rad/s
R = Lωo = (0.005)(2000π) = 31.42 Ω
[b] Re = 31.42‖270 = 28.14 Ω
ωloaded =Re
L= 5628 rad/s
·. . floaded =ωloaded
2π= 895.77 Hz
P 14.3 Note: add the resistor to the cirucit in Fig. 14.4(a).
[a] H(s) =Vo
Vi
=R
sL + R + Rl
=(R/L)
s + (R + Rl)/L
Problems 14–5
[b] H(jω) =(R/L)(
R+Rl
L
)+ jω
|H(jω)| =(R/L)√(
R+Rl
L
)2+ ω2
|H(jω)|max occurs when ω = 0
[c] |H(jω)|max =R
R + Rl
[d] |H(jωc)| =R√
2(R + Rl)=
R/L√(R+Rl
L
)2+ ω2
c
·. . ω2c =
(R + Rl
L
)2
; ·. . ωc = (R + Rl)/L
[e] Note – add 75 Ω resistor in series with the 10 mH inductor.
ωc =127 + 75
0.01= 20,200 rad/s
H(jω) =12,700
20,200 + jω
H(j0) = 0.6287
H(j20,200) =0.6287√
2/− 45 = 0.4446/− 45
H(j6060) =12,700
20,200 + j6060= 0.6022/− 16.70
H(j60,600) =12,700
20,200 + j60,600= 0.1988/− 71.57
P 14.4 [a] ωc =1
RC=
1(103)(100 × 10−9)
= 10 krad/s
fc =ωc
2π= 1591.55 Hz
[b] H(jω) =ωc
s + ωc
=10,000
s + 10,000
H(jω) =10,000
10,000 + jω
H(jωc) =10,000
10,000 + j10,000= 0.7071/− 45
14–6 CHAPTER 14. Introduction to Frequency-Selective Circuits
H(j0.1ωc) =10,000
10,000 + j1000= 0.9950/− 5.71
H(j10ωc) =10,000
10,000 + j100,000= 0.0995/− 84.29
[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(1/sC)RL + 1/sC
=RL
RLCs + 1
Then H(s) =Z
Z + R
=RL
RRLCs + R + RL
=(1/RC)
s +(
R + RL
RRLC
)
[b] |H(jω)| =(1/RC)√
ω2 + [(R + RL)/RRLC]2
|H(jω)| is maximum at ω = 0
[c] |H(jω)|max =RL
R + RL
[d] |H(jωc)| =RL√
2(R + RL)=
(1/RC)√ω2
c + [(R + RL)/RRLC]2
·. . ωc =R + RL
RRLC=
1RC
(1 + (R/RL))
[e] ωc =1
(103)(10−7)[1 + (103/104)] = 10,000(1 + 0.1) = 11,000 rad/s
H(j0) =10,00011,000
= 0.9091/0
Problems 14–7
H(jωc) =10,000
11,000 + j11,000= 0.6428/− 45
H(j0.1ωc) =10,000
11,000 + j1100= 0.9046/− 5.71
H(j10ωc) =10,000
11,000 + j110,000= 0.0905/− 84.29
P 14.6 [a] fc =ωc
2π=
50,0002π
=502π
× 103 = 7957.75 Hz
[b]1
RC= 50 × 103
R =1
(50 × 103)(0.5 × 10−6)= 40 Ω
[c] ωc =1
RC
(1 +
R
RL
)
·. .R
RL
= 0.05 ·. . RL = 20R = 800 Ω
[d] H(j0) =RL
R + RL
=800840
= 0.9524
P 14.7 [a]1
RC=
1(50 × 103)(5 × 10−9)
= 4000 rad/s
fc =40002π
= 636.62 Hz
[b] H(s) =s
s + ωc
·. . H(jω) =jω
4000 + jω
H(jωc) = H(j4000) =j4000
4000 + j4000= 0.7071/45
H(j0.2ωc) = H(j800) =j800
4000 + j800= 0.1961/78.69
H(j5ωc) = H(j20, 000) =j20,000
4000 + j20,000= 0.9806/11.31
[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
14–8 CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.8 [a] H(s) =Vo
Vi
=R
R + Rc + (1/sC)
=R
R + Rc
· s
[s + (1/(R + Rc)C)]
[b] H(jω) =R
R + Rc
· jω
jω + (1/(R + Rc)C)
|H(jω)| =R
R + Rc
· ω√ω2 + 1
(R+Rc)2C2
The magnitude will be maximum when ω = ∞.
[c] |H(jω)|max =R
R + Rc
[d] |H(jωc)| =Rωc
(R + Rc)√
ω2c + [1/(R + Rc)C]2
·. . |H(jω)| =R√
2(R + Rc)when
·. . ω2c =
1(R + Rc)2C2
or ωc =1
(R + Rc)C
[e] ωc =1
(62.5 × 103)(5 × 10−9)= 3200 rad/s
R
R + Rc
=50
62.5= 0.8
·. . H(jω) =0.8jω
3200 + jω
H(jωc) =(0.8)j3200
3200 + j3200= 0.5657/45
H(j0.2ωc) =(0.8)j640
3200 + j640= 0.1569/78.69
H(j5ωc) =(0.8)j16,000
3200 + j16,000= 0.7845/11.31
P 14.9 [a] ωc =1
RC= 2π(300) = 600π rad/s
·. . R =1
ωcC=
1(600π)(100 × 10−9)
= 5305.16 Ω
Problems 14–9
[b] Re = 5305.16‖47,000 = 4767.08 Ω
ωc =1
ReC=
1(4767.08)(100 × 10−9)
= 2097.7 rad/s
fc =ωc
2π=
2097.72π
= 333.86 Hz
P 14.10 [a] ωc =R
Lso R = ωcL = (25 × 103)(5 × 10−3) = 125 Ω
[b] ωc(loaded) =R
L· RL
R + RL
= 24,000 rad/s
·. .RL
R + RL
=ωc(loaded)
ωc(unloaded)=
24,00025,000
= 0.96
RL = 0.96(R + RL) ·. . 0.04RL = 0.96R = (0.96)(125)
·. . RL =(0.96)(125)
0.04= 3 kΩ
P 14.11 By definition Q = ωo/β therefore β = ωo/Q. Substituting this expression into Eqs.14.34 and 14.35 yields
ωc1 = − ωo
2Q+
√√√√( ωo
2Q
)2
+ ω2o
ωc2 =ωo
2Q+
√√√√( ωo
2Q
)2
+ ω2o
Now factor ωo out to get
ωc1 = ωo
− 1
2Q+
√√√√1 +(
12Q
)2
ωc2 = ωo
1
2Q+
√√√√1 +(
12Q
)2
P 14.12 ωo =√
ωc1ωc2 =√
(121)(100) = 110 krad/s
fo =ωo
2π= 17.51 kHz
β = 121 − 100 = 21 krad/s or 3.34 kHz
Q =ωo
β=
11021
= 5.24
14–10 CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.13 β =ωo
Q=
50,0004
= 12.5 krad/s;12,500
2π= 1.99 kHz
ωc2 = 50,000
1
8+
√1 +
(18
)2 = 56.64 krad/s
fc2 =56.64 k
2π= 9.01 kHz
ωc1 = 50,000
−1
8+
√1 +
(18
)2 = 44.14 krad/s
fc1 =44.14 k
2π= 7.02 kHz
P 14.14 [a] ω2o =
1LC
so L =1
[8000(2π)]2(5 × 10−9)= 79.16 mH
R =ωoL
Q=
8000(2π)(79.16 × 10−3)2
= 1.99 kΩ
[b] fc1 = 8000
−1
4+
√1 +
116
= 6.25 kHz
[c] fc2 = 8000
1
4+
√1 +
116
= 10.25 kHz
[d] β = fc2 − fc1 = 4 kHz
or
β =fo
Q=
80002
= 4 kHz
P 14.15 [a] ω2o =
1LC
=1
(10 × 10−3)(10 × 10−9)= 1010
ωo = 105 rad/s = 100 krad/s
[b] fo =ωo
2π=
105
2π= 15.92 kHz
[c] Q = ωoRC = (100 × 103)(8000)(10 × 10−9) = 8
[d] ωc1 = ωo
− 1
2Q+
√√√√1 +(
12Q
)2 = 105
− 1
16+
√1 +
1256
= 93.95 krad/s
[e] ·. . fc1 =ωc1
2π= 14.96 kHz
Problems 14–11
[f] ωc2 = ωo
1
2Q+
√√√√1 +(
12Q
)2 = 105
1
16+
√1 +
1256
= 106.45 krad/s
[g] ·. . fc2 =ωc2
2π= 16.94 kHz
[h] β =ωo
Q=
105
8= 12.5 krad/s or 1.99 kHz
P 14.16 [a] L =1
ω2oC
=1
(50 × 10−9)(20 × 103)2 = 50 mH
R =Q
ωoC=
5(20 × 103)(50 × 10−9)
= 5 kΩ
[b] ωc2 = ωo
1
2Q+
√√√√1 +(
12Q
)2 = 20,000
1
10+
√1 +
1100
= 22.10 krad/s ·. . fc2 =ωc2
2π= 3.52 kHz
ωc1 = ωo
− 1
2Q+
√√√√1 +(
12Q
)2 = 20,000
− 1
10+
√1 +
1100
= 18.10 krad/s ·. . fc1 =ωc1
2π= 2.88 kHz
[c] β =ωo
Q=
20,0005
= 4000 rad/s or 636.62 Hz
P 14.17 [a] ω2o =
1LC
=1
(40 × 10−3)(40 × 10−9)= 625 × 106
ωo = 25 × 103 rad/s = 25 krad/s
fo =25,000
2π= 3978.87 Hz
[b] Q =ωoL
R + Ri
=(25 × 103)(40 × 10−3)
200= 5
[c] ωc1 = ωo
− 1
2Q+
√√√√1 +(
12Q
)2 = 25,000
− 1
10+
√1 +
1100
= 22.62 krad/s or 3.60 kHz
[d] wc2 = ωo
1
2Q+
√√√√1 +(
12Q
)2 = 25,000
1
10+
√1 +
1100
= 27.62 krad/s or 4.40 kHz
14–12 CHAPTER 14. Introduction to Frequency-Selective Circuits
[e] β = ωc2 − ωc1 = 27.62 − 22.62 = 5 krad/sor
β =ωo
Q=
25,0005
= 5 krad/s or 795.77 Hz
P 14.18 [a] H(s) =(R/L)s
s2 + (R+Ri)L
s + 1LC
For the numerical values in Problem 14.17 we have
H(s) =4500s
s2 + 5000s + 625 × 106
·. . H(jω) =4500jω
(625 × 106 − ω2) + j5000ω
H(jωo) =j4500(25 × 103)j5000(25 × 103)
= 0.9/0
·. . 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)
(113.12 + j113.12) × 106 = 0.6364/45
·. . 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)
(−138.12 + j138.12) × 106 = 0.6364/− 45
·. . vo(t) = 500(0.6364) cos(27,620t − 45) = 318.2 cos(27,620t − 45) mV
P 14.19 [a]
[b] L =1
ω2oC
=1
(50 × 103)2(20 × 10−9)= 20 mH
R =ωoL
Q=
(50 × 103)(20 × 10−3)6.25
= 160 Ω
Problems 14–13
[c] Re = 160‖480 = 120 Ω
Re + Ri = 120 + 80 = 200 Ω
Qsystem =ωoL
Re + Ri
=(50 × 103)(20 × 10−3)
200= 5
[d] βsystem =ωo
Qsystem=
50 × 103
5= 10 krad/s
βsystem(Hz) =10,000
2π= 1591.55 Hz
P 14.20 [a]Vo
Vi
=Z
Z + Rwhere Z =
1Y
and Y = sC +1sL
+1
RL
=LCRLs2 + sL + RL
RLLs
H(s) =RLLs
RLRLCs2 + (R + RL)Ls + RRL
=(1/RC)s
s2 +[(
R+RL
RL
) (1
RC
)]s + 1
LC
=
(RL
R+RL
) (R+RL
RL
) (1
RC
)s
s2 +[(
R+RL
RL
) (1
RC
)]s + 1
LC
=Kβs
s2 + βs + ω2o
, K =RL
R + RL
[b] βL =(
R + RL
RL
) 1RC
[c] βU =1
RC
·. . βL =(
R + RL
RL
)βU =
(1 +
R
RL
)βU
[d] QL =ωo
β=
ωoRC(R+RL
RL
)[e] QU = ωoRC
·. . QL =(
RL
R + RL
)QU =
1[1 + (R/RL)]
QU
[f] H(jω) =Kjωβ
ω2o − ω2 + jωβ
H(jωo) = K
14–14 CHAPTER 14. Introduction to Frequency-Selective Circuits
Let ωc represent a cutoff frequency. Then
|H(jωc)| =K√2
=Kωcβ√
(ω2o − ω2
c )2 + ω2cβ
2
·. .1√2
=ωcβ√
(ω2o − ω2
c )2 + ω2cβ
2
Squaring both sides leads to
(ω2o − ω2
c )2 = ω2
cβ2 or (ω2
o − ω2c ) = ±ωcβ
·. . ω2c ± ωcβ − ω2
o = 0
or
ωc = ∓β
2±√
β2
4+ ω2
o
The two positive roots are
ωc1 = −β
2+
√β2
4+ ω2
o and ωc2 =β
2+
√β2
4+ ω2
o
where
β =(1 +
R
RL
) 1RC
and ω2o =
1LC
P 14.21 [a] ω2o =
1LC
=1
(5 × 10−3)(200 × 10−12)= 1012
ωo = 1 Mrad/s
[b] β =R + RL
RL
· 1RC
=(
500 × 103
400 × 103
)(1
(100 × 103)(200 × 10−12)
)= 62.5 krad/s
[c] Q =ωo
β=
106
62.5 × 103 = 16
[d] H(jωo) =RL
R + RL
= 0.8/0
·. . vo(t) = 250(0.8) cos(106t) = 200 cos 106t mV
[e] β =(1 +
R
RL
) 1RC
=(1 +
100RL
)(50 × 103) rad/s
ωo = 106 rad/s
Q =ωo
β=
201 + (100/RL)
where RL is in kilohms
Problems 14–15
[f]
P 14.22 ω2o =
1LC
=1
(2 × 10−6)(50 × 10−12)= 1016
ωo = 100 Mrad/s
Qu = ωoRC = (100 × 106)(2.4 × 103)(50 × 10−12) = 12
·. .(
RL
R + RL
)12 = 7.5; ·. . RL =
7.54.5
R = 4 kΩ
P 14.23 [a] In analyzing the circuit qualitatively we visualize vi as a sinusoidal voltage andwe seek the steady-state nature of the output voltage vo.At zero frequency the inductor provides a direct connection between the inputand the output, hence vo = vi when ω = 0.At infinite frequency the capacitor provides the direct connection, hencevo = vi when ω = ∞.At the resonant frequency of the parallel combination of L and C theimpedance of the combination is infinite and hence the output voltage will bezero when ω = ωo.At frequencies on either side of ωo the amplitude of the output voltage will benonzero 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 + 1/sC
=sL
s2LC + 1
Then
H(s) =Vo
Vi
=R
Z + R=
R(s2LC + 1)sL + R(s2LC + 1)
=[s2 + (1/LC)]
s2 +(
1RC
)s +
(1
LC
)
14–16 CHAPTER 14. Introduction to Frequency-Selective Circuits
H(s) =s2 + ω2
o
s2 + βs + ω2o
[c] From part (b) we have
H(jω) =ω2
o − ω2
ω2o − ω2 + jωβ
It follows that H(jω) = 0 when ω = ωo
·. . ωo =1√LC
[d] |H(jω)| =ω2
o − ω2√(ω2
o − ω2)2 + ω2β2
|H(jω)| =1√2
when ω2β2 = (ω2o − ω2)2
or ± ωβ = ω2o − ω2, thus
ω2 ± βω − ω2o = 0
The two positive roots of this quadratic are
ωc1 =−β
2+
√√√√(β
2
)2
+ ω2o
ωc2 =β
2+
√√√√(β
2
)2
+ ω2o
Also note that since β = ωo/Q
ωc1 = ωo
−1
2Q+
√√√√1 +(
12Q
)2
ωc2 = ωo
1
2Q+
√√√√1 +(
12Q
)2
[e] It follows from the equations derived in part (b) that
β = 1/RC
[f] By definition Q = ωo/β = ωoRC
Problems 14–17
P 14.24 [a] ω2o =
1LC
=1
(50 × 10−6)(20 × 10−9)= 1012
·. . ωo = 1 Mrad/s
[b] fo =ωo
2π= 159.15 kHz
[c] Q = ωoRC = (106)(750)(20 × 10−9) = 15
[d] ωc1 = ωo
− 1
2Q+
√√√√1 +(
12Q
)2 = 106
− 1
30+
√1 +
1900
= 967.22 krad/s
[e] fc1 =ωc1
2π= 153.94 kHz
[f] ωc2 = ωo
1
2Q+
√√√√1 +(
12Q
)2 = 106
1
30+
√1 +
1900
= 1.03 Mrad/s
[g] fc2 =ωc2
2π= 164.55 kHz
[h] β = fc2 − fc1 = 10.61 kHz
P 14.25 [a] ωo = 2πfo = 8π krad/s
L =1
ω2oC
=1
(8000π)2(0.5 × 10−6)= 3.17 mH
R =Q
ωoC=
5(8000π)(0.5 × 10−6)
= 397.89 Ω
[b] fc2 = fo
1
2Q+
√√√√1 +(
12Q
)2 = 4000
1
10+
√1 +
1100
= 4.42 kHz
fc1 = fo
− 1
2Q+
√√√√1 +(
12Q
)2 = 4000
− 1
10+
√1 +
1100
= 3.62 kHz
[c] β = fc2 − fc1 = 800 Hz
or
β =fo
Q=
40005
= 800 Hz
14–18 CHAPTER 14. Introduction to Frequency-Selective Circuits
P 14.26 [a] Re = 397.89‖1000 = 284.63 Ω
Q = ωoReC = (8000π)(284.63)(0.5 × 10−6) = 3.58
[b] β =fo
Q=
40003.58
= 1.12 kHz
[c] fc2 = 4000
1
7.15+
√1 +
17.152
= 4.60 kHz
[d] fc1 = 4000
− 1
7.15+
√1 +
17.152
= 3.48 kHz
P 14.27 [a] Let Z =RL(sL + (1/sC))RL + sL + (1/sC)
Z =RL(s2LC + 1)
s2LC + RLCs + 1
Then H(s) =Vo
Vi
=s2RLCL + RL
(R + RL)LCs2 + RRLCs + R + RL
Therefore
H(s) =(
RL
R + RL
)· [s2 + (1/LC)][
s2 +(
RRL
R+RL
)sL
+ 1LC
]
=K(s2 + ω2
o)s2 + βs + ω2
o
where K =RL
R + RL
; ω2o =
1LC
; β =(
RRL
R + RL
) 1L
[b] ωo =1√LC
[c] β =(
RRL
R + RL
) 1L
[d] Q =ωo
β=
ωoL
[RRL/(R + RL)]
[e] H(jω) =K(ω2
o − ω2)(ω2
o − ω2) + jβω
H(jωo) = 0
[f] H(j0) =Kω2
o
ω2o
= K
Problems 14–19
[g] H(jω) =K[(ωo/ω)2 − 1
][
(ωo/ω)2 − 1]+ jβ/ω
H(j∞) =−K
−1= K
[h] H(jω) =K(ω2
o − ω2)(ω2
o − ω2) + jβω
H(j0) = H(j∞) = K
Let ωc represent a corner frequency. Then
|H(jωc)| =K√2
·. .K√2
=K(ω2
o − ω2c )√
(ω2o − ω2
c )2 + ω2cβ
2
Squaring both sides leads to
(ω2o − ω2
c )2 = ω2
cβ2 or (ω2
o − ω2c ) = ±ωcβ
·. . ω2c ± ωcβ − ω2
o = 0
or
ωc = ∓β
2±√
β2
4+ ω2
o
The two positive roots are
ωc1 = −β
2+
√β2
4+ ω2
o and ωc2 =β
2+
√β2
4+ ω2
o
where
β =RRL
R + RL
· 1L
and ω2o =
1LC
P 14.28 [a] ω2o =
1LC
=1
(10−6)(4 × 10−12)= 0.25 × 1018 = 25 × 1016
ωo = 5 × 108 = 500 Mrad/s
β =RRL
R + RL
· 1L
=(30)(150)
180· 110−6 = 25 Mrad/s = 3.98 MHz
Q =ωo
β=
500 M25 M
= 20
14–20 CHAPTER 14. Introduction to Frequency-Selective Circuits
[b] H(j0) =RL
R + RL
=150180
= 0.8333
H(j∞) =RL
R + RL
= 0.8333
[c] fc2 =250π
1
40+
√1 +
11600
= 81.59 MHz
fc2 =250π
− 1
40+
√1 +
11600
= 77.61 MHz
Check: β = fc2 − fc1 = 3.98 MHz.
[d] Q =ωo
β=
500 × 106
RRL
R+RL· 1
L
=500(R + RL)
RRL
=503
(1 +
30RL
)where RL is in ohms.
[e]
P 14.29 [a] ω2o =
1LC
= 1012
·. . L =1
(1012)(400 × 10−12)= 2.5 mH
RL
R + RL
= 0.96; ·. . 0.04RL = 0.96R
·. . RL = 24R ·. . R =36,000
24= 1.5 kΩ
[b] β =(
RL
R + RL
)R · 1
L= 576 × 103
Q =ωo
β=
106
576 × 103 = 1.74
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 − 2ζ2 = 132,287.57 rad/s
·. . ω = 0
ω = 132,287.57 rad/s
[b] ωp = ωn
√1 − 2ζ2 = 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
Thus β = ωc2 − ωc1 = 1533.10 rad/s
Calculate inductance using Eq. (14.32) and capacitance using Eq. (14.31):
L =R
β=
6001533.10
= 0.39 H
C =1
Lωc1ωc2=
1(0.39)(4379.38)(5912.48)
= 0.10 µF
[b] At the outermost two frequencies in the low-frequency group (687 Hz and 941Hz) the amplitudes are
|V697 Hz| = |V941 Hz| =|Vpeak|√
2= 0.707|Vpeak|
because these are cutoff frequencies. We calculate the amplitudes at the othertwo low frequencies using Eq. (14.32):
|V | = (|Vpeak|)(|H(jω)|) = |Vpeak| ωβ√(ω2
o − ω2)2 + (ωβ)2
Therefore
|V770 Hz| = |Vpeak| (4838.05)(1533.10)√(5088.522 − 4838.052)2 + [(4838.05)(1533.10)]2
= 0.948|Vpeak|and
|V852 Hz| = |Vpeak| (5353.27)(1533.10)√(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 frequenciesin both bands of the DTMF system were carefully chosen to produce this typeof predictable behavior with linear filters. In other words, the frequencies werechosen to be equally far apart with respect to the response produced by a linearfilter. 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 theDTMF tones remind us of musical notes! Unlike musical scales, DTMFfrequencies were selected to be harmonically unrelated, to lower the risk ofmisidentifying 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. Theamplitude of a tone with this frequency is
|V1209 Hz| = |Vpeak| =(7596.37)(1533.10)√
(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 cutofffrequencies, 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 andcapacitance values are
L =R
β=
6002664
= 0.225 H
C =1
ωc1ωc2L= 0.057 µF
At the highest of the low-band frequencies, 941 Hz, the amplitude is
|Vω| = |Vpeak| ωβ√(ω2
o − ω2)2 + ω2β2
where ωo = √ωc1ωc2 . Thus,
|Vω| =|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 thelow-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 is0.948|Vpeak|. The response to the 20 Hz tone is
|V20 Hz| =|Vpeak|(125.6)(1533)
[(50892 − 125.62)2 + [(125.6)(1533)]2]1/2
= 0.00744|Vpeak|
·. .0.00744|Vring-peak|0.948|VDTMF-peak| = 0.5
·. . |Vring-peak| = 63.7|VDTMF-peak|
Thus, the 20 Hz signal can be 63.7 times as large as the DTMF tones.
15Active Filter Circuits
Assessment Problems
AP 15.1 H(s) =−(R2/R1)ss + (1/R1C)
1R1C
= 1 rad/s; R1 = 1 Ω, ·. . C = 1 F
R2
R1= 1, ·. . R2 = R1 = 1 Ω
·. . Hprototype(s) =−s
s + 1
AP 15.2 H(s) =−(1/R1C)
s + (1/R2C)=
−20,000s + 5000
1R1C
= 20,000; C = 5 µF
·. . R1 =1
(20,000)(5 × 10−6)= 10 Ω
1R2C
= 5000
·. . R2 =1
(5000)(5 × 10−6)= 40 Ω
15–1
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
kfkm
·. . 0.5 × 10−6 =1
kfkm
·. . km =1
(0.5 × 10−6)(62,831.85)= 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
s2 +(
2R2C
)s +
(1
R1R2C2
)
Equate the transfer functions. For C = 1F,
2R2C
=√
2, ·. . R2 =√
2 = 1.414 Ω
1R1R2C2 = 1, ·. . R1 =
1√2
= 0.707 Ω
AP 15.5 Q = 8, K = 5, ωo = 1000 rad/s, C = 1 µF
For the circuit in Fig 15.26
H(s) =−( 1
R1C
)s
s2 +( 2
R3C
)s +
(R1 + R2
R1R2R3C2
)
=Kβs
s2 + βs + ω2o
β =2
R3C, ·. . R3 =
2βC
β =ωo
Q=
10008
= 125 rad/s
Problems 15–3
·. . R3 =2 × 106
(125)(1)= 16 kΩ
Kβ =1
R1C
·. . R1 =1
KβC=
15(125)(1 × 10−6)
= 1.6 kΩ
ω2o =
R1 + R2
R1R2R3C2
106 =(1600 + R2)
(1600)(R2)(16,000)(10−6)2
Solving for R2,
R2 =(1600 + R2)106
256 × 105 , 246R2 = 16,000, R2 = 65.04 Ω
AP 15.6 ωo = 1000 rad/s; Q = 4;
C = 2 µF
H(s) =s2 + (1/R2C2)
s2 +[4(1 − σ)
RC
]s +
( 1R2C2
)
=s2 + ω2
o
s2 + βs + ω2o
; ωo =1
RC; β =
4(1 − σ)RC
R =1
ωoC=
1(1000)(2 × 10−6)
= 500 Ω
β =ωo
Q=
10004
= 250
·. .4(1 − σ)
RC= 250
4(1 − σ) = 250RC = 250(500)(2 × 10−6) = 0.25
1 − σ =0.254
= 0.0625; ·. . σ = 0.9375
15–4 CHAPTER 15. Active Filter Circuits
Problems
P 15.1 Summing the currents at the inverting input node yields
0 − Vi
Zi
+0 − Vo
Zf
= 0
·. .Vo
Zf
= −Vi
Zi
·. . H(s) =Vo
Vi
= −Zf
Zi
P 15.2 [a] Zf =R2(1/sC2)
[R2 + (1/sC2)]=
R2
R2C2s + 1
=(1/C2)
s + (1/R2C2)Likewise
Zi =(1/C1)
s + (1/R1C1)
·. . H(s) =−(1/C2)[s + (1/R1C1)][s + (1/R2C2)](1/C1)
= −C1
C2
[s + (1/R1C1)][s + (1/R2C2)]
[b] H(jω) =−C1
C2
[jω + (1/R1C1)jω + (1/R2C2)
]
H(j0) =−C1
C2
(R2C2
R1C1
)=
−R2
R1
[c] H(j∞) = −C1
C2
(j
j
)=
−C1
C2
[d] As ω → 0 the two capacitor branches become open and the circuit reduces to aresistive inverting amplifier having a gain of −R2/R1.As ω → ∞ the two capacitor branches approach a short circuit and in this casewe encounter an indeterminate situation; namely vn → vi but vn = 0 becauseof the ideal op amp. At the same time the gain of the ideal op amp is infinite sowe have the indeterminate form 0 · ∞. Although ω = ∞ is indeterminate wecan reason that for finite large values of ω H(jω) will approach −C1/C2 invalue. In other words, the circuit approaches a purely capacitive invertingamplifier with a gain of (−1/jωC2)/(1/jωC1) or −C1/C2.
Problems 15–5
P 15.3 [a] Zf =(1/C2)
s + (1/R2C2)
Zi = R1 +1
sC1=
R1
s[s + (1/R1C1)]
H(s) = − (1/C2)[s + (1/R2C2)]
· s
R1[s + (1/R1C1)]
= − 1R1C2
s
[s + (1/R1C1)][s + (1/R2C2)]
[b] H(jω) = − 1R1C2
jω(jω + 1
R1C1
) (jω + 1
R2C2
)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 andtherefore vo = 0.
P 15.4 [a] K = 10(10/20) = 3.16 =R2
R1
R2 =1
ωcC=
1(2π)(103)(750 × 10−9)
= 212.21 Ω
R1 =R2
K=
212.213.16
= 67.11 Ω
[b]
P 15.5 [a] R1 =1
ωcC=
1(2π)(8 × 103)(3.9 × 10−9)
= 5.10 kΩ
K = 10(14/20) = 5.01 =R2
R1
·. . R2 = 5.01R1 = 25.57 kΩ
15–6 CHAPTER 15. Active Filter Circuits
[b]
P 15.6 For the RC circuit
H(s) =Vo
Vi
=(1/RC)
s + (1/RC)
R′ = kmR; C ′ =C
kmkf
·. . R′C ′ = kmRC
kmkf
=1kf
RC =1kf
1R′C ′ = kf
H ′(s) =(1/R′C ′)
s + (1/R′C ′)=
kf
s + kf
H ′(s) =1
(s/kf ) + 1
For the RL circuit
H(s) =V0
Vi
=R/L
s + (R/L)
R′ = kmR; L′ =km
kf
L
R′
L′ =kmRkm
kfL
= kf
(R
L
)= kf
H ′(s) =(R′/L′)
s + (R′/L′)=
kf
s + kf
H ′(s) =1
(s/kf ) + 1
Problems 15–7
P 15.7 For the RC circuit
H(s) =Vo
Vi
=s
s + (1/RC)
R′ = kmR; C ′ =C
kmkf
·. . R′C ′ =RC
kf
=1kf
;1
R′C ′ = kf
H ′(s) =s
s + (1/R′C ′)=
s
s + kf
=(s/kf )
(s/kf ) + 1
For the RL circuit
H(s) =s
s + (R/L)
R′ = kmR; L′ =kmL
kf
R′
L′ = kf
(R
L
)= kf
H ′(s) =s
s + (R′/L′)=
s
s + kf
=(s/kf )
(s/kf ) + 1
P 15.8 H(s) =(R/L)s
s2 + (R/L)s + (1/LC)=
βs
s2βs + ω2o
For the prototype circuit ωo = 1 and β = ωo/Q = 1/Q.For the scaled circuit
H ′(s) =(R′/L′)s
s2 + (R′/L′)s + (1/L′C ′)
where R′ = kmR; L′ =km
kf
L; and C ′ =C
kfkm
·. .R′
L′ =kmRkm
kfL
= kf
(R
L
)= kfβ
1L′C ′ =
kfkm
km
kfLC
=k2
f
LC= k2
f
15–8 CHAPTER 15. Active Filter Circuits
Q′ =ω′
o
β′ =kfωo
kfβ= Q
therefore the Q of the scaled circuit is the same as the Q of the unscaled circuit.Also note β′ = kfβ.
·. . H ′(s) =
(kf
Q
)s
s2 +(
kf
Q
)s + k2
f
H ′(s) =
(1Q
) (skf
)[(
skf
)2+ 1
Q
(skf
)+ 1
]
P 15.9 [a] L = 1 H; C = 1 F
R =1Q
=120
= 0.05 Ω
[b] kf =ω′
o
ωo
= 40,000; km =R′
R=
50000.05
= 100,000
Thus,
R′ = kmR = (0.05)(100,000) = 5 kΩ
L′ =km
kf
L =100,00040,000
(1) = 2.5 H
C ′ =C
kmkf
=1
(40,000)(100,000)= 250 pF
[c]
P 15.10 [a] Since ω2o = 1/LC and ωo = 1 rad/s,
C =1L
=1Q
[b] H(s) =(R/L)s
s2 + (R/L)s + (1/LC)
H(s) =(1/Q)s
s2 + (1/Q)s + 1
Problems 15–9
[c] In the prototype circuit
R = 1 Ω; L = 16 H; C =1L
= 0.0625 F
·. . km =R′
R= 10,000; kf =
ω′o
ωo
= 25,000
Thus
R′ = kmR = 10 kΩ
L′ =km
kf
L =10,00025,000
(16) = 6.4 H
C ′ =C
kmkf
=0.0625
(10,000)(25,000)= 250 pF
[d]
[e] H ′(s) =116
(s
25,000
)(
s25,000
)2+ 1
16
(s
25,000
)+ 1
H ′(s) =1562.5s
s2 + 1562.5s + 625 × 106
P 15.11 [a] Using the first prototype
ωo = 1 rad/s; C = 1 F; L = 1 H; R = 25 Ω
km =R′
R=
40,00025
= 1600; kf =ω′
o
ωo
= 50,000
Thus,
R′ = kmR = 40 kΩ; L′ =km
kf
L =1600
50,000(1) = 32 mH;
C ′ =C
kmkf
=1
(1600)(50,000)= 12.5 nF
Using the second prototype
ωo = 1 rad/s; C = 25 F
L =125
= 40 mH; R = 1 Ω
15–10 CHAPTER 15. Active Filter Circuits
km =R′
R= 40,000; kf =
ω′o
ωo
= 50,000
Thus,
R′ = kmR = 40 kΩ; L′ =km
kf
L =40,00050,000
(0.04) = 32 mH;
C ′ =C
kmkf
=25
(40,000)(50,000)= 12.5 nF
[b]
P 15.12 For the scaled circuit
H ′(s) =s2 +
(1
L′C′
)s2 +
(R′L′
)s +
(1
L′C′
)
L′ =km
kf
L; C ′ =C
kmkf
·. .1
L′C ′ =k2
f
LC; R′ = kmR
·. .R′
L′ = kf
(R
L
)
It follows then that
H ′(s) =s2 +
(k2
f
LC
)
s2 +(
RL
)kfs +
k2f
LC
=
(skf
)2+(
1LC
)[(
skf
)2+(
RL
) (skf
)+(
1LC
)]
= H(s)|s=s/kf
P 15.13 For the circuit in Fig. 15.31
H(s) =s2 +
(1
LC
)s2 + s
RC+(
1LC
)
Problems 15–11
It follows that
H ′(s) =s2 + 1
L′C′
s2 + sR′C′ + 1
L′C′
where R′ = kmR; L′ =km
kf
L;
C ′ =C
kmkf
·. .1
L′C ′ =k2
f
LC
1R′C ′ =
kf
RC
H ′(s) =s2 +
(k2
f
LC
)
s2 +(
kf
RC
)s +
k2f
LC
=
(skf
)2+ 1
LC(skf
)2+(
1RC
) (skf
)+ 1
LC
= H(s)|s=s/kf
P 15.14 [a] For the circuit in Fig. P15.14(a)
H(s) =Vo
Vi
=s +
1s
1Q
+ s +1s
=s2 + 1
s2 +(
1Q
)s + 1
For the circuit in Fig. P15.14(b)
H(s) =Vo
Vi
=Qs + Q
s
1 + Qs + Qs
=Q(s2 + 1)
Qs2 + s + Q
H(s) =s2 + 1
s2 +(
1Q
)s + 1
15–12 CHAPTER 15. Active Filter Circuits
[b] k=f
ω′o
ωo
= 104; Q = 8;
Replace s with s/kf .
H ′(s) =
(s
104
)2+ 1(
s104
)2+ 1
8
(s
104
)+ 1
=s2 + 108
s2 + 1250s + 108
P 15.15 For prototype circuit (a):
H(s) =Vo
Vi
=Q
Q + 1s+ 1
s
=Q
Q + ss2+1
H(s) =Q(s2 + 1)
Q(s2 + 1) + s=
s2 + 1
s2 +(
1Q
)s + 1
For prototype circuit (b):
H(s) =Vo
Vi
=1
1 + (s/Q)(s2+1)
=s2 + 1
s2 +(
1Q
)s + 1
P 15.16 From the solution to Problem 14.15, ωo = 100 krad/s and β = 12.5 krad/s. Computethe two scale factors:
kf =ω′
o
ωo
=2π(200 × 103)
100 × 103 = 4π
km =1kf
C
C ′ =14π
10 × 10−9
2.5 × 10−9 =1π
Thus,
R′ = kmR =8000
π= 2546.48 Ω L′ =
km
kf
L =1/π4π
(10 × 10−3) = 253.303 µH
Calculate the cutoff frequencies:
ω′c1 = kfωc1 = 4π(93.95 × 103) = 1180.6 krad/s
ω′c2 = kfωc2 = 4π(106.45 × 103) = 1337.7 krad/s
To check, calculate the bandwidth:
β′ = ω′c2 − ω′
c1 = 157.1 krad/s = 4πβ (Checks!)
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 =ω′
o
ωo
=50 × 103
106 = 0.05
km =kfL
′
L=
0.05(200 × 10−6)50 × 10−6 = 0.2
Thus,
R′ = kmR = (0.2)(750) = 150 Ω C ′ =C
kmkf
=20 × 10−9
(0.2)(0.05)= 2 µF
Calculate the bandwidth:
β′ = kfβ = (0.05)[2π(10.61 × 103)] = 3333 rad/s
To check, calculate the quality factor:
Q =ωo
β=
106
2π(10.61 × 103)= 15
Q′ =ω′
o
β′ =50 × 103
3333= 15 (Checks)
P 15.18 [a] km =R′
R=
10001
= 1000; kf =C
kmC ′ =1
(1000)(200 × 10−9)= 5000
L′ =km
kf
(L) =10005000
(1) = 200 mH
[b]V − 10/s
1000+
V
0.2s+
V
1000 + (5 × 106/s)= 0
V( 1
1000+
5s
+s
1000s + 5 × 106
)=
1100s
V =10(s + 5000)
2s2 + 10,000s + 25 × 106 =5(s + 5000)
s2 + 5000s + 12.5 × 106
15–14 CHAPTER 15. Active Filter Circuits
Io =V
0.2s=
25(s + 5000)s(s2 + 5000s + 12.5 × 106)
=K1
s+
K2
s + 2500 − j2500+
K∗2
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 thecurrent is reduced by a factor of 1000. Since kf = 5000 the coefficients of t aremultiplied by 5000.
P 15.19 km =R′
R=
500050
= 100; kf =ω′
o
ωo
= 5000
C ′ =C
kmkf
=4 × 10−3
(100)(5000)= 8 nF
50 Ω → 5 kΩ; 700 Ω → 70 kΩ
L′ =km
kf
L =1005000
(20) = 0.4 H
0.05vφ → 0.05100
vφ = 5 × 10−4vφ
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,
io(t) = 17.28 + 28.80e−105t cos(75,000t + 126.87) mA
Problems 15–15
P 15.20 [a] From Eq 15.1 we have
H(s) =−Kωc
s + ωc
where K =R2
R1, ωc =
1R2C
·. . H ′(s) =−K ′ω′
c
s + ω′c
where K ′ =R′
2
R′1
ω′c =
1R′
2C′
By hypothesis R′1 = kmR1; R′
2 = kmR2,
and C ′ =C
kfkm
. It follows that
K ′ = K and ω′c = kfωc, therefore
H ′(s) =−Kkfωc
s + kfωc
=−Kωc(skf
)+ ωc
[b] H(s) =−K
(s + 1)
[c] H ′(s) =−K(skf
)+ 1
=−Kkf
s + kf
P 15.21 [a] From Eq. 15.4
H(s) =−Ks
s + ωc
where K =R2
R1and
ωc =1
R1C
·. . H ′(s) =−K ′ss + ω′
c
where K ′ =R′
2
R′1
and ω′c =
1R′
1C′
By hypothesis
R′1 = kmR1; R′
2 = kmR2; C ′ =C
kmkf
It follows that
K ′ = K and ω′c = kfωc
·. . H ′(s) =−Ks
s + kfωc
=−K(s/kf )(
skf
)+ ωc
15–16 CHAPTER 15. Active Filter Circuits
[b] H(s) =−Ks
(s + 1)
[c] H ′(s) =−K(s/kf )(
skf
+ 1) =
−Ks
(s + kf )
P 15.22 [a] Hhp =−s
s + 1; kf =
ω′o
ω=
1000(2π)1
= 2000π
·. . H ′hp =
−s
s + 2000π1
RHCH
= 2000π; ·. . RH =1
(2000π)(0.1 × 10−6)= 1.59 kΩ
Hlp =−1
s + 1; kf =
ω′o
ω=
5000(2π)1
= 10,000π
·. . H ′lp =
−10,000πs + 10,000π
1RLCL
= 10,000π; ·. . RL =1
(10,000π)(0.1 × 10−6)= 318.3 Ω
[b] H ′(s) =−s
s + 2000π· −10,000πs + 10,000π
=10,000πs
(s + 2000π)(s + 10,000π)
[c] ωo =√
ωc1ωc1 =√
(2000π)(10,000π) = 1000π√
20 rad/s
H ′(jωo) =(10,000π)(j1000π
√20)
(2000π + j1000π√
20)(10,000π + j1000π√
20)
=j10
√20
(2 + j√
20)(10 + j√
20)= 0.8333/0
Problems 15–17
[d] G = 20 log10(0.8333) = −1.58 dB
[e]
P 15.23 [a] For the high-pass section:
kf =ω′
o
ω=
4000(2π)1
= 8000π
H ′(s) =−s
s + 8000π
·. .1
R1(10 × 10−9)= 8000π; R1 = 3.98 kΩ ·. . R2 = 3.98 kΩ
For the low-pass section:
kf =ω′
o
ω=
400(2π)1
= 800π
H ′(s) =−800π
s + 800π
·. .1
R2(10 × 10−9)= 800π; R2 = 39.8 kΩ ·. . R1 = 39.8 kΩ
0 dB gain corresponds to K = 1. In the summing amplifier we are free tochoose Rf and Ri so long as Rf/Ri = 1. To keep from having many differentresistance values in the circuit we opt for Rf = Ri = 39.8 kΩ.
15–18 CHAPTER 15. Active Filter Circuits
[b]
[c] H ′(s) =s
s + 8000π+
800πs + 800π
=s2 + 1600πs + 64 × 105π2
(s + 800π)(s + 8000π)
[d] ωo =√
(8000π)(800π) = 800π√
10
H ′(j800π√
10) =−(800π
√10)2 + 1600π(j800π
√10) + 64 × 105π2
(800π + j800π√
10)(8000π + j800π√
10)
=j128 × 104
√10π2
(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/sC)
R + (1/sC)=
(1/RC)s + (1/RC)
H(jω) =(1/RC)
jω + (1/RC)
|H(jω)| =(1/RC)√
ω2 + (1/RC)2
|H(jω)|2 =(1/RC)2
ω2 + (1/RC)2
[b] Let Va be the voltage across the capacitor, positive at the upper terminal. Then
Va − Vi
R1+ sCVa +
Va
R2 + sL= 0
Solving for Va yields
Va =(R2 + sL)Vi
R1LCs2 + (R1R2C + L)s + (R1 + R2)
But
Vo =sLVa
R2 + sL
Therefore
Vo =sLVi
R1LCs2 + (L + R1R2C)s + (R1 + R2)
H(s) =sL
R1LCs2 + (L + R1R2C)s + (R1 + R2)
H(jω) =jωL
[(R1 + R2) − R1LCω2] + jω(L + R1R2C)
15–20 CHAPTER 15. Active Filter Circuits
|H(jω)| =ωL√
[R1 + R2 − R1LCω2]2 + ω2(L + R1R2C)2
|H(jω)|2 =ω2L2
(R1 + R2 − R1LCω2)2 + ω2(L + R1R2C)2
=ω2L2
R21L
2C2ω4 + (L2 + R21R
22C
2 − 2R21LC + 2R1R2LC)ω2 + (R1 + R2)2
[c] Let Va be the voltage across R2 positive at the upper terminal. Then
Va − Vi
R1+
Va
R2+ VasC + VasC = 0
(0 − Va)sC + (0 − Va)sC +0 − Vo
R3= 0
·. . Va =R2Vi
2R1R2Cs + R1 + R2
and Va = − Vo
2R3Cs
It follows directly that
H(s) =Vo
Vi
=−2R2R3Cs
2R1R2Cs + (R1 + R2)
·. . H(jω) =−2R2R3C(jω)
(R1 + R2) + jω(2R1R2C)
|H(jω)| =2R2R3Cω√
(R1 + R2)2 + ω24R21R
22C
2
|H(jω)|2 =4R2
2R23C
2ω2
(R1 + R2)2 + 4R21R
22C
2ω2
P 15.25 ωo = 2πfo = 400π rad/s
β = 2π(1000) = 2000π rad/s
·. . ωc2 − ωc1 = 2000π
√ωc1ωc2 = ωo = 400π
Solve for the cutoff frequencies:
ωc1ωc2 = 16 × 104π2
Problems 15–21
ωc2 =16 × 104π2
ωc1
·. .16 × 104π2
ωc1
− ωc1 = 2000π
or ω2c1
+ 2000πωc1 − 16 × 104π2 = 0
ωc1 = −1000π ±√
106π2 + 0.16 × 106π2
ωc1 = 1000π(−1 ±√
1.16) = 242.01 rad/s
·. . ωc2 = 2000π + 242.01 = 6525.19 rad/s
Thus, fc1 = 38.52 Hz and fc2 = 1038.52 Hz
Check: β = fc2 − fc1 = 1000Hz
ωc2 =1
RLCL
= 6525.19
RL =1
(6525.19)(5 × 10−6)= 30.65 Ω
ωc1 =1
RHCH
= 242.01
RH =1
(242.01)(5 × 10−6)= 826.43 Ω
P 15.26 ωo = 1000 rad/s; GAIN = 6
β = 4000 rad/s; C = 0.2 µF
β = ωc2 − ωc1 = 4000
ωo =√
ωc1ωc2 = 1000
Solve for the cutoff frequencies:
·. . ω2c1
+ 4000ωc1 − 106 = 0
ωc1 = −2000 ± 1000√
5 = 236.07 rad/s
15–22 CHAPTER 15. Active Filter Circuits
ωc2 = 4000 + ωc1 = 4236.07 rad/s
Check: β = ωc2 − ωc1 = 4000 rad/s
ωc1 =1
RLCL
·. . RL =1
(0.2 × 10−6)(236.07)= 21.81 kΩ
1RHCH
= 4236.07
RH =1
(0.2 × 10−6)(4236.07)= 1.18 kΩ
Rf
Ri
= 6
If Ri = 1 kΩ Rf = 6Ri = 6 kΩ
P 15.27 [a] y = 20 log101√
1 + ω2n= −10 log10(1 + ω2n)
From the laws of logarithms we have
y =(−10
ln 10
)ln(1 + ω2n)
Thus
dy
dω=(−10
ln 10
) 2nω2n−1
(1 + ω2n)
x = log10 ω =ln ω
ln 10·. . ln ω = x ln 10
1ω
dω
dx= ln 10,
dω
dx= ω ln 10
dy
dx=(
dy
dω
)(dω
dx
)=
−20nω2n
1 + ω2ndB/decade
at ω = ωc = 1 rad/s
dy
dx= −10n dB/decade.
Problems 15–23
[b] y = 20 log101
[√
1 + ω2]n= −10n log10(1 + ω2)
=−10nln 10
ln(1 + ω2)
dy
dω=
−10nln 10
( 11 + ω2
)2ω =
−20nω
(ln 10)(1 + ω2)
As before
dω
dx= ω(ln 10); ·. .
dy
dx=
−20nω2
(1 + ω2)
At the corner ωc =√
21/n − 1 ·. . ω2c = 21/n − 1
dy
dx=
−20n[21/n − 1]21/n
dB/decade.
[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 amplitudecharacteristic at the cutoff frequency decreases at a much faster rate for theButterworth filter.Hence the transition region of the Butterworth filter will be much narrowerthan that of the cascaded sections.
P 15.28 [a] n ∼= (−0.05)(−30)log10(7000/2000)
∼= 2.76
·. . n = 3
[b] Gain = 20 log101√
1 + (7000/2000)6= −32.65 dB
P 15.29 [a] For the scaled circuit
H ′(s) =1/(R′)2C ′
1C′2
s2 + 2R′C′
1s + 1
(R′)2C′1C′
2
where
R′ = kmR; C ′1 = C1/kfkm; C ′
2 = C2/kfkm
15–24 CHAPTER 15. Active Filter Circuits
It follows that
1(R′)2C ′
1C′2
=k2
f
R2C1C2
2R′C ′
1=
2kf
RC1
·. . H ′(s) =k2
f/R2C1C2
s2 + 2kf
RC1s +
k2f
R2C1C2
=1/R2C1C2(
skf
)2+ 2
RC1
(skf
)+ 1
R2C1C2
P 15.30 [a] H(s) =1
(s + 1)(s2 + s + 1)[b] fc = 2000 Hz; ωc = 4000π rad/s; kf = 4000π
H ′(s) =1
( skf
+ 1)[( skf
)2 + skf
+ 1]
=k3
f
(s + kf )(s2 + kfs + k2f )
=(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 =R′
R=
10001
= 1000; kf =ω′
o
ωo
=2π(2000)
1= 4000π
R′ = kmR = 1000 Ω; C ′ =C
kmkf
=1
(1000)(4000π)= 79.58 nF
In the second-order circuit R = 1 Ω, 2/C1 = 1 so C1 = 2 F, andC2 = 1/C1 = 0.5 F. Therefore in the scaled second-order circuit
R′ = kmR = 1000 Ω; C ′1 =
C1
kmkf
=2
(1000)(4000π)= 159.15 nF
C ′2 =
C2
kmkf
=0.5
(1000)(4000π)= 39.79 nF
Problems 15–25
[b]
P 15.32 [a] n =(−0.05)(−48)
log10(2000/500)= 3.99 ·. . n = 4
From Table 15.1 the transfer function of the first section is
H1(s) =s2
s2 + 0.765s + 1For the prototype circuit
2R2
= 0.765; R2 = 2.61 Ω; R1 =1R2
= 0.383 Ω
The transfer function of the second section is
H2(s) =s2
s2 + 1.848s + 1For the prototype circuit
2R2
= 1.848; R2 = 1.082 Ω; R1 =1R2
= 0.9240 Ω
The scaling factors are:
kf =ω′
o
ωo
=2π(2000)
1= 4000π
C ′ =C
kmkf
·. . 10 × 10−9 =1
4000πkm
·. . km =1
4000π(10 × 10−9)= 7957.75
Therefore in the first section
R′1 = kmR1 = 3.04 kΩ; R′
2 = kmR2 = 20.80 kΩ
In the second section
R′1 = kmR1 = 7.35 kΩ; R′
2 = kmR2 = 8.61 kΩ
15–26 CHAPTER 15. Active Filter Circuits
[b]
P 15.33 n = 5: 1 + (−1)5s10 = 0; s10 = 1
s10 = 1/0 + 36k
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 15–27
n = 6: 1 + (−1)6s12 = 0 s12 = −1
s12 = 1/15 + 36k
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) =s2
s2 + 2kmR2(C/kmkf )s + 1
kmR1kmR2(C2/k2mk2
f)
H ′(s) =s2
s2 + 2kf
R2Cs +
k2f
R1R2C2
=(s/kf )2
(s/kf )2 + 2R2C
(skf
)+ 1
R1R2C2
15–28 CHAPTER 15. Active Filter Circuits
P 15.35 [a] n =(−0.05)(−48)log10(32/8)
= 3.99 ·. . n = 4
From Table 15.1 the transfer function is
H(s) =1
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
The capacitor values for the first stage prototype circuit are
2C1
= 0.765 ·. . C1 = 2.61 F
C2 =1C1
= 0.38 F
The values for the second stage prototype circuit are
2C1
= 1.848 ·. . C1 = 1.08 F
C2 =1C1
= 0.92 F
The scaling factors are
km =R′
R= 1000; kf =
ω′o
ωo
= 16,000π
Therefore the scaled values for the components in the first stage are
R1 = R2 = R = 1000 Ω
C1 =2.61
(16,000π)(1000)= 52.01 nF
C2 =0.38
(16,000π)(1000)= 7.61 nF
The scaled values for the second stage are
R1 = R2 = R = 1000 Ω
C1 =1.08
(16,000π)(1000)= 21.53 nF
C2 =0.92
(16,000π)(1000)= 18.38 nF
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)
·. . H ′hp(s) =
(s/4000π)4
[(s/4000π)2 + 0.765(s/4000π) + 1]
· 1[(s/4000π)2 + 1.848(s/4000π) + 1]
=s4
(s2 + 3060πs + 16 × 106π2)(s2 + 7392πs + 16 × 106π2)
For the low pass section kf = 16,000π
Hlp(s) =1
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
·. . H ′lp(s) =
1[(s/16,000π)2 + 0.765(s/16,000π) + 1]
· 1[(s/16,000π)2 + 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) = H ′hp(s)H
′lp(s)
15–30 CHAPTER 15. Active Filter Circuits
For convenience let
D1 = s2 + 3060πs + 16 × 106π2
D2 = s2 + 7392πs + 16 × 106π2
D3 = s2 + 12,240πs + 256 × 106π2
D4 = s2 + 29,568πs + 256 × 106π2
Then
H ′(s) =65,536 × 1012π4s4
D1D2D3D4
[d] ωo = 2π(4000) = 8000π rad/s
s = j8000π
s4 = 4096 × 1012π4
D1 = (16 × 106π2 − 64 × 106π2) + j(8000π)(3060π)
= 106π2(−48 + j24.48) = 106π2(53.88/152.98)
D2 = (16 × 106π2 − 64 × 106π2) + j(8000π)(7392π)
= 106π2(−48 + j59.136) = 106π2(76.16/129.07)
D1 = (256 × 106π2 − 64 × 106π2) + j(8000π)(12,240π)
= 106π2(192 + j97.92) = 106π2(215.53/27.02)
D1 = (256 × 106π2 − 64 × 106π2) + j(8000π)(29,568π)
= 106π2(192 + j236.544) = 106π2(304.66/50.93)
H ′(jωo) =(65,536)(4096)π8 × 1024
(π8 × 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 theprototype bandpass circuit
R1 =Q
K=
1610
= 1.6 Ω
R2 =Q
2Q2 − K=
16502
Ω
R3 = 2Q = 32 Ω
Problems 15–31
The scaling factors are
kf =ω′
o
ωo
= 2π(6400) = 12,800π
km =C
C ′kf
=1
(20 × 10−9)(12,800π)= 1243.40
Therefore,
R′1 = kmR1 = (1.6)(1243.40) = 1.99 kΩ
R′2 = kmR2 = (16/502)(1243.40) = 39.63 Ω
R′3 = kmR3 = 32(1243.40) = 39.79 kΩ
[b]
P 15.38 From Eq 15.58 we can write
H(s) =−(
2R3C
) (R3C
2
) (1
R1C
)s
s2 + 2R3C
s + R1+R2R1R2R3C2
or
H(s) =−(
R32R1
) (2
R3Cs)
s2 + 2R3C
s + R1+R2R1R2R3C2
Therefore
2R3C
= β =ωo
Q;
R1 + R2
R1R2R3C2 = ω2o ;
and K =R3
2R1
By hypothesis C = 1 F and ωo = 1 rad/s
·. .2R3
=1Q
or R3 = 2Q
15–32 CHAPTER 15. Active Filter Circuits
R1 =R3
2K=
Q
K
R1 + R2
R1R2R3= 1
Q
K+ R2 =
(Q
K
)(2Q)R2
·. . R2 =Q
2Q2 − K
P 15.39 [a] First we will design a unity gain filter and then provide the passband gain withan inverting amplifier. For the high pass section the cut-off frequency is 500Hz. The order of the Butterworth is
n =(−0.05)(−20)log10(500/200)
= 2.51
·. . n = 3
Hhp(s) =s3
(s + 1)(s2 + s + 1)
For the prototype first-order section
R1 = R2 = 1 Ω, C = 1 F
For the prototype second-order section
R1 = 0.5 Ω, R2 = 2 Ω, C = 1 F
The scaling factors are
kf =ω′
o
ωo
= 2π(500) = 1000π
km =C
C ′kf
=1
(15 × 10−9)(1000π)=
106
15π
In the scaled first-order section
R′1 = R′
2 = kmR1 =106
15π(1) = 21.22 kΩ
C ′ = 15 nF
In the scaled second-order section
R′1 = 0.5km = 10.61 kΩ
R′2 = 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 theButterworth filter is
n =(−0.05)(−20)
log10(11,250/4500)= 2.51; ·. . n = 3
Hlp(s) =1
(s + 1)(s2 + s + 1)
For the prototype first-order section
R1 = R2 = 1 Ω, C = 1 F
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′
R= 104; kf =
ω′o
ωo
= (4500)(2π) = 9000π
For the scaled first-order section
R′1 = R′
2 = 10 kΩ; C ′ =C
kfkm
=1
(9000π)(104)= 3.54 nF
For the scaled second-order section
R′1 = R′
2 = 10 kΩ
C ′1 =
C1
kfkm
=2
(9000π)(104)= 7.07 nF
C ′2 =
C2
kfkm
=0.5
(9000π)(104)= 1.77 nF
GAIN AMPLIFIER
20 log10 K = 20 dB, ·. . K = 10
Since we are using 10 kΩ resistors in the low-pass stage, we will useRf = 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
Hhp(s) =s3
(s + 1)(s2 + s + 1)
The frequency scaling factor is kf = (ω′o/ωo) = 1000π. Therefore the scaled
transfer function is
H ′hp(s) =
(s/1000π)3(s
1000π+ 1
) [(s
1000π
)2+ s
1000π+ 1
]
=s3
(s + 1000π)[s2 + 1000πs + 106π2]
Unscaled low-pass stage
Hlp(s) =1
(s + 1)(s2 + s + 1)
The frequency scaling factor is kf = (ω′o/ωo) = 9000π. Therefore the scaled
transfer function is
H ′lp(s) =
1(s
9000π+ 1
) [(s
9000π
)2+(
s9000π
)+ 1
]
=(9000π)3
(s + 9000π)(s2 + 9000πs + 81 × 106π2)
Thus the transfer function for the filter is
H ′(s) = 10H ′hp(s)H
′lp(s) =
729 × 1010π3s3
D1D2D3D4
Problems 15–35
where
D1 = s + 1000π
D2 = s + 9000π
D3 = s2 + 1000πs + 106π2
D4 = s2 + 9000πs + 81 × 106π2
[b] At f = 200 Hz ω = 400π rad/s
D1(j400π) = 400π(2.5 + j1)
D2(j400π) = 400π(22.5 + j1)
D3(j400π) = 4 × 105π2(2.1 + j1.0)
D4(j400π) = 4 × 105π2(202.1 + j9)
Therefore
D1D2D3D4(j400π) = 256π61014(28,534.82/52.36)
H ′(j400π) =(729π3 × 1010)(64 × 106π3)
256π6 × 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) = 106π2(−8 + j3)
D4(j3000π) = 9 × 106π2(8 + j3)
H ′(j3000π) =(729 × π3 × 1010)(27 × 109π3)
27 × 1018π6(730/270 )
= 9.99/90
·. . 20 log10 |H ′(j3000π)| = 19.99 dB
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 frequencywe would expect the high-pass stage of the filter to predict the loss in gain at200 Hz. For a 3nd order Butterworth
GAIN = 20 log101√
1 + (500/200)6= −23.89 dB.
1500 Hz is in the passband for this bandpass filter, and is in fact the centerfrequency. 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 functionconfirms 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(
1s2 + 0.518
(1s
)+ 1
) (1s2 +
√2(
1s
)+ 1
) (1s2 + 1.932
(1s
)+ 1
)
Hhp(s) =s6
(s2 + 0.518s + 1)(s2 +√
2s + 1)(s2 + 1.932s + 1)
P 15.42 [a] kf = 25,000
H ′hp(s) =
(s/25,000)6
[(s/25,000)2 + 0.518(s/25,000) + 1]
· 1[(s/25,000)2 + 1.414s/25,000 + 1][(s/25,000)2 + 1.932s/25,000 + 1]
=s6
(s2 + 12,950s + 625 × 106)(s2 + 35,350s + 625 × 106)
· 1(s2 + 48,300s + 625 × 106)
[b] H ′(j25,000) =−(25,000)6
[12,950(j25,000)][35,350(j25,000)][48,300(j25,000)]
=−(25,000)3
(12,950)(25,350)(48,300)j3
= 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 opamp is ideal the current in R3 is zero. Therefore at low frequencies the circuitbehaves as an inverting amplifier with a gain of R2/R1. At very highfrequencies the capacitor branches are short circuits and hence the outputvoltage is zero.
[b] Let the node where R1, R2, R3, and C2 join be denoted as a, then
(Va − Vi)G1 + VasC2 + (Va − Vo)G2 + VaG3 = 0
−VaG3 − VosC1 = 0
or
(G1 + G2 + G3 + sC2)Va − G2Vo = G1Vi
Va =−sC1
G3Vo
Solving for Vo/Vi yields
H(s) =−G1G3
(G1 + G2 + G3 + sC2)sC1 + G2G3
=−G1G3
s2C1C2 + (G1 + G2 + G3)C1s + G2G3
=−G1G3/C1C2
s2 +[
(G1+G2+G3)C2
]s + G2G3
C1C2
=−G1G2G3
G2C1C2
s2 +[
(G1+G2+G3)C2
]s + G2G3
C1C2
=−Kbo
s2 + b1s + bo
where K =G1
G2; bo =
G2G3
C1C2
and b1 =G1 + G2 + G3
C2
[c] Equating coefficients we see that
G1 = KG2
G3 =boC1C2
G2=
boC1
G2
since by hypothesis C2 = 1 F
b1 =G1 + G2 + G3
C2= G1 + G2 + G3
15–38 CHAPTER 15. Active Filter Circuits
·. . b1 = KG2 + G2 +boC1
G2
b1 = G2(1 + K) +boC1
G2
Solving this quadratic equation for G2 we get
G2 =b1
2(1 + K)±√
b21 − boC14(1 + K)
4(1 + K)2
=b1 ±
√b21 − 4bo(1 + K)C1
2(1 + K)
For G2 to be realizable
C1 <b21
4bo(1 + K)
[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 = boC1/G2
P 15.44 [a] In the second order section of a third order Butterworth filter bo = b1 = 1Therefore,
C1 ≤ b21
4bo(1 + K)=
1(4)(1)(5)
= 0.05 F
·. . C1 = 0.05 F (limiting value)
[b] G2 =1
2(1 + 4)= 0.1 S
G3 =1
0.1(0.05) = 0.5 S
G1 = 4(0.1) = 0.4 S
Therefore,
R1 =1
G1= 2.5 Ω; R2 =
1G2
= 10 Ω; R3 =1
G3= 2 Ω
Problems 15–39
[c] kf =ω′
o
ωo
= 2π(2500) = 5000π
km =C2
C ′2kf
=1
(10 × 10−9)kf
= 6366.2
C ′1 =
0.05kfkm
= 0.5 × 10−9 = 500 pF
R′1 = (2.5)(6366.2) = 15.92 kΩ
R′2 = (10)(6366.2) = 63.66 kΩ
R′3 = (2)(6366.2) = 12.73 kΩ
[d] R′1 = R′
2 = (6366.2)(1) = 6.37 kΩ
C ′ =C
kfkm
=1
108 = 10 nF
[e]
P 15.45 [a] By hypothesis the circuit becomes:
For very small frequencies the capacitors behave as open circuits and thereforevo is zero. As the frequency increases, the capacitive branch impedancesbecome small compared to the resistive branches. When this happens thecircuit becomes an inverting amplifier with the capacitor C2 dominating thefeedback path. Hence the gain of the amplifier approaches(1/jωC2)/(1/jωC1) or C1/C2. Therefore the circuit behaves like a high-passfilter 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
VaG2 + (Va − Vi)sC1 + (Va − Vo)sC2 + VasC3 = 0
or
Va[G2 + s(C1 + C2 + C3)] − VosC2 = sC1Vi
Summing the currents away from the inverting input terminal gives
(0 − Va)sC3 + (0 − Vo)G1 = 0
or
sC3Va = −G1Vo; Va =−G1Vo
sC3
Therefore we can write
−G1Vo
sC3[G2 + s(C1 + C2 + C3)] − sC2Vo = sC1Vi
Solving for Vo/Vi gives
H(s) =Vo
Vi
=−C1C3s
2
[C2C3s2 + G1(C1 + C2 + C3)s + G1G2]
=−C1C2
s2[s2 + G1
C2C3(C1 + C2 + C3)s + G1G2
C2C3
]
=−Ks2
s2 + b1s + bo
Therefore the circuit implements a second-order high-pass filter with apassband gain of C1/C2.
[c] C1 = K:
b1 =G1
(1)(1)(K + 2) = G1(K + 2)
·. . G1 =b1
K + 2; R1 =
(K + 2
b1
)
bo =G1G2
(1)(1)= G1G2
·. . G2 =bo
G1=
bo
b1(K + 2)
·. . R2 =b1
bo(K + 2)
Problems 15–41
[d] From Table 15.1 the transfer function of the second-order section of athird-order high-pass Butterworth filter is
H(s) =Ks2
s2 + s + 1
Therefore b1 = bo = 1
Thus
C1 = K = 8 F
R1 =8 + 2
1= 10 Ω
R2 =1
1(8 + 2)= 0.10 Ω
P 15.46 [a] Low-pass filter:
n =(−0.05)(−30)
log10(1000/400)= 3.77; ·. . n = 4
In the first prototype second-order section: b1 = 0.765, bo = 1, C2 = 1 F
C1 ≤ b21
4bo(1 + K)≤ (0.765)2
(4)(2)≤ 0.0732
choose C1 = 0.03 F
G2 =0.765 ±
√(0.765)2 − 4(2)(0.03)
4=
0.765 ± 0.5884
Arbitrarily select the larger value for G2, then
G2 = 0.338 S; ·. . R2 =1
G2= 2.96 Ω
G1 = KG2 = 0.338 S; ·. . R1 =1
G1= 2.96 Ω
G3 =boC1
G2=
(1)(0.03)0.338
= 0.089 ·. . R3 = 1/G3 = 11.3 Ω
Therefore in the first second-order prototype circuit
R1 = R2 = 2.96 Ω; R3 = 11.3 Ω
C1 = 0.03 F; C2 = 1 F
In the second second-order prototype circuit: b1 = 1.848, b0 = 1, C2 = 1 F
·. . C1 ≤ (1.848)2
8≤ 0.427
15–42 CHAPTER 15. Active Filter Circuits
choose C1 = 0.30 F
G2 =1.848 ±
√(1.848)2 − 8(0.3)
4=
1.848 ± 1.0084
Arbitrarily select the larger value, then
G2 = 0.7139 S; ·. . R2 =1
G2= 1.4008 Ω
G1 = KG2 = 0.7139 S; ·. . R1 =1
G1= 1.4008 Ω
G3 =boC1
G2=
(1)(0.30)0.7139
= 0.4202 S ·. . R3 = 1/G3 = 2.3796 Ω
In the low-pass section of the filter
kf =ω′
o
ωo
= 2π(400) = 800π
km =C
C ′kf
=1
(10 × 10−9)kf
=125,000
π
Therefore in the first scaled second-order section
R′1 = R′
2 = 2.96km = 118 kΩ
R′3 = 11.3km = 450 kΩ
C ′1 =
0.03kfkm
= 300 pF
C ′2 = 10 nF
In the second scaled second-order section
R′1 = R′
2 = 1.4008km = 55.74 kΩ
R′3 = 2.3796km = 94.68 kΩ
C ′1 =
0.3kfkm
= 3 nF
C ′2 = 10 nF
High-pass filter section
n =(−0.05)(−30)
log10(6400/2560)= 3.77; n = 4.
In the first prototype second-order section: b1 = 0.765; bo = 1; C2 = C3 = 1 F
C1 = K = 1 F
Problems 15–43
R1 =K + 2
b1=
30.765
= 3.92 Ω
R2 =b1
bo(K + 2)=
0.7653
= 0.255 Ω
In the second prototype second-order section: b1 = 1.848; bo = 1;C2 = C3 = 1 F
C1 = K = 1 F
R1 =K + 2
b1=
31.848
= 1.623 Ω
R2 =b1
bo(K + 2)=
1.8483
= 0.616 Ω
In the high-pass section of the filter
kf =ω′
o
ωo
= 2π(6400) = 12,800π
km =C
C ′kf
=1
(10 × 10−9)(12,800π)=
7812.5π
In the first scaled second-order section
R′1 = 3.92km = 9.75 kΩ
R′2 = 0.255km = 634 Ω
C ′1 = C ′
2 = C ′3 = 10 nF
In the second scaled second-order section
R′1 = 1.623km = 4.04 kΩ
R′2 = 0.616km = 1.53 kΩ
C ′1 = C ′
2 = C ′3 = 10 nF
In the gain section, let Ri = 10 kΩ and Rf = 10 kΩ.
15–44 CHAPTER 15. Active Filter Circuits
[b]
P 15.47 [a] The prototype low-pass transfer function is
Hlp(s) =1
(s2 + 0.765s + 1)(s2 + 1.848s + 1)
The low-pass frequency scaling factor is
kflp= 2π(400) = 800π
The scaled transfer function for the low-pass filter is
H ′lp(s) =
1[(s
800π
)2+ 0.765s
800π+ 1
] [(s
800π
)2+ 1.848s
800π+ 1
]
=4096 × 108π4
[s2 + 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
H ′hp(s) =
(s/12,800π)4[(s
12,800π
)2+ 0.765s
12,800π+ 1
] [(s
12,800π
)2+ 1.848s
12,800π+ 1
]
=s4
[s2 + 9792πs + (12,800π)2][s2 + 23,654.4πs + (12,800π)2]
The transfer function for the filter is
H ′(s) =[H ′
lp(s) + H ′hp(s)
]
[b] fo =√
fc1fc2 =√
(400)(6400) = 1600 Hz
ωo = 2πfo = 3200π rad/s
(jωo)2 = −1024 × 104π2
(jωo)4 = 1,048,576 × 108π4
H ′lp(jωo) =
4096 × 108π4
[−960 × 104π2 + j612(3200π2)]×
1[−960 × 104π2 + j1478.4(3200π2)]
=40,000
(−3000 + j612)(−3000 + j1478.4)
= 3906.2 × 10−6/37.76
H ′hp(jωo) =
1,048,576 × 108π4
[15,360 × 104π2 + j9792(3200π2)]
1[15,360 × 104π2 + j23,654.4(3200π2)]
=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 highfrequencies the capacitor branches are short circuits and the output voltage iszero. 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. Observethis will also be the voltage at the left-hand terminal of R2. The s-domainequations are
(Va − Vi)G1 + (Va − Vo)sC1 = 0
(Vo − Va)G2 + sC2Vo = 0
or
(G1 + sC1)Va − sC1Vo = G1Vi
−G2Va + (G2 + sC2)Vo = 0
·. . Va =G2 + sC2Vo
G2
·. .
[(G1 + sC1)
(G2 + sC2)G2
− sC1
]Vo = G1Vi
·. .Vo
Vi
=G1G2
(G1 + sC1)(G2 + sC2) − C1G2s
which reduces to
Vo
Vi
=G1G2/C1C2
s2 + G1C1
s + G1G2C1C2
=bo
s2 + b1s + bo
[c] There are four circuit components and two restraints imposed by H(s);therefore there are two free choices.
[d] b1 =G1
C1
·. . G1 = b1C1
bo =G1G2
C1C2
·. . G2 =bo
b1C2
[e] No, all physically realizeable capacitors will yield physically realizeableresistors.
[f] From Table 15.1 we know the transfer function of the prototype 4th orderButterworth filter is
H(s) =1
(s2 + 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
0.765(1) = 1.307 S
Problems 15–47
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
1.848
)(1) = 0.541 S
R2 = 1/G2 = 1.848 Ω
P 15.49 [a] kf =ω′
o
ωo
= 2π(3000) = 6000π
km =C
C ′kf
=1
(4.7 × 10−9)(6000π)=
106
28.2π
In the first section
R′1 = 1.307km = 14.75 kΩ
R′2 = 0.765km = 8.64 kΩ
In the second section
R′1 = 0.541km = 6.11 kΩ
R′2 = 1.848km = 20.86 kΩ
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 outputvoltage is zero. As the frequency increases the capacitor branches approachshort 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 + VoG2 = 0
It follows that
Va(G1 + sC1) − G1Vo = sC1Vi
and Va =(G2 + sC2)Vo
sC2
Thus[(G2 + sC2)
sC2
](G1 + sC1) − G1
Vo = sC1Vi
Vos2C1C2 + sC1G2 + G1G2 = s2C1C2Vi
Problems 15–49
H(s) =Vo
Vi
=s2(
s2 +G2
C2s +
G1G2
C1C2
)
=Vo
Vi
=s2
s2 + b1s + bo
[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.
[d] bo =G1G2
C1C2; b1 =
G2
C2
·. . G2 = b1C2; R2 =1
b1C2
G1 =bo
b1C1
·. . R1 =b1
boC1
[e] No, all realizeable capacitors will produce realizeable resistors.
[f] The second-order section in a 3rd-order Butterworth high-pass filter iss2/(s2 + s + 1). Therefore bo = b1 = 1 and
R1 =1
(1)(1)= 1 Ω.
R2 =1
(1)(1)= 1 Ω.
P 15.51 [a] kf =ω′
o
ωo
= 104π
km =C
C ′kf
=1
(75 × 10−9)(104π)=
105
75π
C1 = C2 = 75 nF; R′1 = R′
2 = kmR = 424.4 Ω
[b] R = 424.4 Ω; C = 75 nF
15–50 CHAPTER 15. Active Filter Circuits
[c]
[d] Hhp(s) =s3
(s + 1)(s2 + s + 1)
H ′hp(s) =
(s/104π)3
[(s/104π) + 1][(s/104π)2 + (s/104π) + 1]
=s3
(s + 104π)(s2 + 104πs + 108π2)
[e] H ′hp(j104π) =
(j104π)3
(j104π + 104π)[(j104π)2 + 104π(j104π) + 108π2]= 0.7071/135
·. . |H ′hp| = 0.7071 = −3 dB
P 15.52 [a] It follows directly from Eq 15.64 that
H(s) =s2 + 1
s2 + 4(1 − σ)s + 1
Now note from Eq 15.69 that (1 − σ) equals 1/4Q, hence
H(s) =s2 + 1
s2 + 1Qs + 1
[b] For Example 15.13, ωo = 5000 rad/s and Q = 5. Therefore kf = 5000 and
H ′(s) =(s/5000)2 + 1
(s/5000)2 +15
(s
5000
)+ 1
=s2 + 25 × 106
s2 + 1000s + 25 × 106
P 15.53 [a] ωo = 2000π rad/s
·. . kf =ω′
o
ωo
= 2000π
Problems 15–51
km =C
C ′kf
=1
(15 × 10−9)(2000π)=
105
3π
R′ = kmR =105
3π(1) = 10,610 Ω
R′
2= 5,305 Ω
σ = 1 − 14Q
= 1 − 14(20)
= 0.9875
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
(s/2000π)2 + 120(s/2000π) + 1
=s2 + 4 × 106π2
s2 + 100πs + 4 × 106π2
P 15.54 To satisfy the gain specification of 20 dB at ω = 0 and α = 1 requires
R1 + R2
R1= 10 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
2π(40)(105)= 39.79 nF
15–52 CHAPTER 15. Active Filter Circuits
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
(111.111.1
)α=1
= 20.01 dB
When C1 = 39 nF the frequency 1/R2C1 is
1R2C1
=109
105(39)= 256.41 rad/s = 40.81 Hz
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||11.1 × 103 + j256.41(11.1)(100)(39)10−3| = 7.11
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
)= 13.98
·. .R1 + R2
R1= 5; ·. . R2 = 4R1
Choose R1 = 100 kΩ. Then R2 = 400 kΩ
1R2C1
= 100π rad/s; ·. . C1 =1
(100π)(400 × 103)= 7.96 nF
P 15.56 [a] |H(j0)| =R1 + αR2
R1 + (1 − α)R2=
11.1 + α(100)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
sC1+
α
sC1=
1sC1
which is identical to the impedance of the capacitor in Fig. P15.53(a).
[b]
Vx =α/sC1
(1 − α)/sC1 + α/sC1V = α
Vy =αR2
(1 − α)R2 + αR2= α = Vx
[c] Since x and y are both at the same potential, they can be shorted together, andthe 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 noeffect on the ratio Vo/Vs, as this feedback path is not involved in the nodalequation that defines the voltage ratio. Thus, the circuit in Fig. 15.53(c) can besimplified into the form of Fig. 15.2, where the input impedance is theequivalent impedance of R1 in series with the parallel combination of(1 − α)/sC1 and (1 − α)R2, and the feedback impedance is the equivalentimpedance of R1 in series with the parallel combination of α/sC1 and αR2:
Zi = R1 +(1−α)sC1
· (1 − α)R2
(1 − α)R2 + (1−α)sC1
=R1 + (1 − α)R2 + R1R2C1s
1 + R2C1s
Zf = R1 +α
sC1· αR2
αR2 + αsC1
=R1 + αR2 + R1R2C1s
1 + R2C1s
P 15.58 As ω → 0
|H(iω)| → 2R3 + R4
2R3 + R4= 1
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
|H(j∞)|β=1∼= Ro
R3> 1
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
|H(j∞)|β=0∼= R3
Ro
< 1
Thus, when β = 0 we have attenuation or “cut”.Also note that when β = 0.5
|H(jω)|β=0.5 =(0.5R4 + Ro)(0.5R4 + R3)(0.5R4 + R3)(0.5R4 + Ro)
= 1
Thus, the transition from amplification to attenuation occurs at β = 0.5. If β > 0.5we 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 15–55
P 15.59 [a] |H(j∞)|β=1 =Ro(R4 + R3)R3(R4 + Ro)
=(65.9)(505.9)(5.9)(565.9)
= 9.99
·. . maximum boost = 20 log10 9.99 = 19.99 dB
[b] |H(j∞)|β=0 =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/R3C2)|β=1 =
∣∣∣∣∣∣(2R3 + R4) + j Ro
R3(R4 + R3)
(2R3 + R4) + j(R4 + Ro)
∣∣∣∣∣∣
=∣∣∣∣∣511.8 + j 65.9
5.9 (505.9)511.8 + j565.9
∣∣∣∣∣= 7.44
20 log10 |H(j/R3C2)|β=1 = 20 log10 7.44 = 17.43 dB
[e] When β = 0
|H(j/R3C2)|β=0 =(2R3 + R4) + j(R4 + Ro)
(2R3 + R4) + jRo
R3(R4 + R3)
Note this is the reciprocal of |H(j/R3C2)|β=1.
·. . 20 log10 |H(j/R3C2)|β+0 = −17.43 dB
[f] The frequency 1/R3C2 is very nearly where the gain is 3 dB off from itsmaximum boost or cut. Therefore for frequencies higher than 1/R3C2 thecircuit 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]
16Fourier Series
Assessment Problems
AP 16.1 av =1T
∫ 2T/3
0Vm dt +
1T
∫ T
2T/3
(Vm
3
)dt =
79Vm = 7π V
ak =2T
[∫ 2T/3
0Vm cos kω0t dt +
∫ T
2T/3
(Vm
3
)cos kω0t dt
]
=( 4Vm
3kω0T
)sin
(4kπ
3
)=(6
k
)sin
(4kπ
3
)
bk =2T
[∫ 2T/3
0Vm sin kω0t dt +
∫ T
2T/3
(Vm
3
)sin kω0t dt
]
=( 4Vm
3kω0T
) [1 − cos
(4kπ
3
)]=(6
k
) [1 − cos
(4kπ
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
[c] ω0 =(2π
T
)= 50 rad/s
[d] f3 = 3f0 = 23.87 Hz
[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.
vg(t) =(6Vm
T
)t, 0 ≤ t ≤ T/6; av = 0, ak = 0 for all k
16–1
16–2 CHAPTER 16. Fourier Series
bk = 0 for k even
bk =8T
∫ T/4
0f(t) sin kω0t dt, k odd
=8T
∫ T/6
0
(6Vm
T
)t sin kω0t dt +
8T
∫ T/4
T/6Vm sin kω0t dt
=(12Vm
k2π2
)sin
(kπ
3
)
vg(t) =12Vm
π2
∞∑n=1,3,5
1n2 sin
nπ
3sin nω0t V
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
A3 = 0; A4 = −1.3 − j2.25 = 2.6/− 120
A5 = 1.04 − j1.8 = 2.1/− 60
θ1 = −120; θ2 = −60; θ3 not defined;
θ4 = −120; θ5 = −60
[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
vi =8Aπ2
∞∑n=1,3,5
( 1n2 sin
nπ
2
)sin nω0(t + T/4)
=8Aπ2
∞∑n=1,3,5
( 1n2 sin2 nπ
2
)cos nω0t
=8Aπ2
∞∑n=1,3,5
1n2 cos nω0t
8Aπ2 =
8(281.25π2)π2 = 2250 mV
ω0 =2πT
=2π
200π× 103 = 10
Problems 16–3
·. . vi = 2250∞∑
n=1,3,5
1n2 cos 10nt mV
From the circuit we have
Vo =Vi
R + (1/jωC)· 1jωC
=Vi
1 + jωRC
Vo =1/RC
1/RC + jωVi =
100100 + jω
Vi
Vi1 = 2250/0 mV; ω0 = 10 rad/s
Vi3 =2250
9/0 = 250/0 mV; 3ω0 = 30 rad/s
Vi5 =225025
/0 = 90/0 mV; 5ω0 = 50 rad/s
Vo1 =100
100 + j10(2250/0) = 2238.83/− 5.71 mV
Vo3 =100
100 + j30(250/0) = 239.46/− 16.70 mV
Vo5 =100
100 + j50(90/0) = 80.50/− 26.57 mV
·. . vo = 2238.33 cos(10t − 5.71) + 239.46 cos(30t − 16.70)
+ 80.50 cos(50t − 26.57) + . . . mV
AP 16.6 [a] The Fourier series of the input voltage is
vg =4Aπ
∞∑n=1,3,5
1n
sin nω0(t + T/4)
= 42∞∑
n=1,3,5
[ 1n
sin(
nπ
2
)]cos 2000nt V
From the circuit we have
VosC +Vo
sL+
Vo − Vg
R= 0
·. .Vo
Vg
= H(s) =s/RC
s2 + (s/RC) + (1/LC)
16–4 CHAPTER 16. Fourier Series
Substituting in the numerical values yields
H(s) =500s
s2 + 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) =500(j2000)
108 − 4 × 106 + 500(j2000)=
j196 + j1
= 0.01042/89.40
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 outputvoltage. The circuit is a bandpass filter with a center frequency of 10,000 rad/sand 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 harmonicterms in the output signal.
AP 16.7 ω0 =2π × 103
2094.4= 3 rad/s
jω0k = j3k
Problems 16–5
VR =2
2 + s + 1/s(Vg) =
2sVg
s2 + 2s + 1
H(s) =(
VR
Vg
)=
2ss2 + 2s + 1
H(jω0k) = H(j3k) =j6k
(1 − 9k2) + j6k
vg1 = 25.98 sin ω0t V; Vg1 = 25.98/0 V
H(j3) =j6
−8 + j6= 0.6/− 53.13; VR1 = 15.588/− 53.13 V
P1 =(15.588/
√2)2
2= 60.75 W
vg3 = 0, therefore P3 = 0 W
vg5 = −1.04 sin 5ω0t V; Vg5 = 1.04/180
H(j15) =j30
−224 + j30= 0.1327/− 82.37
VR5 = (1.04/180)(0.1327/− 82.37) = 138/97.63 mV
P5 =(0.138/
√2)2
2= 4.76 mW; therefore P ∼= P1
∼= 60.75 W
AP 16.8 Odd function with half- and quarter-wave symmetry, therefore av = 0, ak = 0 for allk, bk = 0 for k even; for k odd we have
bk =8T
∫ T/8
02 sin kω0t dt +
8T
∫ T/4
T/88 sin kω0t dt
=( 8
πk
) [1 + 3 cos
(kπ
4
)], k odd
Therefore Cn =(−j4
nπ
) [1 + 3 cos
(nπ
4
)], n odd
16–6 CHAPTER 16. Fourier Series
AP 16.9 [a] Irms =
√2T
[(2)2
(T
8
)(2) + (8)2
(3T8
− T
8
)]=
√34 = 5.831 A
[b] C1 =−j12.5
π; C3 =
j1.5π
; C5 =j0.9π
;
C7 =−j1.8
π; C9 =
−j1.4π
; C11 =j0.4π
Irms =
√√√√I2dc + 2
∞∑n=1,3,5
|Cn|2 ∼=√
2π2 (12.52 + 1.52 + 0.92 + 1.82 + 1.42 + 0.42)
∼= 5.777 A
[c] % Error =5.777 − 5.831
5.831× 100 = −0.93%
[d] Using just the terms C1 – C9,
Irms =
√√√√I2dc + 2
∞∑n=1,3,5
|Cn|2 ∼=√
2π2 (12.52 + 1.52 + 0.92 + 1.82 + 1.42)
∼= 5.774 A
% Error =5.774 − 5.831
5.831× 100 = −0.98%
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)
− j4
nπ
(1 + 3 cos
nπ
4
)ejnω0(t−T/4)
=4π
∞∑n=−∞
n(odd)
1n
(1 + 3 cos
nπ
4
)e−j(n+1)(π/2)ejnω0t
Problems 16–7
Problems
P 16.1 [a] ωoa =2π
200 × 10−6 = 31, 415.93 rad/s
ωob =2π
40 × 10−6 = 157.080 krad/s
[b] foa =1T
=1
200 × 10−6 = 5000 Hz; fob =1
40 × 10−6 = 25,000 Hz
[c] ava = 0; avb =100(10 × 10−6)
40 × 10−6 = 25 V
[d] The periodic function in Fig. P16.1(a) has half-wave symmetry. Therefore,
ava = 0; aka = 0 for k even; bka = 0 for k even
For k odd,
aka =4T
∫ T/4
040 cos
2πkt
Tdt +
4T
∫ T/2
T/480 cos
2πkt
Tdt
=160T
T
2πksin
2πkt
T
∣∣∣∣T/4
0+
320T
T
2πksin
2πkt
T
∣∣∣∣T/2
T/4
=80πk
sinπk
2+
160πk
(sin πk − sin
πk
2
)
= − 80πk
sinπk
2, k odd
bka =4T
∫ T/4
040 sin
2πkt
Tdt +
4T
∫ T/2
T/480 sin
2πkt
Tdt
=−160
T
T
2πkcos
2πkt
T
∣∣∣∣T/4
0− 320
T
T
2πkcos
2πkt
T
∣∣∣∣T/2
T/4
=−80πk
(0 − 1) − 160πk
(−1 − 0)
=240πk
The periodic function in Fig. P16.1(b) is even; therefore, bk = 0 for all k. Also,
avb = 25 V
akb =4T
∫ T/8
0100 cos
2πkt
Tdt
=400T
T
2πksin
2πk
Tt∣∣∣∣T/8
0
=200πk
sinπk
4
16–8 CHAPTER 16. Fourier Series
[e] For the periodic function in Fig. P16.1(a),
v(t) =80π
∞∑n=1,3,5
(− 1
nsin
nπ
2cos nωot +
3n
sin nωot)
V
For the periodic function in Fig. P16.1(b),
v(t) = 25 +200π
∞∑n=1
( 1n
sinnπ
4cos nωot
)V
P 16.2 In studying the periodic function in Fig. P16.2 note that it can be visualized as thecombination 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 isgiven as the answer to Problem 16.3(c).
v1(t) =100π
+ 50 sin ωot − 200π
∞∑n=2,4,6
cos nωot
(n2 − 1)V
v2(t) =60π
+ 30 sin ωo(t − T/2) − 120π
∞∑n=2,4,6
cos nωo(t − T/2)(n2 − 1)
V
Observe the following, noting that n is even:
sin ωo(t − T/2) = sin(ωot − 2π
T
T
2
)= sin(ωot − π) = − sin ωot
Problems 16–9
cos nωo(t − T/2) = cos(nωot − 2πn
T
T
2
)= cos(nωot − nπ) = cos nωot
Using the observations above,
v2(t) =60π
− 30 sin ωot − 120π
∞∑n=2,4,6
cos(nωot)(n2 − 1)
V
Thus,
v(t) = v1(t) + v2(t) =160π
+ 20 sin ωot − 320π
∞∑n=2,4,6
cos(nωot)(n2 − 1)
V
P 16.3 [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
bk =8T
∫ T/4
0Vm sin kω0t dt =
4Vm
kπ, k odd
and v(t) =4Vm
π
∞∑n=1,3,5
1n
sin nω0t V
[b] Even function: bk = 0 for k
av =2T
∫ T/2
0Vm sin
π
Tt dt =
2Vm
π
ak =4T
∫ T/2
0Vm sin
π
Tt cos kω0t dt =
2Vm
π
( 11 − 2k
+1
1 + 2k
)
=4Vm/π
1 − 4k2
and v(t) =2Vm
π
[1 + 2
∞∑n=1
11 − 4n2 cos nω0t
]V
[c] av =1T
∫ T/2
0Vm sin
(2πT
)t dt =
Vm
π
ak =2T
∫ T/2
0Vm sin
2πT
t cos kω0t dt =Vm
π
(1 + cos kπ
1 − k2
)
Note: ak = 0 for k-odd, ak =2Vm
π(1 − k2)for k even,
bk =2T
∫ T/2
0Vm sin
2πT
t sin kω0t dt = 0 for k = 2, 3, 4, . . .
For k = 1, we have b1 =Vm
2; therefore
v(t) =Vm
π+
Vm
2sin ω0t +
2Vm
π
∞∑n=2,4,6
11 − n2 cos nω0t V
16–10 CHAPTER 16. Fourier Series
P 16.4 Starting with Eq. (16.2),
f(t) sin kω0t = av sin kω0t +∞∑
n=1an cos nω0t sin kω0t +
∞∑n=1
bn sin nω0t sin kω0t
Now integrate both sides from to to to + T. All the integrals on the right-hand sidereduce to zero except in the last summation when n = k, therefore we have∫ to+T
tof(t) sin kω0t dt = 0 + 0 + bk
(T
2
)or bk =
2T
∫ to+T
tof(t) sin kω0t dt
P 16.5 [a] I6 =∫ to+T
tosin mω0t dt = − 1
mω0cos mω0t
∣∣∣∣to+T
to
=−1mω0
[cos mω0(to + T ) − cos mω0to]
=−1mω0
[cos mω0to cos mω0T − sin mω0to sin mω0T − cos mω0to]
=−1mω0
[cos mω0to − 0 − cos mω0to] = 0 for all m,
I7 =∫ to+T
tocos mω0to dt =
1mω0
[sin mω0t]∣∣∣∣to+T
to
=1
mω0[sin mω0(to + T ) − sin mω0to]
=1
mω0[sin mω0to − sin mω0to] = 0 for all m
[b] I8 =∫ to+T
tocos mω0t sin nω0t dt =
12
∫ to+T
to[sin(m + n)ω0t − sin(m − n)ω0t] dt
But (m + n) and (m − n) are integers, therefore from I6 above, I8 = 0 for allm, n.
[c] I9 =∫ to+T
tosin mω0t sin nω0t dt =
12
∫ to+T
to[cos(m − n)ω0t − cos(m + n)ω0t] dt
If m = n, both integrals are zero (I7 above). If m = n, we get
I9 =12
∫ to+T
todt − 1
2
∫ to+T
tocos 2mω0t dt =
T
2− 0 =
T
2
[d] I10 =∫ to+T
tocos mω0t cos nω0t dt
=12
∫ to+T
to[cos(m − n)ω0t + cos(m + n)ω0t] dt
If m = n, both integrals are zero (I7 above). If m = n, we have
I10 =12
∫ to+T
todt +
12
∫ to+T
tocos 2mω0t dt =
T
2+ 0 =
T
2
Problems 16–11
P 16.6 av =1T
∫ to+T
tof(t) dt =
1T
∫ 0
−T/2f(t) dt +
∫ T/2
0f(t) dt
Let t = −x, dt = −dx, x =T
2when t =
−T
2
and x = 0 when t = 0
Therefore1T
∫ 0
−T/2f(t) dt =
1T
∫ 0
T/2f(−x)(−dx) = − 1
T
∫ T/2
0f(x) dx
Therefore av = − 1T
∫ T/2
0f(t) dt +
1T
∫ T/2
0f(t) dt = 0
ak =2T
∫ 0
−T/2f(t) cos kω0t dt +
2T
∫ T/2
0f(t) cos kω0t dt
Again, let t = −x in the first integral and we get
2T
∫ 0
−T/2f(t) cos kω0t dt = − 2
T
∫ T/2
0f(x) cos kω0x dx
Therefore ak = 0 for all k.
bk =2T
∫ 0
−T/2f(t) sin kω0t +
2T
∫ T/2
0f(t) sin kω0t dt
Using the substitution t = −x, the first integral becomes
2T
∫ T/2
0f(x) sin kω0x dx
Therefore we have bk =4T
∫ T/2
0f(t) sin kω0t dt
P 16.7 bk =2T
∫ 0
−T/2f(t) sin kω0t dt +
2T
∫ T/2
0f(t) sin kω0t dt
Now let t = x − T/2 in the first integral, then dt = dx, x = 0 when t = −T/2 andx = T/2 when t = 0, also sin kω0(x − T/2) = sin(kω0x − kπ) = sin kω0x cos kπ.Therefore
2T
∫ 0
−T/2f(t) sin kω0t dt = − 2
T
∫ T/2
0f(x) sin kω0x cos kπ dx and
bk =2T
(1 − cos kπ)∫ T/2
0f(x) sin kω0t dt
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
bk =4T
∫ T/2
0f(t) sin kω0t dt when k is odd
16–12 CHAPTER 16. Fourier Series
P 16.8 Because the function is even and has half-wave symmetry, we have av = 0, ak = 0for k even, bk = 0 for all k and
ak =4T
∫ T/2
0f(t) cos kω0t dt, 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 =4T
∫ T/4
0f(t) cos kω0t dt +
4T
∫ T/2
T/4f(t) cos kω0t dt
Now let t = (T/2 − x) in the second integral, then dt = −dx, x = T/4 whent = T/4 and x = 0 when t = T/2. Therefore we get
4T
∫ T/2
T/4f(t) cos kω0t dt = − 4
T
∫ T/4
0f(x) cos kπ cos kω0x dx
Therefore we have
ak =4T
(1 − cos kπ)∫ T/4
0f(t) cos kω0t dt
But k is odd, hence
ak =8T
∫ T/4
0f(t) cos kω0t dt, k odd
P 16.9 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
bk =4T
∫ T/2
0f(t) sin kω0t dt
The function also has quarter-wave symmetry, therefore f(t) = f(T/2 − t) in theinterval T/4 ≤ t ≤ T/2. Thus we have
bk =4T
∫ T/4
0f(t) sin kω0t dt +
4T
∫ T/2
T/4f(t) sin kω0t dt
Now let t = (T/2 − x) in the second integral and note that dt = −dx, x = T/4when t = T/4 and x = 0 when t = T/2, thus
4T
∫ T/2
T/4f(t) sin kω0t dt = − 4
Tcos kπ
∫ T/4
0f(x)(sin kω0x) dx
But k is odd, therefore the expression becomes
bk =8T
∫ T/4
0f(t) sin kω0t dt
Problems 16–13
P 16.10 [a] f =1T
=1
16 × 10−3 = 62.5 Hz
[b] no, because f(3 ms) = 10 mA but f(−3 ms) = −10 mA.
[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 =8T
∫ T/4
0f(t) sin kωot, k odd
=8T
∫ T/8
05t sin kωot dt +
∫ T/4
T/80.01 sin kωot dt
=8T
Int1 + Int2
Int1 = 5∫ T/8
0t sin kωot dt
= 5[
1k2ω2
o
sin kωot − t
kωo
cos kωot
∣∣∣∣T/8
0
]
=5
k2ω2o
sinkπ
4− 0.625T
kωo
coskπ
4
Int2 = 0.01∫ T/4
T/8sin kωot dt =
−0.01kωo
cos kωot
∣∣∣∣T/4
T/8=
0.01kωo
coskπ
4
Int1 + Int2 =5
k2ω2o
sinkπ
4+(0.01
kωo
− 0.625Tkωo
)cos
kπ
4
0.625T = 0.625(16 × 10−3) = 0.01
·. . Int1 + Int2 =5
k2ω2o
sinkπ
4
bk =[ 8T
· 54π2k2 · T 2
]sin
kπ
4=
0.16π2k2 sin
kπ
4, k odd
i(t) =160π2
∞∑n=1,3,5
sin(nπ/4)n2 sin nωot mA
16–14 CHAPTER 16. Fourier Series
P 16.11 [a] T = 1; ωo =2πT
= 2π rad/s
[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
ak =8T
∫ T/4
0Vm cos kω0t dt =
4Vm
πksin
(kπ
2
)
v(t) =4Vm
π
∞∑n=1,3,5
[ 1n
sinnπ
2
]cos nω0t V
[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 =8T
∫ T/4
0
(4Vp
Tt − Vp
)cos kω0t dt = − 8Vp
π2k2
Therefore v(t) = −8Vp
π2
∞∑n=1,3,5
1n2 cos nω0t V
P 16.13 [a] i(t) is even, therefore bk = 0 for all k.
av =12
· T
4· Im · 2 · 1
T=
Im
4A
ak =4T
∫ T/4
0
(Im − 4Im
Tt)
cos kωot dt
=4Im
T
∫ T/4
0cos kωot dt − 16Im
T 2
∫ T/4
0t cos kωot dt
= Int1 − Int2
Int1 =4Im
T
∫ T/4
0cos kωot dt =
2Im
πksin
kπ
2
Int2 =16Im
T 2
∫ T/4
0t cos kωot dt
=16Im
T 2
1
k2ω2o
cos kωot +t
kωo
sin kωot
∣∣∣∣T/4
0
=4Im
π2k2
(cos
kπ
2− 1
)+
2Im
kπsin
kπ
2
Problems 16–15
·. . ak =4Im
π2k2
(1 − cos
kπ
2
)A
·. . i(t) =Im
4+
4Im
π2
∞∑n=1
1 − cos(nπ/2)n2 cos nωot A
[b] Shifting the reference axis to the left is equivalent to shifting the periodicfunction to the right:
cos nωo(t − T/2) = cos nπ cos nωot
Thus
i(t) =Im
4+
4Im
π2
∞∑n=1
(1 − cos(nπ/2)) cos nπ
n2 cos nωot A
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, ak = 0, for k even (half-wave symmetry)
bk = 0, for all k (function is even)
Because of the quarter-wave symmetry, the expression for ak is
ak =8T
∫ T/4
0f(t) cos kω0t dt, k odd
=88
∫ 2
04t2 cos kω0t dt = 4
[2t
k2ω20
cos kω0t +k2ω2
0t2 − 2
k3ω30
sin kω0t
]2
0
16–16 CHAPTER 16. Fourier Series
kω0(2) = k(2π
8
)(2) =
kπ
2
cos(kπ/2) = 0, since k is odd
·. . ak = 4[0 +
4k2ω20 − 2
k3ω30
sin(kπ/2)]
=16k2ω2
0 − 8k3ω3
0sin(kπ/2)
ω0 =2π8
=π
4; ω2
0 =π2
16; ω3
0 =π3
64
ak =(
k2π2 − 8k3π3
)(64) sin(kπ/2)
f(t) = 64∞∑
n=1,3,5
[n2π2 − 8
π3n3
]sin(nπ/2) cos(nω0t)
[e] cos nω0(t − 2) = cos(nω0t − π/2) = sin nω0t sin(nπ/2)
f(t) = 64∞∑
n=1,3,5
[n2π2 − 8
π3n3
]sin2(nπ/2) sin(nω0t)
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 interval0 < t < 4.
[d] av = 0, (half-wave symmetry); ak = 0, for all k (function is odd)
bk = 0, for k even (half-wave symmetry)
bk =8T
∫ T/4
0f(t) sin kω0t dt, k odd
=88
∫ 2
0t3 sin kω0t dt
Problems 16–17
=[
3t2
k2ω20
sin kω0t − 6k4ω4
0sin kω0t − t3
kω0cos kω0t +
6tk3ω3
0cos kω0t
]2
0
kω0(2) = k(2π
8
)(2) =
kπ
2
cos(kπ/2) = 0, since k is odd
·. . bk =[
12k2ω2
0sin(kπ/2) − 6
k4ω40
sin(kπ/2)]
kω0 = k(2π
8
)=
kπ
4; k2ω2
0 =k2π2
16; k4ω4
0 =k4π4
256
·. . bk =192π2k2
[1 − 8
π2k2
]sin(kπ/2), k odd
f(t) =192π2
∞∑n=1,3,5
[ 1n2
(1 − 8
π2n2
)sin(nπ/2)
]sin nω0t
[e] sin nω0(t − 2) = sin(nω0t − π/2) = − cos nω0t sin(nπ/2)
f(t) =−192π2
∞∑n=1,3,5
[ 1n2
(1 − 8
π2n2
)sin2(nπ/2)
]cos nω0t
P 16.16 [a]
[b] av = 0; ak = 0, for k even; bk = 0, for all k
ak =8T
∫ T/4
0f(t) cos kω0t dt, for k odd
=8T
∫ T/8
0
120tT
cos kω0t dt +8T
∫ T/4
T/8
(10 +
40T
t)
cos kω0t dt
=960T 2
∫ T/8
0t cos kω0t dt +
80T
∫ T/4
T/8cos kω0t dt +
320T 2
∫ T/4
T/8t cos kω0t dt
16–18 CHAPTER 16. Fourier Series
=960T 2
[cos kω0t
k2ω20
+t sin kω0t
kω0
]T/8
0
+80T
sin kω0t
kω0
∣∣∣∣T/4
T/8
+320T 2
[cos kω0t
k2ω20
+t sin kω0t
kω0
]T/4
T/8
kω0T
4=
kπ
2; kω0
T
8=
kπ
4
bk =960T 2
[cos(kπ/4)
k2ω20
+T
8kω0sin(kπ/4) − 1
k2ω20
]+
80kω0T
[sin(kπ/2) − sin(kπ/4)]
+320T 2
[cos(kπ/2)
k2ω20
+T
4sin(kπ/2)
kω0− cos(kπ/4)
k2ω20
− T sin(kπ/4)8kω0
]
=640
(kω0T )2 cos(kπ/4) +160
kω0T 2 sin(kπ/2) − 960(kω0T )2
kω0T = 2kπ; (kω0T )2 = 4k2π2
ak =160π2k2 cos(kπ/4) +
80πk
sin(kπ/2) − 240π2k2
[c] ak =80
π2k2 [2 cos(kπ/4) + πk sin(kπ/2) − 3]
a1 =80π2 [2 cos(π/4) + πk sin(π/2) − 3] ∼= 12.61
a3 =809π2 [2 cos(3π/4) + πk sin(3π/2) − 3] ∼= −12.46
a5 =80
25π2 [2 cos(5π/4) + πk sin(5π/2) − 3] ∼= 3.66
f(t) = 12.61 cos(ω0t) − 12.46 cos(3ω0t) + 3.66 cos(5ω0t) + . . .
[d] t =T
4; ω0t =
2πT
· T
4=
π
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 beenspecified 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
ak =4T
∫ T/6
0
(−2Vm
3
)cos kωotdt = − 4
T
2Vm
31
kωo
sin kωot∣∣∣∣T/6
0
= − 8Vm
3k2πsin
(kπ
3
)= −4Vm
3kπsin
(kπ
3
)
Therefore, v2(t − T/6) = −2Vm
9− 4Vm
3π
∞∑n=1
1n
sin(
nπ
3
)cos nωot
and v2(t) = −2Vm
9− 4Vm
3π
∞∑n=1
1n
sin(
nπ
3
)cos nωo(t + T/6)
Then, v(t) = v1(t) + v2(t). Simplifying,
v(t) =7Vm
9− 4Vm
3π
∞∑n=1
1n
[sin
(nπ
3
)cos
(nπ
3
)]cos nωot
+4Vm
3π
∞∑n=1
1n
[sin2
(nπ
3
)]sin nωot V
If Vm = 9π then av = 7π = 21.99 (Checks)
ak = −(12
n
)sin
(nπ
3
)cos
(nπ
3
)= −
(12n
)(12
)sin
(2nπ
3
)=( 6
n
)sin
(4nπ
3
)
bk =(12
n
)sin2
(nπ
3
)=(12
n
)(12
) [1 − cos
(2nπ
3
)]=( 6
n
) [1 − cos
(4nπ
3
)]
a1 = 6 sin(4π/3) = −5.2; b1 = 6[1 − cos(4π/3)] = 9
a2 = 3 sin(8π/3) = 2.6; 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!
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 =4T
∫ T/2
0
(Im − 2Im
Tt)
cos kωot dt
=4T
∫ T/2
0Im cos kω0t dt − 8Im
T 2
∫ T/2
0t cos kω0t dt
=4Im
T
sin kω0t
kω0
∣∣∣∣T/2
0−8Im
T 2
[cos kωot
k2ω20
+t
kω0sin kω0T
]T/2
0
= 0 − 8Im
T 2
[cos kπ
k2ω20
− 1k2ω2
0
]
=(8Im
T 2
)( 1k2ω2
0
)(1 − cos kπ)
=4Im
π2k2 =20k2 , for k odd
bk =4T
∫ T/2
0
(Im − 2Im
Tt)
sin kωot dt
=4Im
T
∫ T/2
0sin kω0t dt − 8Im
T 2
∫ T/2
0t sin kω0t dt
=4Im
T
[− cos kω0t
kω0
]T/2
0
− 8Im
T 2
[sin kωot
k2ω20
− t
kω0cos kω0t
]T/2
0
=4Im
T
[1 − cos kπ
kω0
]− 8Im
T 2
[−T cos kπ
2kω0
]
=8Im
kω0T
[1 +
12
cos kπ]
=2Im
πk=
10πk
, for k odd
ak − jbk =20k2 − j
10πk
=10k
(2k
− jπ)
=10k2
√π2k2 + 4/− θk
where tan θk =πk
2
i(t) = 10∞∑
n=1,3,5
√(nπ)2 + 4
n2 cos(nω0t − θn)
Problems 16–21
[b] A1 = 10√
4 + π2 ∼= 37.24 A tan θ1 =π
2θ1
∼= 57.52
A3 =109
√4 + 9π2 ∼= 10.71 A tan θ3 =
3π2
θ3∼= 78.02
A5 =1025
√4 + 25π2 ∼= 6.33 A tan θ5 =
5π2
θ5∼= 82.74
A7 =1049
√4 + 49π2 ∼= 4.51 A tan θ7 =
7π2
θ7∼= 84.80
A9 =1081
√4 + 81π2 ∼= 3.50 A tan θ9 =
9π2
θ9∼= 85.95
i(t) ∼= 37.24 cos(ωot − 57.52) + 10.71 cos(3ωot − 78.02)
+ 6.33 cos(5ωot − 82.74) + 4.51 cos(7ωot − 84.80)
+ 3.50 cos(9ωot − 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:
i(
T
4
)=
12(5π2) ∼= 24.67 A
P 16.19 The function has half-wave symmetry, thus ak = bk = 0 for k-even, av = 0; fork-odd
ak =4T
∫ T/2
0Vm cos kω0t dt − 8Vm
ρT
∫ T/2
0e−t/RC cos kω0t dt
where ρ =[1 + e−T/2RC
].
Upon integrating we get
ak =4Vm
T
sin kω0t
kω0
∣∣∣∣T/2
0
−8Vm
ρT· e−t/RC
(1/RC)2 + (kω0)2 ·[− cos kω0t
RC+ kω0 sin kω0t
] ∣∣∣∣T/2
0
=−8VmRC
T [1 + (kω0RC)2]
16–22 CHAPTER 16. Fourier Series
bk =4T
∫ T/2
0Vm sin kω0t dt − 8Vm
ρT
∫ T/2
0e−t/RC sin kω0t dt
= −4Vm
T
cos kω0t
kω0
∣∣∣∣T/2
0
−8Vm
ρT· −e−t/RC
(1/RC)2 + (kω0)2 ·[sin kω0t
RC+ kω0 cos kω0t
] ∣∣∣∣T/2
0
=4Vm
πk− 8kω0VmR2C2
T [1 + (kω0RC)2]
P 16.20 [a] a2k + b2
k = a2k +
(4Vm
πk+ kω0RCak
)2
= a2k [1 + (kω0RC)2] + 8Vm
πk
[2Vm
πk+ kω0RCak
]
But ak =−8VmRC
T [1 + (kω0RC)2]
Therefore a2k =
64V 2mR2C2
T 2[1 + (kω0RC)2]2, thus we have
a2k + b2
k =64V 2
mR2C2
T 2[1 + (kω0RC)2]+
16V 2m
π2k2 − 64V 2mkω0R
2C2
πkT [1 + (kω0RC)2]
Now let α = kω0RC and note that T = 2π/ω0, thus the expression for a2k + b2
k
reduces to a2k + b2
k = 16V 2m/π2k2(1 + α2). It follows that
√a2
k + b2k =
4Vm
πk√
1 + (kω0RC)2
[b] bk = kω0RCak +4Vm
πk
Thusbk
ak
= kω0RC +4Vm
πkak
= α − 1 + α2
α= − 1
α
Thereforeak
bk
= −α = −kω0RC
Problems 16–23
P 16.21 Since av = 0 (half-wave symmetry), Eq. 16.38 gives us
vo(t) =∞∑
1,3,5
4Vm
nπ
1√1 + (nω0RC)2
cos(nω0t − θn) where tan θn =bn
an
But from Eq. 16.57, we have tan βk = kω0RC. It follows from Eq. 16.72 thattan βk = −ak/bk or tan θn = − cot βn. Therefore θn = 90 + βn andcos(nω0t − θn) = cos(nω0t − βn − 90) = sin(nω0t − βn), thus ourexpression for vo becomes
vo =4Vm
π
∞∑n=1,3,5
sin(nω0t − βn)
n√
1 + (nω0RC)2
P 16.22 [a] e−x ∼= 1 − x for small x; therefore
e−t/RC ∼=(1 − t
RC
)and e−T/2RC ∼=
(1 − T
2RC
)
vo∼= Vm − 2Vm[1 − (t/RC)]
2 − (T/2RC)=(
Vm
RC
) [ 2t − (T/2)2 − (T/2RC)
]
∼=(
Vm
RC
)(t − T
4
)=(
Vm
RC
)t − VmT
4RCfor 0 ≤ t ≤ T
2
[b] ak =( −8
π2k2
)Vp =
( −8π2k2
)(VmT
4RC
)=
−4Vm
πω0RCk2
P 16.23 [a] Express vg as a constant plus a symmetrical square wave. The constant is Vm/2and the square wave has an amplitude of Vm/2, is odd, and has half- andquarter-wave symmetry. Therefore the Fourier series for vg is
vg =Vm
2+
2Vm
π
∞∑n=1,3,5
1n
sin nω0t
The dc component of the current is Vm/2R, and withsin nω0t = cos(nω0t − 90) the kth harmonic phase current is
Ik =2Vm/kπ
R + jkω0L/− 90 =2Vm
kπ√
R2 + (kω0L)2/− 90 − θk
where θk = tan−1
(kω0L
R
)
Thus the Fourier series for the steady-state current is
i =Vm
2R+
2Vm
π
∞∑n=1,3,5
sin(nω0t − θn)
n√
R2 + (nω0L)2A
16–24 CHAPTER 16. Fourier Series
[b]
The steady-state current will alternate between I1 and I2 in exponential tracesas shown. Assuming t = 0 at the instant i increases toward (Vm/R), we have
i =Vm
R+(I1 − Vm
R
)e−t/τ for 0 ≤ t ≤ T
2
and i = I2e−[t−(T/2)]/τ for T/2 ≤ t ≤ T, where τ = L/R. Now we solve for I1
and I2 by noting that
I1 = I2e−T/2τ and I2 =
Vm
R+(I1 − Vm
R
)e−T/2τ
These two equations are now solved for I1. Letting x = T/2τ, we get
I1 =(Vm/R)e−x
1 + e−x
Therefore the equations for i become
i =Vm
R−[
Vm
R(1 + e−x)
]e−t/τ for 0 ≤ t ≤ T
2and
i =[
Vm
R(1 + e−x)
]e−[t−(T/2)]/τ for
T
2≤ t ≤ T
A check on the validity of these expressions shows they yield an average valueof (Vm/2R):
Iavg =1T
∫ T/2
0
[Vm
R+(I1 − Vm
R
)e−t/τ
]dt +
∫ T
T/2I2e
−[t−(T/2)]/τ dt
=1T
VmT
2R+ τ(1 − e−x)
(I1 − Vm
R+ I2
)
=Vm
2Rsince I1 + I2 =
Vm
R
Problems 16–25
P 16.24 vi =4Aπ
∞∑n=1,3,5
1n
sin nω0(t + T/4)
=4Aπ
∞∑n=1,3,5
( 1n
sinnπ
2
)cos nω0t
ω0 =2π4π
× 103 = 500 rad/s;4Aπ
= 60
vi = 60∞∑
n=1,3,5
( 1n
sinnπ
2
)cos 500nt V
From the circuit
Vo =Vi
R + jωL· jωL =
jω
R/L + jωVi =
jω
1000 + jωVi
Vi1 = 60/0 V; ω = 500 rad/s
Vi3 = −20/0 = 20/180 V; 3ω = 1500 rad/s
Vi5 = 12/0 V; 5ω = 2500 rad/s
Vo1 =j500
1000 + j500(60/0) = 26.83/63.43 V
Vo3 =j1500
1000 + j1500(20/180) = 16.64/− 146.31 V
Vo5 =j2500
1000 + j2500(12/0) = 11.14/21.80 V
·. . 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 inputvoltage is
vg = 42∞∑
n=1,3,5
[ 1n
sin(
nπ
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
R+
Vo − Vg
sL+ (Vo − Vg)sC = 0
·. .Vo
Vg
= H(s) =s2 + 1/LC
s2 + (s/RC) + (1/LC)
Substituting in the numerical values gives
H(s) =s2 + 108
s2 + 500s + 108
H(j2000) =96
96 + j1= 0.9999/− 0.60
H(j6000) =64
64 + j3= 0.9989/− 2.68
H(j10,000) = 0
H(j14,000) =96
96 − j7= 0.9974/4.17
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 eliminatedfrom the output voltage by the circuit, which is a bandreject filter with a centerfrequency of 10,000 rad/s.
P 16.26 [a] Note – find io(t)
V0 − Vg
16s+ V0(12.5 × 10−6s) +
V0
1000= 0
V0
[ 116s
+ 12.5 × 10−6s +1
1000
]=
Vg
16s
Problems 16–27
V0(1000 + 0.2s2 + 16s) = 1000Vg
V0 =5000Vg
s2 + 80s + 5000
I0 =V0
1000=
5Vg
s2 + 80s + 5000
H(s) =I0
Vg
=5
s2 + 80s + 5000
H(njω0) =5
(5000 − n2ω20) + j80nω0
ω0 =2πT
= 240π; ω20 = 57,600π2; 80ω0 = 19,200π
H(jnω0) =5
(5000 − 57,600π2n2) + 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
vg =680π
− 1360π
[13
cos ω0t +115
cos 2ω0t +135
cos 3ω0t +163
cos 4ω0t + . . .]
i0 =680π
× 10−3 − 13603π
(8.82 × 10−6) cos(ω0t − 173.89)
− 136015π
(2.20 × 10−6) cos(2ω0t − 176.96)
− 136035π
(9.78 × 10−7) cos(3ω0t − 177.97)
− 136063π
(5.5 × 10−7) cos(4ω0t − 178.48) − . . .
= 216.45 × 10−3 − 1.27 × 10−3 cos(ω0t − 173.89)
− 6.35 × 10−5 cos(2ω0t − 176.96)
− 1.21 × 10−5 cos(3ω0t − 177.97)
− 3.8 × 10−6 cos(4ω0t − 178.48) − . . .
16–28 CHAPTER 16. Fourier Series
i0 ∼= 216.45 − 1.27 cos(ω0t − 173.89) mA
Note that the sinusoidal component is very small compared to the dccomponent, so
i0 ∼= 216.45 mA (a dc current)
[b] Yes, the solution makes sense. The circuit is a low-pass filter which nearlyeliminates 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
bk =8T
∫ T/4
0f(t) sin kωot, k odd
=8T
∫ T/10
0500t sin kωot dt +
∫ T/4
T/10sin kωot dt
=8T
Int1 + Int2
Int1 = 500∫ T/10
0t sin kωot dt
= 500[
1k2ω2
o
sin kωot − t
kωo
cos kωot
∣∣∣∣T/10
0
]
=500k2ω2
o
sinkπ
5− 50T
kωo
coskπ
5
Int2 =∫ T/4
T/10sin kωot dt =
−1kωo
cos kωot
∣∣∣∣T/4
T/10=
1kωo
coskπ
5
Int1 + Int2 =500k2ω2
o
sinkπ
5+( 1
kωo
− 50Tkωo
)cos
kπ
5
50T = 50(20 × 10−3) = 1
·. . Int1 + Int2 =500k2ω2
o
sinkπ
5
bk =[ 8T
· 5004π2k2 · T 2
]sin
kπ
5=
20π2k2 sin
kπ
5, k odd
Problems 16–29
i(t) =20π2
∞∑n=1,3,5
sin(nπ/5)n2 sin nωot A
From the circuit,
H(s) =Vo
Ig
= Zeq
Yeq =1R1
+1
R2 + sL+ sC
Zeq =1/C(s + R2/L)
s2 + s(R1R2C + L)/R1LC + (R1 + R2)/R1LC
Therefore,
H(s) =320 × 104(s + 32 × 104)
s2 + 32.8 × 104s + 28.8 × 108
We want the output for the third harmonic:
ω0 =2πT
=2π
20 × 10−3 = 100π; 3ω0 = 300π
Ig3 =209π2 sin
3π5 sin 3ω0t
= 0.214/− 90
H(j300π) =320 × 104(j300π + 32 × 104)
(j300π)2 + 32.8 × 104(j300π) + 28.8 × 108 = 353.6/− 5.96
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 =2πT
=2π10π
× 106 = 200 krad/s
·. . n =3 × 106
0.2 × 106 = 15; n =5 × 106
0.2 × 106 = 25
H(s) =Vo
Vg
=(1/RC)s
s2 + (1/RC)s + (1/LC)
16–30 CHAPTER 16. Fourier Series
1RC
=1012
(250 × 103)(4)= 106;
1LC
=(103)(1012)
(10)(4)= 25 × 1012
H(s) =106s
s2 + 106s + 25 × 1012
H(jω) =jω × 106
(25 × 1012 − ω2) + j106ω
15th harmonic input:
vg15 = (150)(1/15) sin(15π/2) cos 15ωot = −10 cos 3 × 106t V
·. . Vg15 = 10/− 180 V
H(j3 × 106) =j3
16 + j3= 0.1843/79.38
Vo15 = (10)(0.1843)/− 100.62 V
vo15 = 1.84 cos(3 × 106t − 100.62) V
25th harmonic input:
vg25 = (150)(1/25) sin(25π/2) cos 5 × 106t = 6 cos 5 × 106t V
·. . Vg25 = 6/0 V
H(j5 × 106) =j5
0 + j5= 1/0
Vo25 = 6/0 V
vo25 = 6 cos 5 × 106t V
P 16.29 [a] av =T
2
[12
(T
2
)Im +
T
2Im
]=
3Im
4
i(t) =2Im
Tt, 0 ≤ t ≤ T/2
i(t) = Im, T/2 ≤ t ≤ T
ak =2T
∫ T/2
0
2Im
Tt cos kωot dt +
2T
∫ T
T/2Im cos kωot dt
Problems 16–31
=Im
π2k2 (cos kπ − 1)
bk =2T
∫ T/2
0
2Im
Tt sin kωot dt +
2T
∫ T
T/2Im sin kωot dt
=−Im
πk
av =3Im
4, a1 =
−2Im
π2 , a2 = 0
a3 =−2Im
9π2
b1 =−Im
π, b2 =
−Im
2π
·. . Irms = Im
√916
+2π4 +
12π2 +
18π2 = 0.8040Im (Eq. 16.81)
Irms = 192.95 mA
P = (0.19295)2(1000) = 37.23 W
[b] Area under i2:
A =∫ T/2
0
4I2m
T 2 t dt + I2m
T
2
=4I2
m
T 2
t3
3
∣∣∣∣T/2
0+I2
m
T
2
= I2mT
[16
+36
]=
23TI2
m
Irms =
√1T
· 23TI2
m =
√23Im = 195.96 mA
P = (0.19596)21000 = 38.4 W
[c] Error =(37.23
38.40− 1
)(100) = −3.05%
P 16.30 vg = 10 +80π2
∞∑n=1,3,5
1n2 cos nωot V
ωo =2πT
=2π4π
× 103 = 500 rad/s
vg = 10 +80π2 cos 500t +
809π2 cos 1500t + . . .
16–32 CHAPTER 16. Fourier Series
Vo − Vg
sL+ sCVo +
Vo
R= 0
Vo(RLCs2 + Ls + R) = RVg
H(s) =Vo
Vg
=1/LC
s2 + s/RC + 1/LC
1LC
=106
(0.1)(10)= 106
1RC
=106
(50√
2)(10)= 1000
√2
H(s) =106
s2 + 1000√
2s + 106
H(jω) =106
106 − ω2 + j1000ω√
2
H(j0) = 1
H(j500) = 0.9701/− 43.31
H(j1500) = 0.4061/− 120.51
vo = 10(1) +80π2 (0.9701) cos(500t − 43.31)
+809π2 (0.4061) cos(1500t − 120.51) + . . .
vo = 10 + 7.86 cos(500t − 43.31) + 0.3658 cos(1500t − 120.51) + . . .
Problems 16–33
Vrms∼=√√√√102 +
(7.86√
2
)2
+(
0.3658√2
)2
= 11.44 V
P ∼= V 2rms
50√
2= 1.85 W
Note – the higher harmonics are severely attenuated and can be ignored. Forexample, the 5th harmonic component of vo is
vo5 = (0.1580)( 80
25π2
)cos(2500t − 146.04) = 0.0512 cos(2500t − 146.04) V
P 16.31 [a] av =2(
12
T4 Vm
)T
=Vm
4
ak =4T
∫ T/4
0
[Vm − 4Vm
Tt]cos kωot dt
=4Vm
π2k2
[1 − cos
kπ
2
]
bk = 0, all k
av =604
= 15 V
a1 =240π2
a2 =2404π2 (1 − cos π) =
120π2
Vrms =
√√√√(15)2 +12
[(240π2
)2
+(120
π2
)2]
= 24.38 V
P =(24.38)2
10= 59.46 W
[b] Area under v2; 0 ≤ t ≤ T/4
v2 = 3600 − 28,800T
t +57,600
T 2 t2
A = 2∫ T/4
0
[3600 − 28,800
Tt +
57,600T 2 t2
]dt = 600T
Vrms =
√1T
600T =√
600 = 24.49 V
P =√
6002/10 = 60 W
16–34 CHAPTER 16. Fourier Series
[c] Error =(59.46
60.00− 1
)100 = −0.9041%
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
P = (15)(2) +12(400)(5) cos(30) +
12(100)(3) cos(−75) = 934.85 W
[b] Vrms =
√√√√(15)2 +(
400√2
)2
+(
100√2
)2
= 291.93 V
[c] Irms =
√√√√(2)2 +(
5√2
)2
+(
3√2
)2
= 4.58 A
P 16.33 [a] Area under v2 = A = 4∫ T/6
0
36V 2m
T 2 t2 dt + 2V 2m
(T
3− T
6
)
=2V 2
mT
9+
V 2mT
3
Therefore Vrms =
√√√√ 1T
(2V 2
mT
9+
V 2mT
3
)= Vm
√29
+13
= 74.5356 V
[b] vg = 105.30 sin ω0t − 4.21 sin 5ω0t + 2.15 sin 7ω0t + · · · V
Therefore Vrms∼=√
(105.30)2 + (4.21)2 + (2.15)2
2= 74.5306 V
P 16.34 [a] v(t) =480π
sin ωot +13
sin 3ωot +15
sin 5ωot +17
sin 7ωot +19
sin 9ωot + · · ·
Vrms∼= 480
π
√√√√( 1√2
)2
+(
13√
2
)2
+(
15√
2
)2
+(
17√
2
)2
+(
19√
2
)2
=480π√
2
√1 +
19
+125
+149
+181
∼= 117.55 V
[b] % error =(117.55
120− 1
)(100) = −2.04%
[c] v(t) =960π2
sin ωot +
19
sin 3ωot +125
sin 5ωot
+149
sin 7ωot +181
sin 9ωot − · · ·
Vrms∼= 960
π2√
2
√1 +
181
+1
625+
12401
+1
6561
∼= 69.2765 V
Problems 16–35
Vrms =120√
3= 69.2820 V
% error =(69.2765
69.2820− 1
)(100) = −0.0081%
P 16.35 [a] v(t) ≈ 340π
− 680π
13
cos ωot +115
cos 2ωot + · · ·
Vrms ≈√√√√√(340
π
)2
+(680
π
)2(
13√
2
)2
+(
115
√2
)2
=340π
√1 + 4
( 118
+1
450
)= 120.0819 V
[b] Vrms =170√
2= 120.2082
% error =(120.0819
120.2082− 1
)(100) = −0.11%
[c] v(t) ≈ 170π
+ 85 sin ωot − 3403π
cos 2ωot
Vrms ≈√√√√(170
π
)2
+(
85√2
)2
+(
3403√
2π
)2
≈ 84.8021 V
Vrms =1702
= 85 V
% error = −0.23%
P 16.36 [a] Half-wave symmetry av = 0, ak = bk = 0, even k. For k odd,
ak =4T
∫ T/4
0
4Im
Tt cos kω0t dt =
16Im
T 2
∫ T/4
0t cos kω0t dt
=16Im
T 2
cos kω0t
k2ω20
+t
kω0sin kω0t
∣∣∣∣T/4
0
=16Im
T 2
0 +
T
4kω0sin
kπ
2− 1
k2ω20
ak =2Im
πk
[sin
(kπ
2
)− 2
πk
],
bk =4T
∫ T/4
0
4Im
Tt sin kω0t dt =
16Im
T 2
∫ T/4
0t sin kω0t dt
=16Im
T 2
sin kω0t
k2ω20
− t
kω0cos kω0t
∣∣∣∣T/4
0=
4Im
π2k2 sin(
kπ
2
)
16–36 CHAPTER 16. Fourier Series
[b] ak − jbk =2Im
πk
[sin
(kπ
2
)− 2
πk
]−[j
2πk
sin(
kπ
2
)]
a1 − jb1 =2Im
π
(1 − 2
π
)− j
2π
= 0.47Im/− 60.28
a3 − jb3 =2Im
3π
(−1 − 2
3π
)+ j
( 23π
)= 0.26Im/170.07
a5 − jb5 =2Im
5π
(1 − 2
5π
)− j
( 25π
)= 0.11Im/− 8.30
a7 − jb7 =2Im
7π
(−1 − 2
7π
)+ j
( 27π
)= 0.10Im/175.23
ig = 0.47Im cos(ω0t − 60.28) + 0.26Im cos(3ω0t + 170.07)
+ 0.11Im cos(5ω0t − 8.30) + 0.10Im cos(7ω0t + 175.23) + · · ·
[c] Ig =
√√√√ ∞∑n=1,3,5
(A2
n
2
)
∼= Im
√(0.47)2 + (0.26)2 + (0.11)2 + (0.10)2
2= 0.39Im
[d] Area under i2g = 2∫ T/4
0
(4Im
Tt)2
dt =(
32I2m
T 2
)(t3
3
) ∣∣∣∣T/4
0=
I2mT
6
Ig =
√√√√ 1T
(I2mT
6
)=
Im√6
= 0.41Im
[e] % error =(
estimatedexact
− 1)
100 =(
0.3927Im
(Im/√
6)− 1
)100 = −3.8%
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
bk =8T
∫ T/4
0f(t) sin kωot dt, k-odd
=8T
∫ T/8
0
Vm
4sin kωot dt +
∫ T/4
T/8Vm sin kωot dt
=8Vm
4T
[−cos kωot
kωo
∣∣∣∣T/8
0
]+
8Vm
T
[−cos kωot
kωo
∣∣∣∣T/4
T/8
]
=8Vm
4kωoT
[1 − cos
kπ
4
]+
8Vm
Tkωo
[cos
kπ
4− 0
]
Problems 16–37
=8Vm
kωoT
14
− 14
coskπ
4+ cos
kπ
4
=4Vm
πk
14
+ 0.75 coskπ
4
=
1k[10 + 30 cos(kπ/4)]
b1 = 10 + 30 cos(π/4) = 31.21
b3 =13[10 + 30 cos(3π/4)] = −3.74
b5 =15[10 + 30 cos(5π/4)] = −2.24
b7 =17[10 + 30 cos(7π/4)] = 4.46
V (rms) ≈ Vm
√31.212 + 3.742 + 2.242 + 4.462
2= 22.51
[b] Area under v2 = 2[2(2.5π)2
(T
8
)+ 100π2
(T
4
)]= 53.125π2T
V (rms) =
√1T
(53.125π2)T =√
53.125π = 22.90
[c] % Error =(22.51
22.90− 1
)(100) = −1.7%
P 16.38 [a] From Problem 16.16,
The area under v2:
A = 4[∫ T/8
0
14,400T 2 t2 dt +
∫ T/4
T/8
(10 +
40tT
)2
dt
]
16–38 CHAPTER 16. Fourier Series
=57,600
T 2
t3
3
∣∣∣∣T/8
0+ 400t
∣∣∣∣T/4
T/8+
3200T
t2
2
∣∣∣∣T/4
T/8+
6400T 2
t3
3
∣∣∣∣T/4
T/8
=57,6001536
T + 400T
8+ 1600
3T64
+ 64007T
1536=
5753
T
Vrms =
√1T
(5753
T)
=
√5753
= 13.84 V
[b] P =V 2
rms
15= 12.78 W
[c] From Problem 16.16,
b1 =80π2 (2 cos 45 + π sin 90 − 3) = 12.61 V
vg∼= 12.61 sin ω0t V
P =(19.57/
√2)2
15= 5.30 W
[d] % error =( 5.30
13.84− 1
)(100) = −61.71%
P 16.39 Figure P16.39(b): ta = 0.2 s; tb = 0.6 s
v = 50t 0 ≤ t ≤ 0.2
v = −50t + 20 0.2 ≤ t ≤ 0.6
v = 25t − 25 0.6 ≤ t ≤ 1.0
Area 1 under v2 = A1 =∫ 0.2
02500t2 dt =
203
Area 2 = A2 =∫ 0.6
0.2100(4 − 20t + 25t2) dt =
403
Area 3 = A3 =∫ 1.0
0.6625(t2 − 2t + 1) dt =
403
A1 + A2 + A3 =1003
Vrms =
√11
(1003
)=
10√3
V.
Problems 16–39
Figure P16.39(c): ta = tb = 0.4 s
v(t) = 25t 0 ≤ t ≤ 0.4
v(t) =503
(t − 1) 0.4 ≤ t ≤ 1
A1 =∫ 0.4
0625t2 dt =
403
A2 =∫ 1.0
0.4
25009
(t2 − 2t + 1) dt =603
A1 + A2 =1003
Vrms =
√1T
(A1 + A2) =
√11
(1003
)=
10√3
V.
Figure P16.39 (d): ta = tb = 1
v = 10t 0 ≤ t ≤ 1
A1 =∫ 1
0100t2 dt =
1003
Vrms =
√11
(1003
)=
10√3
V.
P 16.40 cn =1T
∫ T/4
0Vme−jnωot dt =
Vm
T
[e−jnωot
−jnωo
∣∣∣∣T/4
0
]
=Vm
Tnωo
[j(e−jnπ/2 − 1)] =Vm
2πnsin
nπ
2+ j
Vm
2πn
(cos
nπ
2− 1
)
=Vm
2πn
[sin
nπ
2− j
(1 − cos
nπ
2
)]
v(t) =∞∑
n=−∞cne
jnωot
co = av =1T
∫ T/4
0Vm dt =
Vm
4
16–40 CHAPTER 16. Fourier Series
or
co =Vm
2πlimn→0
[sin(nπ/2)
n− j
1 − cos(nπ/2)n
]
=Vm
2πlimn→0
[(π/2) cos(nπ/2)
1− j
(π/2) sin(nπ/2)1
]
=Vm
2π
[π
2− j0
]=
Vm
4
Note it is much easier to use co = av than to use L’Hopital’s rule to find the limit of0/0.
P 16.41 co = av =VmT
2· 1T
=Vm
2
cn =1T
∫ T
0
Vm
Tte−jnωot dt
=Vm
T 2
[e−jnω0t
−n2ω20(−jnω0t − 1)
]T
0
=Vm
T 2
[e−jn2πT/T
−n2ω20
(−jn
2πT
T − 1)
− 1−n2ω2
0(−1)
]
=Vm
T 2
[1
n2ω20(1 + jn2π) − 1
n2ω20
]
= jVm
2nπ, n = ±1, ±2, ±3, . . .
P 16.42 [a] Vrms =
√1T
∫ T
0v2 dt =
√1T
∫ T
0
(Vm
T
)2
t2 dt
=
√V 2
m
T 3
t3
3
∣∣∣∣T0
=
√V 2
m
3=
Vm√3
P =(120/
√3)2
10= 480 W
[b] From the solution to Problem 16.41
c0 =1202
= 60 V; c4 = j1208π
= j15π
c1 = j1202π
= j60π
; c5 = j12010π
= j12π
Problems 16–41
c2 = j1204π
= j30π
; c6 = j12012π
= j10π
c3 = j1206π
= j20π
; c7 = j12014π
= j8.57π
Vrms =
√√√√c2o + 2
∞∑n=1
|cn|2
=√
602 + 2π2 (602 + 302 + 202 + 152 + 122 + 102 + 8.572)
= 68.58 V
[c] P =(68.58)2
10= 470.29 W
% error =(470.29
480− 1
)(100) = −2.02%
P 16.43 [a] Co = av =(1/2)(T/2)Vm
T=
Vm
4
Cn =1T
∫ T/2
0
2Vm
Tte−jnωot dt
=2Vm
T 2
[e−jnωot
−n2ω2o
(−jnωot − 1)]T/2
0
=Vm
2n2π2 [e−jnπ(jnπ + 1) − 1]
Since e−jnπ = cos nπ we can write
Cn =Vm
2π2n2 (cos nπ − 1) + jVm
2nπcos nπ
[b] Co =544
= 13.5 V
C−1 =−54π2 + j
27π
= 10.19/122.48 V
C1 = 10.19/− 122.48 V
C−2 = −j13.5π
= 4.30/− 90 V
C2 = 4.30/90 V
C−3 =−6π2 + j
9π
= 2.93/101.98 V
C3 = 2.93/− 101.98 V
16–42 CHAPTER 16. Fourier Series
C−4 = −j6.75π
= 2.15/− 90 V
C4 = 2.15/90 V
[c]
Vo
250+
Vo
sL+ VosC +
Vo − Vg
62.5 × 103 = 0
·. . (250LCs2 + 1.004sL + 250)Vo = 0.004sLVg
Vo
Vg
= H(s) =(1/62, 500C)s
s2 + 1/249C + 1/LC
H(s) =16s
s2 + 1/249Cs + 4 × 1010
ωo =2πT
=2π10π
× 106 = 2 × 105 rad/s
H(j0) = 0
H(j2 × 105k) =jk
12, 500(1 − k2) + j251k
Therefore,
H−1 = 0.0398/0; H1 = 0.0398/0
H−2 =−j2
−37, 500 − j20= 5.33 × 10−5/86.23; H2 = 5.33 × 10−5/− 89.23
H−3 =−3j
−10−5 − j753= 3.00 × 10−5/89.57; H2 = 3.00 × 10−5/− 89.57
H−4 =−4j
−187, 500 − j1004= 2.13 × 10−5/89.69; H2 = 2.13 × 10−5/− 89.69
The output voltage coefficients:
C0 = 0
C−1 = (10.19/122.48)(0.00398/0) = 0.0406/122.48 V
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∼=√√√√C2
o + 24∑
n=1|Cn|2 ∼=
√√√√24∑
n=1|Cn|2
∼=√
2(0.04062 + (2.29 × 10−4)2 + (8.79 × 10−5)2 + (4.58 × 10−5)2 ∼= 0.0574 V
P =(0.0574)2
250= 13.2 µW
P 16.44 [a] Vrms =
√1T
∫ T/2
0
(2Vm
Tt)2
dt
=
√√√√ 1T
[4V 2
m
T 2
t3
3
]T/2
0
=
√√√√ 4V 2m
(3)(8)=
Vm√6
Vrms =54√
6= 22.05 V
[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
[c] % Error =(21.29
22.05− 1
)(100) = −3.44%
16–44 CHAPTER 16. Fourier Series
P 16.45 [a] From Example 16.3 we have:
av =404
= 10 V, ak =40πk
sin(
kπ
2
)
bk =40πk
[1 − cos
(kπ
2
)], Ak/− θ
k = ak − jbk
A1 = 18.01 V θ1 = −45, A2 = 12.73 V, θ2 = −90
A3 = 6 V, θ3 = −135, A4 = 0, A5 = 3.6 V, θ5 = −45
A6 = 4.24 V, θ6 = −90, A7 = 2.57 V, θ7 = −135
[b] Cn =an − jbn
2, C−n =
an + jbn
2= C∗
n
C0 = av = 10 V C3 = 3/135 V C6 = 2.12/90 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
Problems 16–45
P 16.46 [a] From the solution to Problem 16.29 we have
Ak = ak − jbk =Im
π2k2 (cos kπ − 1) + jIm
πk
A0 = 0.75Im = 180 mA
A1 =240π2 (−2) + j
240π
= 90.56/122.48 mA
A2 = j2402π
= 38.20/90 mA
A3 =2409π2 (−2) + j
2403π
= 26.03/101.98 mA
A4 = j2404π
= 19.10/90 mA
A5 =24025π2 (−2) + j
2405π
= 15.40/97.26 mA
A6 = j2406π
= 12.73/90 mA
[b] C0 = A0 = 180 mA
C1 =12A1/− θ1 = 45.28/122.48 mA
C−1 = 45.28/− 122.48 mA
C2 =12A2/− θ2 = 19.1/90 mA
C−2 = 19.1/− 90 mA
16–46 CHAPTER 16. Fourier Series
C3 =12A3/− θ3 = 13.02/101.98 mA
C−3 = 13.02/− 101.98 mA
C4 =12A4/− θ4 = 9.55/90 mA
C−4 = 9.55/− 90 mA
C5 =12A5/− θ5 = 7.70/97.26 mA
C−5 = 7.70/− 97.26 mA
C6 =12A6/− θ6 = 6.37/90 mA
C−6 = 6.37/− 90 mA
P 16.47 [a] v = A1 cos(ωot + 90) + A3 cos(3ωot − 90)
+A5 cos(5ωot + 90) + A7 cos(7ωot − 90)
v = −A1 sin ωot + A3 sin 3ωot − A5 sin 5ωot + A7 sin 7ωot
[b] v(−t) = A1 sin ωot − A3 sin 3ωot + A5 sin 5ωot − A7 sin 7ωot
·. . v(−t) = −v(t); odd function
[c] v(t − T/2) = −A1 sin(ωot − π) + A3 sin(3ωot − 3π)
−A5 sin(5ωot − 5π) + A7 sin(7ωot − 7π)
= A1 sin ωot − A3 sin 3ωot + A5 sin 5ωot − A7 sin 7ωot
·. . 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 − ωot) + A3 sin(3π − 3ωot)
A5 sin(5π − 5ωot) + A7 sin(7π − 7ωot)
= −A1 sin ωot + A3 sin 3ωot − A5 sin 5ωot + A7 sin 7ωot
·. . 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), Function is even
[c] Yes, A0 = 0, An = 0 for n even
[d] Irms =
√11,0252 + 12252 + 4412 + 2252
2= 7.85 mA
[e] A1 = 11,025/0 µA; C1 = 5512.50/0 µA
A3 = 1225/180 µA; C3 = 612.5/180 µA
A5 = 441/180 µA; C5 = 220.5/180 µA
A7 = 225/0 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−j180e−j50,000t + 612.5e−j180
e−j30,000t
+ 5512.5e−j10,000t + 5512.5ej10,000t + 612.5ej180ej30,000t
+ 220.5ej180ej50,000t + 112.5ej70,000t µA
16–48 CHAPTER 16. Fourier Series
[f]
θn
(krad/s)10 30 50 70
−70 −50 −30 −10
−180˚
180˚
P 16.49 From Table 15.1 we have
H(s) =1
(s + 1)(s2 + s + 1)
After scaling we get
H ′(s) =106
(s + 100)(s2 + 100s + 104)
ωo =2πT
=2π5π
× 103 = 400 rad/s
·. . H ′(jnωo) =1
(1 + j4n)[(1 − 16n2) + j4n]
It follows that
H(j0) = 1/0
Problems 16–49
H(jωo) =1
(1 + j4)(−15 + j4)= 0.0156/− 241.03
H(j2ωo) =1
(1 + j8)(−63 + j8)= 0.00195/− 255.64
vg(t) =A
π+
A
2sin ωot − 2A
π
∞∑n=2,4,6,
cos nωot
n2 − 1
= 54 + 27π sin ωot − 36 cos 2ωot − · · · 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 forvg(t). We get
vg(t) = 100 +800π2
∞∑n=1,3,5,
1n2 cos nωot
The transfer function of the prototype second-order low pass Butterworth filter is
H(s) =1
s2 +√
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
(j5000)2 + 2000√
2(j5000)2 + 4 × 106= 0.1580/− 146.04
H(j15,000) =4 × 106
(j15,000)2 + 2000√
2(j15,000)2 + 4 × 106= 0.0178/− 169.13
Vdc = 100 V
Vg1 =800π2 /0 V
Vg3 =8009π2 /0 V
16–50 CHAPTER 16. Fourier Series
Vodc = 100(1) = 100 V
Vo1 =800π2 (0.1580/− 146.04) = 12.81/− 146.04 V
Vo3 =8009π2 (0.0178/− 169.13) = 0.16/− 169.13 V
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 − Vg
R1+
Va
R2+ VasC2 + (Va − Vo)sC1 = 0
(0 − Va)sC2 +0 − Vo
R3= 0
Solving for Vo in terms of Vg yields
Vo
Vg
= H(s) =−1
R1C1s
s2 + 1R3
(1
C1+ 1
C2
)s + R1+R2
R1R2R3C1C2
It follows that
ω2o =
R1 + R2
R1R2R3C1C2
β =1R3
( 1C1
+1C2
)
Ko =R3
R1
(C2
C1 + C2
)
Note that
H(s) =−R3
R1
(C2
C1+C2
)1
R3
(1
C1+ 1
C2
)s
s2 + 1R3
(1
C1+ 1
C2
)s +
(R1+R2
R1R2R3C1C2
)[b] For the given values of R1, R2, R3, C1, and C2 we have
−R3
R1
(C2
C1 + C2
)= − R3
2R1= −400
313
1R3
( 1C1
+1C2
)= 2000
R1 + R2
R1R2R3C1C2= 0.16 × 1010 = 16 × 108
Problems 16–51
H(s) =−(400/313)(2000)s
s2 + 2000s + 16 × 108
ωo =2πT
=2π50π
× 106 = 4 × 104 rad/s
H(jnωo) =−(400/313)(2000)jnωo
16 × 108 − n2ω2o + j2000nωo
=−j(20/313)n
(1 − n2) + j0.05n
H(jωo) =−j(20/313)
j(0.050)= −400
313= −1.28
H(j3ωo) =−j(20/313)(3)
−8 + j0.15= 0.0240/91.07
H(j5ωo) =−j(100/313)−24 + j0.25
= 0.0133/90.60
vg(t) =4Aπ
∞∑n=1,3,5
1n
sin(nπ/2) cos nωot
A = 15.65π V
vg(t) = 62.60 cos ωot − 20.87 cos 3ωot + 12.52 cos 5ωot − · · ·
vo(t) = −80 cos ωot − 0.50 cos(3ωot + 91.07)
+ 0.17 cos(5ωot + 90.60) − · · · V
17The Fourier Transform
Assessment Problems
AP 17.1 [a] F (ω) =∫ 0
−τ/2(−Ae−jωt) dt +
∫ τ/2
0Ae−jωt dt
=A
jω[2 − ejωτ/2 − e−jωτ/2]
=2Ajω
[1 − ejωτ/2 + e−jωτ/2
2
]
=−j2A
ω[1 − cos
ωτ
2]
[b] F (ω) =∫ ∞
0te−ate−jωt dt =
∫ ∞
0te−(a+jω)t dt =
1(a + jω)2
AP 17.2 f(t) =12π
∫ −2
−34ejtω dω +
∫ 2
−2ejtω dω +
∫ 3
24ejtω dω
=1
j2πt4e−j2t − 4e−j3t + ej2t − e−j2t + 4ej3t − 4ej2t
=1πt
[3e−j2t − 3ej2t
j2+
4ej3t − 4e−j3t
j2
]
=1πt
(4 sin 3t − 3 sin 2t)
AP 17.3 [a] F (ω) = F (s) |s=jω= Le−at sin ω0ts=jω
=ω0
(s + a)2 + ω20
∣∣∣∣s=jω
=ω0
(a + jω)2 + ω20
[b] F (ω) = Lf(−t)s=−jω =[
1(s + a)2
]s=−jω
=1
(a − jω)2
17–1
17–2 CHAPTER 17. The Fourier Transform
[c] f+(t) = te−at, f−(t) = teat
Lf+(t) =1
(s + a)2 , Lf−(−t) =−1
(s + a)2
Therefore F (ω) =1
(a + jω)2 − 1(a − jω)2 =
−j4aω
(a2 + ω2)2
AP 17.4 [a] f ′(t) =2Aτ
,−τ
2< t < 0; f ′(t) =
−2Aτ
, 0 < t <τ
2
·. . f ′(t) =2Aτ
[u(t + τ/2) − u(t)] − 2Aτ
[u(t) − u(t − τ/2)]
=2Aτ
u(t + τ/2) − 4Aτ
u(t) +2Aτ
u(t − τ/2)
·. . f ′′(t) =2Aτ
δ(t +
τ
2
)− 4A
τδ(t) +
2Aτ
δ(t − τ
2
)
[b] Ff ′′(t) =[2A
τejωτ/2 − 4A
τ+
2Aτ
e−jωτ/2]
=4Aτ
[ejωτ/2 + e−jωτ/2
2− 1
]=
4Aτ
[cos
(ωτ
2
)− 1
]
[c] Ff ′′(t) = (jω)2F (ω) = −ω2F (ω); therefore F (ω) = − 1ω2Ff ′′(t)
Thus we have F (ω) = − 1ω2
4Aτ
[cos
(ωτ
2
)− 1
]
AP 17.5 v(t) = Vm
[u(t +
τ
2
)− u
(t − τ
2
)]
Fu(t +
τ
2
)=[πδ(ω) +
1jω
]ejωτ/2
Fu(t − τ
2
)=[πδ(ω) +
1jω
]e−jωτ/2
Therefore V (ω) = Vm
[πδ(ω) +
1jω
] [ejωτ/2 − e−jωτ/2
]
= j2Vmπδ(ω) sin(
ωτ
2
)+
2Vm
ωsin
(ωτ
2
)
=(Vmτ) sin(ωτ/2)
ωτ/2
Problems 17–3
AP 17.6 [a] Ig(ω) = F10sgn t =20jω
[b] H(s) =Vo
Ig
Using current division and Ohm’s law,
Vo = −I2s = −[ 44 + 1 + s
](−Ig)s =
4s5 + s
Ig
H(s) =4s
s + 5, H(ω) =
j4ω5 + jω
[c] Vo(ω) = H(ω) · Ig(ω) =(
j4ω5 + jω
)(20jω
)=
805 + jω
[d] vo(t) = 80e−5tu(t) V
[e] Using current division,
i1(0−) =15ig =
15(−10) = −2 A
[f] i1(0+) = ig + i2(0+) = 10 + i2(0−) = 10 + 8 = 18 A
[g] Using current division,
i2(0−) =45(10) = 8 A
[h] Since the current in an inductor must be continuous,
i2(0+) = i2(0−) = 8 A
[i] Since the inductor behaves as a short circuit for t < 0,
vo(0−) = 0 V
[j] vo(0+) = 1i2(0+) + 4i1(0+) = 80 V
AP 17.7 [a] Vg(ω) =1
1 − jω+ πδ(ω) +
1jω
H(s) =Va
Vg
=0.5‖(1/s)
1 + 0.5‖(1/s)=
1s + 3
, H(ω) =1
3 + jω
Va(ω) = H(ω)Vg(ω)
=1
(1 − jω)(3 + jω)+
1jω(3 + jω)
+πδ(ω)3 + jω
=1/4
1 − jω+
1/43 + jω
+1/3jω
− 1/33 + jω
+πδ(ω)3 + jω
=1/4
1 − jω+
1/3jω
− 1/123 + jω
+πδ(ω)3 + jω
=1/4
1 − jω+
1/3jω
− 1/123 + jω
+ πδ(ω)
17–4 CHAPTER 17. The Fourier Transform
Therefore va(t) =[14etu(−t) +
16
sgn t − 112
e−3tu(t) +16
]V
[b] va(0−) =14
− 16
+ 0 +16
=14
V
va(0+) = 0 +16
− 112
+16
=14
V
va(∞) = 0 +16
+ 0 +16
=13
V
AP 17.8 v(t) = 4te−tu(t); V (ω) =4
(1 + jω)2
Therefore |V (ω)| =4
1 + ω2
W1Ω =1π
∫ √3
0
[4
(1 + ω2)
]2
dω
=16π
12
[ω
ω2 + 1+ tan−1 ω
1
]√3
0
= 16[√
38π
+16
]= 3.769 J
W1Ω(total) =8π
[ω
ω2 + 1+ tan−1 ω
1
]∞0
=8π
[0 +
π
2
]= 4 J
Therefore % =3.769
4(100) = 94.23%
AP 17.9 |V (ω)| = 6 −( 6
2000π
)ω, 0 ≤ ω ≤ 2000π
|V (ω)|2 = 36 −( 72
2000π
)ω +
( 364π2 × 106
)ω2
W1Ω =1π
∫ 2000π
0
[36 − 72ω
2000π+
36 × 10−6
4π2 ω2
]dω
=1π
[36ω − 72ω2
4000π+
36 × 10−6ω3
12π2
]2000π
0
=1π
[36(2000π) − 72
4000π(2000π)2 +
36 × 10−6(2000π)3
12π2
]
Problems 17–5
= 36(2000) − 72(2000)2
4000+
36 × 10−6(2000)3
12
= 24 kJ
W6kΩ =24 × 103
6 × 103 = 4 J
17–6 CHAPTER 17. The Fourier Transform
Problems
P 17.1 [a] F (ω) =∫ 2
−2
[A sin
(π
2
)t]e−jωt dt =
−j4πA
π2 − 4ω2 sin 2ω
[b] F (ω) =∫ 0
−τ/2
(2Aτ
t + A)
e−jωt dt +∫ τ/2
0
(−2Aτ
t + A)
e−jωt dt
=4Aω2τ
[1 − cos
(ωτ
2
)]
P 17.2 [a] F (ω) =∫ τ/2
−τ/2
2Aτ
te−jωt dt
=2Aτ
[e−jωt
−ω2 (−jωt − 1)]τ/2
−τ/2
=2Aω2τ
[e−jωτ/2
(jωτ
2+ 1
)− ejωτ/2
(−jωτ
2+ 1
)]
F (ω) =2Aω2τ
[e−jωτ/2 − ejωτ/2 + j
ωτ
2
(e−jωτ/2 + ejωτ/2
)]
F (ω) = j2Aτ
[ωτ cos(ωτ/2) − 2 sin(ωτ/2)
ω2
]
[b] Using L’Hopital’s rule,
F (0) = limω→0
j2A[ωτ(τ/2)(− sin ωτ/2) + τ cos ω(τ/2) − 2(τ/2) cos(ωτ/2)
2ωτ
]
= limω→0
j2A[−ωτ(τ/2) sin(ωτ/2)
2ωτ
]
= limω→0
j2A[−τ sin(ωτ/2)
4
]= 0
·. . F (0) = 0
[c] When A = 1 and τ = 1
F (ω) = j2[ω cos(ω/2) − 2 sin(ω/2)
ω2
]
|F (ω)| =∣∣∣∣∣2ω cos(ω/2) − 4 sin(ω/2)
ω2
∣∣∣∣∣F (0) = 0
Problems 17–7
|F (2)| =∣∣∣∣4 cos 1 − 4 sin 1
4
∣∣∣∣ = 0.30
|F (4)| =∣∣∣∣8 cos 2 − 4 sin 2
16
∣∣∣∣ = 0.44
|F (6)| =∣∣∣∣12 cos 3 − 4 sin 3
36
∣∣∣∣ = 0.35
|F (8)| =∣∣∣∣16 cos 4 − 4 sin 4
64
∣∣∣∣ = 0.12
|F (9)| =∣∣∣∣18 cos 4.5 − 4 sin 4.5
81
∣∣∣∣ ∼= 0
|F (10)| =∣∣∣∣20 cos 5 − 4 sin 5
100
∣∣∣∣ = 0.10
|F (12)| =∣∣∣∣24 cos 6 − 4 sin 6
144
∣∣∣∣ = 0.17
|F (14)| =∣∣∣∣28 cos 7 − 4 sin 7
196
∣∣∣∣ = 0.09
|F (15.5)| =∣∣∣∣31 cos 7.75 − 4 sin 7.75
240.25
∣∣∣∣ ∼= 0
P 17.3 [a] F (ω) = A +2Aωo
ω, −ωo/2 ≤ ω ≤ 0
F (ω) = A − 2Aωo
ω, 0 ≤ ω ≤ ωo/2
F (ω) = 0 elsewhere
17–8 CHAPTER 17. The Fourier Transform
f(t) =12π
∫ 0
−ωo/2
(A +
2Aωo
ω)
ejtω dω
+12π
∫ ωo/2
0
(A − 2A
ωo
ω)
ejtω dω
f(t) =12π
[ ∫ 0
−ωo/2Aejtω dω +
∫ 0
−ωo/2
2Aωo
ωejtω dω
+∫ ωo/2
0Aejtω dω −
∫ ωo/2
0
2Aωo
ωejtω dω]
=12π
[ Int1 + Int2 + Int3 − Int4 ]
Int1 =∫ 0
−ωo/2Aejtω dω =
A
jt(1 − e−jtωo/2)
Int2 =∫ 0
−ωo/2
2Aωo
ωejtω dω =2Aωot2
(1 − jtωo
2e−jtωo/2 − e−jtωo/2)
Int3 =∫ ωo/2
0Aejtω dω =
A
jt(ejtωo/2 − 1)
Int4 =∫ ωo/2
0
2Aωo
ωejtω dω =2Aωot2
(−jtωo
2ejtωo/2 + ejtωo/2 − 1)
Int1 + Int3 =2At
sin(ωot/2)
Int2 − Int4 =4Aωot2
[1 − cos(ωot/2)] − 2At
sin(ωot/2)
·. . f(t) =12π
[ 4Aωot2
(1 − cos(ωot/2))]
=2A
πωot2
[2 sin2(ωot/4)
]
=4ωoA
πω2ot
2 sin2(ωot/4)
=ωoA
4π
[sin(ωot/4)(ωot/4)
]2
[b] f(0) =ωoA
4π(1)2 = 79.58 × 10−3ωoA
Problems 17–9
[c] A = 20π; ωo = 2 rad/s
f(t) = 10[sin(t/2)(t/2)
]2
P 17.4 [a] F (s) = Lte−at =1
(s + a)2
F (ω) = F (s)∣∣∣∣s=jω
+ F (s)∣∣∣∣s=−jω
F (ω) =[
1(a + jω)2
]+[
1(a − jω)2
]
=2(a2 − ω2)
(a2 − ω2)2 + 4a2ω2 =2(a2 − ω2)(a2 + ω2)2
[b] F (s) = Lt3e−at =6
(s + a)4
F (ω) = F (s)∣∣∣∣s=jω
+ F (s)∣∣∣∣s=−jω
F (ω) =6
(a + jω)4 − 6(a − jω)4 = −j48aω
a2 − ω2
(a2 + ω2)4
[c] F (s) = Le−at cos ω0t =s + a
(s + a)2 + ω20
=0.5
(s + a) − jω0+
0.5(s + a) + jω0
F (ω) = F (s)∣∣∣∣s=jω
+ F (s)∣∣∣∣s=−jω
F (ω) =0.5
(a + jω) − jω0+
0.5(a + jω) + jω0
+0.5
(a − jω) − jω0+
0.5(a − jω) + jω0
=a
a2 + (ω − ω0)2 +a
a2 + (ω + ω0)2
17–10 CHAPTER 17. The Fourier Transform
[d] F (s) = Le−at sin ω0t =ω0
(s + a)2 + ω20
=−j0.5
(s + a) − jω0+
j0.5(s + a) + jω0
F (ω) = F (s)∣∣∣∣s=jω
− F (s)∣∣∣∣s=−jω
F (ω) =−ja
a2 + (ω − ω0)2 +ja
a2 + (ω + ω0)2
[e] F (ω) =∫ ∞
−∞δ(t − to)e−jωt dt = e−jωto
(Use the sifting property of the Dirac delta function.)
P 17.5 Fsin ω0t = F
ejω0t
2j
− F
e−jω0t
2j
=12j
[2πδ(ω − ω0) − 2πδ(ω + ω0)]
= jπ[δ(ω + ω0) − δ(ω − ω0)]
P 17.6 f(t) =12π
∫ ∞
−∞[A(ω) + jB(ω)][cos tω + j sin tω] dω
=12π
∫ ∞
−∞[A(ω) cos tω − B(ω) sin tω] dω
+j
2π
∫ ∞
−∞[A(ω) sin tω + B(ω) cos tω] dω
But f(t) is real, therefore the second integral in the sum is zero.
P 17.7 By hypothesis, f(t) = −f(−t). From Problem 17.6, we have
f(−t) =12π
∫ ∞
−∞[A(ω) cos tω + B(ω) sin tω] dω
For f(t) = −f(−t), the integral∫∞−∞ A(ω) cos tω dω must be zero. Therefore, if
f(t) is real and odd, we have
f(t) =−12π
∫ ∞
−∞B(ω) sin tω dω
P 17.8 F (ω) =−j2ω
; therefore B(ω) =−2ω
; thus we have
f(t) = − 12π
∫ ∞
−∞
(−2ω
)sin tω dω =
1π
∫ ∞
−∞sin tω
ωdω
Butsin tω
ωis even; therefore f(t) =
2π
∫ ∞
0
sin tω
ωdω
Problems 17–11
Therefore,
f(t) =2π
· π
2= 1 t > 0
f(t) =2π
·(−π
2
)= −1 t < 0
from a table of definite integrals
Therefore f(t) = sgn t
P 17.9 From Problem 17.4[c] we have
F (ω) =ε
ε2 + (ω − ω0)2 +ε
ε2 + (ω + ω0)2
Note that as ε → 0, F (ω) → 0 everywhere except at ω = ±ω0. At ω = ±ω0,F (ω) = 1/ε, therefore F (ω) → ∞ at ω = ±ω0 as ε → 0. The area under eachbell-shaped curve is independent of ε, that is∫ ∞
−∞εdω
ε2 + (ω − ω0)2 =∫ ∞
−∞εdω
ε2 + (ω + ω0)2 = π
Therefore as ε → 0, F (ω) → πδ(ω − ω0) + πδ(ω + ω0)
P 17.10 A(ω) =∫ ∞
−∞f(t) cos ωt dt
=∫ 0
−∞f(t) cos ωt dt +
∫ ∞
0f(t) cos ωt dt
= 2∫ ∞
0f(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 = −
∫ ∞
0f(t) sin ωt dt
P 17.11 A(ω) =∫ 0
−∞f(t) cos ωt dt +
∫ ∞
0f(t) cos ωt dt = 0
since f(t) cos ωt is an odd function.
B(ω) = −2∫ ∞
0f(t) sin ωt dt, since f(t) sin ωt is an even function.
P 17.12 [a] F
df(t)dt
=∫ ∞
−∞df(t)dt
e−jωt dt
Let u = e−jωt, then du = −jωe−jωt; let dv = [df(t)/dt] dt, then v = f(t).
Therefore F
df(t)dt
= f(t)e−jωt
∣∣∣∣∞−∞−∫ ∞
−∞f(t)[−jωe−jωt dt]
= 0 + jωF (ω)
17–12 CHAPTER 17. The Fourier Transform
[b] Fourier transform of f(t) exists, i.e., f(∞) = f(−∞) = 0.
[c] To find F
d2f(t)dt2
, let g(t) =
df(t)dt
Then F
d2f(t)dt2
= F
dg(t)dt
= jωG(ω)
But G(ω) = F
df(t)dt
= jωF (ω)
Therefore we have F
d2f(t)dt2
= (jω)2F (ω)
Repeated application of this thought process gives
F
dnf(t)dtn
= (jω)nF (ω).
P 17.13 [a] F∫ t
−∞f(x) dx
=∫ ∞
−∞
[∫ t
−∞f(x) dx
]e−jωt dt
Now let u =∫ t
−∞f(x) dx, then du = f(t)dt
Let dv = e−jωt dt, then v =e−jωt
−jω
Therefore,
F∫ t
−∞f(x) dx
=
e−jωt
−jω
∫ t
−∞f(x) dx
∣∣∣∣∞−∞−∫ ∞
−∞
[e−jωt
−jω
]f(t) dt
= 0 +F (ω)jω
[b] We require∫ ∞
−∞f(x) dx = 0
[c] No, because∫ ∞
−∞e−axu(x) dx =
1a
= 0
P 17.14 [a] Ff(at) =∫ ∞
−∞f(at)e−jωt dt
Let u = at, du = a dt, u = ±∞ when t = ±∞Therefore,
Ff(at) =∫ ∞
−∞f(u)e−jωu/a
(du
a
)=
1aF(
ω
a
), a > 0
Problems 17–13
[b] Fe−|t| =1
1 + jω+
11 − jω
=2
1 + ω2
Therefore Fe−a|t| =(1/a)2
(ω/a)2 + 1
Therefore Fe−0.5|t| =4
4ω2 + 1, Fe−|t| =
2ω2 + 1
Fe−2|t| = 1/[0.25ω2 + 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] Ff(t − a) =∫ ∞
−∞f(t − a)e−jωt dt
Let u = t − a, then du = dt, t = u + a, and u = ±∞ when t = ±∞.Therefore,
Ff(t − a) =∫ ∞
−∞f(u)e−jω(u+a) du
= e−jωa∫ ∞
−∞f(u)e−jωu du = e−jωaF (ω)
[b] Fejω0tf(t) =∫ ∞
−∞f(t)e−j(ω−ω0)t dt = F (ω − ω0)
[c] Ff(t) cos ω0t = F
f(t)[ejω0t + e−jω0t
2
]
=12F (ω − ω0) +
12F (ω + ω0)
P 17.16 Y (ω) =∫ ∞
−∞
[∫ ∞
−∞x(λ)h(t − λ) dλ
]e−jωt dt
=∫ ∞
−∞x(λ)
[∫ ∞
−∞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) =∫ ∞
−∞x(λ)
[∫ ∞
−∞h(u)e−jω(u+λ) du
]dλ
=∫ ∞
−∞x(λ)
[e−jωλ
∫ ∞
−∞h(u)e−jωu du
]dλ
=∫ ∞
−∞x(λ)e−jωλH(ω) dλ = H(ω)X(ω)
P 17.17 Ff1(t)f2(t) =∫ ∞
−∞
[ 12π
∫ ∞
−∞F1(u)ejtudu
]f2(t)e−jωt dt
=12π
∫ ∞
−∞
[∫ ∞
−∞F1(u)f2(t)e−jωtejtu du
]dt
=12π
∫ ∞
−∞
[F1(u)
∫ ∞
−∞f2(t)e−j(ω−u)t dt
]du
=12π
∫ ∞
−∞F1(u)F2(ω − u) du
P 17.18 [a] F (ω) =∫ ∞
−∞f(t)e−jωt dt
dF
dω=∫ ∞
−∞d
dω
[f(t)e−jωt
]dt = −j
∫ ∞
−∞tf(t)e−jωt dt = −jFtf(t)
Therefore jdF (ω)
dω= Ftf(t)
d2F (ω)dω2 =
∫ ∞
−∞(−jt)(−jt)f(t)e−jωt dt = (−j)2Ft2f(t)
Note that (−j)n =1jn
Thus we have jn
[dnF (ω)
dωn
]= Ftnf(t)
[b] (i) Fe−atu(t) =1
a + jω= F (ω);
dF (ω)dω
=−j
(a + jω)2
Therefore j
[dF (ω)
dω
]=
1(a + jω)2
Therefore Fte−atu(t) =1
(a + jω)2
Problems 17–15
(ii) F|t|e−a|t| = Fte−atu(t) − Fteatu(−t)
=1
(a + jω)2 − jd
dω
(1
a − jω
)
=1
(a + jω)2 +1
(a − jω)2
(iii) Fte−a|t| = Fte−atu(t) + Fteatu(−t)
=1
(a + jω)2 + jd
dω
(1
a − jω
)
=1
(a + jω)2 − 1(a − jω)2
P 17.19 [a] f1(t) = cos ω0t, F1(u) = π[δ(u + ω0) + δ(u − ω0)]
f2(t) = 1, −τ/2 < t < τ/2, and f2(t) = 0 elsewhere
Thus F2(u) =τ sin(uτ/2)
uτ/2
Using convolution,
F (ω) =12π
∫ ∞
−∞F1(u)F2(ω − u) du
=12π
∫ ∞
−∞π[δ(u + ω0) + δ(u − ω0)]τ
sin[(ω − u)τ/2](ω − u)(τ/2)
du
=τ
2
∫ ∞
−∞δ(u + ω0)
sin[(ω − u)τ/2](ω − u)(τ/2)
du
+τ
2
∫ ∞
−∞δ(u − ω0)
sin[(ω − u)τ/2](ω − u)(τ/2)
du
=τ
2· sin[(ω + ω0)τ/2]
(ω + ω0)(τ/2)+
τ
2· sin[(ω − ω0)τ/2]
(ω − ω0)τ/2
[b] As τ increases, the amplitude of F (ω) increases at ω = ±ω0 and at the sametime 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
2
∫ ∞
−∞sin[(ω − ω0)(τ/2)]
(ω − ω0)(τ/2)dω =
∫ ∞
−∞sin[(ω − ω0)(τ/2)]
(ω − ω0)(τ/2)[(τ/2) dω] = π
Therefore as t → ∞,
f1(t)f2(t) → cos ω0t and F (ω) → π[δ(ω − ω0) + δ(ω + ω0)]
17–16 CHAPTER 17. The Fourier Transform
P 17.20 [a] vg = 100u(t)
Vg(ω) = 100[πδ(ω) +
1jω
]
H(s) =10
5s + 10=
2s + 2
H(ω) =2
jω + 2
Vo(ω) = H(ω)Vg(ω) =200πδ(ω)jω + 2
+200
jω(jω + 2)
= V1(ω) + V2(ω)
v1(t) =12π
∫ ∞
−∞200πejtω
jω + 2δ(ω) dω =
12π
(200π2
)= 50 (sifting property)
V2(ω) =K1
jω+
K2
jω + 2=
100jω
− 100jω + 2
v2(t) = 50sgn(t) − 100e−2tu(t)
vo(t) = v1(t) + v2(t) = 50 + 50sgn(t) − 100e−2tu(t)
= 100u(t) − 100e−2tu(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) =200jω
Then,
Vo(ω) = H(ω)Vg(ω) =400
jω(jω + 2)=
K1
jω+
K2
jω + 2=
200jω
− 200jω + 2
·. . vo(t) = 100sgn(t) − 200e−2tu(t) V
[b]
P 17.22 [a] Find the Thévenin equivalent with respect to the terminals of the capacitor:
vTh =56vg; RTh = 60‖12 = 10 kΩ
Io =VTh
10,000 + 106/2s=
2sVTh
20,000s + 106
H(s) =Io
VTh=
10−4s
s + 50; H(ω) =
jω × 10−4
jω + 50
17–18 CHAPTER 17. The Fourier Transform
vTh =56vg = 30 sgn(t); VTh =
60jω
Io = H(ω)VTh(ω) =(
60jω
)(jω × 10−4
jω + 50
)=
6 × 10−3
jω + 50
io(t) = 6e−50tu(t) mA
[b] At t = 0− the circuit is
At t = 0+ the circuit is
ig(0+) =30 + 36
12= 5.5 mA
i60k(0+) =3060
= 0.5 mA
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 RThC Thus,
τ = (10,000)(2 × 10−6) = 20 ms; ·. .1τ
= 50,
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 =VTh
104 + (106/2s)· 106
2s
H(s) =Vo
VTh=
50s + 50
H(jω) =50
jω + 50
VTh(ω) =60jω
Vo(ω) = H(jω)VTh(ω) =(
60jω
)50
jω + 50
=3000
(jω)(jω + 50)=
60jω
− 60jω + 50
vo(t) = 30sgn(t) − 60e−50tu(t) V
[b] vo(0−) = −30 V
vo(0+) = 30 − 60 = −30 V
This makes sense because there cannot be an instantaneous change in thevoltage 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
Vg
= H(s) =4/s
0.5 + 0.01s + 4/s
H(s) =400
s2 + 50s + 400=
400(s + 10)(s + 40)
H(jω) =400
(jω + 10)(jω + 40)
Vg(ω) =6jω
17–20 CHAPTER 17. The Fourier Transform
Vo(ω) = Vg(ω)H(jω) =2400
jω(jω + 10)(jω + 40)
Vo(ω) =K1
jω+
K2
jω + 10+
K3
jω + 40
K1 =2400400
= 6; K2 =2400
(−10)(30)= −8
K3 =2400
(−40)(−30)= 2
Vo(ω) =6jω
− 8jω + 10
+2
jω + 40
vo(t) = 3sgn(t) − 8e−10tu(t) + 2e−40tu(t) V
[b] vo(0−) = −3 V
[c] vo(0+) = 3 − 8 + 2 = −3 V
[d] For t ≥ 0+:
Vo − 3/s0.5 + 0.01s
+(Vo + 3/s)s
4= 0
Vo
[ 100s + 50
+s
4
]=
300s(s + 50)
− 0.75
Vo =1200 − 3s2 − 150ss(s + 10)(s + 40)
=K1
s+
K2
s + 10+
K3
s + 40
K1 =1200400
= 3; K2 =1200 − 300 + 1500
(−10)(30)= −8
K3 =1200 − 4800 + 6000
(−40)(−30)= 2
vo(t) = (3 − 8e−10t + 2e−40t)u(t) V
[e] Yes.
Problems 17–21
P 17.25 [a] Io =Vg
0.5 + 0.01s + 4/s
H(s) =Io
Vg
=100s
s2 + 50s + 400=
100s(s + 10)(s + 40)
H(ω) =100(jω)
(jω + 10)(jω + 40)
Vg(ω) =6jω
Io(ω) = H(ω)Vg(ω) =600
(jω + 10)(jω + 40)
=20
jω + 10− 20
jω + 40
io(t) = (20e−10t − 20e−40t)u(t) A
[b] io(0−) = 0
[c] io(0+) = 0
[d]
Io =6/s
0.5 + 0.01s + 4/s=
600s2 + 50s + 400
=600
(s + 10)(s + 40)=
20s + 10
− 20s + 40
io(t) = (20e−10t − 20e−40t)u(t) A
[e] Yes.
P 17.26 [a] Io =IgR
R + 1/sC=
RCsIg
RCs + 1; H(s) =
Io
Ig
=s
s + 1/RC
1RC
=106
25 × 103 = 40; H(ω) =jω
jω + 40
ig = 200sgn(t) µA; Ig = (200 × 10−6)(
2jω
)=
400 × 10−6
jω
17–22 CHAPTER 17. The Fourier Transform
Io = Ig[H(ω)] =400 × 10−6
jω· jω
jω + 40=
400 × 10−6
jω + 40
io(t) = 400e−40tu(t) µA
[b] Yes, at the time the source current jumps from −200 µA to +200 µA thecapacitor is charged to (200)(50) × 10−3 = 10 V, positive at the lowerterminal. The circuit at t = 0− is
At t = 0+ the circuit is
The time constant is (50 × 103)(0.5 × 10−6) = 25 ms.
·. .1τ
= 40 ·. . for t > 0, io = 400e−40t µA
P 17.27 [a] Vo =IgR(1/sC)R + (1/sC)
=IgR
RCs + 1
H(s) =Vo
Ig
=1/C
s + (1/RC)=
2 × 106
s + 40
H(ω) =2 × 106
40 + jω; Ig(ω) =
400 × 10−6
jω
Vo(ω) = H(ω)Ig(ω) =(
400 × 10−6
jω
)(2 × 106
40 + jω
)
=800
jω(40 + jω)=
20jω
− 2040 + jω
vo(t) = 10sgn(t) − 20e−40tu(t) V
[b] Yes, at the time the current source jumps from −200 to +200 µA the capacitoris charged to −10 V. That is, at t = 0−,vo(0−) = (50 × 103)(−200 × 10−6) = −10 V.
Problems 17–23
At t = ∞ the capacitor will be charged to +10 V. That is,vo(∞) = (50 × 103)(200 × 10−6) = 10 VThe time constant of the circuit is (50 × 103)(0.5 × 10−6) = 25 ms, so1/τ = 40. The function vo(t) is plotted below:
P 17.28 [a] ig = 3e−5|t|
·. . Ig(ω) =3
jω + 5+
3−jω + 5
=30
(jω + 5)(−jω + 5)
Vo
10+
Vos
10= Ig
·. .Vo
Ig
= H(s) =10
s + 1; H(ω) =
10jω + 1
Vo(ω) = Ig(ω)H(ω) =300
(jω + 1)(jω + 5)(−jω + 5)
=K1
jω + 1+
K2
jω + 5+
K3
−jω + 5
K1 =300
(4)(6)= 12.5
K2 =300
(−4)(10)= −7.5
K3 =300
(6)(10)= 5
Vo(ω) =12.5
jω + 1− 7.5
jω + 5+
5−jω + 5
vo(t) = [12.5e−t − 7.5e−5t]u(t) + 5e5tu(−t) V
17–24 CHAPTER 17. The Fourier Transform
[b] vo(0−) = 5 V
[c] vo(0+) = 12.5 − 7.5 = 5 V
[d] ig = 3e−5tu(t), t ≥ 0+
Ig =3
s + 5; H(s) =
10s + 1
vo(0+) = 5 V; γC = 0.5
Vo
10+
Vos
10= Ig + 0.5
Vo(s + 1) =30
s + 5+ 5
Vo =30
(s + 5)(s + 1)+
5s + 1
=−7.5s + 5
+7.5
s + 1+
5s + 1
=12.5s + 1
− 7.5s + 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
sL1+
Vo
sL2+
Vo
R= 0
·. . Vo =RVg
L1
[s + R
( 1L1
+1L2
)]
Problems 17–25
Io =Vo
sL2
·. .Io
Vg
= H(s) =R/L1L2
s(s + R[(1/L1) + (1/L2)])
R
L1L2= 12 × 105
R( 1
L1+
1L2
)= 3 × 104
·. . H(s) =12 × 105
s(s + 3 × 104)
H(ω) =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)
io(t) =1500π × 105
2π
∫ ∞
−∞[δ(ω + 4 × 104) + δ(ω − 4 × 104)]ejtω
jω(jω + 3 × 104)dω
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
e−j40,000t
−j(3 − j4)+
ej40,000t
j(3 + j4)
=75400
e−j40,000t
5/− 143.13 +ej40,000t
5/143.13
= 0.075 cos(40,000t − 143.13) A
io(t) = 75 cos(40,000t − 143.13) mA
17–26 CHAPTER 17. The Fourier Transform
[b] In the phasor domain:
Vo − 125j200
+Vo
j800+
Vo
120= 0
12Vo − 1500 + 3Vo + j20Vo = 0
Vo =1500
15 + j20= 60/− 53.13 V
Io =Vo
j800= 75 × 10−3/− 143.13 A
io(t) = 75 cos(40,000t − 143.13) mA
P 17.30 [a]
Vo =Vgs
25 + (100/s) + s=
Vgs2
s2 + 25s + 100
H(s) =Vo
Vg
=s2
(s + 5)(s + 20); H(ω) =
(jω)2
(jω + 5)(jω + 20)
vg = 25ig = −450e10tu(−t) − 450e−10tu(t) V
Vg = − 450−jω + 10
− 450jω + 10
Vo(ω) = H(ω)Vg =−450(jω)2
(−jω + 10)(jω + 5)(jω + 20)
+−450(jω)2
(jω + 10)(jω + 5)(jω + 20)
=K1
−jω + 10+
K2
jω + 5+
K3
jω + 20+
K4
jω + 5+
K5
jω + 10+
K6
jω + 20
Problems 17–27
K1 =450(100)(15)(30)
= −100 K4 =−450(25)(5)(15)
= −150
K2 =450(25)(15)(15)
= −50 K5 =−450(100)(−5)(10)
= 900
K3 =450(400)(30)(−15)
= 400 K6 =−450(400)(−15)(−10)
= −1200
Vo(ω) =−100
−jω + 10+
−200jω + 5
+−800
jω + 20+
900jω + 10
vo = −100e10tu(−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 Ωresistor change instantaneously, and the capacitor will not let the voltageacross 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 consistentwith the solution for vo obtained in part (c).
17–28 CHAPTER 17. The Fourier Transform
P 17.31
Vo − Vg
25+
100Vo
s+
Vos
100s + 125 × 104 = 0
·. . Vo =s(100s + 125 × 104)Vg
125(s2 + 12,000s + 25 × 106)
Io =sVo
100s + 125 × 104
H(s) =Io
Vg
=s2
125(s2 + 12,000s + 25 × 106)
H(ω) =−8 × 10−3ω2
(25 × 106 − ω2) + j12,000ω
Vg(ω) = 300π[δ(ω + 5000) + δ(ω − 5000)]
Io(ω) = H(ω)Vg(ω) =−2.4πω2[δ(ω + 5000) + δ(ω − 5000)]
(25 × 106 − ω2) + j12,000ω
io(t) =−2.4π
2π
∫ ∞
−∞ω2[δ(ω + 5000) + δ(ω − 5000)](25 × 106 − ω2) + j12,000ω
ejtω dω
= −1.2
25 × 106e−j5000t
−j(12,000)(5000)+
25 × 106ej5000t
j(12,000)(5000)
=612
e−j5000t
−j+
ej5000t
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 beforet = 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 AAt t = 0+ the circuit is
io(0+) =30 − (−10)
5=
405
= 8 A
[c] The s-domain circuit is
Vo =[
Vg
5 + (10/s)
] (10s
)=
2Vg
s + 2
Vo
Vg
= H(s) =2
s + 2
17–30 CHAPTER 17. The Fourier Transform
H(ω) =2
jω + 2
Vg(ω) = 5(
2jω
)− 5[2πδ(ω)] +
30jω + 5
=10jω
− 10πδ(ω) +30
jω + 5
Vo(ω) = H(ω)Vg(ω) =2
jω + 2
[10jω
− 10πδ(ω) +30
jω + 5
]
=20
jω(jω + 2)− 20πδ(ω)
jω + 2+
60(jω + 2)(jω + 5)
=K0
jω+
K1
jω + 2+
K2
jω + 2+
K3
jω + 5− 20πδ(ω)
jω + 2
K0 =202
= 10; K1 =20−2
= −10; K2 =603
= 20; K3 =60−3
= −20
Vo(ω) =10jω
+10
jω + 2− 20
jω + 5− 20πδ(ω)
jω + 2=
10jω
+10
jω + 2+
20jω + 5
− 10πδ(ω)
vo(t) = 5sgn(t) + [10e−2t − 20e−5t]u(t) − 5 V
P 17.33 [a]
(Vo − Vg)s106 +
Vo
4s+
Vo
800= 0
·. . Vo =s2Vg
s2 + 1250s + 25 × 104
Vo
Vg
= H(s) =s2
(s + 250)(s + 1000)
H(ω) =(jω)2
(jω + 250)(jω + 1000)
vg = 45e−500|t|; Vg(ω) =45,000
(jω + 500)(−jω + 500)
·. . Vo(ω) = H(ω)Vg(ω) =45,000(jω)2
(jω + 250)(jω + 500)(jω + 1000)(−jω + 500)
=K1
jω + 250+
K2
jω + 500+
K3
jω + 1000+
K4
−jω + 500
Problems 17–31
K1 =45,000(−250)2
(250)(750)(750)= 20
K2 =45,000(−500)2
(−250)(500)(1000)= −90
K3 =45,000(−1000)2
(−750)(−500)(1500)= 80
K4 =45,000(500)2
(750)(1000)(1500)= 10
·. . vo(t) = [20e−250t − 90e−500t + 80e−1000t]u(t) + 10e500tu(−t) V
[b] vo(0−) = 10 V; Vo(0+) = 20 − 90 + 80 = 10 V
vo(∞) = 0 V
[c] IL =Vo
4s=
0.25sVg
(s + 250)(s + 1000)
H(s) =IL
Vg
=0.25s
(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)
=K1
jω + 250+
K2
jω + 500+
K3
jω + 1000+
K4
−jω + 500
K4 =(0.25)(500)(45,000)(750)(1000)(1500)
= 5 mA
iL(t) = 5e500tu(−t); ·. . iL(0−) = 5 mA
K1 =(0.25)(−250)(45,000)
(250)(750)(750)= −20 mA
K2 =(0.25)(−500)(45,000)(−250)(500)(1000)
= 45 mA
K3 =(0.25)(−1000)(45,000)(−750)(−500)(1500)
= −20 mA
·. . iL(0+) = K1 + K2 + K3 = −20 + 45 − 20 = 5 mA
Checks, i.e., iL(0+) = iL(0−) = 5 mA
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 Laplacetransform. Our circuit becomes
Vo
800+
Vo
4s+
(Vo − Vg)s106 + 35 × 10−6 +
5 × 10−3
s= 0
·. . (s2 + 1250s + 25 × 104)Vo = s2Vg − (35s + 5000)
vg(t) = 45e−500tu(t) V; Vg =45
s + 500
·. . (s + 250)(s + 1000)Vo =45s2 − (35s + 5000)(s + 500)
(s + 500)
·. . Vo =10s2 − 22,500s − 250 × 104
(s + 250)(s + 500)(s + 1000)
=20
s + 250− 90
s + 500+
80s + 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
4 − jω− 36
4 + jω=
72jω(4 − jω)(4 + jω)
Problems 17–33
Vo(s) =(16/s)
10 + s + (16/s)Vg(s)
H(s) =Vo(s)Vg(s)
=16
s2 + 10s + 16=
16(s + 2)(s + 8)
H(ω) =16
(jω + 2)(jω + 8)
Vo(ω) = H(ω) · Vg(ω) =1152jω
(4 − jω)(4 + jω)(2 + jω)(8 + jω)
=K1
4 − jω+
K2
4 + jω+
K3
2 + jω+
K4
8 + jω
K1 =1152(4)
(8)(6)(12)= 8
K2 =1152(−4)(8)(−2)(4)
= 72
K3 =1152(−2)(6)(2)(6)
= −32
K4 =1152(−8)
(12)(−4)(−6)= −32
·. . Vo(jω) =8
4 − jω+
724 + jω
− 322 + jω
− 328 + jω
·. . vo(t) = 8e4tu(−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 changeinstantaneously across a capacitor.
P 17.35 Vo(s) =10s
+30
s + 20− 40
s + 30=
600(s + 10)s(s + 20)(s + 30)
Vo(s) = H(s) · 15s
·. . 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
·. . Vo(ω) =30jω
· 40(jω + 10)(jω + 20)(jω + 30)
=1200(jω + 10)
jω(jω + 20)(jω + 30)
vo(ω) =20jω
+60
jω + 20− 80
jω + 30
vo(t) = 10sgn(t) + [60e−20t − 80e−30t]u(t) V
P 17.36 [a] f(t) =12π
∫ 0
−∞eωejtω dω +
∫ ∞
0e−ωejtω dω
=
1/π1 + t2
[b] W = 2∫ ∞
0
(1/π)2
(1 + t2)2 dt =2π2
∫ ∞
0
dt
(1 + t2)2 =12π
J
[c] W =1π
∫ ∞
0e−2ω dω =
1π
e−2ω
−2
∣∣∣∣∞0
=12π
J
[d]1π
∫ ω1
0e−2ω dω =
0.92π
, 1 − e−2ω1 = 0.9, e2ω1 = 10
ω1 = (1/2) ln 10 ∼= 1.15 rad/s
P 17.37
Io =IgR
R + (1/sC)=
RCsIg
RCs + 1
H(s) =Io
Ig
=s
s + (1/RC)
RC = (100 × 103)(1.25 × 10−6) = 125 × 10−3;1
RC=
10.125
= 8
H(s) =s
s + 8; H(ω) =
jω
jω + 8
Ig(ω) =30 × 10−6
jω + 2
Io(ω) = H(ω)Ig(ω) =30 × 10−6jω
(jω + 2)(jω + 8)
Problems 17–35
|Io(ω)| =ω(30 × 10−6)
(√
ω2 + 4)(√
ω2 + 64)
|Io(ω)|2 =900 × 10−12ω2
(ω2 + 4)(ω2 + 64)=
K1
ω2 + 4+
K2
ω2 + 64
K1 =(900 × 10−12)(−4)
(60)= −60 × 10−12
K2 =(900 × 10−12)(−64)
(−60)= 960 × 10−12
|Io(ω)|2 =960 × 10−12
ω2 + 64− 60 × 10−12
ω2 + 4
W1Ω =1π
∫ ∞
0|Io(ω)|2 dω =
960 × 10−12
π
∫ ∞
0
dω
ω2 + 64− 60 × 10−12
π
∫ ∞
0
dω
ω2 + 4
=120 × 10−12
πtan−1 ω
8
∣∣∣∣∞0
−30 × 10−12
πtan−1 ω
2
∣∣∣∣∞0
=(120
π· π
2− 30
π· π
2
)× 10−12 = (60 − 15) × 10−12 = 45 pJ
Between 0 and 4 rad/s
W1Ω =[120
πtan−1 1
2− 30
πtan−1 2
]× 10−12 = 7.14 pJ
% =7.1445
(100) = 15.86%
P 17.38 [a] Vg(ω) =60
(jω + 1)(−jω + 1)
H(s) =Vo
Vg
=0.4
s + 0.5; H(ω) =
0.4(jω + 0.5)
Vo(ω) =24
(jω + 1)(jω + 0.5)(−jω + 1)
Vo(ω) =−24
jω + 1+
32jω + 0.5
+8
−jω + 1
vo(t) = [−24e−t + 32e−t/2]u(t) + 8etu(−t) V
17–36 CHAPTER 17. The Fourier Transform
[b] |Vg(ω)| =60
(ω2 + 1)
[c] |Vo(ω)| =24
(ω2 + 1)√
ω2 + 0.25
[d] Wi = 2∫ ∞
0900e−2t dt = 1800
e−2t
−2
∣∣∣∣∣∞
0
= 900 J
[e] Wo =∫ 0
−∞64e2t dt +
∫ ∞
0(−24e−t + 32e−t/2)2 dt
= 32 +∫∞0 [576e−2t − 1536e−3t/2 + 1024e−t] dt
= 32 + 288 − 1024 + 1024 = 320 J
Problems 17–37
[f] |Vg(ω)| =60
ω2 + 1, |V 2
g (ω)| =3600
(ω2 + 1)2
Wg =3600
π
∫ 2
0
dω
(ω2 + 1)2
=3600
π
12
(ω
ω2 + 1+ tan−1 ω
) ∣∣∣∣20
=1800
π
(25
+ tan−1 2)
= 863.53 J
·. . % =(863.53
900
)× 100 = 95.95%
[g] |Vo(ω)|2 =576
(ω2 + 1)2(ω2 + 0.25)
=1024
ω2 + 0.25− 768
(ω2 + 1)2 − 1024(ω2 + 1)
Wo =1π
1024 · 2 · tan−1 2ω
∣∣∣∣20
−768(1
2
)(ω
ω2 + 1+ tan−1 ω
)2
0
−1024 tan−1 ω
∣∣∣∣20
=2048
πtan−1 4 − 384
π
(25
+ tan−1 2)
− 1024π
tan−1 2
= 319.2 J
% =319.2320
× 100 = 99.75%
P 17.39 Io =0.5sIg
0.5s + 25=
sIg
s + 50
H(s) =Io
Ig
=s
s + 50
H(ω) =jω
jω + 50
I(ω) =12
jω + 10
Io(ω) = H(ω)I(ω) =12(jω)
(jω + 10)(jω + 50)
17–38 CHAPTER 17. The Fourier Transform
|Io(ω)| =12ω√
(ω2 + 100)(ω2 + 2500)
|Io(ω)|2 =144ω2
(ω2 + 100)(ω2 + 2500)
=−6
ω2 + 100+
150ω2 + 2500
Wo(total) =1π
∫ ∞
0
150dω
ω2 + 2500− 1
π
∫ ∞
0
6dω
ω2 + 100
=3π
tan−1(
ω
50
) ∣∣∣∣∞0
−0.6π
tan−1(
ω
10
) ∣∣∣∣∞0
= 1.5 − 0.3 = 1.2 J
Wo(0–100 rad/s) =3π
tan−1(2) − 0.6π
tan−1(10)
= 1.06 − 0.28 = 0.78 J
Therefore, the percent between 0 and 100 rad/s is
0.781.2
(100) = 64.69%
P 17.40 [a] |Vi(ω)|2 =4 × 104
ω2 ; |Vi(100)|2 =4 × 104
1002 = 4; |Vi(200)|2 =4 × 104
2002 = 1
Problems 17–39
[b] Vo =ViR
R + (1/sC)=
RCVi
RCs + 1
H(s) =Vo
Vi
=s
s + (1/RC);
1RC
=10610−3
(0.5)(20)=
100010
= 100
H(ω) =jω
jω + 100
|Vo(ω)| =200|ω| · |ω|√
ω2 + 104=
200√ω2 + 104
|Vo(ω)|2 =4 × 104
ω2 + 104 , 100 ≤ ω ≤ 200 rad/s; |Vo(ω)|2 = 0, elsewhere
|Vo(100)|2 =4 × 104
104 + 104 = 2; |Vo(200)|2 =4 × 104
5 × 104 = 0.8
[c] W1Ω =1π
∫ 200
100
4 × 104
ω2 dω =4 × 104
π
[− 1
ω
]200
100
=4 × 104
π
[ 1100
− 1200
]=
200π
∼= 63.66 J
[d] W1Ω =1π
∫ 200
100
4 × 104
ω2 + 104 dω =4 × 104
π· tan−1 ω
100
∣∣∣∣200
100
=400π
[tan−1 2 − tan−1 1] ∼= 40.97 J
17–40 CHAPTER 17. The Fourier Transform
P 17.41 [a] Vi(ω) =A
a + jω; |Vi(ω)| =
A√a2 + ω2
H(s) =s
s + α; H(ω) =
jω
α + jω; |H(ω)| =
ω√α2 + ω2
Therefore |Vo(ω)| =ωA√
(a2 + ω2)(α2 + ω2)
Therefore |Vo(ω)|2 =ω2A2
(a2 + ω2)(α2 + ω2)
WIN =∫ ∞
0A2e−2at dt =
A2
2a; when α = a we have
WOUT(a) =A2
π
∫ a
0
ω2 dω
(ω2 + a2)2 =A2
π
∫ a
0
dω
a2 + ω2 −∫ a
0
a2 dω
(a2 + ω2)2
=A2
4aπ
(π
2− 1
)
WOUT(total) =A2
π
∫ ∞
0
[ω2
(a2 + ω2)2
]dω =
A2
4a
ThereforeWOUT(a)
WOUT(total)= 0.5 − 1
π= 0.1817 or 18.17%
[b] When α = a we have
WOUT(α) =1π
∫ α
0
ω2A2dω
(a2 + ω2)(α2 + ω2)
=A2
π
∫ α
0
[K1
a2 + ω2 +K2
α2 + ω2
]dω
where K1 =a2
a2 − α2 and K2 =−α2
a2 − α2
Therefore
WOUT(α) =A2
π(a2 − α2)
[a tan−1
(α
a
)− απ
4
]
WOUT(total) =A2
π(a2 − α2)
[aπ
2− α
π
2
]=
A2
2(a + α)
ThereforeWOUT(α)
WOUT(total)=
2π(a − α)
·[a tan−1
(α
a
)− απ
4
]
For α = a√
3, this ratio is 0.2723, or 27.23% of the output energy lies in thefrequency 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 thefrequency band between 0 and a/
√3.