Solutions Manual for Introduction to Electric Circuits 8ed - Dorf

Prévia do material em texto

1 
Circuit Variables 
Assessment Problems 
AP 1.1 To solve this problem we use a product of ratios to change units from 
dollars/year to dollars/millisecond. We begin by expressing $10 billion in 
scientific notation: 
$100 billion = $100 x 109 
Now we determine the number of milliseconds in one year, again using a 
product of ratios: 
1 year 1 day 1 hour 1 min 1 sec 1 year . . . . 
365.25 days 24 hours 60 mins 60 secs 1000 ms 31.5576 x 109 ms 
Now we can convert from dollars/year to dollars/millisecond, again with a 
product of ratios: 
$100 x 109 
1 year 
1 year 100 
31.5576 x 109 ms - 31.5576 = $3.17 /ms 
AP 1.2 First, we recognize that 1ns=10-9 s. The question then asks how far a 
signal will travel in 10-9 s if it is traveling at 80% of the speed of light. 
Remember that the speed of light c = 3 x 108 m/s. Therefore, 80% of c is 
(0.8)(3 x 108) = 2.4 x 108 m/s. Now, we use a product of ratios to convert 
from meters/second to inches/nanosecond: 
2.4 x 108 ID 1 s 100 CID 1 in (2.4 x 108)(100) 9.45 in 
- -
ls 109 ns 1 m 2.54 cm (109 )(2.54) 1 ns 
Thus, a signal traveling at 80% of the speed of light will travel 9.45" in a 
nanosecond. 
1-1 
1-2 CHAPTER 1. Circuit Variables 
AP 1.3 Remember from Eq. (1.2), current is the time rate of change of charge, or 
i = 1ff In this problem, we are given the current and asked to find the total 
charge. To do this, we must integrate Eq. (1.2) to find an expression for 
charge in terms of current: 
q(t) =lot i(x) dx 
We are given the expression for current, i, which can be substituted into the 
above expression. To find the total charge, we let t -+ oo in the integral. Thus 
we have 
q =Joo 20e-5ooox dx = 20 e-5oooxjoo = 20 (e-oo - eo) 
total o -5000 o -5000 
20 20 
- _ 5000 (0-1) = 5000 = 0.004 C = 4000
p,C 
AP 1.4 Recall from Eq. (1.2) that current is the time rate of change of charge, or 
i = ~. In this problem we are given an expression for the charge, and asked to 
find the maximum current. First we will find an expression for the current 
using Eq. (1.2): 
i = dq = !!:_ [~ - (! + ~) e-at] 
dt dt a 2 a a2 
d ( 1 ) d ( t -at) d ( 1 -at) 
= dt a 2 - dt ~e - dt a2 e 
( 1 -at t -at) ( 1 -at) = 0 - a e - a a e - -a a 2 e 
Now that we have an expression for the current, we can find the maximum 
value of the current by setting the first derivative of the current to zero and 
solving for t: 
di d - = -(te-at) = e-at + t(-a)eat = (1 - at)e-at = 0 
dt dt 
Since e-at never equals 0 for a finite value of t, the expression equals O only 
when (1 - at) = 0. Thus, t = l/a will cause the current to be maximum. For 
this value of t, the current is 
· 1 -a/a 1 -1 
i = -e = -e 
a a 
Problems 1-3 
Remember in the problem statement, a:= 0.03679. Using this value for a:, 
1 
i = 0.03679 e-i ':::'. 10 A 
AP 1.5 Start by drawing a picture of the circuit described in the problem statement: 
-:CJ 20V + 2 
-+4A 
Also sketch the four figures from Fig. 1.6: 
~ ~ 
+ + 
v v 
- -
(a) (bl 
~ ~ 
- -
v v 
+ + 
{c) (d) 
[al Now we have to match the voltage and current shown in the first figure 
with the polarities shown in Fig. 1.6. Remember that 4A of current 
entering Terminal 2 is the same as 4A of current leaving Terminal 1. We 
get 
(a) v = -20V, i = -4A; (b) v = -20V, i = 4A 
(c)v=20V, i=-4A; (d)v=20V, i=4A 
[b] Using the reference system in Fig. 1.6(a) and the passive sign convention, 
p =vi= (-20)(-4) =SOW. Since the power is greater than 0, the box is 
absorbing power. 
[cl From the calculation in part (b), the box is absorbing 80 W. 
AP 1.6 Applying the passive sign convention to the power equation using the voltage 
and current polarities shown in Fig. 1.5, p =vi. From Eq. (1.3), we know that 
power is the time rate of change of energy, or p = ~~. If we know the power, 
we can find the energy by integrating Eq. (1.3). To begin, find the expression 
for power: 
p =vi= (lO,OOOe-5000t)(2oe-5000t) = 200,oooe-10•000t = 2 x 105e-10•000t W 
1 ~4 CHAPTER 1. Circuit Variables 
Now find the expression for energy by integrating Eq. (1.3): 
w(t) = J:p(x) dx 
Substitute the expression for power, p, above. Note that to find the total 
energy, we let t -+ oo in the integral. Thus we have 
w = Joo 2 x 105e-10,000x dx = 2 x 105 e-10,000xloo 
0 -10,000 0 
2x105 0 2x105 2x105 
= -10,000(e-oo - e) = -10,000(0- l) = 10,000 = 20 J 
AP 1. 7 At the Oregon end of the line the current is leaving the upper terminal, and 
thus entering the lower terminal where the polarity marking of the voltage is 
negative. Thus, using the passive sign convention, p = -vi. Substituting the 
values of voltage and current given in the figure, 
p = -(800 x 103)(1.8 x 103 ) = -1440 x 106 = -1440 MW 
Thus, because the power associated with the Oregon end of the line is 
negative, power is being generated at the Oregon end of the line and 
transmitted by the line to be delivered to the California end of the line. 
Chapter Problems 
p 1.1 
(250 x 106)( 440) 
)9 = 110 giga-watt hours H· 
p 1.2 
. 5280 ft 2526 lb 1 kg 
(4 cond.) · (845 m1) · 1 mi · lOOO ft · 2.2 lb = 20.5 x 10
6 kg 
p 1.3 
1000 songs x songs 
[a] (32)(24)(2.1) mm3 - 1 mm3 
(1000)(1) . . 
x = (32)(24)(2.l) = 0.62 3-mmute songs, or about 111.6 seconds of
 music 
p 1.4 
Problems 1-5 
4 x 109 bytes x x 106 MB 
[b) (32)(24)(2.1) mm3 - (0.1)3 mm3 
(4 x 109)(0.001) 
x = (32)(24)(2.l) = 2480 bytes 
(320)(240) pixels 2 bytes 30 frames 4 608 106 b , / ------ · · = . x ytes sec 
1 frame 1 pixel 1 sec 
(4.608 x 106 bytes/sec)(x secs)= 30 x 109 bytes 
30 x 109 
x = 4.tiOS x 106 = 6510 sec = 108.5 min of video 
P 1.5 [a] We can set up a ratio to determine how long it takes the bamboo to grow 
10 p,m First, recall that 1 mm= 103µm. Let's also express the rate of 
growth of bamboo using the units mm/s instead of mm/day. Use a 
product of ratios to perform this conversion: 
250 mm 1 day 1 hour 1 min 250 10 / 
1 day 24 hours. 60 min. 60 sec - (24)(60)(60) = 3456 mm s 
Use a ratio to determine the time it takes for the bamboo to grow 10 µm: 
10/3456 x 10-3 m 10 x 10-6 m 10 x 10-6 
1 s x s so x = 10/3456 x 10-3 = 3.455 s 
[h) 1 cell · 3600 s · (24)(7) hr = 175 000 cells/week 
3.456 s 1 hr 1 week ' 
P 1.6 Volume = area x thickness 
p 1.7 
Convert values to millimeters, noting that 10 m2 = 106 mm2 
106 = (10 x 106)(thickness) 
106 
==?- thickness = 10 x 106 = 0.10 mm 
C/rn3 = 1.6022 x 10-19 C x 1029 elec:rons = 1.6022 x 1010 C/rn3 
1 electron 1 m 
Cross-sectional area of wire = 7rr2 = 7r(l.5 x 10-3 m)2 = 7.07 x 10-6 m2 
C/m = (1.6022 x 101°C/m3)(7.07 x 10-6rn2) = 113.253 x 103 C/m 
Therefore, i (s~c) = (113.253 x 103) (~) x avg vel (7) 
. i 1200 
Thus, average velocity = 113_253 x 103 = 113.253 x 103 = 0.0106 m/s 
1-6 
p 1.8 
p 1.9 
CHAPTER 1. Circuit Variables 
35 x 10-6 C/s 2 18 1014 1 / n= =. x eecs 
1.6022 x 10-19 C/elec 
First we use Eq. (1.2) to relate current and charge: 
dq 
i = dt = 24 cos 4000t 
Therefore, dq = 24 cos 4000t dt 
To find the charge, we can integrate both sides of the last equation. Note that 
we substitute x for q on the left side of the integral, and y for t on the right 
side of the integral: 
1q(t) Int dx = 24 cos 4000y dy 
q(O) 0 
We solve the integral and make the substitutions for the limits of the integral, 
remembering that sin 0 = 0: 
sin 4000y It 24 24 . 24 . 
q(t) - q(O) = 24 4000 0 = 4000 sin4000t - 4000 sm4000(0) = 4000 sm4000t 
But q(O) = 0 by hypothesis, i.e., the current passes through its maximum 
value at t = 0, so q(t) = 6 x 10-3 sin4000tC = 6sin4000tmC 
P 1.10 w = qV = (1.6022 x 10-19)(6) = 9.61 x 10-19 = 0.961 aJ 
P 1.11 p = (9)(100 x 10-3) = 0.9 W; 5 h . 3600 s = 18 000 r lhr ' s 
w(t) = fot pdt {18,000 w(18,000) = lo 0.9 dt = 0.9(18,000) = 16.2 kJ 
P 1.12 Assume we are standing at box A looking toward box B. Then, using the 
passive sign convention p = vi, since the current i is flowing into the + 
terminal of the voltage v. Now we just substitute the values for v and i into 
the equation forpower. Remember that if the power is positive, Bis absorbing 
power,.so the power must be flowing from A to B. If the power is negative, B 
is generating power so the power must be flowing from B to A. 
[a] p = (120)(5) = 600 w 600 W from A to B 
[b] p = (250)(-8) = -2000 w 
[c] p = (-150)(16) = -2400 W 
[d] p = (-480)(-'-10) = 4800 w 
2000 W from B to A 
2400 W from B to A 
4800 W from A to B 
P 1.13 [a] 
p 1.14 
p 1.15 
p =vi= (40)(-10) = -400 W 
Power is being delivered by the box. 
[b] Leaving 
[c] Gaining 
[a] p = vi = (-60)(-10) = 600 W, so power is being absorbed by the box. 
[b] Entering 
[c] Losing -lOA 
- 1-:o:vJ 
[a] In Car A, the current i is in the direction of the voltage drop across the 12 
V battery( the current i flows into the + terminal of the battery of Car 
A). Therefore using the passive sign convention, 
p =vi= (30)(12) = 360 W. 
Since the power is positive, the battery in Car A is absorbing power, so 
Car A must have the "dead" battery. 
[b] w(t) =lat pdx; 1 min= 60 s 
[60 
w(60) = lo 360 dx 
w = 360(60 - 0) = 360(60) = 21,600 J = 21.6 kJ 
t 
P 1.16 p =vi; w = f pdx lo 
Since the energy is the area under the power vs. time plot, let us plot p vs. t. 
p 
l80x10·3 
120x10·3 
0 288 ks t 
Note that in constructing the plot above, we used the fact that 80 hr 
= 288,000 s = 288 ks 
p(O) = (9)(20 x 10-3 ) = 180 x 10-3 W 
1-8 CHAPTER 1. Circuit Variables 
p(288 ks) = (6)(20 x 10-3) = 120 x 10-3 w 
w = (120 x 10-3)(288 x 103) + ~(180 x 10-3 - 120 x 10-3)(288 x 103) = 43.2 kJ 
p 1.17 [al P =vi= 30e-so0t _ 30e-1soot _ 40e-1ooot + soe-2000t _ ioe-3000t 
p(l ms)= 3.1 mW 
(b] w(t) = ht (30e-5oox _ 30e-1500x _ 4oe-1ooox+ 
50e-2ooox _ ioe-3000x)dx 
= 21.67 - 60e-5oot + 20e-15o0t + 40e-1oo0t_ 
25e-2000t + 3.33e-3000t µJ 
w(l ms) = l.24µJ 
[c] Wtotal = 21.67µJ 
P 1.18 [a] v(lO ms)= 40oe-1 sin2 = 133.8 V 
i(lO ms)= 5e-1 sin2 = 1.67 A 
p(lO ms)= vi= 223.79 W 
[h] p - vi= 2000e-200t sin2 200t 
- 2oooe-200t (~ - ~ cos40ot] 
- 10ooe-200t - 1oooe-200t cos 400t 
w - rxi 10ooe-200t dt - f 00 10ooe-200t cos400t dt 
lo .lo 
e-200t loo 
- 1000 -200 0 
-1000 { (200)~-:~400)2 [-200cos400t + 400sin400t]} 1
00 
200 ° 
- 5 - 1000 [ 4 x 104 + 16 x 1Q4] = 5 - 1 
w - 4 J 
Problems 1-9 
P 1.19 [a] 0 s :::; t < 4 s: 
v = 2.5t V; 
4 s < t < 8 s: 
v = 10 V; 
8 s < t < 16 s: 
i = lµA; p = 2.5tµW 
i =0 A; p=OW 
v = -2.5t + 30 V; i = -lµA; p = 2.5t-30µW 
16 s < t:::; 20 s: 
v = -10 V; 
20 s:::; t < 36 s: 
v = t-30 V; 
36 s < t:::; 46 s: 
v=6V; 
46 s:::; t < 50 s: 
i =0 A; p=OW 
i = 0.4µA; p = 0.4t-12µW 
i =0 A; p=OW 
v = -l.5t + 75 V; i = -0.6µA; p = 0.9t-45µW 
t > 50 s: 
v=OV; i =O A; p=OW 
10 
-10 
t(•J 
52 
[b] Calculate the area under the curve from zero up to the desired time: 
w(4) - !(4)(10)=20µJ 
w(12) w(4) - ~(4)(10) = 0 J 
w(36) - w(12) + !(4)(10) - H10)(4) + ~(6)(2.4) = 7.2 µJ 
w(50) - w(36) - ~(4)(3.6) = 0 J 
1-10 CHAPTER 1. Circuit Variables 
P 1.20 [a] p =vi= (0.05e-1000t)(75 - 75e-1000t) = (3.75e-1000t - 3.75e-2000t) W 
dp = -3750e-1000t + 7500e-2000t = 0 
dt 
2 = elOOOt so ln2 = lOOOt 
Pmax = p(693.15µs) = 937.5 mW 
so 
thus pis maximum at t = 693.15 p,s 
(b) W= foo[3.75e-1000t_3,75e-2000t)dt= [ 3.75 e-lOOOt_ 3.75 e-2000tlooJ 
lo -1000 -2000 o 
= 3. 75 _ 3. 75 = 1.875 rnJ 
1000 2000 
P 1.21 [a] p =vi = 900 sin(2007rt) cos(2007rt) = 450 sin( 4007rt) W 
Therefore, Pmax = 450 W 
[h] Pmax( extracting) = 450 W 
5x10-3 
[c] Pavg 200 h 450sin(4007rt) dt 
- 9 X 104 [-cos 4001l't 15x10-s = 225 [1 - cos 27r] = 0 
4007r 0 1l' 
180 180 
[d] Pavg = -[1- cos2.57r] = - = 57.3 W 
7r 7r 
P 1.22 [a] q area under i vs. t plot 
- [H5)(4) + (10)(4) + H8)(4) + (8)(6) + ~(3)(6)] x 103 
[10 + 40 + 16 + 48 + 9]103 = 123,000 c 
[b] w - j pdt = j vi dt 
v - 0.2 x 10-3t + 9 
0:::; t < 4000s 
i - 15 - 1.25 x 10-3t 
0 < t ~ 15 ks 
p - 135 - 8.25 x 10-3t - 0.25 x 10-6t2 
rooo 
lo (135 - 8.25 x 10-3t - 0.25 x 10-6t2) dt 
- (540 - 66 - 5.3333)103 = 468.667 kJ 
4000 :::; t :::; 12 '000 
i - 12 - 0.5 x 10-3t 
p - 108 - 2.1 x 10-3t - 0.1x10-6t2 
112,000 W2 - (108 - 2.1 X 10-3t - 0.1 X l0-6t2) dt 
4000 
- (864 - 134.4 - 55.467)103 = 674.133 kJ 
p 1.23 [a] 
ProbleIDB 1-11 
12,000 < t < 15,000 
'l -
p -
W3 -
-
WT -
p -
-
-
dp 
-
dt 
30- 2 x 10-3t 
270 - 12 x 10-3t - 0.4 x 10-6t2 
{15,000 
11 (270 - 12 x 10-3t - 0.4 x 10-6t2) dt 
12,000 
(810 - 486 - 219.6)103 = 104.4 kJ 
W1 + W2 + W3 = 468.667 + 674.133 + 104.4 = 1247.2 kJ 
vi= [16,000t + 20)e-8o0t][(128t + 0.16)e-800t] 
2048 x 103t2e-1600t + 5120te-1600t + 3.2e-1600t 
3.2e-1600t[640,000t2 + 1600t + 1] 
3.2{ e-1600t[1280 x 103t + 1600] - l60oe-1600t[640,000t2 + 1600t + 1]} 
-3.2e-1600t[128 x 104(800t2 + t)] = -409.6 x 104e-1600tt(800t + 1) 
dp 
Therefore, dt = 0 when t = 0 
so Pma:x occurs at t == 0. 
[b] Pma:x - 3.2e-0[0 + 0 + 1] 
[c] 
- 3.2W 
w - lotpdx 
w 0t t t 
3_2 - lo 640,000x2e-1600x dx + lo l600xe-1600x dx + lo e-1600x dx 
640,000e-1600x [256 x 104x2 3200x 2] It 
- -4096 x 106 + + + 
0 
1600e-1600x It e-1600x It 
256 x 104 (-l500x - l) 0 + -1600 0 
When t -4 oo all the upper limits evaluate to zero, hence 
w (640,000)(2) 1600 1 
3.2 = 4096 x 106 + 256 x 104 + 1600 
w = 10-3 + 2 x 10-3 + 2 x 10-3 = 5 mJ. 
P 1.24 [a] We can find the time at which the power is a maximum by writing an 
expression for p(t) = v(t)i(t), taking the first derivative of p(t) and 
setting it to zero, then solving for t. The calculations are shown below: 
1-12 CHAPTER 1. Circuit Variables 
p 
p 
dp 
a$ 
dt 
t 
-
-
-
-
-
0 t < 0, p = 0 t > 3 s 
vi= t(3 - t)(6 - 4t) = 18t - 18t2 + 4t3 mW 
18 - 36t + 12t2 = 12(t2 - 3t + 1.5) 
0 when t2 - 3t + 1.5 = 0 
3±yi9=6 3±J3 
2 2 
o::;t::;3s 
ti - 3/2 - v'3/2 = 0.634 s; t2 = 3/2 + v'3/2 = 2.366 s 
p(t1) -
p(t2) -
18(0.634) - 18(0.634)2 + 4(0.634)3 = 5.196 mW 
18(2.366) -18(2.366)2 + 4(2.366)3 = -5.196 mW 
Therefore, maximum power is being delivered at t = 0.634 s. 
[b] The maximum power was calculated in part (a) to determine the time at 
which the power is maximum: Pmax = 5.196 mW (delivered) 
[c] As we saw in part (a), the other "maximum" power is actually a 
minimum, or the maximum negative power. As we calculated in part (a), 
maximum power is being extracted at t = 2.366 s. 
[d] This maximum extracted power was calculated in part (a) to determine 
the time at which power is maximum: Pmax = 5.196 mW (extracted) 
rt r 
[e] w = lo pdx = lo (18x - 18x2 + 4x3)dx = 9t2 - 6t3 + t4 
w ( 0) = 0 mJ w( 2) = 4 mJ 
w(l) = 4 mJ w(3) = 0 mJ 
To give you a feel for the quantities of voltage, current, power, and energy 
and their relationships among one another, they are plotted below: 
2.5 
2.0 
1.5 
2 ,. 
-0.5 
1.00 2.00 3.00 
6 
4 
«("" 
2 
§. 0 
·- -2 
3.00 
-4 
-6 
t(s) 
Problems 1-13 
6 
§' 
.§, 
<>.. -2 
1.00 
-4 
-6 
6 
5 
"" 
4 
.§, 3 
~ 2 
1 
0 
0 1.00 2.00 3.00 
t(s) 
P 1.25 [a] p vi 
- 400 x 103t2e-800t + 700te-BOOt + 0.25e-so0t 
- e-800t[400,000t2 + 700t + 0.25] 
dp 
dt 
{ e-800t[800 x 103t + 700) - 80oe-800t[400,000t2 + 700t + 0.25]} 
- [-3,200,000t2 + 2400t + 5}10oe-800t 
Therefore, 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-BOO(.OOl68) 
- 666 mW 
[c] w - fntpdx 
0t t t 
w - lo 400,000x2e-800x dx + lo 700xe-BDOx dx + lo 0.25e-800x dx 
400 oooe-BOOx It 
- -~12 x 106 [64 x 104x2 + l600x + 2] 0 + 
700e-soox It e-soox It 
64 x 1Q4 (-800x - 1) o + 0.25 -800 o 
When t = oo all the upper limits evaluate to zero, hence 
w = (400,000)(2) 700 0.25 = 2 97 J 
512 x 106 + 64 x 104 + 800 · m · 
P 1.26 We use the passive sign convention to determine whether the power equation 
is p = vi or p = -vi and substitute into the power equation the values for v 
and i, as shown below: 
1-14 CHAPTER 1. Circuit Variables 
Pa - Vaia = (0.150)(0.6) = 90 mW 
Pb vbib = (0.150)(-1.4) = -210 mW 
Pc - -vcic = -(0.100)(-0.8) = 80 mW 
Pd - vdid = (0.250)(-0.8) = -200 mW 
Pe - -veie = -(0.300)(-2) = 600 mWPf - vrir = (-0.300)(1.2) = -360 mW 
Remember that if the power is positive, the circuit element is absorbing 
power, whereas is the power is negative, the circuit element is developing 
power. We can add the positive powers together and the negative powers 
together - if the power balances, these power sums should be equal: 
LPdev = 210 + 200 + 360 = 770 mW; 
LPabs = 90 + 80 + 600 = 770 mW 
Thus, the power balances and the total power developed in the circuit is 770 
mW. 
P 1.27 [a] From the diagram and the table we have 
Pa - -Vaia = -(5000)(-0.150) = 750 W 
Pb - vbib = (2000)(0.250) = 500 W 
Pc - -Vcic = -(3000)(0.200) = -600 W 
Pd - Vdid = (-5000)(0.400) = -2000 W 
Pe - -Veie = -(1000)(-0.050) = 50 W 
Pr - vrir = (4000)(0.350) = 1400 W 
Pg - -vgig = -(-2000)(0.400) = 800 W 
Ph - -vhih = -(-6000)(-0.350) = -2100 W 
LPdeI - 600 + 2000 + 2100 = 4700 w 
LPabs - 750 + 500 + 50 + 1400 + 800 = 3500 w 
Therefore, LPdeI # LPabs and the sub9rdinate engineer is correct. 
[b] The difference between the power delivered to the circuit and the power 
absorbed by the circuit is 
-4700+ 3500 = 1200 w 
One-half of this difference is 600 W, so it is likely that Pc 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 
Ve should be -3.0 kV, not 3.0 kV!) If the sign of Pc is changed from 
negative to positive, we can recalculate the power delivered and the 
power absorbed as follows: 
p 1.28 
p 1.29 
Problems 1-15 
LPdel = 2000 + 2100 = 4100 W 
LPabs = 750 + 500 + 600 + 50 + 1400 + 800 = 4100 W 
Pa 
Pb 
Pc 
Pd 
Pe 
Pf 
Pg 
Ph 
Pj 
Now the power delivered equals the power absorbed and the power 
balances for the circuit. 
- -vaia = -(36)(250 X 10-6) = -9 mW 
- vbib = (44)(-250 x 10-6 ) = -11 mW 
- Vcic = (28)(-250 X 10-6) = -7 mW 
- vdid = (-108)(100 x 10-6) = -10.8 mW 
- Veie = (-32)(150 x 10-6) = -4.8 mW 
- -vrif = -(60)(-350 x 10-6) = 21 mW 
vgig = (-48)(-200 x 10-6) = 9.6 mW 
- vhih = (80)(-150 x 10-6 ) = -12 mW 
- -vjij = -(80)(-300 x 10-6 ) = 24 mW 
Therefore, 
LPabs = 21+9.6 + 24 = 54.6 mW 
_LP<lel = 9 + 11 + 7 + 10.8 + 4.8 + 12 = 54.6 W 
Thus, the interconnection satisfies the power check 
Pa - -vaia = -(1.6)(0.080) = -128 mW 
Pb -vbib = -(2.6)(0.060) = -156 mW 
Pc - Vcic = (-4.2)(-0.050) = 210 mW 
Pd - -vdid = -(1.2)(0.020) = -24 mW 
Pe - Veie = (1.8)(0.030) = 54 mW 
Pf - -vrir = -(-1.8)(-0.040) = -72 mW 
Pg - Vgig = ( -3.6) ( -0.030) = 108 ID W 
Ph - vhih = (3.2)(-0.020) = -64 mW 
Pj -vJij = -(-2.4)(0.030) = 72 mW 
LPdel = 128 + 156 + 24 + 72 + 64 = 444 ID W 
L::Pabs = 210 + 54 + 108 + 72 = 444 ID W 
1-16 CHAPTER 1. Circuit Variables 
Therefore, 2:.::Pdel = l:Pabs = 444 mW 
Thus, the interconnection satisfies the power check 
P 1.30 [a) From an examination of reference polarities, elements a, b, e, and f absorb 
power, while elements c, d, g, and h supply power. 
[b] Pa - Vaia = (0.300)(25 X 10-6) = 7.5 µW 
Pb - -vbib = -(-0.100)(10 X 10-6) = l /LW 
Pc - Vcic = (-0.200)(15 X 10-6) = -3 µW 
Pd - -vdid = -(-0.200)(-35 x 10-
6) = -7 µW 
Pe - -Veie = -(0.350)(-25 X 10-6) = 8.75µW 
Pf - Vfif = (0.200)(10 x 10-6) = 2 µW 
Pg - Vgig = (-0.250)(35 x 10-6 ) = -8.75 µW 
Ph - vhih = (0.050)(-10 x 10-6 ) = -0.5 µW 
LPabs = 7.5+1+8.75 + 2 = 19.25µW 
LP<lel = 3 + 7 + 8.75 + 0.5 = 19.25µW 
Thus, 19.25 /LW of power is delivered and 19.25 p,W of power is absorbed, 
and the :power balances 
-----2 
Circuit Elements 
Assessment Problems 
AP 2.1 
lb 
lb + + vg vi SA 
4 
[a) Note that the current ib is in the same circuit branch as the 8 A current 
source; however, ib is defined in the opposite direction of the current 
source. Therefore, 
ib =-SA 
Next, note that the dependent current source and the independent 
current source are in parallel with the same polarity. Therefore, their 
voltages are equal, and 
ib -8 
Vg= 4 = 4 = -2V 
[b) To find the power associated with the 8 A source, we need to find the 
voltage drop across the source, vi. Note that the two independent sources 
are in parallel, and that the voltages vg and v1 have the same polarities, 
so these voltages are equal: 
Vi= Vg = -2V 
Using the passive sign convention, 
Ps = (8A)(vi) = (8A)(-2V) = -16W 
Thus the current source generated 16 W of power. 
2-1 
2-2 CHAPTER 2. Circuit Elements 
AP2.2 
AP 2.3 
+ 
[a] Note from the circuit that Vx = -25 V. To find a note that the two 
current sources are in the same branch of the circuit but their currents 
flow in opposite directions. Therefore 
avx = -15A 
Solve the above equation for a and substitute for vx, 
a= -15A = -15A = 0.6 A/V 
Vx -25V 
[b] To find the power associated with the voltage source we need to know the 
current, iv. Note that this current is in the same branch of the circuit as 
the dependent current source and these two currents flow in the same 
direction. Therefore, the current iv is the same as the current of the 
dependent source: 
iv= O:Vx = (0.6)(-25) = -15A 
Using the passive sign convention, 
Ps = -(iv)(25V) = -(-15A)(25V) = 375W. 
Thus the voltage source dissipates 375 W. 
[a] The resistor and the voltage source are in parallel and the resistor voltage 
and the voltage source have the same polarities. Therefore these two 
voltages are the same: 
VR = Vg = lkV 
Problems 2~3 
Note from the circuit that the current through the resistor is ig = 5 mA. 
Use Ohm's law to calculate the value of the resistor: 
R = vn = 1 kV = 200 kn 
ig 5rnA 
Using the passive sign convention to calculate the power in the resistor, 
PR= (vR)(i9) = (1 kV)(5 mA) = 5 W 
The resistor is dissipating 5 W of power. 
(b] Note from part (a) the VR = v9 and iR = i 9 . The power delivered by the 
source is thus 
Psource = -Vg'tg so V = _Psource = - -3W = 40V 
9 i 75mA g 
Since we now have the value of both the voltage and the current for the 
resistor, we can use Ohm's law to calculate the resistor value: 
Vg 40V 
R = iu = 75 mA = 533.33f! 
The power absorbed by the resistor must equal the power generated by 
the source. Thus, 
PR= -Psource = -(-3W) = 3W 
[c] Again, note the in= i 9 . The power dissipated by the resistor can be 
determined from the resistor's current: 
PR= R(in)2 = R(i9)2 
Solving for i 9 , 
i 2 = Pr _ 480 mW = O l6 
9 R 300f! .OO 
Then, since vn = Vg 
so i 9 = V0.0016 = 0.04A = 40mA 
vR = Rin = Ri9 = (300f!)(40mA) = 12V so Vg = 12V 
2-4 CHAPTER 2. Circuit Elements 
AP 2.4 
+ 
G 
[a] Note from the circuit that the current through the conductance G is ig, 
flowing from top to bottom, because the current source and the 
conductance are in the same branch of the circuit so must have the same 
current. The voltage drop across the current source is vg, positive at the 
top, because the current source and the conductance are also in parallel 
so must have the same voltage. From a version of Ohm's law, 
ig 0.5A 
119 = G = 50 mS = 10 V 
Now that we know the voltage drop across the current source, we can 
find the power delivered by th,is source: 
Psource = -Vgig = -(10)(0.5) = -5 W 
Thus the current source delivers 5 W to the circuit. 
[b] We can find the value of the conductance using the power, and the value 
of the current using Ohm's law and the conductance value: 
so 
p 9 
G = _!!_ = - = 0.048 = 40mS 
v2 152 g 
ig = Gvg = (40mS)(15V) = 0.6A 
[c] We can find the voltage from the power and the conductance, and then 
use the voltage value in Ohm's law to find the current: 
pg= cv; so 2 p9 8W vg = G = 200 µS = 40,000 
Thus Vg = V 40,000 = 200 V 
ig = Gvg = (200pS)(200V) = 0.04A = 40mA 
Problems 2~5 
AP 2.5 [a] Redraw the circuit with all of the voltages and currents labeled for every 
circuit element. 
Write a KVL equation clockwise around the circuit, starting below the 
voltage source: 
-24 V + V2 + V5 - V1 = Q 
Next, use Ohm's law to calculate the three unknown voltages from the 
three currents: 
A KCL equation at the upper right node gives i 2 = i 5 ; a KCL equation at 
thebottom right node gives i 5 = -i1 ; a KCL equation at the upper left 
node gives is = -i2 . Now replace the currents i 1 and i 2 in the Ohm's law 
equations with i5: 
V5 = 7i5; 
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 
[b) V1 = -2i5 = -2(2) = -4 V 
(c] v2 = 3i5 = 3(2) = 6V 
(d] V5 = 7i5 = 7(2) = 14 V 
[e] A KCL equation at the lower left node gives is:;:::: i 1 . Since i 1 = -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. 
2-6 CHAPTER 2. Circuit Elements 
AP 2.6 Redraw the circuit labeling all voltages and currents: 
R 
200V 
+ 
120V 
We can find the value of the unknown resistor if we can find the value of its 
voltage and its current. To start, write a KVL equation clockwise around the 
right loop, starting below the 24 n resistor: 
-120V+v3 =0 
Use Ohm's law to calculate the voltage across the 8 n resistor in terms of its 
current: 
V3 = 8i3 
Substitute the expression for v3 into the first equation: 
-120V + 8i3 = 0 so . - 120 -15A 'l3 - -
8 
Also use Ohm's law to calculate the value of the current thrm:igh the 24 n 
resistor: 
Now write a KCL equation at the top middle node, summing the currents 
leaving: 
so ii = i2 + i3 = 5 + 15 = 20 A 
Write a KVL equation clockwise around the left loop, starting below the 
voltage source: 
-200V + V1+120V = 0 so V1 = 200 - 120 = 80 V 
Now that we know the values of both the voltage and the current for the 
unknown resistor, we can use Ohm's law to calculate the resistance: 
AP 2.7 [a] Plotting a graph of Vt versus it gives 
vtM 
25 
20 
15 
10 
5 
~(A) 
0.1 0.2 0.3 
Problems 2-7 
Note that when it = 0, Vt = 25 V; therefore the voltage source must be 25 
V. Since the plot is a straight line, its slope can be used to calculate the 
value of resistance: 
Llv 25 - 0 25 
R = D..i = 0.25 - o = 0.25 =won 
A circuit model having the same v - i characteristic is a 25 V source in 
series with a lOOr! resistor, as shown below: 
25V 
[bl Draw the circuit model from part (a) and attach a 25 n resistor: 
100 n . 
-..it 
+ 
25V 
To find the power delivered to the 25 n resistor we must calculate the 
current through the 25 n resistor. Do this by first using KCL to recognize 
that the current in each of the components is it, fl.owing in a clockwise 
direction. Write a KVL equation in the clockwise direction, starting 
below the voltage source, and using Ohm's law to express the voltage 
drop across the resistors in the direction of the current it fl.owing through 
the resistors: 
-25 V + lOOit + 25it = 0 so 125it = 25 
Thus, the power delivered to the 25 n resistor is 
P2s = (25)i; = (25)(0.2)2 = 1 W. 
so it= 25 = 0.2A 
125 
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 
2-8 CHAPTER 2. Circuit Elements 
is a straight line, its slope can be used to calculate the value of resistance: 
R = Av = 25 - 0 = ~ = lOO n 
Ai 0.25 - 0 0.25 
A circuit model having the same v - i characteristic is a 0.25 A current 
source in parallel with a lOOn resistor, as shown below: 
+ 
0.25A 
(b] Draw the circuit model from part (a) and attach a 25 n resistor: 
0.25A 
Note that by writing a KVL equation around the right loop we see that 
the voltage drop across both resistors is Vt. Write a KCL equation at the 
top center node, summing the currents leaving the node. Use Ohm's law 
to specify the currents through the resistors in terms of the voltage drop 
across the resistors and the value of the resistors. 
Vt Vt 
-0.25 + 100 + 25 = 0, 
112 
P25= 2;=1 W. 
so 5Vt = 25, thus Vt= 5V 
AP 2.9 First note that we know the current through all elements in the circuit except 
the 6 k!l resistor (the current in the three elements to the left of the 6 kn 
resistor is i 1; the current in the three elements to the right of the 6 kn resistor 
is 30i1). To find the current in the 6 kn resistor, write a KCL equation at the 
top node: 
We can then use Ohm's law to find the voltages across each resistor in terms 
Problems 2-9 
of i 1 . The results are shown in the figure below: 
+ 54,000i1 -
lV +v-
- 54,000i1 + 
54kQ 1.8kQ 
0 
-~ 'f--
-7· 30i1 30i,~ 11 
+ 
5V 
1 
186,000il 6kQ : 8V 
11 
i -i 31i1 30i1 
[a] To find ii, write a KVL equation around the left-hand loop, summing 
voltages in a clockwise direction starting below the 5V source: 
-5 v + 54,000i1 - 1 v + l86,000i1 = 0 
Solving for i 1 
54,00lli1 + l86,000i1 = 6 v so 
Thus, 
6 -25 A ii = 240 000 - µ 
' 
240,000ii = 6 v 
[b] Now that we have the value of i 1 , we can calculate the voltage for each 
component except the dependent source. Then we can write a KVL 
equation for the right-hand loop to find the voltage v of the dependent 
source. Sum the voltages in the clockwise direction, starting to the left of 
the dependent source: 
+v - 54,000i1 + 8 V - 186,000i1 = 0 
Thus, 
v = 240,000i1 - 8V = 240,000(25 x 10-6 ) - 8V = 6V - 8V = -2V 
We now know the values of voltage and current for every circuit element. 
2-10 CHAPTER 2. Circuit Elements 
Let's construct a power table: 
Element Current Voltage Power Power 
(ltA) (V) Equation (µW) 
5V 25 5 p= -vi -125 
54kn 25 1.35 p= Ri2 33.75 
lV 25 1 p= -vi -25 
6kn 775 4.65 p= Ri2 3603.75 
Dep. source 750 -2 p= -vi 1500 
i.skn 750 1.35 p= Ri2 1012.5 
8V 750 8 p= -vi -6000 
[c] The total power generated in the circuit is the surn of the negative power 
values in the power table: 
-125µW + -25J1,W + -6000µW = -6150µW 
Thus, the total power generated in the circuit is 6150 p,W. 
[d] The total power absorbed in the circuit is the surn of the positive power 
values in the power table: 
33.75 µW + 3603.75 JLW + 1500 µW + 1012.5 µW = 6150 µW 
Thus, the total power absorbed in the circuit is 6150 µW. 
AP 2.10 Given that i1 = 2 A, we know the current in the dependent source is 
2i1 = 4 A. We can write a KCL equation at the left node to find the current in 
the 10 n resistor. Surnrning the currents leaving the node, 
-5 A + 2 A + 4 A+ i10n = 0 so i10n = 5 A - 2 A - 4 A = -1 A 
Thus, the current in the 10 n resistor is 1 A, flowing right to left, as seen in 
the circuit below. 
10 0 
5A 
Problems 2-11 
[a] To find Vs, write a KVL equation, summing the voltages counter-clockwise 
around the lower right loop. Start below the voltage source. 
-vs+ (1 A)(10r2) + (2A)(30r2) = 0 so Vs = 10 V + 60 V = 70 V 
[b] The current in the voltage source can be found by writing a KCL equation 
at the right-hand node. Sum the currents leaving the node 
-4 A + 1 A +iv = 0 so iv=r4A-1A=3A 
The current in the voltage source is 3 A, flowing top to bottom. The 
power associated with this source is 
p =vi= (70V)(3A) = 210W 
Thus, 210 W are absorbed by the voltage source. 
[ c] The voltage drop across the independent current source can be found by 
writing a KVL equation around the left loop in a clockwise direction: 
-v5A + (2 A)(30 0) = 0 so V5A = 60V 
The power associated with this source is 
p = -V5Ai = -(60V)(5A) = -300W 
This source thus delivers 300 W of power to the circuit. 
[d] The voltage across the controlled current source can be found by writing a 
KVL equation around the upper right loop in a clockwise direction: 
+v4A + (10r2)(1 A)= 0 so V4A = -lOV 
The power associated with this source is 
p = V4Ai = (-10V)(4A) = -40W 
This source thus delivers 40 W of power to the circuit. 
[e] The total power dissipated by the resistors is given by 
(i300)2 (3on) + (i10n)2(10n) = (2)2(30n) + (1)2(100) = 120+ 10 = 130W 
2-12 CHAPTER 2. Circuit Elements 
Problems 
P 2.1 [a] Yes, independent voltage sources can carry the 8 A current required by the 
connection; independent current source can support any voltage required 
by the connection, in this case 20 V, positive at the top. 
[b] 30 V source:absorbing 
10 V source: delivering 
8 A source: delivering 
[c] P3ov - (30)(8) = 240 W (abs) 
P10v - -(10)(8) = -80 W (del) 
PsA - -(20)(8) = -160 W (del) 
LPabs = LPdel = 240 W 
[d] The interconnection is valid, but in this circuit the voltage drop across the 
8 A current source is 40 V, positive at the top; 30 V source is absorbing, 
the 10 V source is absorbing, and the 8 A source is delivering 
P3ov - (30)(8) = 240 W (abs) 
P10v - (10)(8) = 80 W (abs) 
PsA - -(40)(8) = -320 W (del) 
LPabs = LPdel = 320 W 
P 2.2 The interconnection is valid. The 10 A current source has a voltage drop of 
100 V, positive at the top, because the 100 V source supplies its voltage drop 
across a pair of terminals shared by the 10 A current source. The right hand 
branch of the circuit must also have a voltage drop of 100 V from the left 
terminal of the 40 V source to the bottom terminal of the 5 A current source, 
because this branch shares the same terminals as the 100 V source. This 
means that the voltage drop across the 5 A current source is 140 V, positive at 
the top. Also, the two voltage sources can carry the current required of the 
interconnection. This is summarized in the figure below: 
+ 
lOOV 
40V 
lOOV 
Problems 2~ 13 
+ 
140V 
From the values of voltage and current in the figure, the power supplied by the 
currents sources is calculated as follows: 
P10A = -(100)(10) = -1000 W (dev) 
PsA = -(140)(5) = -700 W (dev) 
LPdev = 1700 W 
P 2.3 The interconnection is not valid. Note that both current sources in the right 
hand branch supply current through the 100 V source. If the interconnection 
was valid, these two current sources would supply the same current in the 
same direction, which they do not. 
P 2 .4 The interconnect is valid since the voltage sources can all carry 5 A of current 
supplied by the current source, and the current source can carry the voltage 
drop required by the interconnection. Note that the branch containing the 10 
V, 40 V, and 5 A sources must have the same voltage drop as the branch 
containing the 50 V source, so the 5 A current source must have a voltage 
drop of 20 V, positive at the right. The voltages and currents are summarize 
in the circuit below: 
lOV 
- + 
50V + 
5A 40V -
- 20V + 
Psov - (50)(5) = 250 w (abs) 
P10v - (10)(5) = 50 w (abs) 
P4ov -(40)(5) = -200 w (dev) 
PsA - -(20)(5) = -100 w (dev) 
2--14 CHAPTER 2. Circuit Elements 
P 2.5 The interconnection is valid, since the voltage sources can carry the 10 A 
current supplied by the current source, and the current sources can carry 
whatever voltage drop is required by the interconnection. In particular, note 
the the voltage drop across the three sources in the right hand branch must be 
the same as the voltage drop across the 20 A current source in the middle 
branch, since the middle and right hand branch are connected between the 
same two terminals. In particular, this means that 
p 2.6 
p 2.7 
v1 (the voltage drop across the middle branch) 
= lOOV - 50V - v2 (the voltage drop across the right hand branch) 
Hence any combination of v1 and v2 such that v1 + v2 = 50 V is a valid 
solution. 
6V 4V 
8V 12V 
The interconnection is invalid. The voltage drop between the top terminal and 
the bottom terminal on the left hand side is due to the 6 V and 8 V sources, 
giving a total voltage drop between these terminals of 14 V. But the voltage 
drop between the top terminal and the bottom terminal on the right hand side 
is due to the 4 V and 12 V sources, giving a total voltage drop between these 
two terminals of 16 V. The voltage drop between any two terminals in a valid 
circuit must be the same, so the interconnection is invalid. 
[a] Yes, each of the voltage sources can carry the current required by the 
interconnection, and each of the current sources can carry the voltage 
drop required by the interconnection. (Note that if!.= 5 A.) 
[b] No, because the voltage drop between the top terminal and the bottom 
terminal cannot be determined. For example, define v1, v2 , and v3 as 
Problems 2-15 
shown: 
_.__.1.li. 
+ ZOA 
'2 + 
zov 
3i.li. .. 
+ I V3 
"t - lOOV 5A 
The voltage drop across the left branch, the center branch, and the right 
branch must be the same, since these branches are connected at the same 
two terminals. This requires that 
20 + V1 = V2 + 100 = V3 
But this equation has three unknown voltages, so the individual voltages 
cannot be determined, and thus the power of the sources cannot be 
determined. 
P 2.8 The interconnection is invalid. In the middle branch, the value of the current 
iA must be -25 A, since the 25 A cun;ent source supplies current in this 
branch in the direction opposite the direction of the current iA. Therefore, the 
voltage supplied by the dependent voltage source in the left hand branch is 
6(-25) == -150 V. This gives a voltage drop from the top terminal to the 
bottom terminal in the left hand branch of 50 - (-150) = 200 V. But the 
voltage drop between these same terminals in the right hand branch is 250 V, 
due to the voltage source in that branch. Therefore, the interconnection is 
invalid. 
P 2.9 The middle branch has a 4 A current source, so the current iA in that branch 
must also be 4 A, since the two currents are in the same direction. This means 
that the current supplied by the dependent source is 2( 4) = 8 A. Next, 
v0 = 100 V, and this must be the voltage drop across all three branches in the 
circuit, since all three branches connect at the same two terminals. Therefore, 
the voltage drop across the current source in the left hand branch must be 160 
V, positive at the top and the voltage drop across the current source in the 
middle branch must be 180 V, positive at the top. The voltages and currents 
2-16 CHAPTER 2. Circuit Elements 
for all sources are summarized in the figure below: 
12A 
+ 
180V 
- 80V 
+ 
lOOV j 8A 
4A 
From the values of voltage and current in the figure, the power supplied by the 
currents sources is calculated as follows: 
P12A - (160)(12) = 1920 W (abs) 
P6ov - -(60)(12) = -720 W (dev) 
Psov - (80)(4) = 320 W (abs) 
P4A - -(180)(4) = -720 W (dev) 
Pdepsource -(100)(8) = -800 W (dev) 
LPdev = 720 + 720 + 800 = 2240 W 
P 2.10 Since we know the device is a resistor, we can use Ohm's law to calculate the 
resistance. From Fig. P2.10(a), 
v = Ri so v R=-: 
z 
Using the values in the table of Fig. P2.10(b), 
R= -160 = -80 =~=~= 240 = 8kn 
-0.02 -0.01 0.01 0.02 0.03 
P 2.11 Since we know the device is a resistor, we can use the power equation. From 
Fig. P2.ll(a), 
. v2 
p=vi=-
R 
so 
v2 
R=-
P 
Using the values in the table of Fig. P2.ll(b) 
(10)2 R = (-10)2 = (-5)2 - (5)2 -
25 x 10-3 6.25 x 10-3 6.25 x 10-3 25 x 10-3 
~-'-(1_5'---)2_ - (20)2 = 4kn 
56.25 x 10-3 100 x 10-3 
Problems 2-17 
P 2.12 The resistor value is the ratio of the power to the square of the current: 
p 2.13 
R = ~· Using the values for power and current in Fig. P2.12(b), 
'/, 
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 
R91 - resistance of frame between battery and lamp A 
R92 - resistance of frame between lamp A and lamp B 
s - switch 
P 2.14 [a] Plot the v~i characteristic: 
500 
400 
~ 300 
'> 200 
100 
00 4 8 12 16 
it (A) 
From the plot: 
R = !lv = (420 - 100) = 200 
/li (16 - 0) 
When it= 0, Vt= 100 V; therefore the ideal current source must have a 
current of 100/20 = 5 A 
2-18 CHAPTER 2. Circuit Elements 
_lt 
~----+-~ 
+ 
20 0 
[h] We attach a 5 n resistor to the device model developed in part (a): 
20 0 
Write a KCL equation at the top node: 
5 +it= i1 
50 
Write a KVL equation for the right loop, in the direction of the two 
currents, using Ohm's law: 
20ii + 5it = 0 
Combining the two equations and solving,20(5 +it) + 5it = 0 so 25it = -100; thus it= -4A 
Now calculate the power dissipated by the resistor: 
Psn = 5i; = 5(-4)2 =BOW 
P 2.15 [a] Plot the v - i characteristic 
100-.-------------------
80+-:=~~~= 60~ 
Vt (V)40+-------------------
20+-------------------
0-+---~---~---~--~----
0 2 4 6 s 10 
it (A) 
From the plot: 
R = !J.v = ( 90 - 50) = 4 n 
, 1::1i (10 - 0) 
When it= 0, Vt= 50 V; therefore the ideal voltage source has a voltage 
of 50 V. 
[b] 
P 2.16 [a] 
Problems 2-19 
+ 
When Vt= 0, 
-50 
it= 4 = -12.5A 
Note that this result can also be obtained by extrapolating the v - i 
characteristic to Vt= 0. 
!~~ 
0 I 
0 10 20 30 40 50 55 
vsM 
[h] .6.v = 20V; .6.i = 10 mA; 
r - - - - - - - - - - - - - -,--Jlo-i• 
is= 16 mA 
2 kO 3 kO 
2-20 CHAPTER 2. Circuit Elements 
[d] vs(open circuit)= (40 x 10-3)(2 x 103) = 80 V 
[e} The open circuit voltage can be found in the table of values (or from the 
plot) as the value of the voltage Vs when the current is = 0. Thus, 
Vs(open circuit) = 55 V (from the table) 
[f) Linear model cannot predict the nonlinear behavior of the practical 
current source. 
P 2.17 [a] Begin ::(V} 
70 
60 
50 
40 
30 
20 
10 
0 
........ 
0 
-..........__ 
--......... 
75 
--......__ .___ 
---- ----- ------150 2215 300 400 1500 i 8 (mA) 
[b) Since the plot is linear for 0 ::; is ::::; 225 mA amd since R = Av/ Ai, we can 
calculate R from the plotted values as follows: 
R _ Av _ 75 - 30 _ ~ _ 2 n 
- Ai - 0.225 - 0 - 0.225 - OO 
We can determine the value of the ideal voltage source by considering the 
value of Vs when is = 0. When there is no current, there is no voltage 
drop across the resistor, so all of the voltage drop at the output is dne to 
the voltage source. Thus the value of the voltage source must be 75 V. 
The model, valid for 0::::; is ::; 225 mA, is shown below: 
2000 
Wv-
[c] The circuit is shown below: 
2000 -4 
-i:400Q 
Write a KVL equation in the clockwise direction, starting below the 
voltage source. Use Ohm's law to express the voltage drop across the 
resistors in terms of the current i: 
- 75 v + 200i + 400i = 0 so 600i = 75V 
Thus, i = ;~~ = 125mA 
(d) The circuit is shown below: 
2000 
Problems 2-21 
Write a KVL equation in the clockwise direction, starting below the 
voltage source. Use Ohm's law to express the voltage drop across the 
resistors in terms of the current i: 
-75V + 200i = 0 so 200i = 75V 
Thus, 
. 75V 
i=--=375mA 2oon 
(e] The short circuit current can be found in the table of values (or from the 
plot) as the value of the current is when the voltage Vs= 0. Thus, 
isc = 500mA (from table) 
(f] The plot of voltage versus current constructed in part (a) is not linear (it 
is piecewise linear, but not linear for all values of i 8 ). Since the proposed 
circuit model is a linear model, it cannot be used to predict the nonlinear 
behavior exhibited by the plotted data. 
P 2.18 [a] 
50V 
50 - 4ig + 80ib = 20ib + 80ib = lOOib 
ib - 0.5 A, therefore, ia = 2 A and ig = 2.5 A 
[b] ib = 0.5 A 
[c] v0 = 80ib = 40 V 
[d] p4n - i~(4) = 6.25(4) = 25 W 
P2on - i;(20) = (4)(20) = 80 W 
Pson - i~(80) = 0.25(80) = 20 W 
2-22 CHAPTER 2. Circuit Elements 
p 2.19 
[e] p5ov (delivered) = 50ig = 125 W 
Check: 
LP<lis = 25 + 80 + 20 = 125W 
LP<lel = 125W 
300 i2 --=..-
800 \' 900 
+ V30 - + ~ 
90 
[a) Write a KCL equation at the top node: 
so 
Write a KVL equation around the right loop: 
-Vgo + V30 + Vgo = 0 
From Ohm's law, 
Vgo = 80i1, 
Substituting, 
-80ii + 30i2 + 90i2 = 0 so - 80ii + 120i2 = 0 
Solving the two equations for ii and i 2 simultaneously, 
ii= 2.4A and i2 = l.6A 
[h] Write a KVL equation clockwise around the left loop: 
-v0 + Vgo = 0 but vgo = 80ii = 80(2.4) = 192 A 
So v 0 = Vgo = 192V 
[c} Calculate power using p =vi for the source and p = Ri2 for the resistors: 
Psource = -vo(4) = -(192)(4) = -768W 
Psoo = 2.42(80) = 460.8 W 
P3on = 1.62 (30) = 76.8 W 
P9on = 1.62 (90) = 230.4 W 
L Pdev = 768 W l:Pabs = 460.8 + 76.8 + 230.4 = 768 W 
Problems 2-23 
P 2.20 [a] Use KVL for the right loop to calculate the voltage drop across the 
right-hand branch v0 • This is also the voltage drop across the middle 
branch, so once v 0 is known, use Ohm's law to calculate i0 : 
V 0 lOOOia + 4000ia + 3000ia = 8000ia = 8000(0.002) = 16 V 
16 - 2000i0 
16 
'to - 2000 = S mA 
(b) KCL at the top node: i 9 = ia + i 0 = 0.002 + 0.008 = 0.010 A = 10 mA. 
(c) The voltage drop across the source is v0 , seen by writing a KVL equation 
for the left loop. Thus, 
P 2.21 [a] 
[b] 
p9 = -v0 ig = -(16)(0.01) = -0.160 W = -160 mW. 
Thus the source delivers 160 mW. 
180V BO 
.i 
~------..._._ __ ___, '2 
V2 = 180 - 100 = 80V 
. V2 
i2 = 8 = lOA 
i3 + 4 = i2, i3 = 10 - 4 = 6A 
V1 = l0i3 + 8i2 = 10(6) + 8(10) = 140V 
. V1 140 
i1 = 70 = 70 = 2A 
Note also that 
i4 = ii + i3 = 2 + 6 = 8 A 
i9 = i4 + i0 = 8 + 4 = 12 A 
Psn - 82(5) = 320 w 
P2sn - (4)2(25) = 400 w 
Pron - 22(70) = 280 w 
P10n - 62(10) = 360 w 
Psn - 102(8) = 800 w 
[c] 2: P<lis = 320 + 400 + 280 + 360 + 800 = 2160W 
Pdev = 180i9 = 180(12) = 2160W 
2-24 CHAPTER 2. Circuit Elements 
P 2.22 [aJ 
p 2.23 
60 
100 
Va= (9+6)(3) = 45V 
-125 + Va + Vb = 0 SO Vb = 125 - Va = 125 - 45 = 80 V 
ie = vb/(10 + 6) = 80/16 = 5A 
id = ie - 3 = 5 - 3 = 2 A 
Ve = 5id +Vb = 5(2) + 80 = 90 V 
ie = Ve/30 = 90/30 = 3 A 
Vd = 125 - Ve = 125 - 90 = 35 V 
ia = id + ie = 2 + 3 = 5 A 
R = vd/ia = 35/5 = 70. 
[h] i9 = ia + 3 = 5 + 3 = 8 A 
p9 (supplied) = (125)(8) = 1000 W 
450 1' __,.. b 
240V 
18 n 
id= 60/12 = 5A; therefore, vcd = 60 + 18(5) = 150V 
-240 + Vac + Vcd = O; therefore, Vac = 240 - 150 = 90V 
ib = Va.c/45 = 90/45 = 2A; therefore, ic =id - ib = 5- 2 = 3A 
Vbd = lOie + Vcd = 10(3) + 150 = 180 V; 
therefore, ia = Vbd/180 = 180/180 = 1 A 
ie = ia + ic = 1 + 3 = 4 A 
-240 + Vab + Vbd = 0 therefore, Vab = 240 - 180 = 60V 
R = Vab/ie = 60/4 = 150. 
Problems 2-25 
CHECK: i 9 = ib + ie = 2 + 4 = 6 A 
P<lev = (240)(6) = 1440 W 
Lp<lis = 12 (180) + 42(15) + 32(10) + 52(12) + 52 (18) + 22(45) 
= 1440 W (CHECKS) 
P 2.24 (a] 
5kn lOkn 
4kn 
O.lA t 
icd = 500/15,000 = 33.33 mA 
ibd + icd = 0.1 so ibd = 0.1- 0.033 = 66.67mA 
4000ibc + 500 - 7500ibd = 0 SO ibc = (500 - 500)/4000 = 0 
iac = ic<l - ibc = 33.33 - 0 = 33.33 mA 
0.1 = iab + iac SO iab = 0.1- 33.33 = 66.67mA 
Calculate the power dissipated by the resistors using the equation 
PR= Ri~: 
p5ko = (5000)(0.0667)2 = 22.22W p7.5kn = (7500)(0.0667)2 = 33.33W 
Prnkn = (10,000)(0.03333)2 = 11.11 W p15kn = (15,000)(0.0333)2 = 16.67W 
P4kn = (4000)(0)2 =OW 
(b] Calculate the voltage drop across the current source: 
VacI = 5000iab + 7500ibd = 5000(0.0667) + 7500(0.0667) = 833.33 V 
Now that we have both the voltage and the current for the source, we can 
calculate the power supplied by the source: 
p9 = -833.33(0.1) = -83.33W thus Pu (supplied) = 83.33W 
(c] L Pdis = 22.22 + 33.33 + 11.11 + 16.67 + 0 = 83.33 W 
Therefore, 
2-26 CHAPTER 2. Circuit Elements 
P 2.25 [a] 
V2 = 80 + 4(12) = 128 V; V1=128- (8 + 12 + 4)(2) = 80V 
. V1 80 
i1= 6 + 10 = 16 = 5A; i3 = ii - 2 = 5 - 2 = 3 A 
Vg = V1+24i3 = 80 + 24(3) = 152V 
i 4 = 2 + 4 = 6A 
iu = -i4 - i3 = -6 - 3 = -9 A 
[b] Calculate power using the formula p = Ri2 : 
Pan= (8)(2)2 = 32 W; 
P4n = (4)(2)2 = 16W; 
P24n = (24)(3)2 = 216 W; 
P10n = (10)(5)2 = 250W; 
[c] vg = 152V 
P12n = (12)(2)2 = 48W 
P4n = (4)(6)2 = 144 W 
P6n = (6)(5)2 = 150W 
P12n = (12)(4)2 = 192 W 
[d] Sum the power dissipated by the resistors: 
LP<liss = 32+48+16+144+216+150 + 250 + 192 = 1048W 
The power associated with the sources is 
Pvolt-source = (80)(4) = 320W 
Pcurr-source = -Vgig = -(152)(9) = -1368 W 
Thus the total power dissipated is 1048 + 320 = 1368 W and the total 
power developed is 1368 W, so the power balances. 
P 2.26 [a] Start by calculating the voltage drops due to the currents i 1 and i 2 . Then 
use KVL to calculate the voltage drop across and 100 n resistor, and 
Ohm's law to find the current in the lOOnresistor. Finally, KCL at each 
of the middle three nodes yields the currents in the two sources and the 
[b) 
Problems 2~27 
current in the middle 10 n resistor. These calculations are summarized in 
the figure below: 
20 
+ 
130V 
- iooo soov 
460V 
250V 
P13o - -(130)(15) = -1950 W 
P46o - -(460)(30) = -13,800 W 
Lpdis - (15)2(2) + (15)2(10) + (30)2 (2) + (10)2(25) + (25)2(10) + (5)2(100) 
- 450 + 2250 + 1800 + 2500 + 6250 + 2500 = 15, 750 w 
I:Psup - 1950 + 13,800 = 15, 750 W 
Therefore, I:Pdis = I:Psup = 15, 750 W 
p 2.27 iE - iB - ic = 0 
ic=f3iB therefore iE = (1 + (3)iB 
or 
Now replace iE by (1 + (3)iB and solve for iB. Thus 
2-28 CHAPTER 2. Circuit Elements 
P 2.28 First note that we know the current through all elements in the circuit except 
the 200 n resistor (the current in the three elements to the left of the 200 n 
resistor is ifJ; the current in the three elements to the right of the 200 n 
resistor is 29ifJ)· To find the current in the 2000 resistor, write a KCL 
equation at the top node: 
We can then use Ohm's law to find the voltages across each resistor in terms 
of ifJ. The results are shown in the figure below: 
+ 10,000i,e-
lOkQ 0.8V 
- 14,500 i,e + 
5000 0 -7 .--f!l---4---< I'-' 1,e + - Vy+ 29i,e~ 
625V 15.2V 6000i,e 200Q 
1,e 
i j,30i,e 
[a] To find ifJ, write a KVL equation around the left-hand loop, summing 
voltages in a clockwise direction starting below the 15.2V source: 
-15.2 V + 10,000i1 - 0.8 V + 6000ifJ = 0 
Solving for ifJ 
10,000ifJ + 6000ifJ = 16 v so 16,000ifJ = 16 v 
Thus, 
. 16 
ifJ = = lmA 
16,000 
Now that we have the value of ifJ, we can calculate the voltage for each 
component except the dependent source. Then we can write a KVL 
equation for the right-hand loop to find the voltage Vy of the dependent 
source. Sum the voltages in the clockwise direGtion, starting to the left of 
the dependent source: 
-Vy - 14,500ifJ + 25 V - 6000ifJ = 0 
Thus, 
Vy = 25 V - 20,500ifJ = 25 V - 20,500(10-3 ) = 25 V - 20.5 V = 4.5 V 
Problems 2-29 
[b] We now know the values of voltage and current for every circuit element. 
Let's construct a power table: 
Element Current Voltage Power Power 
(mA) (V) Equation (mW) 
15.2V 1 15.2 p= -vi -15.2 
lOkO 1 10 p= Ri2 10 
0.8V 1 0.8 p= -vi -0.8 
2000 30 6 p= Ri2 180 
Dep. source 29 4.5 p=vi 130.5 
5000 29 14.5 p= Ri2 420.5 
25V 29 25 p= -vi -725 
The total power generated in the circuit is the sum of the negative power 
values in the power table: 
-15.2mW +-0.8mW +-725mW = -741mW 
Thus, the total power generated in the circuit is 7 41 mW. The total 
power absorbed in the circuit is the sum of the positive power values in 
the power table: 
10 mW+ 180 mW+ 130.5 mW+ 420.5 mW = 7 41 mW 
Thus, the total power absorbed in the circuit is 7 41 mW and the power in 
the circuit balances. 
P 2.29 [a] i 0 = 0 because no current can exist in a single conductor connecting two 
parts of a circuit. 
[b] 
SOY 
io""'I-
60 = 6000ig i9 =10 mA 
v ti = 5000i9 = 50V 
2000i1 = 500i2 , so i 1 + 4i1 = -300 mA; therefore, i1 = -60 mA 
[c] 300 - 60 + i2 = 0, so i 2 = -240 mA. 
2-30 CHAPTER 2. Circuit Elements 
P 2.30 50i + 0·250 + 0·250 - O· i2 = -0.5 mA 2 50 12.5 - ' 
V1 = lOOi2 = -50 mV 
20i1 + (-~~SO) + ( -0.0005) = O; i 1 = 125 µ,A 
Therefore, v9 = 6.25 mV. 
P 2.31 [a] -50 - 20i17 + 18ia = 0 
Therefore, - 50 - 20i17 + 45i17 = 0, so i17 = 2 A 
18ia = 45iu = 90; so ia = 5 A 
V 0 = 40i17 = 80V 
[b) i 9 = current out of the positive terminal of the 50 V source 
vd = voltage drop across the 8ia source 
Vd = 80 - 20 = 60 V 
I:Pgen - 50i9 + 20iuig = 50(47) + 20(2)(47) = 4230 W 
LP<liss - 18i~ + 5iu(i9 - ia) + 40i; + 8iavd + 8ia(20) 
- (18)(25) + 10(47 - 5) + 4(40) + 40(60) + 40(20) 
- 4230 W; Therefore, 
2.:::Pgen - LP<liss = 4230 W 
P 2.32 Here is Equation 2.25: 
VccR2 = (15)(80) = 12V 
R1 +R2 100 
R1R2 = (20)(80) = 16 kn 
R1 + R2 100 
Problems 2-31 
12 - 0.2 11.8 
iB = 16 + 40(0.1) = 2c) = 0.59 mA 
ic = f3iB = (39)(0.59) = 23.01 mA 
iE = ic + iB = 23 + 0.59 = 23.6 mA 
V3d = (23.6)(0.1) = 2.36V 
Vbd, = Vo + V3d = 2.56V 
. - Vbd, - 2.56 10-3 - 32 A i2 - - x - µ 
R2 80 
ii = i2 + i B = 32 + 590 = 622 p,A 
Vab = 20(0.622) = 12.44V 
ice = ic + ii = 23.01 + 0.622 = 23.632 mA 
Vi3 + 23.01(0.5) + 2.36 = 15 
V13 = l.135V 
P 2.33 [a] 
2 2 
&-+----+---<> 
3 3 
Location 1 Location 
[b] 
2 
e-+----+--..f':> 
3 
2 4 3 
Location Location 2 Location 3 
P 2.34 From the simplified circuit model, using Ohm's law and KVL: 
400i + 50i + 200i - 250 = 0 so i= 250/650= 385 mA 
This current is nearly enough to stop the heart, according to Table 2.1, so a 
warning sign should be posted at the 250 V source. 
2-32 CHAPTER 2. Circuit Elements 
p 2.35 
v + 
P 2.36 [a] p = i 2R 
( 250)
2 
Parm= 650 (400)=59.17W 
( 250)
2 
Pleg = 650 (200) = 29.59 W 
( 250)
2 
Ptrunk = 650 (50) = 7.40W 
[b] (dT) = 2.39 X 10-4Parm = 35.36 X 10_4 o C/s 
dt 4 arm 
tarm = 35~36 x 104 = 1414.23 s or 23.57 min 
(dT) = 2.39 X 10-4 Pieg = 7.07 X 10-40 C/s 
dt leg 10 
5xl~ . 
t1eg = 7.07 = 7,071.13 s or 117.85 mm 
(dT) = 2.39 x 10-4 (7.4) = 0.71 x 10_4 0 C/s 
dt trunk 25 
5 x 104 
itrunk = = 70,422.54 s or 1,173.71 min 0.71 
[c] They are all much greater than a few minutes. 
P 2.37 [a] Rarms = 400 + 400 = 800Q 
i1etgo - 50 mA (minimum) 
Vmin = (800)(50) X 10-3 = 40V 
Problems 2-33 
[b) No, 12/800 = 15 mA. Note this current is sufficient to give a perceptible 
shock. 
P 2.38 Rspoce = 1 MO 
ispace = 3 mA 
V = ispaceRspoce = 3000 V. 
--------3 
Simple Resistive Circuits 
Assessment Problems 
AP 3.1 
7.2 0 6 0 
10 0 
Start from the right hand side of the circuit and make series and parallel 
combinations of the resistors until one equivalent resistor remains. Begin by 
combining the 6 n resistor and the 10 0 resistor in series: 
fin+ 100=160 
Now combine this 16 0 resistor in parallel with the 64 0 resistor: 
1601164!1 = (lfi)(fi4) = 1024 = 12.80 
16 +64 80 
This equivalent 12.8 n resistor is in series with the 7.2 0 resistor: 
12.8n + 1.2n = 200 
Finally, this equivalent 20 0 resistor is in parallel with the 30 0 resistor: 
200ll300 = (20)(30) = 600 = 120 
20 + 30 50 
Thus, the simplified circuit is as shown: 
rEU 
5AL512o 
3-1 
3-2 CHAPTER 3. Simple Resistive Circuits 
AP3.2 
[a] With the simplified circuit we can use Ohm's law to find the voltage across 
both the current source and the 12 n equivalent resistor: 
v = (12f!)(5 A)= 60 V 
[h] Now that we know the value of the voltage drop across the current source, 
we can use the formula p = -vi to find the power associated with the 
source: 
p = -(60 V)(5 A)= -300 W 
Thus, the source delivers 300 W of power to the circuit. 
[cJ 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 n resistor, so we can use Ohm's law to calculate the current 
through this resistor: 
iA = 60 V = 2 A 
30f! 
Now write a KCL equation at the upper left node to find the current is: 
so is= 5 A-iA = 5 A- 2 A= 3 A 
Next, write a KVL equation around the outer loop of the circuit, using 
Ohm's law to express the voltage drop across the resistors in terms of the 
current through the resistors: 
-v + 7.2is + 6ic + lOic = 0 
So 
Thus 
16ic = v - 7.2is = 60 V - (7.2)(3) = 38.4 V 
ic = 38.4 = 2.4 A 
16 
Now that we have the current through the 10 n resistor we can use the 
formula p = Ri2 to find the power: 
Pion= (10)(2.4)2 = 57.6 W 
25kQ 
200V '.!'.' 
+ 
75kQ 
AP 3.3 
Problems 3-3 
[a] We can use voltage division to calculate the voltage v0 across the 75 kn 
resistor: 
75,000 
v0 (no load) = 75,000 + 25,000 (200 V) = 150 V 
[h] When we have a load resistance of 150 kn then the voltage v0 is across the 
parallel combination of the 75 kn resistor and the 150 kO resistor. First, 
calculate the equivalent resistance of the parallel combination: 
5 kOll150 kO = (75,000)(l50,000) = 50 000 0 = 50 kO7 75,000 + 150,000 ' 
Now use voltage division to find v0 across this equivalent resistance: 
Vo = 50,000 (200 V) = 133.3 V 
50,000 + 25,000 
[c] If the load terminals are short-circuited, the 75 kn resistor is effectively 
removed from the circuit, leaving only the voltage source and the 25 kO 
resistor. We can calculate the current in the resistor using Ohm's law: 
i = 200 V = 8 mA 
25 kO 
Now we can use the formula p = Ri2 to find the power dissipated in the 
25 kO resistor: 
P25k = (25,000)(0.008)2 = 1.6 W 
[d) The power dissipated in the 75 kn resistor will be maximum at no load 
since v0 is maximum. In part (a) we determined that the no-load voltage 
is 150 V, so be can use the formula p = v2 / R to calculate the power: 
(150)2 
P75k(max) = 75 000 = 0.3 W 
' 
600 20Ar·-40_0 ___ R~ 
800 
[a] We will write a current division equation for the current throught the 80n 
resistor and use this equation to solve for R: 
R 
ison = R + 40 n + 80 n (20 A) = 4 A so 20R = 4(R + 120) 
Thus 16R = 480 and R= 480 = 300 
16 
~-4 CHAPTER 3. Simple Resistive Circuits 
AP 3.4 
[b] With R = 30 n we can calculate the current through R using current 
division, and then use this current to find the power dissipated by R, 
using the formula p = Ri2 : 
i = 40 + 80 (20 A) = 16 A 
R 40+80+30 
so PR= (30)(16)2 = 7680 W 
[c] Write a KVL equation around the outer loop to solve for the voltage v, 
and then use the formula p = -vi to calculate the power delivered by the 
current source: 
-v + (600)(20 A)+ (300)(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. 
[a] First we need to determine the equivalent resistance to the right of the 
40 n and 70 n resistors: 
Req = 2001!3001!(500 + 100) so 
1 1 1 1 1 
Req = 2on + 300 + 600 =ion 
Thus, Req = 10n 
Now we can use voltage division to find the voltage v0 : 
Vo = 40 + ~~ + 70 ( 60 V) = 20 V 
[b] The current through the 40 n resistor can be found using Ohm's law: 
. V 0 20 V 
i4on = 40 = 40 n = 0.5 A 
This current flows from left to right through the 40 n resistor. To use 
current division, we need to find the equivalent resistance of the two 
parallel branches containing the 20 n resistor and the 50 n and 10 n 
resistors: 
20011(500+100) = (20)(60) = 150 
20+60 
Now we use current division to find the current in the 30 n branch: 
15 
z3on = 15 + 30 (0.5 A) = 0.16667 A= 166.67 rnA 
Problems 3-5 
[c] We can find the power dissipated by the 50 0 resistor if we can find the 
current in this resistor. We can use current division to find this current 
from the current in the 40 0 resistor, but first we need to calculate the 
equivalent resistance of the 20 0 branch and the 30 0 branch: 
AP 3.5 [a] 
[b] 
AP 3.6 [a] 
20011300 = (2o)(3o) = 12 o 
20+30 
Current division gives: 
12 
i5on = 12 + 50 + 10 (0.5 A) = 0.08333 A 
Thus, p500 = (50)(0.08333)2 = 0.34722 W = 347.22 mW 
lV - 1000 
We can find the current i usirtg Ohm's law: 
i = l~OVO = 0.01 A = 10 mA 
5 .5550 ~. 
1. 
lV : 1000 
Rm, = 50 nu5.555 0 = 5 0 
We can use the meter resistance to find the current using Ohm's law: 
lV 
Zmeas = 100 n + 5 n = 0.009524 = 9.524 mA 
15k0 
60Vr + v ?5kQ 
3-6 CHAPTER 3. Simple Resistive Circuits 
[b) 
Use voltage division to find the voltage v: 
75,000 
71 = 75,000 + 15,000 (50 V) = 50 V 
15kQ 
60Vr v 75kQ + 
The meter resistance is a series combination of resistances: 
Rm = 149,950 + 50 = 150,000 n 
We can use voltage division to find v, but first we must calculate the 
equivalent resistance of the parallel combination of the 75 kn resistor and 
the voltmeter: 
75 ooon11150 ooon = (75,000)(150,000) = 50 kn 
' ' 75,000 + 150,000 
Thus, 
50,000 (60 V) - 46 15 V 
Vmeas = 50,000 + 15,000 - . 
AP 3.7 [a] Using the condition for a balanced bridge, the products of the opposite 
resistors must be equal. Therefore, 
lOORx = (1000)(150) so Rx= (1000)(l
50) = 1500n = 1.5 kn 
100 
[b] When the bridge is balanced, there is no current flowing through the 
meter, so the meter acts like an open circuit. This place..s the following 
branches in parallel: The branch with the voltage source, the branch with 
the series combination R1 and R3 and the branch with the series 
combination of R2 and Rx. We can find the current in the latter two 
branches using Ohm's law: 
5V 
'lRi.Ra = 100 n + 150 n = 20 mA; 
5V 
iR2 ,Rx = 1000 + 1500 = 2 mA 
We can calculate the power dissipated by each resistor using the formula 
p = Ri2: 
P1000 = (lOOn)(0.02 A)2 = 40 mW 
p15on = (150n)(0.02 A)2 = 60 mW 
P10oon = (lOOOn)(0.002 A)2 = 4 mW 
P15oon = (1500n)(0.002 A)2 = 6 mW 
Problems 3-7 
Since none of the power dissipation values exceeds 250 rn W, the bridge 
can be left in the balanced state without exceeding the power-dissipating 
capacity of the resistors. 
AP 3.8 Convert the three Y-connected resistors, 20 0, 10 0, and 5 0 to three 
A-connected resistors Ra, Rb, and Re. To assist you the figure below has both 
the Y-connected resistors and the A-connected resistors 
28 Q 
105 Q 
Ra= (5)(10) + (5)(20) + (10)(20) = 17 5 O 
20 . 
Rb= (5)(10) + (5)(20) + (10)(20) = 35 O 
10 
Re= (5)(10) + (5)(20) + (10)(20) = 700 
5 
The circuit with these new A-connected resistors is shown below: 
28 Q 
105 Q 
From this circuit we see that the 70 0 resistor is parallel to the 28 0 resistor: 
70011280 = (70)(28) = 200 
70+28 
Also, the 17.5 0 resistor is parallel to the 105 0 resistor: 
17.50111050 = (l7.5)(105) = 150 
17.5+105 
3-8 CHAPTER 3. Simple Resistive Circuits 
Once the parallel combinations are made, we can see that the equivalent 20 0 
resistor is in series with the equivalent 15 n resistor, giving an equivalent 
resistance of 20 n + 15 n = 35 n. Finally, this equivalent 35 n resistor is in 
parallel with the other 35 n resistor: 
35 Oll35 n = (35)(35) = 11.5 n 
35+35 
Thus, the resistance seen by the 2 A source is 17.5 n, and the voltage can be 
calculated using Ohm's law: 
v = (17.50)(2 A)= 35 V 
Problems 3-9 
Problems 
P 3.1 [a] The 3 kO and 8 kO resistors are in series, as are the 5 kO and 7 kO 
resistors. The simplified circuit is shown below: 
llkQ 
ZmA t lOkQ 6kQ 12kQ 
[b] The 180 0 and 300 n resistors are in series, as are the 140 0 and 200 0 
resistors. The simplified circuit is shown below: 
2400 
480Q 
lOV + 
3400 
'--~~~OM.~~~__, 
[c] The 40 0, 50 0, and 60 n resistors are in series, as are the 45 0 and 30 0 
resistors. The simplified circuit is shown below: 
150Q 50V ~ 75Q 
P 3.2 [a] The 12 0 and 20 n resistors are in parallel, as are the 28 0 and 210 
resistors. The simplified circuit is shown below: 
12Q 
200mA t 7.5Q 18Q 
3-10 CHAPTER 3. Simple Resistive Circuits 
[bJ The 30 0 and 5 0 resistors are in parallel, as are the 9 0 and 18 0 resistors. 
The simplified circuit is shown below: 
4.28570~> 60 
[cJ The 100 kO and 300 kO resistors are in parallel, as are the 75 kn, 50 kO, 
and 150 kO resistors. The simplified circuit is shown below: 
75k0 
0.5V ! 25k0 
25k0 
P 3.3 (a) p4n - i;4 = (12)24 = 576 W 
P3n - (8)23 = 192 W 
Pl8n = ( 4)218 = 288 W 
P6n = (8)26 = 384 W 
[b] P120v( delivered) = 120i8 = 120(12) = 1440 W 
[c] Pdiss = 576 + 288 + 192 + 384 = 1440 W 
P 3.4 [a] From Ex. 3-1: i 1 = 4 A, i2 = 8 A, i 8 = 12 A 
at node b: -12 + 4 + 8 = 0, at node d: 12 - 4 - 8 = 0 
40 30 
d 
(b] V1 = 4i8 = 48 V 
V2 = l8i1 = 72 V V4 = 6i2 = 48 V 
loop abda: -120 + 48 + 72 = 0, 
loop bcdb: - 72 + 24 + 48 = 0, 
loop abcda: -120 + 48 + 24 + 48 = 0 
P 3.5 Always work from the side of the circuit furthest from the source. Remember 
that the current in all series-connected circuits is the same, and that the 
Problems 3~ 11 
voltage drop across all parallel-connected resistors is the same. 
[a) Req = {[(5 k + 7 k)ll6 k] + 3 k + 8 k}ll10 k = [(12 kll6 k) + 11 k]ll10 k 
= (4 k + 11k)[l10k=15 klllO k = 6 kn 
[b) Req = [24011(180+300)]+140 + 200 = (24011480) + 340 = 160 + 340 = 500n 
[c] Req = (40 + 50 + 60)11(30+ 45) = 1501175 = 50n 
P 3.6 Always work from the side of the circuit furthest from the source. Remember 
that the current in all series-connected circuits is the same, and that the 
voltage drop across all parallel-connected resistors is the same. 
[a] Req = 12112011(18 + (281121)] = 12ll20ll(18 + 12) = 1211201130 = 6 n 
[bJ Req = 4 + (9IJ18) + [5ll30ll(20 + 40)] = 4 + 6 + (5l[30ll6o) = 4 + 6 + 4 = 14n 
[c) Req = (100 kll300 k) + (75 kll50 kll150 k) + 25 k = 75 k + 25 k + 25 k = 125 kn 
P 3.7 [a) 12 nll24n = 8 n Therefore, Rab= 8 + 2 + 6 = 16 n 
1 1 1 1 15 1 
[b) Req = 24 kn+ 30 kn+ 20 kn= 120 kn= 8 kn 
Req = 8 kn; Req + 7 = 15 kn 
1 1 1 1 5 1 
Rab= 15 kn+ 30 kn+ 15 kn= 30 kn= 6 kn 
Rab=6 kn 
p 3.8 [a] 601120 = 1200/80 = 15 n 121124 = 288/36 = 8 n 
3011120 = 3600/150 = 24n 15+8+7 = 30n 
Rab = 15 + 24 + 25 = 64 n 
[b) 35+40 = 75n 751150 = 3750/125 = 30n 
3o + 20 =::;: 50 n 50ll75 = 3750/125 = 30 n 
30+ 10 = 40n 401160 + 91118 = 24 + 6 = 30n 
301130 = 15n Rab = 10 + 15 + 5 = 30 n 
[c) 50+30 =son 801120 = 16n 
16+ 14 = 30n 30+ 24 = 54n 
54[[27 = 1sn 18+12 = 30Q 
301130 = 150 Rab = 3 + 15 + 2 = 20 n 
3-12 CHAPTER 3. Simple Resistive Circuits 
P 3.9 [a] For circuit (a) 
Rab= 1511(18+481116) = 100 
For circuit (b) 
5ll10ll15ll10ll(12+18) = 20 
1611(14 + 2) = 8 o 
Rab = 4 + 8 + 12 = 24 0 
For circuit ( c) 
14411(4+12) = 14.40 
14.4 + 5.6 = 200 
201112 = 7.50 
7.5 + 2.5 = 100 
101115 = 60 
14+6+10 = 300 
Rab= 301160 = 200 
202 
(b] Pa = lo = 40 W 
1442 
Pb=27=768W 
Pc= 52(20) = 500 W 
P 3.10 Req = 6ll30jl20 = 4 0 
V30A = vm = (30 A)(40) = 120 V 
Therefore, since the three original resistors are in parallel with the current 
source: 
V3on = V30A = 120 v 
v2 1202 
Thus, P3on = ;~n = W = 480 W 
P 3.11 {a ] 100 120 
1800 
200 
Req = (10 + 40 + 20)11[12 + (2011180)] = 701130 = 21 n 
7Ji2A = 12(21) = 252 V 
40 
Vo= V4on = 10 + 40 + 20 (252) = 144 V 
v20n = 20111so (252) = 18 (252) = 151.2 v 
12 + (2011180) 30 
. = 151.2 = 7 56 A 
'lo 20 . 
[b] P12n = (252/30)2(12) = 846.72 W 
[c] P12A = -(252)(12) = -3024 W 
Problems 3-13 
Thus the power developed by the current source is 3024 W. 
p 3.12 
R2 R 
[a] Req = RllR= 2R = 2 
[h] Req - RllRllRll · · · llR 
- Rll_B_ 
n-1 
R2/(n -1) 
R+R/(n-1) 
[c] One solution: 
(n R's) 
R2 R 
=-= 
nR n 
100 n - 200 n + 500 n 
- 1000/5 + 1000/2 
- 1 kn111 kn111 kn111 kn111kn+1 kn111 kn 
700 Q ___,... 
1 kO 
3-14 CHAPTER 3. Simple Resistive Circuits 
p 3.13 
p 3.14 
[d] One solution: 
5.5 kn 5 kn+o.5 kn 
- 2 kn + 2 kn + 1 kn + o.5 kn 
_ 2 kn + 2 kn + 2 kn + 2 kn 
2 4 
- 2 kn+ 2 kn+ 2 kn112 kn+ 2 kn112 kn112 kn112 kn 
2 kO 
5.5 kq... 
160(3300) 
[a] Vo = ( 4700 + 3300) = 66 V 
[h) i = 160/8000 = 20 mA 
PR1 = (400 x 10-6)(4.7 x 103) = 1.88 w 
PR2 = (400 x 10-6)(3.3 x 103 ) = 1.32 w 
[c) Since R1 and R2 carry the same current and R1 > R2 to satisfy the voltage 
requirement, first pick R1 to meet the 0.5 W specification 
Therefore, (~~) 2 R1 :S 0.5 160- 66 'lR1 = 
942 
Thus, Ri 2 0.5 or R1 2 17,672 n 
Now use the voltage specification: 
R2 
R2 + 17,672 (160) = 66 
Thus, R2 = 12,408 n 
4 = 20R2 
R2+40 
so R2=10n 
3 = 20Re 
40+Re 
so Re= 120 n 
17 
Thus, -=--- so RL = 24n 
p 3.15 
Problems 3-15 
so 
Th 4R = 5 25R = 5.25( 48R2) en, 2 • e 48 + R2 
Thus, R2 = 15 kO and R1 = 4(15 k) = 60 kn 
[b] The resistor that must dissipate the most power is R1 , as it has the largest 
resistance and carries the same current as the parallel combination of R2 
and the load resistor. The power dissipated in R1 will be maximum when 
the voltage across R1 is maximum. This will occur when the voltage 
divider has a resistive load. Thus, 
VR1 = 100 -16 = 84 V 
842 
PR1 = 60 k = 117.6 m W 
Thus the minimum power rating for all resistors should be 1/8 W. 
P 3.16 Refer to the solution to Problem 3.15. The voltage divider will reach the 
maximum power it can safely dissipate when the power dissipated in R1 equals 
0.15 W. Thus, 
l/2 
_BL = 0.15 SO VR1 = 94.87 V 
60 k 
110 = 100 - 94.87 = 5.13 V 
lOORe 
So, 60 k +Re = 5.13 and R-e = 3.25 kn 
Th (l5 k)RL = 3250 
us, 15 k+ RL 
and RL = 4.14 kn 
The minimum value for RL is thus 4.14 kn. 
3~ 16 CHAPTER 3. Simple Resistive Circuits 
P 3.17 [a ] 
20k0 60k0 
p 3.18 
[b ] 
480V 
180 kO + 60 kD = 240 kD 
80 kOll240 kD = 60 kD 
V 0 1 = 
50,000 (480) = 360 V 
(20,000 + 60,000) 
180,000 
V 0 = (240,000) ( V 0 1) = 270 V 
20 kO 
480V 
i = 480 - 48 mA 100,000 - . 
80,000i = 384 v 
= 180,000 (384) = 288 v 
Vo 240,000 
60 kO 
+ 
+ 
[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 
I 80,000 ( ) 
v 01 = (lOO,OOO) 480 = 384 V 
Now note this is the input voltage to the second voltage divider when the 
current controlled voltage source is used. 
(24)2 - 80 
R1 +R2+R3 - ' 
(R1 + R2)24 = 12 
(R1 + R2 + R3) 
Therefore, R1 + R2 + R3 = 7.2 D 
2R3 = 7.2; 
R2 (24) = 5 
Ri +R2 +R3 
4.8R2 = Ri + R2 + 3.6 = 7.2 
Thus, R2 = 1.50; 
P 3.19 [a] At no load: 
At full load: 
Therefore k -
Cl: -
Also, R 
_ (l-k)R 
1 - 2 
k 
[b] Ri - ( 0.05) R = 2.5 kO 
0.68 ° 
( 0.05) f) O.l2 R0 = 14.167 kH 
[c ] 
+ 
60V 
R3 = 3.60 
R _ (1- k)R 1 - 2 
k 
Ri = (l - a:) Re 
a 
Ri = (k- a) Ro 
ak 
Problems 3~17 
Maximum dissipation in R2 occurs at no load, therefore, 
[(60)(0.85)}2 
PR2(max) = 14 167 = 183.6 m w 
' 
Maximum dissipation in R1 occurs at full load. 
p - [60 - 0.80(60)}2 -
Ri(max) - 2500 - 57.60 mW 
3~ 18 CHAPTER 3. Simple Resistive Circuits 
[d ] + 
GOV 
34 kn 
- (50)
2 = 1.44 W = 1440 mW 
2500 
(0)2 
14167 = 0 w , 
short 
P 3.20 [a] Let v0 be the voltage across the parallel branches, positive at the upper 
terminal, then 
ig = v0 G1 + VoG2 + · · · + VoG N = Vo( G1 + G2 + · · · + G N) 
It follows that 
'Lg Vo = ~---;:;...__ __ _ 
( G1 + G2 + ... + G N) 
The current in the kth branch is ik = v0 Gk; Thus, 
. i9Gk 
'lk = 
[G1 + G2 +. + GN] 
[b] z6.25 = 1142(0.l5) = 32 mA 
(4 + 0.4 + 1+0.16 + 0.1+0.05] 
P 3.21 Begin by using the relationships among the branch currents to express all 
branch currents in terms of i 4 : 
ii = 4i2 = 4(8i3) = 5(32i4) 
Now use KCL at the top node to relate the branch currents to the current 
supplied by the source. 
Express the branch currents in terms of i 4 and solve for i 4 : 
so 
Problems 3~ 19 
Since the resistors are in parallel, the same voltage, 1 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 : 
Vg 1 V 
R4 = i4 = (5/206) mA = 41.2 kn 
Calculate i 3 from i 4 and use Ohm's law as above to find R3 : 
• Vg 1 V 
.. R3 = i3 = (25/206) mA = 8240n 
Calculate i 2 from i 4 and use Ohm's law as above to find R2: 
40 . 0.2 A i2 = 'l4 = -
206 
• Vg 1 V 
.. R2 = i2 = (200/206) mA = 1030n 
Calculate i 1 from i 4 and use Ohm's law as above to find R1: 
. 160' 0.8 A i1 = 'l4 = -206 
• Vg 1 V 
. . Ri = ii = (800/206) mA = 257·5 n 
The resulting circuit is shown below: 
+ 
lV 
P 3.22 [a] The equivalent resistance to the right of the 10 kn resistor is 
3 k + 8 k + [6 kll(5 k + 7 k)] = 15 k!l. Therefore, 
. 15 klllO k 6 k 
i10k = 10 k (0.002) = 10 k (0.002) = 1.2 mA 
[b] The voltage drop across the 10 kn resistor can be found using Ohm's law: 
v1ok = (10, OOO)i10k = (10, 000)(0.0012) = 12 V 
[ c] The voltage v10k drops across the 3 kn resistor, the 8 kn resistor and the 
equivalent resistance of the 6 kn and the parallel branch containing the 
5 kn and 7 k!l resistors. Thus, using voltage division, 
6kll(5k+7k) 4 
V6k = 3 k + 8 k + [6 kil(5 k + 7 k)) (12) = 15 (12) = 3.2 V 
[d) The voltage v6k drops across the branch containing the 5 kn and 7 kn 
resistors. Thus, using voltage division, 
5k 
V5k = 5 k+ 7 k(3.2) = 1.33 V 
3-20 CHAPTER 3. Simple Resistive Circuits 
P 3.23 [a] The voltage drop across the 240 n resistor is the same as the voltage drop 
across the parallel combination of the branch containingthe 240 n 
resistor and the branch containing the 180 n and 300, n resistors. Thus 
by voltage division, 
p 3.24 
24011(180 + 300) 160 
V 240 = [24011(180 + 300)] + 140 + 200 (lO) = 500 (lO) = 3·2 V 
[b] The current in the 240 n resistor can be found from its voltage using 
Ohm's law: 
. V240 3.2 
i240 = 240 = 240 = 13.33 mA 
[ c] The current in the 140 n resistor divides between two branches - one 
containing the 180 n and 300 n resistors and the other containing the 
240 n resistor. Using current division, 
. - 24011(180+300) (. ) - 0 01333 
i240 - 240 t140 - . 
1 
(a) Vik = l + 5 (30) = 5 V 
15 
Vi5k = 15 + 60 (30) = 6 V 
Vx = V15k - Vik = 6 - 5 = 1 V 
(b) Vik= ~ (1) = V 8 j6 
V15k = ~; (15) = Vs/5 
so ii40 = 240(0.01333) = 20 mA 
160 
Vx = ( Vs/5) - ( Vs/6) = Vs/30 
P 3.25 60ll30 = 20 n 
i3on = (25)(75) = 15 A 
125 
V2 = (15)(20) = 300 V 
V2 + 30i30 = 750 V 
Vi - 12(25) = 750 
V1 = 1050 V 
P 3.26 i10k = (l8)(l5 k) = 6.75 mA 
40 k 
V15k = -(6.75 m)(15 k) = -101.25 V 
i 3k = 18 m - 6.75 m = 11.25 mA 
V12k = -(12 k)(ll.25 m) = -135 V 
V0 = -101.25- (-135) = 33.75 V 
Problems 3-21 
p 3.27 54011270 = 180; 180 + 20 = 200; 2011 (10 + 15 + 35) = 15 o; 
· - 20 11 60 (15) = 11 25 A· i 0 = 272171
54 (11.25) = 7.5 A 
~n- 20 · ' 
p 3.28 [a] 401110 = 8 0 
8+2=100 
151110 = 60 
6+4=100 
301110 = 7.50 
. 120 
i12ov = ~7 = 16 A .5 
. 7.5 ( ) A 
i40 = 4 + 6 16 = 12 
i2n = 2 ! 8 ( 12) = 7. 2 A 
io = 4~ (7.2) = 1.44 A 
[h] i15o = i4n - i2n = 12 - 7.2 = 4.8 A 
P150 = ( 4.8)2(15) = 345.6 W 
P 3.29 [a] The voltage across the 9 0 resistor is 1(12 + 6) = 18 V. 
The current in the 9 0 resistor is 18/9 = 2 A. The current in the 2 0 
resistor is 1 + 2 = 3 A. Therefore, the voltage across the 24 0 resistor is 
(2)(3) + 18 = 24 v. 
The current in the 24 0 resistor is 1 A. The current in the 3 0 resistor is 
1 + 2 + 1 = 4 A. Therefore, the voltage across the 72 0 resistor is 
24 + 3( 4) = 36 v. 
The current in the 72 0 resistor is 36/72 = 0.5 A. 
The 20 0115 0 resistors are equivalent to a 4 0 resistor. The current in 
this equivalent resistor is 0.5 + 1 + 3 = 4.5 A. Therefore the voltage 
across the 108 0 resistor is 36 + 4.5( 4) = 54 v. 
The current in the 108 0 resistor is 54/108 = 0.5 A. The current in the 
1.2 0 resistor is 4.5 + 0.5 = 5 A. Therefore, 
Vg = (1.2)(5) + 54 = 60 V 
3-22 CHAPTER 3. Simple Resistive Circuits 
[b] The current in the 20 n resistor is 
. = (4.5)(4) = 18 = 09 A 
'l2Q 20 20 . 
Thus, the power dissipated by the 20 n resistor is 
P20 = (0.9)2(20) = 16.2 W 
P 3.30 [a] The model of the ammeter is an ideal ammeter in parallel with a resistor 
whose resistance is given by 
Rm= 1~~V = 500. 
We can calculate the current through the real meter using current 
division: 
(25/12) . 25 . 1 . 
'lm = 50 + (25/12) (imeas) = 625 (imeas) = 25 'lmeas 
(b] At full scale, imeas = 5 A and im = 2 mA so 5 - 0.002 = 4998 mA flows 
throught the resistor RA: 
100 mV 100 
RA = 4998 m A = 4998 O 
(100/4998) . 1 . 
im = 50 + (100/4998) (imeas) = 2500 (imeas) 
[c] Yes 
P 3.31 The measured value is 601130.5 = 20.22 0. 
1~ 00 
'lg = (20.22 + lO) = 5.96 A; imeas = 90_5 (5.96) = 3.95 A 
The true value is 601130 = 20 0. 
180 
ig = (20 + 10) = 6 A; itrue = ~~ ( 6) = 4 A 
%error = [3·:5 - 1) x 100 = -1.283 
P 3.32 Begin by using current division to find the actual value of the current i 0 : 
24 
'ltrue = 24 + 5.5 (20 mA) = 16.27 mA 
24 
'tmeas = 24 + 5.5 + 0.5 (20 mA) = 16 mA 
3 error = [ 6
16 - 1] 100 = -1.663 
1 .27 
P 3.33 For all full-scale readings the total resistance is 
full-scale reading 
Rv + Rrnovement = 10_3 
Rv = 1000 (full-scale reading) - 50 
(a] Rv = 1000(100) - 50 = 99, 950 0 
[b] Rv = 1000(5) - 50 = 4950!1 
[ c] Rv = 100 - 50 = 50 0 
P 3.34 [a] Vmeas = (20 X 10-3)(24IJ5.5IJ4950) = 0.089411 V 
(b] Vtrue = (20 X 10-3)(24115.5) = 0.089492 V 
( 0.089411 ) % error = 0.089492 - 1 x 100 = -0.08998% 
p 3.35 
+ 50mV 
Original meter: Re = 50 x 10-
3 
= 0 005 0 10 . 
Modified meter: R = (0.01
5)(0.005) = 0.00375 0 
e 0.02 
(Jfs)(0.00375) = 50 x 10-3 
Ifs= 13.33 A 
Problems 3-23 
P 3.36 At full scale the voltage across the shunt resistor will be 50 m V; therefore the 
power dissipated will be 
(50 x 10-3 ) 2 
Therefore RA> 0.5 = 5 m!1 
Otherwise the power dissipated in RA will exceed its power rating of 0.5 W 
3-24 CHAPTER 3. Simple Resistive Circuits 
When RA::;::::- 5 mO, the shunt current will be 
i A = 50 x 10-3 = 10 A 
5 x 10-3 
The measured current will be imeas = 10 + 0.001 = 10.001 A 
. ·. Full-scale reading is for practical purposes is 10 A 
P 3.37 The current in the shunt resistor at full-scale deflection is 
iA = iruuscale = 2 x 10-3 A. The voltage across RA at full-scale deflection is 
always 100 mV; therefore, 
100 x 10-3 100 
RA= . 2 lQ-3 -
iru1Iscale - X 1 OOOirunscale - 2 
100 
[a] RA = 5000 _ 2 = 20,008 mO 
100 
[b] RA = 2000 _ 2 = 50.05 mO 
100 
[c] RA = 1000 - 2 = 100.20 mo 
[d] R = lOO :::;::: 2.0830 
A 50-2 
P 3.38 [a 1 
20 kO 
ie 0.6V -------{ + - --
0.2kO 
20 X 103i1 + 80 X 103 (i1 - iB) = 7.5 
80 X 103 (i1 - iB) = 0.6 + 4QiB(0.2 X 103) 
lOOi1 - 80iB = 7 .5 x 10-3 
80i1 - 88iB = 0.6 X 10-3 
Calculator solution yields is = 225 µA 
+ 7.5V 
Problems 3-25 
[b] With the insertion of the ammeter the equations become 
100i1 - 80iB = 7.5 x 10-3 (no change) 
80 X 103(i1 - iB) = 103iB + 0.6 + 40iB(200) 
80i1 - 89iB = 0.6 X 10-3 
Calculator solution yields iB = 216 µA 
( 216 ) [c] % error = 225 -1 100 = -4% 
P 3.39 [a] Vmeter = 180 V 
p 3.40 
(b) Rmeter = (100)(200) = 20 kO 
201170 = 15.56 kO 
180 
Vmeter = 35_56 X 15.56 = 78. 76 V 
[c] 201120 = 10 kO 
180 
Vmeter = 80 (10) = 22.5 V 
[ d] Vmeter a = 180 V 
Vmeter b + Vmeter c = 101.26 V 
No, because of the loading effect. 
[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 ] 
___,.. ~ 
+ 
R,,,n1 = ( 400)(1000) = 400 kO = Rm2 
. . . Rm1 + Rm2 = 800 kO 
. 400 10-3 1 A . 
i1 max = 400 x = m = i2 max 
. ·. imax = 1 mA since meters are in series 
Vmax = 10-3(400 + 400)103 = 800 V 
Thus the meters can be used to measure the voltage 
3-26 CHAPTER 3. Simple Resistive Circuits 
. 504 
[c] 'tm = 800 x 103 = 0.63 mA 
Vml = (0.63)(400) = 252 V = Vm2 
P 3.41 The current in the series-connected voltmeters is 
im = !~~ = 0.82 mA 
v50 kn= (0.82)(50) = 41 V 
Vpower supply = 328 + 328 + 41 = 697 V 
p 3.42 
800 v 
Rmeter = Rm + Rmovement = 1 mA = 800 kn 
vmeas = (300 k011600 knll800 k0)(3.5 mA) = (160 kn)(3.5 mA) = 560 V 
Vtrue = (300 knll600 kn)(3.5 mA) = (200 k0)(3.5 mA) = 700 V 
( 560 ) % error = 700 -1 100 = -20% 
P 3.43 [a] Rrneter = 300 kn+ 600 knll200 kO = 450 kn 
450ll360 = 200 kn 
200 
Vmeter = 240 ( 600) = 500 V 
[b] What is the percent error in the measured voltage? 
360 
True value = 400 ( 600) = 540 V 
( 500 ) % error = 540 - 1 100 = -7.413 
P 3.44 [a] R1 - (50)103 = 50 kO 
R2 - (20)103 = 20 kO 
R3 - (2)103 = 2 kn 
Problems 3-27 
[h] Let ia actual current in the movement 
id - design current in the movement 
Then % error = C: -1) 100 
For the 50 V scale: 
. 50 50 
ia = 50,000 + 100 = 50,100' 
. 50 
id= 50,000 
~a = 50,000 = 0.9980 % error = (0.9980 - 1)100 = -0.20% 
id 50,100 
For the 20 V scale: 
ia = 20,000 = 0.995 % error = (0.995 - 1.0)100 = -0.4975% 
id 20,100 
For the 2 V scale: 
~a = 2000 = 0.9524 % error = (0.9524 - 1.0)100 = -4.76% 
id 2100 
P 3.45 [a] Rmovement = 5 n 
50 
R1 + Rmovement = 2 x 10_3 = 25 kn R 1 = 24,995 n 
100 
R2 + R1 + Rmovement = 2 x 10_3 = 50 kn R2 = 25 kn 
200 
R3 + R2 + R1 + Rmovement = 2 X 10_3 = 100 kn 
:. R3 = 50 kn 
[h] 
200V ....----
~ 
+ t1 
. i 
'move 
v, 500 kO 
imove = ~~~ (2) = 1.88 mA 
3--28 CHAPTER 3. Simple Resistive Circuits 
V1 = (1.88)(50) = 94 V 
. 94 
ii = 500 = 0.188 mA 
i2 = irnove +ii= 1.88 + 0.188 = 2.068 mA 
Vrneas = Vx = 94 + 50i2 = 197.4 V 
{c] v1 - 100 V i2 = 2 + 0.20 = 2.20 mA 
i1 100/500 = 0.20 mA Vrneas = Vx= 100 + 50(2.20) = 210 V 
P 3.46 From the problem statement we have 
80 = \1s(l0) (1) Vs ih mV; Rs in Mn 
lO+Rs 
72 = l1s(5) (2) 
5+Rs 
[a] From Eq (1) 10 +Rs= 0.125\1s 
:. Rs= 0.125Vs - 10 
Substituting into Eq (2) yields 
5Vs 
72 = 0_125l1s _ 5 or Vs= 90 mV 
[b] From Eq (1) 
80 = goo or 80Rs = 100 
lO+Rs 
So Rs = 1250 kn 
P 3.47 Since the bridge is balanced, we can remove the detector without disturbing 
the voltages and currents in the circuit. 
It follows that 
P 3.48 [a ] 
Problems 3~ 29 
R3ig(R2 +Rx) - Rxig(R1 + R3) 
LR - LR 
21 v + 
The condition for a balanced bridge is that the product of the opposite 
resistors must be equal: 
(800)(Rx) = (1200)(600) so R = (1200)(600) = 9000 
x 800 
[b] The source current is the sum of the two branch currents. Each branch 
current can be determined using Ohm's law, since the resistors in each 
branch are in series and the voltage drop across each branch is 21 V: 
. 21 v 21 v 
is= 8000 + 6000 + 12000 + 9000 = 25 mA 
[c) We can use current division to find the current in each branch: 
1200+ 900 
'lJeft = 1200 + 900 + 800 + 600 (25 mA) = 15 mA 
iright - 25 mA - 15 mA = 10 mA 
Now we can use the formula p = Ri2 to find the power dissipated by each 
resistor: 
Psoo = (800)(0.015)2 = 180 mW p600 = (600)(0.015)2 = 135 mW 
p1200 = (1200)(0.010)2 = 120 mW p900 = (900)(0.010)2 = 90 mW 
Thus, the 800 n resistor absorbs the most power; it absorbs 180 mW of 
power. 
3-30 CHAPTER 3. Simple Resistive Circuits 
[d] From the analysis in part (c), the 900n resistor absorbs the least power; it 
absorbs 90 rnW of power. 
P 3.49 Redraw the circuit, replacing the detector branch with a short circuit. 
_....,.. i$ 
+ 
50V -
15 kOll3 kn = 2.5 kn 
g knll45 kn = 7.5 kn 
. 50 
'tg = lO = 5 rnA 
V1 = 5(2.5) = 12.5 V 
V2 = 5(7.5) = 37.5 V 
. 12·5 833 3 A i1 =ls= .. p, 
. 37.5 A 
i2 = -9- = 4166.7 µ 
.l 
It 
.l 
12 
id= i 1 - i2 = -3333.4µA 
+ 
5k0 
3 kO 
V1 
+ 
"2 
P 3.50 Note the bridge structure is balanced, that is 10 x 18 = 30 x 6, hence there is 
no current in the 50n resistor. It follows that the equivalent resistance of the 
circuit is 
Req = 3 + (10+6)11(30+18) = 3+12 = 15n 
The source current is 300/15 = 20 A. 
The current down through the branch containing the 30 n and 18 n resistors is 
i18 = 30 ~ 18 (20) = 5 A 
. . . Pl8 = 18(5)2 = 450 W 
Problems 3-31 
P 3.51 In order that all four decades (1, 10, 100, 1000) that are used to set R3 
contribute to the balance of the bridge, the ratio R2/ R1 should be set to 0.001. 
P 3.52 Begin by transforming the a-connected resistors (10 n, 30 n, 60 n) to 
Y-connected resistors. Both the Y-connected and a-connected resistors are 
shown below to assist in using Eqs. 3.44 - 3.46: 
2BO 
160 600 
BOY 
100 
Now use Eqs. 3.44 - 3.46 to calculate the values of the Y-connected resistors: 
R = (30)(60) = 18 n· 
1 10+30+ 60 ' 
R = (60)(10) = 6 n· 
2 10+30+60 ' 
The transformed circuit is shown below: 
2BO 
BOY 
___,. i1 
16 o 1BO 60 
R = (30)(10) = 3 n 
3 10+ 30+ 60 
The equivalent resistance seen by the 80 V source can be calculated by making 
series and parallel combinations of the resistors to the right of the 24 V source: 
Req = (28 + 6)11(16 + 18) + 3 = 341134 + 3 = 11+3 = 2on 
Therefore, the current i in the 80 V source is given by 
Use current division to calculate the currents i1 and i2. Note that the current 
ii flows in the branch containing the 28 n and 6 n series connected resistors, 
3-32 CHAPTER 3. Simple Resistive Circuits 
while the current i 2 flows in the parallel branch that contains the series 
connection of the 16 0 and 18 n resistors: 
16+18 . 34 
i 1 =16+18+28+6(i) = 68(4 A)= 2 A, 
and i2 = 4 A - 2 A = 2 A 
Now use KVL and Ohm's law to calculate v1. Note that v1 is the sum of the 
voltage drop across the 18 0 resistor, l8i2, and the voltage drop across the 3 0 
resistor, 3i: 
V1 = l8i2 + 3i = 18(2 A)+ 3(4 A)= 36 + 12 = 48 v 
Finally, use KVL and Ohm's law to calculate v2 . Note that v2 is the sum of 
the voltage drop across the 6 0 resistor, 6i1 , and the voltage drop across the 
3 n resistor' 3i: 
V2 = 6i1+3i = 6(2 A)+ 3(4 A)= 12+12 = 24 v 
P 3.53 [a] Calculate the values of the Y-connected resistors that are equivalent to the 
10 0, 30 0, and 600 ~-connected resistors: 
R = (10)(30) = 3 O· 
x 10+30+60 ' 
R = (10)(60) = 60 
z 10 + 30 + 60 
R = (30)(60) = 18 O· 
y 10+30+ 60 ' 
Replacing the R2-R3-R4 delta with its equivalent Y gives 
200 
Now calculate the equivalent resistance Rab by making series and parallel 
combinations of the resistors: 
Rab= 20 + 3 + [(30 + 6)11(18 + 18)} + 9 = 500 
Problems 3-33 
[h] Calculate the values of the A-connected resistors that are equivalent to 
the 10 0, 30 0, and 60 0 Y-connected resistors: 
Rx _ (10)(30) + (30)(60) + (10)(60) = 2700 = 900 
30 30 
(10)(30) + (30)(60) + (10)(60) = 2700 = 2700 
10 10 
Rz _ (10)(30) + (30)(60) + (10)(60) = 2700 = 45 O 
60 60 
Replacing the R2 , R4 , R5 wye with its equivalent A gives 
200 
Make series and parallel combinations of the resistors to find the 
equivalent resistance Rab: 
90011300 = 22.50; 270011180 = 16.8750 
4511(22.5 + 16.875) = 21 o 
Rab= 20 + 21+9 = 500 
(c] Convert the delta connection ~-R5-R6 to its equivalent wye. 
Convert the wye connection R3-R4-R6 to its equivalent delta. 
P 3.54 Replace the upper and lower deltas with the equivalent wyes: 
R = (25)(10) = 5 o· R = (10)(15) = 3 o· R = (25)(15) = 7 5 o 
w w ' w w ' w w . 
R = (125)(25) = 12_5 0 . R = (25)(100) = 10 0 . R = (125)(100) = 50 0 
lL 250 ' 2L 250 ' 3L 250 
The resulting circuit is shown below: 
3-34 CHAPTER 3. Simple Resistive Circuits 
p 3.55 
170 
30 Q 
Now make series and parallel combinations of the resistors: 
(7.5+12.5)11(3 + 17 + 30 + 10) = 201160 = 150 
Rab = 15 + 5 + 15 + 50 + 14 = 99 0 
a 
250 6.250 
GA t 15 Q 
+ 
600 
300 
vx 
c 
25116.25 = 5 0 601130 = 200 
+ 
15 0 
. = (5)(l5) = 2 25 A· 
i1 ( 40) . ' Vx = 20
i1 = 45 V 
Vg = 25i1 = 56.25 V 
V6.25 = Vg - Va; = 11.25 V 
11.252 452 56.252 
Pdevice = 6.25 + 30 + 15 = 298.6875 W 
P 3.56 8 + 12 = 20 o 
201160 = 150 
15+20 = 350 
3511140 = 2s o 
28 + 22 = 500 
501175 = 300 
30+ 10 = 400 
ig = 240 I 40 = 6 A 
i 0 = (6)(50)/125 = 2.4 A 
ii4on = (6 - 2.4)(35)/175 = 0.72 A 
p14on = (0.72)2(140) = 72.576 W 
Problems 3-35 
P 3.57 The top of the pyramid can be replaced by a resistor equal to 
R = (3.6)(1.8) = 1.2 k" 
1 5.4 ,H, 
The lower left and right deltas can be replaced by wyes. Each resistance in the 
wye equals 600 0. Thus our circuit can be reduced to 
1200 0 
Now the 2400 0 in parallel with 1200 0 reduces to 800 0 . 
. ·. Rab = 600 + 800 + 600 = 2000 = 2 kO 
3~36 CHAPTER 3. Simple Resistive Circuits 
P 3.58 [a] Convert the upper delta to a wye. 
R = (80)(200) = 400 
1 400 
R = (80)(120) = 240 
2 400 
R = (120)(200) = 60 n 
3 400 
Convert the lower delta to a wye. 
R = (60)(90) = 18 0 
4 300 
R = (60)(150) = 300 
5 300 
R = (90)(150) = 45 fl 
6 300 
Now redraw the circuit using the wye equivalents. 
a~ 
20 
38 0 15 0 
b ______ _ 
Rab= 2 + 40 + (80~~~20) + 30 = 42 + 48 + 30 = 1200 
[b] When Vab = 600 V 
. 600 
ig = 120 = 5 A 
. _ (5)(80) _ 2 A 
i15 - 200 -
Pt50 = (4)(15) = 60 W 
Problems 3-37 
P 3.59 [a] After the 20 0-100 0-50 0 wye is replaced by its equivalent delta, the 
circuit reduces to 
3200 
--+i1 
80 Q 
4000 
1800 
Now the circuit can be reduced to 
96 
i = :o~ (1000) = 240 rnA 
i 0 = 1~i0 (240) = 96 rnA 
[b] ii = :ooo (240) = 48 mA 
84 Q 
soon 
2400 
[c} Now that i 0 and i 1 are known return to the original circuit 
3200 48mA.., 
20 Q 48mA"' 
+ 
Vg 
V2 = (50)(0.048) + (600)(0.096) = 60 V 
V2 60 
i 2 = 100 = 100 = 600 rnA 
800 Q 
[d] Vg = V2 + 20(0.6 + 0.048) = 60 + 12.96 = 72.96 V 
pg= -(vg)(l) = -72.96 W 
Thus the current source delivers 72.96 W. 
3-38 CHAPTER 3. Simple Resistive Circuits 
P 3.60 [a) Replace the 30--60-10 n delta with a wye equivalent to get 
182 (.l 
500V 27 (.l 
Usingseries/parallel reductions the circuit reduces to 
33 (.l 
500V 
. 500 
zg = 100 = 5 A 
i2 = 200 ( 5) = 4 A 
250 
[b) ii = 33/30 = 1.1 A 
40 Q 
270 
Returning to the original circuit we have 
500V 
+ 33V -
300 
io = 1.1 - 1.0 = 0.1 A 
[c) v = 60i0 = 6 V 
[d) Psupplied = (500)(5.0) = 2500 W 
P 3.61 Subtracting Eq. 3.42 from Eq. 3.43 gives 
+ 
135 v 
Adding this expression to Eq. 3.41 and solving for R1 gives 
p 3.62 
Problems 3-39 
To find R2, subtract Eq. 3.43 from Eq. 3.41 and add this result to Eq. 3.42. 
To find R3, subtract Eq. 3.41 from Eq. 3.42 and add this result to Eq. 3.43. 
Using the hint, Eq. 3.43 becomes 
Ri + R3 = Rb[(R2/R3)Rb + (R2/R1)Rb) = Rb(R1 + R3)R2 
(R2/ R1)Rb +Rb+ (R2/ R3)Rb (R1R2 + R2R3 + R3R1) 
Solving for Rb gives Rb= (R1R2 + R2R3 + R3R1)/ R2· To find Ra: First use 
Eqs. 3.44-3.46 to obtain the ratios (Ri/ R3) =(Re/ Ra) or Re= (Ri/ R3)Ra. 
and (Ri/ R2) = (Rb/ Ra.) or Rb = (R1/ R2)Ra. Now use these relationships to 
eliminate Rb and Re from Eq. 3.42. To find Re, use Eqs. 3.44-3.46 to obtain 
the ratios Rb= (R3/ R2)Rc and Ra.= (R3/ R 1)Re· Now use the relationships to 
eliminate Rb and Ra from Eq. 3.41. 
1 R1 
Ga - - = --------
Ra RiR2 + R2R3 + R3R1 
l/G1 
- (i/G1)(l/G2) + (l/G2)(l/G3) + (1/G3)(1/G1) 
(l/G1)(G1G2G3) G2G3 
- -
G1 + G2 + G3 Gi + G2 + G3 
Similar manipulations generate the expressions for Gb and Ge. 
Therefore 
When Rab= RL, the current into terminal a of the attenuator will be 
vifRL 
Using current division, the current in the RL branch will be 
Vi R2 
RL 2R1 + R2 + RL 
Therefore 
and 
9 x 104 = Ri + RiR2 
Vo= 0.6 = R2 
Vi 2R1 + R2 + 600 
3-40 CHAPTER 3. Simple Resistive Circuits 
: . 1.2R1 + 0.6R2 + 360 = R2 
0.4R2 = l.2R1 + 360 
R2 = 3R1+900 
9 x 104 = Ri + Ri (3R1 + 900) = 4Ri + 900R1 
Ri + 225R1 - 22,500 = 0 
R1 = -112.5 ± j (112.5)2 + 22,500 = -112.5 ± 187.5 
Ri = 75S1 
R2 = 3(75) + 900 = 1125 n 
P 3.64 [a] After making the Y-to-~ transformation, the circuit reduces to 
R 
3R c 
d 
Combining the parallel resistors reduces the circuit to 
0.75R 
a -----t.---
Now note: 
Therefore 
c 
3R 
d 
3RR 
3R+~L 
3RRL 2.25R2 + 3.75RRL 
o. 75R + 3R + RL - 3R + RL 
3R (2.25R
2 + 3.75RRL) 
R 3R + RL 3R(3R + 5RL) 
ab = 3R ( 2.25R
2 + 3. 75RRL) - 15R + 9RL 
+ 3R+Rt 
When Rab = R1, we have 15RRL + 9Rf, = 9R2 + 15RRL 
Therefore Rl, = R2 or RL = R 
[b] When R = RL, the circuit reduces to 
i; 0.75RL jo __,.. c __,.. 
a 
+ + 
vi 3RL v. 
b ______ ,____, 
d 
. ii(3RL) 1 . 1 Vi 
'lo= =-'li=--, 
4.5RL 1.5 1.5 RL 
Therefore Vo = 0.5 
vi 
P 3.65 [a] 3(3R - RL) = 3R + RL 
9R - 1800 = 3R + 600 
6R = 2400, R = 400f! 
0.75RL 
R = 2(400)(600)2 = 2400f! 
2 3(400)2 - (600)2 . 
[b ] i, 2400 Q ----... 
i 400 Q 400 Q 
-4 
180V 
Vo = Vi = 180 = 60 V 
3 3 
i 0 = 60 = 100 mA 
600 
. _ 180 - 60 _ 120 _ 50 A 
ii - 2400 - 2400 - m 
. 180 
i 9 = t300 = 300 mA 
i2 = 300 - 50 = 250 mA 
i3 = 100 - 50 = 50 mA 
i4 = 250 - 50 = 200 mA 
Problems 3--41 
600 Q 
3--42 CHAPTER 3. Simple Resistive Circuits 
P2400 top= (50 x 10-3)2(2400) = 6 W 
P4oo left= (250 x 10-3)2(400) = 25 W 
P4oo right = (50 x 10-3) 2( 400) = 1 W 
P400 vertical = (200 X 10-3 )2 ( 400) = 16 W 
P6oo load= (100 x 10-3) 2(600) = 6 W 
The 400 n resistor carrying i 2 
[c] P400 left = 25 W 
[d] The 400 n resistor carrying i3 
[e] P4oo right = 1 W 
P 3.66 [a ] 
+ 
VinR4 
Va=-----
. Ro+ R4 +AR 
R3 
R4Vin R3 
~=~-~=~+~+AR-~+~~ 
When the bridge is balanced, 
R4 R3 
Ro + R4 Vin = R2 + R3 Vin 
. R4 R3 
. . Ro + R~ - R2 +-R3 
Thus, 
R4Vin R4Vin 
Ro+R4+AR Ro+ R4 
- R4v· [ l - l ] 
m Ro + R4 + AR Ro + R4 
R4vin(-AR) 
(Ro+ R4 + AR)(Ro + R4) 
-(AR)R4vin 
(Ro+ R4)2 ' 
since AR<< R4 
[bl D..R = 0.03R0 
R = R2R4 = (1000)(5000) = lO OOO n 
0 R3 500 ' 
D..R = (0.03)(104) = 3000 
R:i -300(5000)(6) :::;;:: -40 v 
Vo . (15,000)2 . m 
[c] V0 -
(Ro+ R4 + D..R)(Ro + R4) 
-300 ( 5000) ( 6) 
- (15,300)(15,000) 
- -39.2157 m V 
-(t::.R)R4vin 
P 3.67 (a] approx value = (Ro+ R4) 2 
-( t::.R) R4Vin 
true value = (R0 + R4 + D..R)(Ro + R4) 
approx value (R0 + R4 + D..R) 
true value - . .. (R0 + R~) 
Problems 3-43 
[Ro+ R4 + D..R ] D..R 3 error = Ro + R4 - 1 X 100 = Ro + R4 X 100 
p 3.68 
But Ra= R2R4 
R3 
. R3AR 
· · 3 error = R4(R2 + R3) 
(500)(300) 
[b] 3 error = (t>OOO)(l500) x 100 = 23 
t::.R(R3)(100) = 0.5 
(R2 + R3)R4 
D..R(500)(100) = 0.5 
(1500)(5000) 
:. D..R = 750 
75 
3 change = lO,OOO x 100 = 0.753 
3~44 CHAPTER 3. Simple Resistive Circuits 
P 3.69 [a] From Eq 3.64 we have 
· )2 R2 
c~ · = Ri(l: 2u)2 
Substituting into Eq 3.63 yields 
R~ 
R2 = Ri(l + 2u)2R1 
Solving for R2 yields 
R2 = (1+2u)2 R1 
[b] From Eq 3.67 we have 
i1 R2 -=-----
Zb Ri+R2+2Ra 
But R2 = (1+2u)2R1 and Ra= uR1 therefore 
ii (1+2u)2 R1 (1+2u)2 
- = ---------
ib R1 + (1+2u)2R1 + 2uR1 (1+2u) + (1+2u)2 
1+2u 
- 2(1 +a) 
It follows that 
( i1)
2 = (1+2u)2 
ib 4(1 + u)2 
Substituting into Eq 3.66 gives 
R _ (1+2u)2Ra _ (1+2u)2uR1 
b - 4(1 + u)2 - 4(1 + a-)2 
P 3. 70 From Eq 3.69 
ii R2R3 
-=--
i3 D 
Therefore D can be written as 
D = (1+2a-)4Ri 
(l+CT) 
~ 
R2R3(l +CT) 
(1+2CT)4Rt 
(1+2a-)2R1R3(l +a-) 
(1 + 2a-)4 Rt 
(1 + a-)R3 
(l+ 2lT) 2R1 
When this result is substituted into Eq 3.69 we get 
R _ (1 + a-) 2 R5R1 
3 - (1+2<Y)4Rr 
Solving for R3 gives 
P 3. 71 From the dimensional specifications, calculate a and R3 : 
CT = ;; = O.i25 = 0.025; 
Calculate Ri from R3 and a: 
Problems 3-45 
Ra = a-R1 = 0.0259 n Rb= (1 + 2a)2a Ri = 0.0068 n 4(1 + a-) 2 
3-46 CHAPTER 3. Simple Resistive Circuits 
R2 = (1+2o-)2 Ri = 1.1435 0 
Using symmetry, 
Re = Rb = 0.0068 st Rd= Ra= 0.02590 
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, ii, and i 2 : 
i 1 = VdcR2 = 10.7561 A 
D 
i2 = Vdc(R1 + 2Ra) = 10 2439 A D . 
It follows that i~Rb = 3 Wand the power dissipation per meter is 
3/0.025 = 120 W /m. The value of it R1 = 120 W /m. The value of i~R2 = 120 
W /m. Finally, it Ra. = 3 W /m. 
P 3. 72 From the solution to Problem 3. 71 we have ib = 21 A and i3 = 10 A. By 
symmetry ic = 21 A thus the total current supplied by the 12 V source is 
21+21+10 or 52 A. Therefore the total power delivered by the source is p12 v 
(del) = (12)(52) = 624 W. We also have from the solution that 
Pa= Pb= Pc= Pd= 3 W. Therefore the total power delivered to the vertical 
resistors is Pv = (8)(3) = 24 W. The total power delivered to the five 
horizontal resistors is PH = 5(120) = 600 W. 
LPdiss = PH + Pv = 624 W = LPdel 
P 3.73 (a] <J' = 0.05/1.25 = 0.04 
Since the power dissipation is 150 W /m the power dissipated in R3 must 
be 150(1.25) or 187.5 W. Therefore 
122 
R3 = 187.5 = 0.7680 
From Table 3.1 we have 
R = (l + (]') 2 R3 = 0.6106 0 
1 (1+2<J')4 
Ra = (]' R1 = 0.0244 0 
R2 = (1 + 2cr)2 R1 = o. 7122 n 
R = (l + 20")20-Ri = 0.0066 n 
b 4(1+o-)2 
Therefore 
R4 = R2 = 0.7122 n 
Re = Rb = 0.0066 n 
[b} D = 0.4877 
i 1 = Vd_;R2 = 17.52 A 
ii R1 = 187.5 W or 150 W /m 
. - Ri + 2Ra TT - 16 23 A i2 - D Vdc - • 
i~R2 = 187.5 W or 150 W /m 
iiRa = 7.5 W or 150 W/m 
R5 = Ri = 0.6106 n 
Rd= Ra= 0.0244!1 
. - Ri + R2 + 2Ra TT - 33 75 A 
ib - D Vdc - • 
i~Rb = 7.5 W or 150 W/m 
12 
isource = 33.75 + 33.75 + 0.768 = 83.125 A 
Pdel = 12(83.125) = 997.50 W 
PH= 5(187.5) = 937.5 w 
Pv = 8(7.5) = 60 w 
LPdel = LPdiss = 997.50 W 
Problems 3-4 7 
-----4 
Techniques of Circuit Analysis 
Assessment Problems 
AP 4.1 [a] Redraw the circuit, labeling the reference node and the two node voltages: 
2Q 
Place these equations in standard form: 
V1 (2- + 2- + ~) + V2 (-~) 
60 15 5 5 
V1 (-~) 
Solving, v1 = 60 V and v2 = 10 V; 
Therefore, i 1 = ( V1 - v2) /5 = 10 A 
5A 
15 
[h] P15A = -(15 A)v1 = -(15 A)(60 V) = -900 W = 900 W(delivered) 
[c] P5A = (5 A)v2 = (5 A)(lO 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: 
V16Q~ 
+ 
1Q v 
The two node voltage equations are: 
V1 V1 - V2 
-4.5 + 1 + 6 + 2 - 0 
V2 V2 - V1 V2 - 30 
12 + 6 + 2 + 4 - 0 
Place these equations in standard form: 
V1 ( 1 + ~) + V2 ( - ~) - 4.5 
V1 (-~) - 7.5 
Solving, v1 = 6 V v2 = 18 V 
To find the voltage v, first find the current i through the series-connected 6 n 
and 2 n resistors: 
V1 - V2 6 - 18 
i = = = -1.5 A 
6+2 8 
Using a KVL equation, calculate v: 
V = 2i + V2 = 2( -1.5) + 18 = 15 V 
AP 4.3 [a] Redraw the circuit, choosing the node voltages and reference node as 
shown: 
8Q 4Q 
The node voltage equations are: 
V1 - 50 V1 V1 - V2 . 
6 + 8 + 2 - 3i1 - 0 
-a + V2 + V2 - V1 + 3i1 - 0 
4 2 
Problems 4-3 
The dependent source requires the following constraint equation: 
50-vi 
i1 = 6 
Place these equations in standard form: 
V1 (~ + ~ + ~) V2 (-~) ii(-3) 
50 
+ + -6 8 2 6 
V1 (-~) + V2 (~ + ~) + ii(3) - 5 
V1 (!) v2(0) ii(l) 50 + + 6 
Solving, V1 = 32 V; v2 = 16 V; i 1 = 3 A 
Using these values to calculate the power associated with each source: 
P50V = -50i1 - -150 W 
P5A = -5(v2) 
P3i1 = 3i1 ( V2 - v1) 
-80W 
-144 w 
(b] All three sources are delivering power to the circuit because the power 
computed in (a) for each of the sources is negative. 
AP 4.4 Redraw the circuit and label the reference node and the node at which the 
node voltage equation will be written: 
j 1~ 3~_2-'lfOQ111.---"t 
10V : jiil 400 
The node voltage equation is 
V0 V 0 - 10 V0 + 20i~ _ Q 
40 + 10 + 20 -
The constraint equation required by the dependent source is 
. . . 10 - V 0 10 + 20i~ 
i~ = i10n + i3on = 10 + 30 
Place these equations in standard form: 
4-4 CHAPTER 4. Techniques of Circuit Analysis 
( 1 1 1) 
Va 40 + 10 + 20 + ia(l) - 1 
va (:0) 
. ( 20) 10 + ia 1- 30 - 1 + 30 
Solving, ia = -3.2 A and Va= 24 V 
AP 4.5 Red.raw the circuit identifying the three node voltages and the reference node: 
l 7.5Q 10Q 2.50 
ix 
Note that the dependent voltage source and the node voltages v and v2 form a 
supernode. The v1 node voltage equation is 
~ + V1 - V - 4.8 = Q 
7.5 2.5 
The supernode equation is 
The constraint equation due to the dependent source is 
. V1 
i --
x - 7.5 
The constraint equation due to the supernode is 
V +ix= V2 
Place this set of equations in standard form: 
V1 ( 2_ + 2-) 
7.5 2.5 + v (-2-) 2.5 + v2(0) 
V1 (-2-) 
2.5 + v (2~5 + 110) + V2(2~5 +1) 
VI (-2-) 
7.5 
+ v(O) + v2(0) 
v1(0) + v(l) + v2(-l) 
+ ix(O) -
+ ix(O) 
+ ix(l) -
+ ix(l) -
Solving this set of equations gives v 1 = 15 V, v2 = 10 V, ix= 2 A, and 
v=8 V. 
4.8 
12 
0 
0 
Problems 4-5 
AP 4.6 Redraw the circuit identifying the reference node and the two unknown node 
voltages. Note that the right-most node voltage is the sum of the 60 V source 
and the dependent source voltage. 
6i.p 
...----"""'-+;•'>-------, 
60+ 6i.p 
30 
The node voltage equation at v1 is 
The constraint equation due to the dependent source is 
. 60 + 6iq, - V1 
iq, = 
3 
Place these two equations in standard form: 
v1 (!+2-+!) + iq,(-2) - 30+20 
2 24 3 
V1 (~) + iq,(1 - 2) - 20 
Solving, iq, = -4 A and v1 = 48 V 
AP 4. 7 [a] Redraw the circuit identifying the three mesh currents: 
300 
SQ~90Q 
80V ~ 26~ 13 
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 
80 
4-6 CHAPTER 4. Techniques of Circuit Analysis 
Place these equations in standard form: 
3lii - 5i2 - 26i3 - 80 
-5ii + 125i2 - 90i3 - 0 
-26ii - 90i2 + 124i3 - 0 
Solving, 
ii= 5 A; i3 = 2.5 A 
Psov = -(80)i1 = -(80)(5) = -400 W 
Therefore the 80 V source is delivering 400 W to the circuit. 
(b) Psn = (8)i~ = 8(2.5)2 = 50 W, so the 8 0 resistor dissipates 50 W. 
AP 4.8 [a] b = 8, n = 6, b - n + 1 = 3 
[h] Redraw the circuit identifying the three mesh currents: 
-3V.p 
The three mesh-current equations are 
-25+2(ii-i2)+5(i1-i3)+10 - 0 
-(-3v¢) + 14i2 + 3(i2 - i3) + 2(i2 - ii) - 0 
li3-10+5(i3-i1)+3(i3-i2) - 0 
The dependent source constraint equation is 
Vcp = 3(i3 - i2) 
Place these four equations in standard form: 
7i1 - 2i2 - 5i3 +Ovit> - 15 
-2i1+19i2 - 3i3 + 3v4> - 0 
-5i1 - 3i2 + 9i3 + Ovit> - 10 
Oi1 + 3i2 - 3i3 + 1v4> - 0 
Solving 
ii= 4 A; i2 = -1 A; V,p = 12 V 
Problems 4-7 
Pds = -(-3v.p)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: 
2Q 
6Q~8Q 
The mesh current equations are: 
-25 + 6(ia - ib) + 8(ia - ic) - 0 
2ib + 8(ib -ic) + 6(ib - ia) - 0 
5i.p + 8(ic - ia) + 8(ic - ib) - 0 
The dependent source constraint equation is i.p = ia. We can substitute this 
simple expression for i.p into the third mesh equation and place the equations 
in standard form: 
14ia - 6ib - 8ic 25 
-6ia + 16ib - 8ic - 0 
-3ia - Sib + 16ic - 0 
Solving, 
ia = 4 A; ib = 2.5 A; ic = 2 A 
Thus, 
AP 4.10 Redraw the circuit identifying the mesh currents: 
30Vh 2Q ~ ~ tl6A 
~-'1J~---~~•4V'QN il22~i~~-~~J il-~3~i~ 
4--8 CHAPTER 4. Techniques of Circuit Analysis 
Since there is a current source on the perimeter of the i 3 mesh, we know that 
i 3 = -16 A. The remaining two mesh equations are 
-30 + 3ii + 2(ii - i2) + 6ii 0 
8i2 + 5(i2 + 16) + 4i2 + 2(i2 - ii) 0 
Place these equations in standard form: 
llii - 2i2 30 
-2ii + 19i2 - -80 
Solving: ii= 2 A, i2 = -4 A, i3 = -16 A 
The current in the 2 n resistor is i 1 - i 2 = 6 A . ·. 
Thus, the 2 n resistors dissipates 72 W. 
AP 4.11 Redraw the circuit and identify the mesh currents: 
lOA 
+· 
2Q ~ lQ 
+ 
~ v<P 5Q~ 
2v¢ 
5 
P20 = (6)2(2) = 72 W 
There are current sources on the perimeters of both the ib mesh and the ic 
mesh, so we know that 
ib = -10 A; 
. 2vq, 
'lc=-
5 
The remaining mesh current equation is 
-75 + 2(ia + 10) + 5(ia - OAvt/J') = 0 
The dependent source requires the following constraint equation: 
Place the mesh current equation and the dependent source equation is 
standard form: 
7ia - 2vq, 55 
5ia - 3vq, - 0 
Solving: ia = 15 A; ib = -10 A; ic = 10 A; Vq, = 25 V 
Thus, ia = 15 A. 
Problems 4-9 
AP 4.12 Redraw the circuit and identify the mesh currents: 
20 
20~ 
l.c 
10 
The 2 A current source is shared by the meshes ia and ib. Thus we combine 
these meshes to form a supermesh and write the following equation: 
The other mesh current equation is 
The supermesh constraint equation is 
Place these three equations in standard form: 
2ia + 4ib - 4ic - 10 
-2ia - 2ib + 5ic - 6 
ia - ib + Oic 2 
Solving, 
Thus, 
ia = 7 A; ib = 5 A; ic = 6 A 
P1n = i~(l) = (6)2(1) = 36 W 
AP 4.13 Redraw the circuit and identify the reference node and the node voltage v1 : 
15Q V1 10Q 
20Vr~25V 
I l I 
The node voltage equation is 
V1 - 2Q _ 2 V1 - 25 = Q 
15 + 10 
4-10 CHAPTER 4. Techniques of Circuit Analysis 
Rearranging and solving, 
v (!__ !__) - 2 20 25 
1 15 + 10 - + 15 + 10 
V1 = 35 V 
P2A = -35(2) = -70 W 
Thus the 2 A current source delivers 70 W. 
AP 4.14 Redraw the circuit and identify the mesh currents: 
V'IA 
4A 
+ 
.--~~~___,~.~~~~~--. 
There is a current source on the perimeter of the i 3 mesh, so i 3 = 4 A. The 
other two mesh current equations are 
-128 + 4(i1 - 4) + 6(i1 - i2) + 2ii - 0 
30ix + 5i2 + 6(i2 - ii)+ 3(i2 - 4) - 0 
The constraint equation due to the dependent source is 
Substitute the constraint equation into the second mesh equation and place 
the resulting two mesh equations in standard form: 
12i1 - 6i2 144 
24i1+14i2 132 
Solving, 
i2 = -6 A; ix= 9-4 = 5 A 
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: 
36A t 
+ 
60 Va 
1.60 
+ 
80 v 
Transform the 120 V source in series with the 20 n resistor into a 6 A 
source in parallel with the 20 n resistor. Also transform the -60 V source 
in series with the 5 n resistor into a -12 A source in parallel with the 5 0resistor. The result is the following circuit: 
6.AI t 200 12.AI + 5 0 36A t 
+ 
60 Va 
1.60 
+ 
80 v 
Combine the three current sources into a single current source, using 
KCL, and combine the 20 n, 5 n, and 6 n resistors in parallel. The 
resulting circuit is shown on the left. To simplify the circuit further, 
transform the resulting 30 A source in parallel with the 2.4 n resistor into 
a 72 V source in series with the 2.4n resistor. Combine the 2.40 resistor 
in series with the 1.6 n resisor to get a very simple circuit that still 
maintains the voltage v. The resulting circuit is on the right. 
30 t 
1.60 40 
+ 
Va 2.40 80 + v 
~i 
80 + 
v 
Use voltage division in the circuit on the right to calculate v as follows: 
8 
v = 12 (72) = 48 v 
[b] Calculate i in the circuit on the right using Ohm's law: 
i=~= 48 =6A 
8 8 
4-12 CHAPTER 4. Techniques of Circuit Analysis 
Now use i to calculate Va in the circuit on the left: 
Va = 6(1.6 + 8) = 57.6 V 
Returning back to the original circuit, note that the voltage Va is also the 
voltage drop across the series combination of the 120 V source and 20 n 
resistor. Use this fact to calculate the current in the 120 V source, ia: 
. = 120 - Va = 120 - 57.6 = 3.12 A 
'ta 20 20 
P12ov = -(120)ia = -(120)(3.12) = -374.40 W 
Thus, the 120 V source delivers 37 4.4 W. 
AP 4.16 To find RTh, replace the 72 V source with a short circuit: 
12Q 
5Q ~Q 
.----'\IV', y 
Note that the 5 n and 20 0 resistors are in parallel, with an equivalent 
resistance of 5JJ20 = 4r2. The equivalent 4r2 resistance is in series with the 8r2 
resistor for an equivalent resistance of 4 + 8 = 12 n. Finally, the 12 fl 
equivalent resistance is in parallel with the 12 n resistor' so 
RTh = 121112 = 60. 
Use node voltage analysis to find VTh· Begin by redrawing the circuit and 
labeling the node voltages: 
12Q 
5Q 8Q 
20Q 
The node voltage equations are 
V1 - 72 V1 V1 - VTh 
5 + 20 + 8 - 0 
VTh - V1 VTh - 72 
8 + 12 0 
Problems 4-13 
Place these equations in standard form: 
V1 ( ! + __!_ + ! ) VTh (-~) 72 + 5 5 20 8 
V1 (-~) + VTh (! + __!_) - 6 8 12 
Solving, v 1 = 60 V and vTh = 64.8 V. Therefore, the Thevenin equivalent 
circuit is a 64.8 V source in series with a 6 0 resistor. 
AP 4.17 We begin by performing a source transformation, turning the parallel 
combination of the 15 A source and 8 0 resistor into a series combination of a 
120 V source and an 8 n resistor, as shown in the figure on the left. Next, 
combine the 2 n, 8 n and 10 n resistors in series to give an equivalent 20 n 
resistance. Then transform the series combination of the 120 V source and the 
20 0 equivalent resistance into a parallel combination of a 6 A source and a 
20 n resistor, as shown in the figure on the right. 
a 
80 20 
120Vi : r20 6A! t 200 120 
I b 
"""' • HlO 
Finally, combine the 20 0 and 12 0 parallel resistors to give 
RN = 201112 = 7.5 n. Thus, the Norton equivalent circuit is the parallel 
combination of a 6 A source and a 7.5 0 resistor. 
AP 4.18 Find the Thevenin equivalent with respect to A, Busing source 
transformations. To begin, convert the series combination of the -36 V source 
and 12 kO resistor into a parallel combination of a -3 mA source and 12 kO 
resistor. The resulting circuit is shown below: 
15kQ 
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 kll60 k= 10 kO n~sistor. 
Then transform the 15 mA source in parallel with the 10 kO resistor into a 
150 V source in series with a 10 kO re,sistor, and combine this 10 kQ resistor 
in series with the 15 kO resistor. The Thevenin equivalent is thus a 150 V 
4-14 CHAPTER 4. Techniques of Circuit Analysis 
source in series with a 25 krl resistor, as seen to the left of the terminals A,B 
in the circuit below. 
25k0 A 
~ :'8t0kQ 
I - I 
•s 
Now attach the voltmeter, modeled as a 100 kr2 resistor, to the Thevenin 
equivalent and use voltage division to calculate the meter reading VAB: 
v . = lOO,OOO (150) = 120 v 
AB 125 000 
' 
AP 4.19 Begin by calculating the open circuit voltage, which is also VTh, from the 
circuit below: 
20 
80 
Summing the currents away from the node labeled VTh We have 
VTh VTh - 24 
-+4+3ix+ =0 
8 2 
Also, using Ohm's law for the 8 n resistor, 
. VTh 
'lx = -
8 
Substituting the second equation into the first and solving for VTh yields 
VTh = 8 V. 
Now calculate RTh. To do this, we use the test source method. Replace the 
voltage source with a short circuit, the current source with an open circuit, 
Problems 4-15 
and apply the test voltage VT, as shown in the circuit below: 
80 l i 
20 
x 
Write a KCL equation at the middle node: 
Use Ohm's law to determine ix as a function of VT: 
Substitute the second equation into the first equation: 
Thus, 
The Thevenin equivalent is an 8 V source in series with a 1 Q resistor. 
AP 4.20 Begin by calculating the open circuit voltage, which is also VTh, using the 
node volta!Ie method in the circnit below: 
160ih. 
v 
600 800 400 th. 
The node voltage equations are 
V V - (vTh + 160iA) 
60 + 20 - 4 o, 
0 
The dependent source constraint equation is 
4~ 16 CHAPTER 4. Techniques of Circuit Analysis 
Substitute the constraint equation into the node voltage equations and put the 
two equations in standard form: 
V ( 610 + 2~) + VTh (- :0) 4 
v (-2~) 0 
Solving, v = 172.5 V and VTh = 30 V. 
Now use the test source method to calculate the test current and thus RTh. 
Replace the current source with a short circuit and apply the test source to 
get the following circuit: 
600 
Write a KCL equation at the rightmost node: 
The dependent source constraint equation is 
Substitute the constraint equation into the KCL equation and simplify the 
right-hand side: 
Therefore, 
Thus, the Thevenin equivalent is a 30 V source in series with a 10 n resistor. 
AP 4.21 First find the Thevenin equivalent circuit. To find VTh, create an open circuit 
between nodes a and b and use the node voltage method with the circuit 
Problems 4-17 
below: v.p 
100V :!: 
l 
The node voltage equations are: 
VTh - (100 + Vq,) VTh - V1 
4 + 4 - 0 
V1 - 100 V1 - 20 V1 - VTh 
4 + 4 + 4 - () 
The dependent source constraint equation is 
Place these three equations in standard form: 
VTh (~ + 1) + V1 (-1) + vq, (-~) - 25 
VTh (-~) + V1 (! + ! + ! ) 4 4 4 + vq, (0) 30 
VTh (0) + V1 (1) + Vq, (-1) 20 
Solving, VTh = 120 V, V1 = 80 V, and vq, = 60 V. 
Now create a short circuit between nodes a and band use the mesh current 
method with the circuit below: 
V¢i 
- + >---'""'~~ 
40 040 
The mesh current equations are 
-100 + 4(i1 - i2) + Vq, + 20 0 
-V<f> + 4i2 + 4(i2 - isc) + 4(i2 - ii) - 0 
-20 - Vq, + 4(isc - i2) 0 
4-18 CHAPTER 4. Techniques of Circuit Analysis 
The dependent source constraint equation is 
Place these four equations in standard form: 
4i1 - 4i2 + Oise + V<f> - 80 
-4i1 + 12i2 - 4isc - V¢ - 0 
Oi1 - 4i2 + 4isc - V¢ - 20 
4i1 + Oi2 - 4isc - V¢ - 0 
Solving, ii = 45 A, i2 = 30 A, isc = 40 A, and V<f> = 20 V. Thus, 
[a] For maximum power transfer' R = RTh = 3 n 
(b] The Thevenin voltage, VTh = 120 V, splits equally between the Thevenin 
resistance and the load resistance, so 
120 
'Vload = - = 60 V 
2 
Therefore, 
'IJ2 602 
Pmax = load = - = 1200 W 
R1oa<l 3 
AP 4.22 Sustituting the value R = 3 0 into the circuit and identifying three mesh 
currents we have the circuit below: 
30 
The mesh current equations are: 
-100 + 4(i1 - i2) + V¢ + 20 - 0 
-v<P + 4i2 + 4(i2 - ia) + 4(i2 - ii) - 0 
-20-v¢+4(ia-i2)+3ia - 0 
The dependent source constraint equation is 
Place these four equations in standard form: 
4i1 - 4i2 + Oi3 + V¢ 80 
-4i1 + 12i2 - 4i3 - V¢ - 0 
Oi1 - 4i2 + 7i3 - V¢ - 20 
4i1 + Oi2 - 4i3 - v<t> - 0 
Solving, i1 = 30 A, i2 = 20 A, i3 = 20 A, and V¢ = 40 V. 
Problems 4-19 
(a] P10ov = -(100)i1 = -(100)(30) = -3000 W. Thus, the 100 V source is 
delivering3000 W. 
[b] P<lepsource = -11¢i2 = -(40)(20) = -800 W. Thus, the dependent source is 
delivering 800 W. 
[c] From Assessment Problem 4.2l(b), the power delivered to the load resistor 
is 1200 W, so the load power is (1200/3800)100 = 31.58% of the 
combined power generated by the 100 V source and the dependent source. 
4-20 CHAPTER 4. Techniques of Circuit Analysis 
Problems 
P 4.1 [a] There are six circuit components, five resistors and the current source. 
Since the current is known only in the current source, it is unknown in 
the five resistors. Therefore there are five unknown currents. 
[b] There are four essential nodes in this circuit, identified by the dark black 
dots in Fig. P4.4. At three of these nodes you can write KCL equations 
that will be independent of one another. A KCL equation at the fourth 
node would be dependent on the first three. Therefore there are three 
independent KCL equations. 
[c] 
Sum the currents at any three of the four essential nodes a, b, c, and d. 
Using nodes a, b, and c we get 
-i9 + ii + iz = 0 
[d] There are three meshes in this circuit: one on the left with the 
components i 9 , R1 , and R4; one on the top right with components Ri, 
R2 , and R3; and one on the bottom right with components R3, R4 , and 
R5. We cannot write a KVL equation for the left mesh because we don't 
know the voltage drop across the current source. Therefore, we can write 
KVL equations for the two meshes on the right, giving a total of two 
independent KVL equations. 
(e] Sum the voltages around two independent closed paths, avoiding a path 
that contains the independent current source since the voltage across the 
current source is not known. Using the upper and lower meshes formed 
by the five resistors gives 
R1i1 + R3i3 - R2i2 = 0 
p 4.2 
p 4.3 
Problems 4-21 
+ 
18V '! 
[a] 11 branches, 7 branches with resistors, 2 branches with independent 
sources, 2 branches with dependent sources 
[b] The current is unknown in every braI,J.ch except the one containing the 5 
mA current source, so the current is unknown in 10 branches. 
[c] 11 essential branches each containing a single element. 
[d] The current is known only in the essential branch containing the current 
source, and is unknown in the remaining 10 essential branches 
[e] From the figure there are 5 nodes - four identified by rectangular boxes 
and one identified by a triangle. 
[ f] There are 5 essential nodes, four identified with rectangular boxes and one 
identified with a triangle 
[g] A mesh is like a window pane, and as can be seen from the figure there are 
7 window panes or meshes. 
[a] As can be seen from the figure, the circuit has 2 separate parts. 
[h] There are 5 nodes - the four black dots and the node betweem the voltage 
source and the resistor R1 . 
[c) There are 7 branches, each containing one of the seven circuit components. 
4-22 CHAPTER 4. Techniques of Circuit Analysis 
[ d) When a conductor joins the lower nodes of the two separate parts, there is 
now only a single part in the circuit. There would now be 4 nodes, 
because the two lower nodes are now joined as a single node. The 
number of branches remains at 7, where each branch contains one of the 
seven individual circuit components. 
P 4.4 {a] From Problem 4.2( d) there are 10 essential branches were the current is 
unknown, so we need 10 simultaneous equations to describe the circuit. 
p 4.5 
[h) From Problem 4.2(f), there are 5 essential nodes, so we can apply KCL at 
( 5 - 1) = 4 of these essential nodes. There would also be two dependent 
source constraint equations. 
[c) The remaining 4 equations needed to describe the circuit will be derived 
from KVL equations. 
(d) We must avoid using the meshes containing current sources, as we have no 
way of determining the voltage drop across a current source. 
1 2 
3 
[a] At node l: 
At node 2: 
At node 3: 
[h] There are many possible solutions. For example, solve the equation at 
node 1 for i 9 : 
ig =ii+ i2 
Substitute this expression for ig into the equation at node 3: 
so 
Multiply this last equation by -1 to get the equation at node 2: 
so 
P 4.6 Use the lower terminal of the 5 0 resistor as the reference node. 
Solving, V 0 = 10 V 
Problems 4-23 
P 4. 7 [a] From the solution to Problem 4.5 we know v0 = 10 V, therefore 
P3A = 3vo = 30 W 
:. P3A (developed) = -30 W 
[b] The current into the negative terminal of the 60 V source is 
60-10 
i9 = 10 = 5 A 
P60V = -60(5) = -300 W 
. ·. P6ov (developed) = 300 W 
[c] P100 = (5)2 (10) = 250 W 
P5o = (10)2 /5 = 20 W 
LPaev = 300 W 
LP<lis = 250 + 20 + 30 = 300 W 
Vo - 60 v 0 
P 4.8 [a] 10 + 5 + 3 = O; V 0 = 10 V 
[b] Let Vx =voltage drop across 3 A source 
Vx = V 0 - (10)(3) = -20 V 
P3A (developed) = (3)(20) = 60 W 
[c] Let i9 =be the current into the positive terminal of the 60 V source 
i9 = (10- 60)/10 = -5 A 
P60V (developed) = (5)(60) = 300 W 
[d] LP<lis = (5)2(10) + (3)2 (10) + (10) 2 /5 = 360 W 
LP<lis = 300 + 60 = 360 w 
[e] v0 is independent of any finite resistance connected in series with the 3 A 
current source 
p 4 9 2 4 + 2 + V1 - V2 = 0 
. . 125 25 
Solving, v1 = 25 V; 
4-24 CHAPTER 4. Techniques of Circuit Analysis 
CHECK: 
(25)2 
p125n = 125 = 5 W 
P25n = (90 - 25)2 = 169 W 
25 
(90)2 
P2500 = -- = 32.4 w 
250 
(90)2 
p375Q = 375 = 21.6 w 
P2.4A = (25)(2.4) = 60 W 
LPabs = 5 + 169 + 32.4 + 21.6 + 60 = 288 W 
LP<lev = (90)(3.2) = 288 W (CHECKS) 
P 4.10 [a] 
80 
+ 
180 
+ 
0 
- 0 
Solving, v1 = 96 V; v2 = 60 V 
. _ 128 - 96 _ 4 A 
'la - -
8 
ib = 96 = 2 A 
48 
. _96-60_ 2 A 
'le - 18 -
100 
128 
ia = 60 = 3 A 
20 
. _ 60 - 70 __ 1 A 
Ze - 10 -
Problems 4-25 
[b] Pdev = 128(4) + 70(1) = 582 W 
P 4.11 [a] v1 
i 60 J, 
i4 i6 
v2 240 
i 120 
is 
10 
v3 
- 0 
- 0 
0 
- 125 
- 0 
-125 
Solving, v1 =101.24 V; v2 = 10.66 V; v3 = -106.57 V 
Thus, i1 = 125 - vi = 23 76 A 1 . 
. V2 
i2 = 2 = 5.33 A 
i3 = V3 + 125 = 18 43 A 1 . 
. V1 - V2 
i4 = = 15 A 
6 
i 5 = v2 - v\3 = 9. 77 A 
12 
i6 = V1 - V3 = 8.66 A 
24 
[h] LPdev = 125i1+125i3 = 5273.09 W 
Lp<lis = iI(l) + i~(2) + i~(l) + i~(6) + i;(12) + i~(24) = 5273.09 W 
4-26 CHAPTER 4. Techniques of Circuit Anal.rsis 
p 4.12 
p 4.13 
40 
Solving, v1 = 100 V; 
+ 
2 V 0 V0 - 55 O +-+ = 
4 5 
V0 = 20 V 
SO 29V1 - V2 ='= 2880 
SO -V1+17V2 = 240 
V2 = 20 V 
20 
P2A = (20)(2) = 40 W (absorbing) 
P 4.14 [a] 
30 + 
2.50 + 
so 3lv1 - 20v2 + Ov3 = 6400 
so -2v1 + 3v2 - V3 = -64 
Solving,v1 = 380 V; v2 = 269 V; v3 = 111 V, 
[b] iu = 640 - 380 = 52 A 
5 
p9 (del) = (640)(52) = 33,280 W 
p 4.15 
1 
150 
31.250 
2 3 
500 
V1 - ( V2 + 30) V1 - V2 V1 _ 4 _ O 
15 + 31.25 + 25 -
[ 
V1 - ( Vz + 30) l V2 - V3 Vz - V1 
- 15 + 50 + 31.25 = 0 
V3 - Vz V3 l - O 
50 + 50 + -
Solving, V1 = 76 V; V2 = 46 V; V3 = -2 V; i3ov = 0 A 
P4A = -4v1 = -4(76) = -304 W ( del) 
P1A = (1)(-2) = -2 W (del) 
P3ov = (30)(0) = 0 W 
P15n = (0)2(15) = 0 W 
v2 762 
P25n = --1. = - = 231.04 W 
25 25 
(Vt - V2) 2 302 
P31.25n = 31. 25 = 31.25 = 28.8 W 
( V2 - V3)7 482 
p500(1ower) = 50 = 50 = 46.08 W 
v2 4 
P5oo(right) = _i = - = 0.08 W 
50 50 
LP<liss = 0 + 231.04 + 28.8 + 46.08 + 0.08 = 306 W 
LP<lev = 304 + 2 = 306 W (CHECKS) 
Problems 4-27 
4-28 CHAPTER 4. Techniques of Circuit Analysis 
~-~ ~-~ ~-~ ~-~ 
P 4.16 (a] -R- + R + R + · · · + R = 0 
1 
[b] V 0 = 3(150 + 200 - 50) = 100 V 
p 4_17 _ 3 Vo Vo+ 5ia V 0 - 80 = O· 
+ 200 + 10 + 20 ' 
. V 0 - 80 
ia = 
20 
[a] Solving, v0 = 50 V 
[b] . = Vo+ 5ia 'lds 10 
ia = (50 - 80)/20 = -1.5 A 
:. ids= 4.25 A; 5ia = -7.5 V: Pds = (-5iA)(ids) = 31.875 W 
[c] P3A = -3v0 = -3(50) = -150 W (del) 
Psov = 80iA = 80( -1.5) = -120 W ( del) 
LP<lel = 150 + 120 = 270 W 
P 4.18 [a] 
CHECK: 
P20on = 2500/200 = 12.5 W 
P20n = (80 - 50)2 /20 = 900/20 = 45 W 
Prnn = ( 4.25)2(10) = 180.625 W 
LP<liss = 31.875 + 180.625 + 12.5 + 45 = 270 W 
2i 1' 
0 
p 4.19 
Problems 4-29 
so -8v1 + 13v2 -4v3 = 0 
so Ov1 - 4v2 + 29v3 = 192.5 
Solving, V1 = -50 V; v2 = -30 V; V3 = 2.5 V 
. V2 - V3 -30 - 2.5 
{h] zo = = = -0.65 A 
50 50 
. V3 - 5i0 2.5 - 5(-0.65) A 
'l3 = = = 1.15 
5 5 
. _ 38.5 - 2.5 _ 1 8 A 
'lg - 20 - . 
Calculate LP<lev because we don't know if the dependent sources are 
developing or absorbing power. Likewise for the independent source. 
P2i0 = -2ioV1 = -2(-0.65)(-50) = -65 W(dev) 
P5i0 = 5ioi3 = 5(-0.65)(1.15) = -3.7375 W(dev) 
Pg= -38.5(1.8) = -69.30 W(dev) 
LP<lev = 69.3 + 65 + 3.7375 = 138.0375 W 
CHECK 
" 2500 900 400 ( )2( ) ( )2 ( )2( ) ~P<liB - lOO + 200 + 25 + 0.65 50 + 1.15 5 + 1.8 20 
138.0375 w 
LP<lev = LP<lis = 138.0375 W 
50 
+ l 
cr 
V0 - 80 V 0 Vo + 75iu _ O· 
5 + 50 + 25 - ' 
Solving, v0 = 50 V; iu = 1 A 
. Vo 
z =-
0" 50 
4-30 CHAPTER 4. Techniques of Circuit Analysis 
io = 50- (-75)(1) = 5 A 
25 
P75i,,. = 75i,,io = -375 W 
The dependent voltage source delivers 375 W to the circuit. 
P 4.20 [a] 
p 4.21 
Solving, v1 = 30 V; v2 = 15 V; i& = 3 A; 
P5iA = (-15)(1) = -15 W(del) 
PM= -5(30) = -150 W(del) 
.·. Pdev = 165 W 
. _ 15+15 - l A 
'to - -
30 
(b] LPabs = (~°i2 + (~~2 + (~~2 + (3)2 (5) + (1)2 (30) = 165 W 
LPdev = LPabs = 165 W 
8000 
- v + 
0 
v 
50Vi ~ t 750mA 2000 
The two node voltage equations are: 
V1 - 50 V1 V1 - V2 
80 + 50 + 40 - 0 
V2 - V1 V2 V2 - 50 
40 - 0·75 + 200 + 800 - 0 
Place these equations in standard form: 
V1 ( 810 + 5~ + :0) + V2 (-:0) 
V1 (- 4~) 
50 
80 
50 
- 0.75 + 800 
p 4.22 
Solving, V1 = 34 V; v2 = 53.2 V. 
Thus, v0 = V2 - 50 = 53.2 - 50 = 3.2 V. 
POWER CHECK: 
i 9 - (50 - 34)/80 + (50 - 53.2)/800 = 196 m A 
P50V - -(50)(0.196) = -9.8 W 
Pson - (50 - 34)2 /80 = 3.2 W 
Psoon - (50 - 53.2)2 /800 = 12.8 m W 
P40n (53.2 - 34)2 /40 = 9.216 W 
P5on 342 /50 = 23.12 W 
P20on - 53.22 /200 = 14.1512 W 
Po.75A - -(53.2)(0.75) = -39.9 W 
Problems 4-31 
LPabs = 3.2 + .0128 + 9.216 + 23.12 + 14.1512 = 49.7 W = LPdel = 
9.8 + 39.9 = 49.7 
80V + 
so 22v1 - 7v2 = 1120 
so -6v1 + 23v2 = 960 
Solving, v1 = 70 V; V2 = 60 V 
. V1 - V2 
Thus, i 0 = = 1 A 
10 
4--32 CHAPTER 4. Techniques of Circuit Analysis 
P 4.23 [a] 
{b] 
50 
500\11 ~ 
so -22v2 + Ov3 + 37v4 - llv5 = 2000 
Solving, V2 = 300 V; V3 = 180 V; v4 = 280 V; v5 = 160 V 
. - 500 - V4 - 500 - 280 - 20 A isn - 11 - 11 -
Pm= (20)2(5) = 2000 W 
V1 - V2 V1 - V4 
i5oov - 4 + 11 
_ 500 - 300 + 500 - 280 = 50 + 20 = 70 A 
4 11 
Psoov 35,000 W 
P 4.24 [a] 
Problems 
Check: 
70A 20A 
i 
50 
i 60 
50A 40 
-71DA 
20A 
i 
20 J, 50 OVi ~ 30 40 
40A 
-71DA 
30A 
20 
i i 60 30 
30A 40Af-
40A 
H1 
2:Pdis - (50)2( 4) + ( 40)2(3) + (30)2(6) + (20)2 (11) + (10)2(2) 
+(30) 2( 4) + (10)2(2) + ( 40)2( 4) = 35,000 w 
i 
- V lmA+ 
,..---'-.1.-WV---+---4:~·1~--~~-
25k 0 llmA 
V1 - 20 V1 -3 V1 + 10 
25 x 103 + 0.25 x 103 + 11 x 10 + 0.5 x 1Q3 = 0 
V1 = -5 V 
. 20+5 
i1 = 25,000 = 1 mA 
V1 -5 
i2 = - = - = -20 mA 
250 250 
i 5 = - lO + 5 = -10 mA 
500 
-10 
i4 = -- = -10 mA 
1000 
i4 + i3 - 11 + i5 = 0 
: . i3 = 11 - i 4 - i5 = 11 + 10 + 10 = 31 mA 
4-33 
4-34 CHAPTER 4. Techniques of Circuit Analysis 
p 4.25 
[b] P2ov = 20i1 = 20(1x10-3 ) = 20 mW 
Prnv = l0i3 = 10(31 x 10-3) = 310 mW 
V11mA + V1 = -10, V11mA = -10 + 5 = -5 V 
PllmA = -llV11mA = -55 mW (del) 
LP<lev = 20 + 310 = 330 mW 
P25k = 25 x 103ii = 25 mW 
3 2 Po.25k = 0.25 x 10 i2 = 100 mW 
Po.5k = 0.5 x 103i~ = 50 mW 
P1k = 1x103i~ = 100 mW 
LPdiss = 25+100+50+100 + 55 = 330 mW 
LP<liss = LP<lev = 330 m w 
40 
c 
24 ~ 
The two node voltage equations are: 
- lQ + Vb + Vb - Ve O 
6 2 
2v A Ve - Vb Ve - 24 
3+ 2 + 4 - 0 
The constraint equation for the dependent source is: 
VA= Vb 
Place these equations in standard form: 
vb(~+~) 
1 
VA(O) + Ve(-2) + - 10 
1 
v (! + ! ) VA(~) 
24 
vb(--) + + -
2 e 2 4 4 
vb(l) + Ve(O) + VL}.(-1) 0 
Solving, Vb= 18 V,ve = 4V,vA=18 V, and v0 = 24-vc = 20 V 
p 4.26 
Problems 4-35 
20k0 4k0 
This circuit has a supernode includes the nodes v1, v2 and the 40 V source. 
The supernode equation is 
V1 V2 V2 
-0.05 + 8000 + 20 000 + 10 000 = 0 
' ' 
The supernode constraint equation is 
Place these two equations in standard form: 
vi (s:oo) + v2 (20,~oo + 10,~) - 0·05 
v1(-l) + v2(l) - 40 
Solving, V1=160 V and V2 = 200 V, so v0 = V2 = 200 V. 
i4o = 0.05 - 8~~0 = 30 m A 
P40V = -(40)i40 = -(40)(0.03) = -1.2 W 
The 40 V source delivers 1.2 W. 
P 4.27 Place va/5 inside a supernode and use the lower node as a reference. Then 
~-W ~ ~-~~ ~-~~-0 10 + 30 + 39 + 78 -
134v1 - 6v a = 3900; VtJ. = 50-V1 
Solving, V1 = 30 V; va = 20 V; v0 = 30 - va/5 = 30 - 4 = 26 V 
p 4.28 i</> = V3 ~ V4 = 235 ~ 222 = 3.25 A 
30i</> = 30(3.25) = 97.5 v 
4-36 CHAPTER 4. Techniques of Circuit Analysis 
V1 = V4 - 30i¢ = 222 - 97.5 = 124.5 V 
V3 +VD.= 250 
.". V L). = 250 - 235 = 15 V 
3.2VL). = (3.2)(15) = 48 A 
. _ 250 - 124.5 250 - 235 _ 77 75 A 
'lg - 2 + 1 - . 
P25ov = -250ig = -250(77.75) = -19,437.5 W(del) 
i3oi"' = iq, - 222/40 - 48 = 3.25 - 5.55 - 48 = -50.3 A 
P30i"' = (30iq,)i3oi"' = (97.5)(-50.3) = -4904.25 W(dev) 
P3.2v"' = (3.2vD.)(v4) = (48)(222) = 10,656 W(abs) 
.". LP<lev = 19,437.5 + 4904.25 = 24,341.75 W 
VI (124.5)2 
P10n = 10 = 10 = 1550.025 W 
P2n = (250 - 2124.5)2 = 7875.125 W 
Pm = (250 - 235)2 = 225 W 
1 
(235)2 
P20n = 20 = 2761.25 W 
P4n = (3.25)2(4) = 42.25 W 
(222)2 
P40n = 40 = 1232.10 W 
· · · LP<liss = 10,656 + 1550.025 + 7875.125 + 225+ 
2761.250 + 42.25 + 1232.l = 24,341.75 w 
Thus, :~.::>dev = LPdissi Agree with analyst 
Problems 4~37 
p 4.29 -- - - - -- -- - - • - - · -- ·;,: •
5
- ~ - --- - -- - -- - -- - - - - - -; supernode 
~~~~--<+- l I 
~--------------- -----------------, 
I I 
I I 
I I 
I I 
: vz : 
I 
400 
Node equations: 
Constraint equations: 
Solving, V1 = -20.25 V; v2 = 10 V; V3 = -29 V 
I 
I 
I 
I 
I 
I 
I 
I 
I 
I 
Let i 9 be the current delivered by the 20 V source, then 
. _ 20 - (20.25) 20 - 10 _ 30 125 A 
'tg - 2 + 1 - . 
p9 (delivered) = 20(30.125) = 602.5 W 
P 4.30 From Eq. 4.16, is= vc/(1 + {3)RE 
From Eq. 4.17, is= (vb - Vo)/(1 + {3)RE 
4-38 CHAPTER 4. Techniques of Circuit Analysis 
P 4.31 [a] 
2Q 1.5Q 
40 = 50i1 - 45i2 
64 = -45i1 + 50.5i2 
Solving, ii = 9.8 A; i2 = 10 A 
ia =ii= 9.8 A; ib =ii - i2 = -0.2 A; ic = -i2 = -10 A 
(b] If the polarity of the 64 V source is reversed, we have 
40 = 50i1 - 45i2 
P 4.32 [a] 
-64 = -45i1 + 50.5i2 
ii= -1.72 A and i2 = -2.8 A 
110 + 12 = 17i1 - l0i2 - 3i3 
0 = - l0i1 + 28i2 - 12i3 
-12 - 70 = -3i1 - 12i2 + 17i3 
Solving, ii= 8 A; i2 = 2 A; i3 = -2 A 
Pno = -l10i1 = -880 W(del) 
P12 = -12(i1 - i3) = -120 W(del) 
P10 = 70i3 = -140 W(del) 
.". LPdev = 1140 W 
Problems 4-39 
[b) P4n = (8)2 (4) = 256 W 
P 4.33 [a] 
P10n = (6)2(10) = 360 W 
P12n = (-4)2(12) = 192 W 
P2n = (-2)2 (2) = 8 W 
P60 = (2)2(6) = 24 W 
p3n = (10)2(3) = 300 W 
:. LPabs = 1140 W 
jso 
12sv6 -:J 
18 0 100 
la 480 J 200 J 
The three mesh current equations are: 
-128 + 8ia + 48(ia - ic) - 0 
70 + 20(ie - ic) + lOie - 0 
Place these equations in standard form: 
ia(8 + 48) + ic( -48) + ie(O) - 128 
ia( -48) + ic(18 + 20 + 48) + ie(-20) - 0 
ia(O) + ic(-20) + ie(20 + 10) - -70 
Solving, ia = 4 A; ic = 2 A; ie = -1 A 
Now calculate the remaining branch currents: 
ib - ia - ic = 2 A 
[b] Psources = P128V + P70V = -(128)ia + (70)ie 
= -(128)(4) + (70)(-1) = -512- 70 = -582 w 
Thus, the power developed in the circuit is 582 W. Note that the resistors 
cannot develop power! 
4-40 CHAPTER 4. Techniques of Circuit Analysit> 
P 4.34 [a] 
-:-J, J, 60 J, 
ll 
i4 i6 
--:J, 
i6 240 
~ 
13 J, 120 
is 
The three mesh current equations are: 
-125 + li1+6(i1 - i6) + 2(i1 - i3) 0 
Place these equations in standard form: 
- 125 ii(l + 6 + 2) + i3(-2) + i6(-6) 
ii(-6)+i3(-12)+i6(24+12+6) - 0 
Solving, i1 = 23. 76 A; i3 = 18.43 A; i6 = 8.66 A 
Now calculate the remaining branch currents: 
i2 - ii - i3 = 5.33 A 
i4 - ii - i6 = 15.10 A 
(b] Psources = Ptop + Phottom = -(125)(23.76)- (125)(18.43) 
= -2970- 2304 = -5274 w 
Thus, the power developed in the circuit is 5274 W. 
Now calculate the power absorbed by the resistors: 
Pitop = (23.76)2(1) = 564.54 W 
P2 = (5.33) 2 (2) = 56.82 W 
Plbot = (18.43) 2(1) = 339.66 W 
P6 = (15.10) 2 (6) = 1368.06 W 
p 4.35 
P12 = (9.77)2(12) = 1145.43 W 
P24 = (8.66) 2 (24) ±= 1799.89 W 
The power absorbed by the resistors is 
Problems 4--41 
564.54 + 56.82 + 339.66 + 1368.06 + 1145.43 + 1799.89 = 527 4 W so the 
power balances. 
80V + 
The three mesh current equations are: 
Place these equations in standard forpi: 
- 80 
- 0 
Solving, i 1 = 6 A; i 2 = 4 A; 
Thus, i0 = ig - i2 = 1 A. 
i3 = 5 A 
4---42 CHAPTER 4. Techniques of Circuit Analysis 
P 4.36 [a] 
50 
40 ~ 60 
:i 
20 
500 • 30 iJ 40 
20 
60 iJ 30 
10 
The four mesh current equations are: 
-500 + 4(ii - i2) + 3(ii - i3) + 6(ii - i4) - 0 
5i2 + 6i2 + 2(i2 - i3) + 4(i2 - ii) - 0 
4i3 + 2(i3 - i4) + 3(i3 -ii)+ 2(i3 -i2) - 0 
3i4 + li4 + 6(i4 - ii)+ 2(i4 - i3) - 0 
Place these equations in standard form: 
ii(4 + 3 + 6) + i2(-4) + i3(-3) + i4(-6) - 500 
ii(-4) + i2(5 + 6 + 2 + 4) + i3(-2) + i4(0) - 0 
ii (-3) + i2(-2) + i3(2 + 4 + 2 + 3) + i4(-2) - 0 
ii(-6) + i2(0) + i3(-2) + i4(2 + 3 + 1+6) - 0 
Solving, i1 = 70 A; i2 = 20 A; i3 = 30 A; i4 = 40 A 
The power absorbed by the 5 0 resistor is 
p5 = i~(5) = (20)2(5) = 2000 w 
[h] Psoo = -(500)i1 = -(500)(70) = -35 kW 
Problems 4-43 
p 4.37 
40 
80 200 
Solving, ii = 3.5 A 
Psn = (3.5)2(8) = 98 W 
p 4.38 
50 
~ 150 
i 
100 1 ~ 
i3 250 
~ i\ll 
500 
i2 
660 = 30ii - 10i2 - 15i3 
Solving, ii = 42 A; i2 = 27 A; i3 = 22 A; 
20i¢ = 100 v 
4-44 CHAPTER 4. Techniques of Circuit Analysis 
p 4.39 
P2oiq, = -lOOi2 = -100(27) = -2700 W 
. •. P20iq, (developed) = 2700 W 
CHECK: 
P660V = -660(42) = -27,720 W (dev) 
LP<lev - 27, 720 + 2700 = 30,420 W 
Lp<lis - ( 42)2(5) + (22)2(25) + (20)2(15) + (5)2(50)+ 
(15)2(10) 
- 30,420 w 
2. 65v ~ 
~~~~~+->--~~~~ 
250 
+ 
~ 125V 
350 850 
Mesh equations: 
2.65vA + 40i1 - 15i2 - 25i3 = 0 
-15i1 + 150i2 - 100i3 = -125 
-25i1 - 100i2 + 210i3 = 125 
Constraint equations: 
Solving, i 1 = 7 A; i2 = 1.2 A; i3 = 2 A 
VA= 100(i2 - i3) = 100(1.2 - 2) = -80 V 
P2.65vb. = 2.65v6.i1 = -1484 W 
Therefore, the dependent source is developing 1484 W. 
CHECK: 
P12sv = 125i2 = 150 W (left source) 
P12sv = -125i3 = -250 W (right source) 
2.:Pdev = 1484 + 250 = 1734 W 
p3sn = (1.2)2(35) = 50.4 W 
Pssn = (2)2(85) = 340 W 
P1sn = (7 - 1.2)2(15) = 504.6 W 
P2sn = (7 - 2)2(25) = 625 W 
P10on = (1.2 - 2)2(100) = 64 W 
LPdiss = 50.4 + 340 + 504.6 + 625 + 64 + 150 = 1734 W 
P 4.40 [a] 
10 = l8i1 - l6i2 
0 = -16i1+28i2+4iA 
4 = 8iA 
Solving, i 1 = 1 A; iz = 0.5 A; iA = 0.5 A 
Vo= l6(i1 - i2) = 16(0.5) = 8 V 
[b] P4it>. = 4iAi2 = ( 4)(0.5)(0.5) = 1 W (abs) 
.•. P4it>. (deliver) = -1 W 
Problems 4-45 
4-46 CHAPTER 4. Techniques of Circuit Analysis 
p 4.41 aai 
400~ 64oov 
12 
14Q 
12A 
600 = 64i1 - 40i2 - 14i3 
-400 = -40i1 + 50i2 - 2i3 
-12 = i3 
Solving, i1 = 2.9 A; i2 = -6.16 A; i3 = -12 A 
[a) V12A = 2(12 - 6.16) + 14(12 + 2.9) 
= 220.28 v 
P12A = -12v12A = -12(220.28) = -2643.36 W 
Therefore, the 12 A source delivers 2643.36 W. 
[b) P400V = 400(-6.16) = -2464 W 
P6oov = -600i1 = -600(2.9) = -1740 W 
Therefore, the total power delivered is 2643.36+2464+1740 = 6847.36 W 
[c] 2°:Presistors = (2.9)2(10) + (6.16)2(8) + (9.06)2(40) + (14.9)2(14) + (5.84)2(2) 
LPabs = 6847.36 W = LP<lel (CHECKS) 
P 4.42 [a] 
~Q 
8mAcb~0 ~ 
8mA . 
l,e. 
i 
1800Q 
330~ 
il 
4700Q 
The mesh current equation for the right mesh is: 
3300(i1 - 0.008) + 6500i1 + 200(i1 - 0.008) = 0 
Solving, 
Then, 
10,000i1 = 28 . ·. ii = 2.8 mA 
i.6. = i 1 - 0.008 = -5.2 mA 
p 4.43 
[b] V 0 = (Q.008)(980) - (-Q.0052)(3300) = 25 V 
PBmA = -(25)(0.008) = -200 mW 
Thus, the 8 mA source delivers 200 mW 
[c] 200iA = 200( -0.0052) = -1.04 V 
Pdep source= 200iAi1 = (-1.04)(0.0028) = -2.912 mW 
The dependent source delivers 2.912 mW. 
2Q 1Q 
+ vtJ. -
3Q 0.5vtJ. + 
~ ~ . vcs 
: 165V l.3 
Mesh equations: 
7i1 + l(i1 - i3) + 2(i1 - i2) - 0 
,....go+ 2(i2 - ii)+ 3(i2 - i3) + 165 - o 
Constraint equations: 
i3 = 0.5vA; VA= 2(i2 - ii) 
Place these equations in standard form: 
ii(7 + 1+2) + i2(-2) + i3(-1) + VA(O) 
ii(-2) + i2(2 + 3) + i3(-3) + VA(O) 
ii(O) + i2(0) + i3(1) + VA(-0.5) 
ii(-2) +i2(2) +i3(0) +vA(-1) 
-
-
-
-
0 
-75 
0 
0 
Problems 4-4 7 
Solving, i1 = -9 A; i2 = -33 A; i3 = -24 A; VA= -48 V 
Solve the outer loop KVL equation to find vcs: 
-90 + 7i1 + Vcs = O; Vcs = 90 - 7(-9) = 153 V 
Calculate the power for the sources: 
Poov - -(90)(-33) = 2970 W 
P165V = (165) (-33 + 24) = -1485 W 
P<lep source = (153)(0.5(-48)) = -3672 W 
Thus, the total power developed is 1485 + 3672 = 5157 W. 
4-48 CHAPTER 4. Techniques of Circuit Analysis 
CHECK: 
Pm - (9)2 (7) = 567 w 
P2n - (24)2(2) = 1152 w 
P3n - (9)2(3) = 243 w 
Pm - (15)2 (1) = 225 w 
LPabs = 567+1152 + 243 + 225 + 2970 = 5157 W (checks!) 
p 4.44 
+ 
250 
Mesh equations: 
-20i1 + 120i2 - 30i~ - 100i9 = O 
Constraint equations: 
i - 4· g - ' 
Solving, i 1 = 4 A; 
i10on = 4 - i2 = -1 A 
200 
100~ 
J..2 
P4A = -v4Aig = -(-100)(4) = 400 W (abs) 
30il'l 
+ 
Problems 4-49 
V3oi4 = 30iD,. = 30ii = 120 V 
P3Di;:;. = -30iD.i2 = -120(5) = -600 W 
Therefore, the dependent source is developing 600 W, all other elements are 
absorbing power, and the total power developed is thus 600 W. 
CHECK: 
Psn = 16(5) = 80 W 
P25n=OW 
P20n = 1(20) = 20 W 
P10on = 1(100) = 100 W 
P4A = 400 W 
l:Pabs = 80 + 0 + 20 + 100 + 400 = 600 W (CHECKS) 
P 4.45 [a] + v -
0 
~----<~---
~-2vl!. 
+ 
30 
Mesh equations: 
15 = 30ii - 25i2 - 2i3 
-10 = -25ii + 30i2 - i3 
Constraint equations: 
Solving, ii = 10 A; 
i3 
i2n =ii - i3 = 9 - 30 = -20 A 
P2n = (-20)2(2) = 800 W 
10 
: lOV 
40 
i3 = 30 A; Va= 25 V 
4-50 CHAPTER 4. Techniques of Circuit Analysis 
[b) p15y = -15(10) = -150 W(dev) 
P10v = l0i2 = 10(9) = 90 W (abs) 
P 4.46 [a] 
V 0 = (i1 -i3)2+ (i2 -i3)l = -40- 21 = -61 V 
Pl.2v6. = i3Vo = (30)(-61) = -1830 W (dev) 
2.:::Pdev = 1830+150 = 1980 W 
% delivered to 2n = 18i8°0 x 100 = 40.4% 
20 50 
50V: 200 
Mesh equations: 
-50 + 6i1 - 4i2 + 9ia = 0 
-9ia - 4i1 + 29i2 - 20i3 = 0 
Constraint equations: 
'lA = i2; i3 = -l.7va; VA= 2i1 
Solving, ii= -5 A; i2 = 16 A; i3 = 17 A; va = -10 V 
9ia = 9(16) = 144 V 
ia = i2 - ii = 21 A 
ib = i2 - i3 = -1 A 
Vb = 20ib = -20 V 
P5oV = -50i1 = 250 W (absorbing) 
P9i6. = -ia(9ia) = -(21)(144) = -3024 W (delivering) 
Pl.7V = -l.7vavb = i3vb = (17)(-20) = -340 W (delivering) 
[b) LPdev = 3024 + 340 = 3364 W 
Lpdis = 250 + (-5)2(2) + (21)2(4) + (16)2(5) + (-1)2(20) 
= 3364 w 
p 4.47 
p 4.48 
20 60 
Solving, i 1 = -0.6 A; i 2 = 2.4 A 
P1sv = - l8i1 = 10.8 W ( diss) 
P3n = (-0.6)2 (3) = 1.08 W 
P2n = (-0.6)2 (2) = 0.72 W 
pgn = (2.4) 2(9) = 51.84 W 
P6n = (2.4)2(6) = 34.56 W 
LP<liss = 99 W 
V 0 = 15i2 - 15 = 36 - 15 = 21 V 
P3A = -3vo = -63 W (dev) 
p15v = -15i2 = -36 W (dev) 
LP<lev = 99 W = LP<liss 
-100 + 5i1 + 15i2 - 15 = 0 
Problems 4--51 
4~52 CHAPTER 4. Techniques of Circuit Analysis 
:. 20i1 = 70 
ii= 3.5 A; i2 = 6.5 A 
V0 = 15i2 - 15 = 97.5 - 15 = 82.5 V 
P10ov = -lOOii = -350 W(dev) 
P3A = -3vo = -247.5 W(dev) 
Pl5V = -15i2 = -97.5 W(dev) 
2.:Pdev = LPdis = 695 W 
Check: LPdis = (3.5)2(5) + (6.5)2(15) = 695 W 
P 4.49 [aJ Summing around the supermesh used in the solution to Problem 3.27 gives 
-(-10) + 5ii + 15i2 - 15 = 0 
:. ii= -2 A; 
P10v = 10(-2) = -20 W (del) 
V 0 = 15i2 - 15 = 0 V 
P3A = 3vo = 0 W 
Pl5V = -15i2 = -15 W (del) 
LPdiss = (-2)2 (5) + (1)2 (15) = 35 W 
LPdev = 35 W = LPdiss 
[b] With 3 A current source replaced with a short circuit 
ii= -2 A, 
:. Lpdiss = (-2)2(5) + (1)2(15) = 35 W 
[cJ A 3 A source with zero terminal voltage is equivalent to a short circuit 
carrying 3 A. 
P 4.50 [a] 
200 = 85i1 - 25i2 - 50i3 
0 = - 75i1 + 35i2 + 150i3i3 - i2 = 4.3(i1 - i2) 
Solving, i 1 = 4.6 A; 
100 
(supermesh) 
i2 = 5.7 A; i3 = 0.97 A 
ic = i3 = 0.97 A; id= 't1 - i2 = -1.1 A 
Problems 4-53 
[b) lOi2 + V0 + 25(i2 - i1) = 0 
• . . V 0 = -57 - 27.5 = -84.5 V 
P4.3id = -vo(4.3id) = -(-84.5)(4.3)(-1.1) = -399.685 W(dev) 
P20ov = -200(4.6) = -920 W(dev) 
LPdev - 1319.685 W 
2.:Pdis - (5.7)210 + (1.1)2(25) + (0.97)2100 + (4.6)2(10)+ 
(3.63)2(50) 
- 1319.685 w 
LPdev = Lpdis = 1319.685 W 
4-54 CHAPTER 4. Techniques of Circuit Analysis 
P 4.51 [a] 
100 350 
40(i3 - ii)+ 10(i3 - i2) + 35(i4 - i2) + 150 = 0 
35(i2 - i4) + 10(i2 - i3) + 15id = 0 
ii= 30 A 
Solving, ii = 30 A; i 2 = 8 A; i 3 = 24 A; i 4 = 6 A 
ia = 30 - 24 = 6 A; ib = 8 - 24 = -16 A; ic = 8 - 6 = 2 A; 
(b] Va= 40ia = 240 V; Vb = 150 - 35ic = 80 V 
P30A = -30va = -30(240) = -7200 W (gen) 
PI5id = 15idie = 15(6)(8) = 720 W (diss) 
P3ia = 3iavb = 3(6)(80) = 1440 W (diss) 
P1sov = 150id = 150(6) = 900 W (diss) 
P4on = (6)2(40) = 1440 W (diss) 
Pion= (-16)2(10) = 2560 W (diss) 
p 3m = (2)2(35) = 140 W (diss) 
LPgen = 7200 W 
Lpdiss = 720 + 1440 + 900 + 1440 + 2560 + 140 = 7200 W 
Problems 4-55 
P 4.52 [a] Both the mesh-current method and the node-voltage method require three 
equations. The mesh-current method is a bit more intuitive due to the 
presence of the voltage sources. We choose the mesh-current method, 
although technically it is a toss-up. 
[b] 
o .H1 + 
110\11 ~ ~180 v1 
11 . + 
la -~v 54. 6250 
i c + i3 -
110\11 ~ ~10.50 v2 
12 
ib 
0.10 
110 - 18.3i1 - 0.2i2 - l8i3 
110 - -0.2i1 + ll0.8i2 - 110.5i3 
0 - -l8i1 - 110.5i2 + 183.125i3 
Solving, i1 = 10 A; i2 = 5 A; i3 =4 A 
V1 - l8(i1 - i3) = 108 v 
V2 - 110.5( i2 - i3) = 110.5 v 
V3 - 54.625i3 - 218.5 v 
[c] PRl - (ii - i3)2(18) = 648 w 
PR2 - (i2 - i3)2(110.5) = 110.5 w 
PR3 - i~(54.625) = 874 W 
(d] LP<lev = llO(i1 + i2) = 1650 W 
LPioad = 1632.5 W 
o-t d 1. d = 1632•5 x 100 = 98.9407 10 e ivere 1650 10 
4-56 CHAPTER 4. Techniques of Circuit Analysis 
p 4.53 
[e] 
0.10 + 
180 
~ 54. 6250 
+ i2 
v 
110 .50 
0.10 
220 = 128.7i1 - 128.5i2 
0 = -128.5ii + 183.125i2 
Solving, ii = 5. 71 A; 
ii - i2 - 1.7 A 
i2 - 4.01 A 
Vi = (1.7)(18) = 30.6 V 
V2 = (1.7)(110.5) = 187.85 V 
Note v1 is low and v2 is high. Therefore, loads designed for 110 V would 
not function properly, and could be damaged. 
~ 54.6250 
i c 
0 .10 
110 = (R + 0.3)ia - 0.2ib - Ric 
110 = -0.2ia + (R + 0.3)ib - Ric 
(R + 0.3)ia - 0.2ib - Ric= -0.2ia + (R + 0.3)ib - Ric 
(R + 0.3)ia - 0.2ib = -Q.2ia + (R + 0.3)ib 
(R + 0.5)ia = (R + 0.5)ib 
Problems 4-57 
P 4.54 [a] There are three unknown node voltages and only two unknown mesh 
currents. Use the mesh current method to minimize the number of 
simultaneous equations. 
(b] 
The mesh current equations: 
2500( ii - 0.01) + 2000ii + 1000( ii - i2) - 0 
5000(i2 - 0.01) + 1000(i2 - ii)+ 1000i2 - 0 
Place the equations in standard form: 
ii (2500 + 2000 + 1000) + i2(-1000) - 25 
ii (-1000) + i2(5000 + 1000 + 1000) - 50 
Solving, i 1 = 6 mA; i 2 = 8 mA 
Find the power in the 1 kO resistor: 
iik = it - i2 = -2 m A 
P1k = (-0.002)2(1000) = 4 mW 
(c] No, the voltage across the 10 A current source is readily available from the 
mesh currents, and solving two simultaneous mesh-current equations is 
less work than solving three node voltage equations. 
[d] Vg = 2000ii + lOOOi2 = 12 + 8 = 20 v 
P1omA = -(20)(0.01) = -200 mW 
Thus the 10 mA source develops 200 mW. 
P 4.55 [a] There are three unknown node voltages and three unknown mesh currents, 
so the number of simultaneous equations required is the same for both 
methods. The node voltage method has the advantage of having to solve 
the three simultaneous equations for one unknown voltage provided the 
connection at either the top or bottom of the circuit is used as the 
reference node. Therefore recommend the node voltage method. 
[b] 
2 
lOmA t 3 
4-58 CHAPTER 4. Techniques of Circuit Analysis 
The node voltage equations are: 
V1 V1 - V2 V1 - V3 
5000 + 2500 + 1000 
0 
V2 Vz - V1 Vz - V3 
-0.01 + 4000 + 2500 + 2000 0 
V3 - V1 V3 - Vz V3 
1000 + 2000 + 1000 
0 
Put the equations in standard form: 
Vi (50
1
00 + 25~0 + 10~0) + Vz (-2;00) + V3 (-10~0) - O 
Vi (-25
1
00) + Vz (40~0 + 25100 + 2;00) + V3 (-20100) 0.01 
VI (-10~0) + Vz (-2;00) + V 3 (20~0 + 10100 + 1;00) O 
Solving, v1 = 6.67 V; v2 = 13.33 V; v3 = 5.33 V 
Prnm = -(13.33)(0.01) = -133.33 mW 
Therefore, the 10 mA source is developing 133.33 mW 
P 4.56 [a) The node voltage method requires summing the currents at two 
supernodes in terms of four node voltages and using two constraint 
equations to reduce the system of equations to two unknowns. If the 
connection at the bottom of the circuit is used as the reference node, 
then the voltages controlling the dependent sources are node voltages. 
This makes it easy to formulate the constraint equations. The current in 
the 10 V source is obtained by summing the currents at either terminal of 
the source. 
The mesh current method requires summing the voltages around the two 
meshes not containing current sources in terms of four mesh currents. In 
addition the voltages controlling the dependent sources must be 
expressed in terms of the mesh currents. Thus the constraint equations 
are more complicated, and the reduction to two equations and two 
unknowns involves more algebraic manipulation. The current in the 10 V 
source is found by subtracting two mesh currents. 
Because the constraint equations are easier to formulate in the node 
voltage method, it is the preferred approach. 
[b] 
lOV 2 
1 
io-E.- + 
250 vx 
Node voltage equat,ions: 
V1 Vx V2 lO _ O ---+-+ -
25 2 5 
V3 V4 V 
-10+-+-+~=0 
4 2 2 
Constraints: 
Solving, 
3. 
2 <E-
lOA 3 
2ix 
-7· -+ 4 
50 20 
V1 = 50 V; V2 = 40 V; V3 = -20 V; V4 = -10 V; ix= 5 A. 
. V1 Vx 
i = --- =-18A 
0 25 2 
P10v = - lOio = 180 W 
Thus, the 10 V source absorbs 180 W. 
P 4.57 Choose the reference node so that a node voltage is identical to the voltage 
across the 15 A source; thus: 
15A 
·+-
1 PQ 
420¥ 20Q idc 
2 400 3 
Since the 15 A source is developing 3750 W, v1 must be 250 V. 
Since v1 is known, we can sum the currents away from node 1 to find v 2; thus: 
250 - (420 + V2) 250 - V2 250 _ 15 = O 
7.2 + 20 + 15 
4-60 CHAPTER 4. Techniques of Circuit Analysis 
V2 = -50 V 
Now that we know v2 we sum the currents away from node 2 to find 113 ; thus: 
112 + 420 - 250 V2 - 250 V2 - 113 _ O 
7.2 + 20 + 40 -
.°. 113 = 50/3 V 
Now that we know v3 we sum the currents away from node 3 to find idc; thus: 
V3 V3 + 50 . 
50 + 40 = idc 
P 4.58 [a] 
150 
If i 0 = 0 then V1 = 113; therefore, 
V1 - V2 V1 - 230 
5 + 20 = 0 
'02 - V1 V2 - V3 112 - 115 _ O 
5 + 15 + 10 -
Solving, 111 = 170 V = v3; v2 = 155 V 
.". 170 - 155 + 170 - Vdc = O 
15 25 
Solving, Vdc = 195 V 
[b] ia = 230 - 170 = 3 A 
20 
ib = 115 - 155 = _ 4 A 
10 
. _ 195 - 170 _ 1 A 
ic - 25 -
id = 170 - 155 = 3 A 
5 
. ~ 155 - 170 __ 1 A 
'le - 15 -
P230V = -230ia = -690 W ( dev) 
Pll5V = -115ib = 460 W (abs) 
Pvdc = -Vdcic = -195 W ( dev) 
P2on = i~(20) = 180 W 
p5rz = i~(5) = 45 W 
P10n = i~(lO) = 160 W 
p15rz = i;(15) = 15 W 
P25rz = i~(25) = 25 W 
LPdiss = 460+180 + 45+160+15 + 25 = 885 W 
LPdev = 690+195 = 885 W (CHECJ(S) 
Problems 4-61 
P 4.59 [a] Apply source transformations to both current sources to get 
2 .3k0 2. 7k0 1k0 
13.8Vl._ _____ i_o~-------'r·2V 
13.8+4.2 3 A 
io = 2700 + 2300 + 1000 = ID 
[b] 
2.7k0 
The node voltage equations: 
3 V1 V1 - V2 
6 x rn- + 2300 + 2700 - 0 
V2 Vz - V1 -3 
1000 + 2700 - 4·2 x 10 - 0 
4-62 CHAPTER 4. Techniques of Circuit Analysis 
P 4.60 [a] 
Place these equations in standard form: 
Vi (27~0 + 2;00) + V 2 (-2;00) -
Vi ( - 2;00) + V 2 (10100 + 2;00) -
Solving, v1 = -6.9 V; V2 = 1.2 V 
V2 -V1 
'lo= 
2700 
lOOV ; 
=3mA 
20k0 3k0 
80k0 
lkO 
4k0 
16k0 
-6 x 10-3 
4.2 x 10-3 
5k0 
60k0 
5k0 
60k0 
20k0 5k0 
5k0 
15k0 l 
io 
r--2okO12ov6 l 
I i: 
lOkO 
'lo= 120 = 4 mA 
30,000 
60k0 
lOkO 
l lOkO 
i 
0 
t lOkO 
l. 
0 
lOkO 
[b] 
Va 
'la -
'lb -
Vb -
'lg -
P1oov -
Check: 
P12mA 
LP<lev 
LP<lis 
20k0 3k0 
+ 80k0 
vb 
i' 12mA + 60k0 
v 
lkO 
(15,000)(0.004) = 60 v 
Va 
60,000 = 1 mA 
12 - 1 - 4 = 7 mA 
60 - (0.007)(4000) = 32 v 
32 
0.007 - 80 OOO = 6.6 mA 
' 
a . - Ja 
-(100)(6.6 x 10-3 ) = -660 mW 
- -(60)(12 x 10-3 ) = -720 mW 
- 660 + 720 = 1380 m w 
Problems 4-63 
l lOkO 
4mA 
- (20,000)(6.6 x 10-3) 2 + (80,000)(0.4 x 10-3) 2 + (4000)(7 x 10-3) 2 
+(60,000)(1x10-3) 2 + (15,000)(4 x 10-3) 2 
- 1380 mW 
P 4.61 [a] First remove the 8 n and 80 0 resistors: 
~~ 
50 + 
200 450 V 0 
100 
Next use a source transformation to convert the 5 A current source and 
20 0 resistor: 
+ 
100 
4-64 CHAPTER 4. Techniques of Circuit Analysis 
which simplifies to 
350 + 
240V ,. 
45 
Vo= SO (-240) = -135 V; 
-135 
io= -- = -3 A 
45 
[b] Return to the original circuit with v0 = -135 V and i0 = -3 A: 
i 9 = 
340 - (-3) = 7.25 A 
80 
100 
P340V = -(340)(7.25) = -2465 W 
Therefore, the 340 V source is developing 2465 W. 
[c] V1 = 340 + 60i0 = 340 - 180 = 160 V 
Vg = V1 + 5(8) = 160 + 40 = 200 V 
P5A = -(5)(200) = -1000 W 
Therefore the 5 A source is developing 1000 W. 
( d] LPdev = 2465 + 1000 = 3465 W 
LPdiss = (5)2(8) + (8)2 (20) + (4.25)2(80) + (3) 2(60) = 3465 W 
·. · LPdiss = LPdev 
P 4.62 (a] Applying a source transformation to each current source yields 
40 10 lOV 
20V ~ 20 
Problems 4-65 
Now combine the 20 V and 10 V sources into a single voltage source and 
the 5 n. 4 n and 1 n re8i8tors into a single resistor to get 
100 
lOV ~ 20 
Now use a source transformation on each voltage source, thus 
[h] 
which can be reduced to 
1.25A 80 
io = (1.25)(8) = 1 A 
10 
20 
SQ ~ 
1 a 
50ia - 40ib = 20 - 10 - 10 = 0 
-40ia + 42b = 10 
S 1 . . Nb 1 A . o v1ng, ib = ~ = = 'lo 
60 
p 4.63 VTh = 50 (40) = 48 V 
RTh = s + (4o)(l0) = 16 n 
50 
20 
4~66 CHAPTER 4. Techniques of Circuit Analysis 
p 4.64 
SA 
,--~~~~----1-~·~~~~~~--. 
120 
V1 - 12 Vt _ 8 = O 
12 + 6 
V1 = 36 V 
VTh = V1 + (2)(8) = 52 V 
RTh = 2 + (l2)(5) = 60 
18 
1 
20 
+ 
b 
P 4. 65 After making a source transformation the circuit becomes 
300 = 48i1 - 40i2 
Problems 4-67 
-450 = -40i1 + 200i2 
:. i1 = 5.25 A and i2 = -1.2 A 
VTh = 8i1 + l0i2 = 30 V 
RTh = (40118+10)11150 = 15Q 
.--~V>A- •a I 1s o 
30vQ 
'~ ---•b 
P 4.66 First we make the observation that the 8-mA current source and the 20 kQ 
resistor will have no influence on the behavior of the circuit with respect to 
the terminals a,b. This follows because they are in parallel with an ideal 
voltage source. Hence our circuit can be simplified to 
or 
Therefore the Norton equivalent is 
..-----+----• a 
10k0 
....._ __ ......_ ____ b 
P 4.67 [a] First, find the Thevenin equivalent with respect to a,b using a succession 
of source transformations. 
4-68 CHAPTER 4. Techniques of Circuit Analysis 
4kQ 3kQ 
BkQ 
3kQ 
12kQ lOkQ 
6 Q 
•a 
'VTh = 48 V 
6kQ 
48r,--+-----. lOOkQ 
v 
meas 
100 
'Vmeas = 106 ( 48) = 45.28 V 
( 45.28 -
48) 
[b] %error = 48 x 100 = -5.673 
lOkQ 
a 
b 
a 
b 
Problems 4-69 
P 4.68 [a] Open circuit: 
200 
1.BA 
--~--'Vvi.-~-.--~-~·~~--w-~-~ 
50 + 
9V : 
V2 - 9 V2 - 1 8 = 0 
20 + 70 . 
V2 = 35 V 
60 
VTh = 70 V2 = 30 V 
Short circuit: 
200 ~. la 
1.8A 
--~--'Vvi.-~-.--~-~~·-----"'1----W 
50 
9V : 
V2 - 9 V2 -1.8 = O 
20 + 10 
:. V2 = 15 V 
. - 9 - 15 - -0 3 A 
ia - 20 - . 
isc = 1.8 - 0.3 = 1.5 A 
30 
RTh = - = 200 
1.5 
600 
4--70 CHAPTER 4. Techniques of Circuit Analysis 
[b) 
200 
50 
250 
100 
RTh = (20+101160 = 2on (CHECKS) 
p 4.69 
12.5 = VTh - 2RTh 
11.7 = VTh - lSRTh 
Solving the above equations for VTh and RTh yields 
VTh = 12.6v, RTh = 50 mn 
: . IN = 252 A, RN = 50 mO 
P 4. 70 First, find the Thevenin equivalent with respect to R0 • 
p 4.71 
Ra 'lo Vo 
0 1.6 0 
2 1.5 3 
6 1.33 8 
10 1.2 12 
15 1.067 16 
400 
60 
240 
Ra io 
20 0.96 
30 0.8 
50 0.6 
60 0.533 
70 0.48 
600 
Va 
19.2 
24 
30 
32 
33.6 
9800 
OPEN CIRCUIT 
V2 = -4Qib 40 X 103 = -16 X 105ib 
5 X 10-5V2 = -80ib 
60 
980ib + 5 X 10-5V2 = 980ib - 80ib = 900ib 
Problems 4-71 
~------a 
4-72 CHAPTER 4. Techniques of Circuit Analysis 
So 900ib is the voltage across the 100 n resistor. 
From KCL at the top left node, 
900i 
540 µA = lOOb + ib = lOib 
540 x 10-6 = 54 A 
Zb = lQ µ 
VTh = -16 X 105 (54 X 10-6) = -86.40 V 
SHORT CIRCUIT 
Zb =;: 54 x 10-3 = 54 x 10-6 = 50 A 
1080 1.08 µ 
isc = -40(50) = -2000 µA= -2 mA 
RTh = - 86 .4 x 103 = 43.2 krl . -2 
.---_,"./IA--• a 
86.4J 43.2k0 
l~---•b 
p 4.72 
3 
lOkO 
50k0 40k0 
The node voltage equations are: 
Problems 4-73 
V1 - 40 V1 V1 - V2 
2000 + 20,000 + 5000 0 
V2 - VJ V2 V2 - V3 O V1 
5000 + 50,000 + 10,000 + 3 20,000 - 0 
V3 - V2 V3 V1 
10,000 + 40,000 - 30 20,000 0 
In standard form: 
( 1 1 1 ) ( 1 ) 
40 
VI 2000 + 20,000 + 5000 + V2 -5000 + v3 (0) = 2000 
vi ( - 5;00 + 2o~ioo) + v2 ( 5o1oo + 50,~oo + 10,~oo) + v3 ( - 10,~oo) = 0 
711 ( - 20~~00) + 712 ( - 10,~oo) + 713 ( 10,~oo + 40,~oo) = 0 
Solving, v1 = 24 V; v 2 = -10 V; v3 = 280 V 
VTh = V3 = 280 v 
2k0 Sk 0 
The mesh current equations are: 
-40 + 2000i1 + 20,000( i 1 - i2) 
lOkO 
50k0' 
isc 
l 
isc 
0 
5000i2 + 50,000( i2 - isc) + 20,000( i2 - ii) - 0 
50,000(isc - i2) + 10,000(isc - 30i~) - 0 
The constraint equation is: 
Put these equations in standard form: 
4-7 4 CHAPTER 4. Techniques of Circuit Analysis 
ii(22,000) + i2(-20,000) + isc(O) + iA(O) 40 
ii (-20,000) + i2(75,000) + isc(-50,000) + iA(O) - 0 
ii(O) + i2(-50,000) + isc(60,000) + iA(-300,000) 0 
ii(-1) +i2(l) +isc(O) +iA(l) - 0 
Solving, 
ii = 13.6 m A; i2 = 12.96 m A; isc = 14 m A; iA = 640 µ A 
280 
RTh=~=20kn 
.---Wv---• a I 2Dk0 
2sov6 
'~-----•b 
P 4.73 [a] Use source transformations to simplify the left side of the circuit. 
2.5kQ 
~ 
lDV 6 ... 
7.5V 
. 10- 7.5 1 A ib= = m 
2.5 
Let R0 = RmeterlllO kn= 7.5/0.8 = 9.375 kn 
:. (Rmeter)(lO) = 9.375; 
Rmetere + 10 
R = (9·375)(10) = 150 kO meter 0.625 
[b) Actual value of ve: 
10 
ib = 2.5 + (0.8)(10) = 0.9524 mA 
Ve - 0.8ib(lQ) = 7.62 V 
[J1 (7.5 - 7.62) (J1 
70 error = 7_62 x 100 = -1.57 70 
P 4.74 [a] Find the Thevenin equivalent with respect to the terminals of the 
ammeter. This is most easily done by first finding the Thevenin with 
respect to the terminals of the 50 n resistor. 
Thevenin voltage: note i<P is zero. 
a 
b 
VTh VTh VTh VTh - 40 
80 + 40 + 240 + 16 = 0 
Solving, vTh = 24 V. 
Short-circuit current: 
24 
RTh = -.- = -480 
-0.5 
2Q 
~ 
24v6 
24 
Rtotal = lO = 2.4 0 
b 
isc = -0.5 A 
x 
y 
Problems 4-75 
144Q 
y 
4-76 CHAPTER 4. Techniques of Circuit Analysis 
Rmeter = 2.4 - 2 = 0.400 
[b) Actual current: 
20 
x 
y 
24 
iactual = 2 = 12 A 
10-12 
3 error = 12 x 100 = -16.673 
p 4.75 
45V 15k0 
i 1 = 45/15,000 = 3 mA 
VTh = 45 + 0.003RTh 
+ 
25V 5k0 
i2 = 25/5000 = 5 mA 
25 = VTh - 0.005RThi VTh = 25 + 0.005RTh 
.'. 45 + 0.003RTh = 25 + 0.005RTh so 
VTh = 45 + 30 = 75 V 
p 4.76 
~---'W'v~--• a 
15J) 10k0 
.__I ---•b 
120¥ ~ 
600 
V1=-(120)=48 V 
1500 
800 
V2 = 2004 (120) = 47.9042 V 
1204Q 
b 
+ 
VTh = V1 - V2 = 48 - 47.9042 = 95.8 mV 
R = (900)(600) (1204)(800) = 2,105,800 = 840_64 n 
Th 1500 + 2004 2505 
840.64Q 
a 
95.Bmr 300 
b 
95.8 x 10-3 
iga1 = 0.87064 x 103 = 110.03 µA 
Problems 4-77 
4-78 CHAPTER 4. Techniques of Circuit Analysis 
P 4. 77 VTh = 0, since circuit contains no independent sources. 
lOi 
. VT - V1 
'lA=---
12 
100 
In standard form: 
60 
120 a 
+ 
b 
V1 (-:2) +VT ( 1~ + ~) + iA (- 16°) = 1 
Solving, 
V1=2 V; VT= 8 V; iA = 0.5 A 
80 
Problems 4-79 
P 4. 78 VTh = 0 since there are no independent sources in the circuit. To find RTh we 
first find Ra'b'. 
+ 
65iA = 2.6VT 
VT 
-. = 50/6.25 = 8 n = Ra'b' 
'lT 
: . RTh = 12 + s = 20 n 
200 
100 120+ 
b 
4-80 CHAPTER 4. Techniques of Circuit Analysis 
P 4. 79 We begin by finding the Thevenin equivalent with respect to R0 • After making 
a couple of source transformations the circuit simplifies to 
80V 
r ooo 
80 + 60ix 
'lx = 140 ix= 1 A 
VTh = 40ix - 60ix = -20ix = -20 V 
+ 
Using the test-source method to find the Thevenin resistance gives 
ix VT 
500 600 
2000 
Use the node voltage method: 
VT - V + 'UT + 60ix _ l = Q 
60 40 
Solving, VT= 50 V. 
VT 
RTh= -=50!1 
1 A 
+ 
1' lA 
Thus our problem is reduced to analyzing the circuit shown below. 
Problems 4~-81 
500 
( -20 )
2 
50 + R0 Ro = 1.5 
400Ro = 1.5 
R~ + 100R0 + 2500 
l.5R~ - 250R0 + 3750 = 0 
.·. R0 = 16.67D; R0 = 150D 
P 4.80 [a] Find the Thevenin equivalent with respect to the terminals of R1 . 
Open circuit voltage: 
40 
its ~ 
4 i2 
60 80 
48 0\11 : ~ 400 ~ 
il i3 
40 
The mesh current equations are: 
480 + 6(ii ~ i2) + 40(ii - i3) + 4ii - 0 
4i2 + 8(i2 - i3) + 6(i2 - ii) 0 
-20i,s + 2i3 + 40(i3 - ii)+ 8(i3 - i2) - 0 
The dependent source constraint equation is: 
i,s =ii - i2 
Place these equations in standard form: 
ii(6 + 40 + 4) + i2(-6) + i3(-40) + i,s(O) 
ii ( -6) + i2( 4 + 8 + 6) + i3(-8) + i,s(O) 
20i/3 
+ 
- -480 
- 0 
ii(-40) + i2(-8) + i3(8 + 2 + 40) + i(3(-20) - 0 
0 
4--82 CHAPTER 4. Techniques of Circuit Analysis 
Solving, ii= -99.6 A; i2 = -78 A; i3 = -100.8 A; i/J = -21.6 A 
V.rh = 40(ii - i3) = 48 V 
Short-circuit current: 
60 
48 0\11 ; 
40 
The mesh current equations are: 
480 + 6( ii - i2) + 4ii - 0 
4i2 + 8(i2 - i3) + 6(i2 - ii) = 0 
-20i/J + 2i3 + 8(i3 - i2) - 0 
80 
The dependent source constraint equation is: 
i/J =ii - i2 
Place these equations in standard form: 
ii(6 + 4) + i2(-6) + i3(0) + i/J(O) 
ii (-6) + i2( 4 + 8 + 6) + i3(-8) + i/3(0) 
ii (0) + i2(-8) + i3(8 + 2) + i/J(-20) 
ii(-1) +i2(l) +i3(0) +i/3(1) 
-
~ 
-
-
-480 
0 
0 
0 
Solving, ii = -92 A; i2 = - 73.33 A; i3 = -96 A; i/J = -18.67 A 
. . . 4 A RTh -- v;_Th = 48 = 12" Zsc = i1 - i3 = ; J.G 
'lsc 4 
RL = RTh = 12!1 
242 
(b] Pmax = J:2 = 48 W 
120 + 
24V 120 
Problems 4~83 
P 4.81 (a] v 
6k0 1.25k0 
+ 
18VI ~ 
I • 
VTh - 18 VTh - 50 _ O 
6000 + 10,000 -
Solving, VTh = 30 V 
4k0 l .25k0 
2k0 lOkO 
RTh = 1250 + 10,000ll (2000 + 4000) = 5 kn 
[b) 
Pmax = (3 x 10-3)2(5000) = 45 ro W 
4-84 CHAPTER 4. Techniques of Circuit Analysis 
P 4.82 Write KCL equations at each of the labeled nodes, place them in standard 
form, and solve: 
V1 4k0 Vz 1.25k0 
9mA f 2k0 5k0 
_ X l -3 ~ V1 - V2 _ 
9 0 + 2000 + 4000 - 0 
V2 - V1 V2 - 50 V2 
4000 + 10,000 + 6250 = 0 
Standard form: 
VI (20~0 + 40100) + V2 ( - 40~0) = 0.009 
( 1 ) ( 1 1 1 ) 50 
VI - 4000 + V2 4000 + 10,000 + 6250 = 10,000 
Calculator solution: 
V1 = 18.25 V V2 = 18.75 V 
Calculate currents: 
50- V2 
i2 = 10000 = 3.125 m A 
' 
Calculate power delivered by the sources: 
P9mA = (9 X 10-3)v1 = (9 X 10-3)(18.25) = 164.25 mW 
p5ov = i2(50) = (3.125 x 10-3 )(50) = 156.25 mW 
Problems 4-85 
P<lelivere<l = 164.25 + 156.25 = 320.5 mW 
From Problem 4.81, 
P5k = 45 mW 
% delivered to R0 : 3:i.5 (100) = 14.04% 
P 4.83 [a] From the solution of Problem 4.72 we have RTh = 20 kn and VTh = 280 V. 
Therefore 
Ro = RTh = 20 kn 
(140)2 
[b] p = 20 OOO = 980 mW 
' 
[c] 
The node voltage equations are: 
V1 - 40 V1 V1 - V2 
2000 + 20,000 + 5000 
lOkO 
SOk 0 40k0 
- 0 
V2 - V1 V2 V2 - V3 . 
5000 + 50 000 + 10 00() + 30zA - O 
' ' 
V3 - V2 V3 . V3 
10 000 + 40 000 - 30zA + 20 000 
' ' ' 
0 
The dependent source constraint equation is: 
. V1 
'lil = 20 000 
' Place these equations in standard form: 
3 
a 
20k0 
b 
( 1 1 1) ( 1) . 40 Vi 2000 + 20,000 + 5000 + V2 -5000 + v3 (0) + iA(O) = 2000 
vi (- 4;oo) +v2 ( 4o
1
oo + 50,~oo + 10,~oo) +v3 (-10,~oo) +i6 (3o) = 0 
Vi(O) + V2 (-10,~00) + V3 ( 10,~00 + 40,~00 + 20,~00) + iA(-30) = () 
4-86 CHAPTER 4. Techniques of Circuit Analysis 
V1 ( 20~0100) + v2(0) + v3(0) + iA(l) = 0 
Solving, v1 = 18.4 V; v2 = -31 V; v3 = 140 V; it:.. = 920 p,A 
Calculate the power: 
. = 40 - 18.4 = 10.8 mA 
'tg 2000 
P40V = -( 40)(10.8 X 10-3) = -432 mW 
Pdep source= (v2 - V3)(30it:..) = -4719.6 ill W 
LP<lev = 432 + 4719.6 = 5151.6 ill W 
980 x 10-3 
% delivered = 5151.6 x 10_3 x 100 = 19.02% 
P 4.84 [a] From the solution to Problem 4.70 we have 
Ro(O) 
[b] 
p0 (W) 
25 
20 
15 
10 
5 
0 
2 
6 
10 
15 
Po(W) Ro(n) Po(W) 
0 20 18.432 
4.5 30 19.2 
10.67 50 18 
14.4 60 17.067 
17.067 70 16.128 
R0 (Q) 
o..-~~~~~~~~~~~~~~~~~~~~~-----, 
0 10 20 30 40 50 60 70 
(c} R0 = 300, P0 (max) = 19.2 W 
Problems 4~87 
P 4.85 Find the Thevenin equivalent with respect to the terminals of R0 • 
Open circuit voltage: 
a b 
50 
~ 
20 ib 30 
+Vt:,-
46. ff\11 ~ ~ 
10 
~ 
i ~ 42 .4V ic a 
( 46.8 - 42.4) = 3ia - 2ib - ic 
0 = -2ia + lOib - 3ic 
Solving, ib = 74.8 A 
.". VTh = 5ib = 37 4 V 
Short circuit current: 
i s 
20 30 
+Vt:;-
46.8\11 ~ ~ 
10 
~ 
i ~ 42. 4V ic a 
46.8 - 42.4 = 3ia - 2isc - ic 
4-88 CHAPTER 4. Techniques of Circuit Analysis 
0 = -2ia + 5isc - 3ic 
Solving, isc = 6.8 A; ia = 8 A; ic = 6 A; Vti. = 2.4 V 
RTh = VTh/isc = 374/6.8 = 55 n 
Ro= 55n 
With R0 equal to 55 n the circuit becomes 
550 
+ 187V 
50 
20 ~ 30 lb 
+vi::.-
46. 8\li : ~ 
10 
~ 
i : 42. 4V l a c 
46.8 - 42.4 = 3ia - 2ib - 2.5(2)(ia - ib) 
187 + 3ib - 3(2.5)(2)(ia - ib) + 2ib - 2ia = 0 
Solving, ia = 59 A; ib = 40.8 A 
Vti. = 2(59 - 40.80) = 36.4 V 
ic = 91 A 
Thus we have 
+ 187V 
50 
20 30 
-7 18.2A -750 .2A 
46. 8\11 ~ 
59A 
t 
Ve = 42.4 - 32 - 150.6 = -140.20 V 
'2:Pctev = 46.8(59) + 42.4(32) + 140.20(91) = 16,876.20 W 
CHECK: 
'2:Pctis = (18.2)2 (2) + (50.2)2(3) + (32)2(1) 
+ 187(3.4) + 187(37.4) = 16,876.20 w 
. (55)(3.4)2 (100) 
% delivered = 16,876.2 = 3. 77% 
Problems 4-89 
P 4.86 [a] We begin by finding the Thevenin equivalent with respect to the terminals 
of R0 • 
Open circuit voltage 
14il!. 
~~~~~·+->-~~~~~ 
10 20 
2 0 0\11 : 200 
: lOOV 
i 
40 l!. 
4-90 CHAPTER 4. Techniques of Circuit Anal.Ysis 
0 = -i1 + 3i2 - 2i3 + 14iA 
100 = - 20i1 - 2i2 + 25i3 
. . . 
'lA = 'l1 - 'l3 
Solving, ii = -2.5 A; i2 = 37.5 A; i3 = 5 A; iA = -7.5 A 
VTh = 20(i1 - i3) = 20(-7.5) = -150 V 
Now find the short-circuit current. 
10 
2 0 0\11 : : lOOV 
40 30 
Note with the short circuit from a to b that it:;.. is zero, hence 14it:;.. is also 
zero. 
-200 = 5i1 - li2 + Oi3 
0 = - li1 + 3i2 - 2i3 
100 = Oi1 - 2i2 + 5i3 
Solving, i 1 = -40 A; 
isc =ii - i3 = -60 A 
RTh = (-150)/(-60) = 2.50 
2.50 
75V 2.50 
+ 
i3 = 20 A 
For maximum power transfer Ro = RTh = 2.5 n 
752 
[b] Pmax = - = 2250 W 
2.5 
Problems 4-91 
P 4.87 From the solution of Problem 4.86 we know that when R0 is 2.5 0, the voltage 
across R0 is 75 V, positive at the lower terminal. Therefore our problem 
reduces to the analysis of the following circuit. In constructing the circuit we 
have used the fact that ifl is -3.75 A, and hence 14ifl is -52.5 V. 
52 .5V ids 
.---~~~~~~~~+>--~~~~---'""~~~ 
10 
75V 
+ 
Using the node voltage method to find v1 and v2 yields 
- 75 - V1 - 75 - V2 
-33.75+ + = 0 
1 2 
V1+52.5 = V2 
Solving, v1 = -115 V; v2 = -62.5 V.It follows that 
. - -115 + 200 - 21 25 A 
'lg1 - 4 - . 
. _ -62.5 + 100 _ 12 5 A 
'lg2 - 3 - . 
iz = -62.5 + 75 = 6 25 A 
2 . 
ids = -6.25 - 12.5 = -18. 75 A 
P20ov = -200ig1 = -4250 W(dev) 
P10ov = - l00ig2 = -1250 W ( dev) 
Pds = 52.5ids = -984.375 W(dev) 
LP<lev = 4250 + 1250 + 984.375 = 6484.375 W 
3 delivered - 64~~~~75 ( 100) = 34. 73 
34. 73 of developed power is delivered to load 
4-92 CHAPTER 4. Techniques of Circuit Anal.rsis 
P 4.88 [a] Open circuit voltage 
20 
100\1! ~ 
Node voltage equation: 
V1 - 100 V1 - 13i .6. V1 - V2 _ O 
2 + 5 + 4 -
Constraint equations: 
100- VI 
'l,6. = 
2 
40 
Solving, V2 = 90 V = VTh; V1 = 90 V; VD.= 0 V; iD. = 5 A 
Short circuit current: 
20 40 
+ Vt, -
10 OVi ~ 
V1 - 100 V1 - 13iA V1 _ O 
2 + 5 +4-
100 - V1 
iD. = ---2 
Solving, v1 = 80 V = v.6.; iA = 10 A 
. V1 
'lsc = 4 + VA = 20 + 80 = 100 A 
VTh 90 
RTh = isc = lOO = 0.90 
Ra= RTh = 0.90 
[b] 
(c] 
Problems 4-93 
0.90 + 
45V 0.90 
(45)2 
Pmax = {).9 = 2250 W 
20 1 50A 
+ Vt, -
+ 
100\11: 45V 0.90 
'1.11 - 100 '1.11 - 13iA '1.11 - 45 
2 + 5 + 4 =O 
100 - V1 
iA=---
2 
Solving, v1 = 85 V; iA = 7.5 A; VA= '1.11 - V2 = 85 - 45 = 40 V 
i10ov = iA = 7.5 A 
P10ov (dev) = 100(7.5) = 750 W 
i12 = VA/4 = 40/4 = 10 A 
ii = i12 - iA - 10 - 7.5 = 2.5 A 
Pl3ia.. (dev) = (97.5)(2.5) = 243.75 W 
Pva.. (dev) = (45)(40) = 1800 W 
LPdev = 750+243.75+1800 = 2793.75 W 
3 delivered = 2~~:.~5 x 100 = 80.543 
4~94 CHAPTER 4. Techniques of Circuit Analysis 
P 4.89 [a] First find the Thevenin equivalent with respect to R0 • 
Open circuit voltage: i</> = O; 50i</> = 0 
~-------~-------------------------------------------, 
I 
I 
I 
ov 
r---------------------------------------1 1 
I I I I 
t : I I 
: v1 + vf::. - a v2: 
I ---.~-A,AJ\,..---"t~~--'v\.AJ:=.---.A-~'W,,-~--;----_.;...,~~~ 
_2 100 50 200 ~-----1 
1000 
~ 280V 4000 
b 
V1 = 210 V; VA= 14 V 
VTh = 280 - V £). = 280 - 14 = 266 v 
Short circuit current 
r----------------------------------------------------1 I 
I 
50ilf> + -
r-------·--------------------------~----1 
I 
I 
I 
I 
I 
+ Vf::. -
100 50 200 
1000 
~ 280V 
b 
~ V] - 280 V2 V2 0.5125(280) = 0 
100 + 10 + 20 + 400 + 
V,6. = 280 V 
V2 + 50i.p = V1 
. 280 V2 
i</> = S + 20 = 56 + 0.05v2 
I 
v• 2• 
l 
L-----"" 
[bJ 
[c] 
V2 = -968 V; V1 = -588 V 
i¢ = isc = 56 + 0.05( -968) = 7.6 A 
RTh = VTh/isc = 266/7.6 = 350 
:. Ro= 350 
350 + 
133V 350 
Pmax = (133)2/35 = 505.4 W 
Problems 4~95 
r----------------------------------------------------1 
I t 
I t 
I + t 
50i\D - : 
r---------------------------------------1 I I 
I I t 
I + I I 
IV I v. - I V 1 
I 1 : U a : 21 
: .._,.__-'111/'v---t~~-"JV\r-----~-w<.~-r--e--::--~---, 
I I 
l. - - - .... ..! 100 
1000 
50 
+ 
b 
i\D 
350 
200 I I L-----..1 
4000 
~ V1 - 280 V2 - 133 V2 0.5125(280 - 133) = 0 
100 + 10 + 20 + 400 + 
v2 + 50i¢ = v1 ; i¢ = 133/35 = 3.8 A 
Therefore, v1 = -189 V and v2 = -379 V; thus, 
. _ 280 - 133 280 + 189 _ 76 30 A 
ig - 5 + 10 - . 
P2sov (dev) = (280)(76.3) = 21,364 W 
P 4.90 [a] Since 0:::; R0 $ oo maximum power will be delivered to the 6 0 resistor 
when R0 = 0. 
[h] p - 3~2 = 150 w 
4-96 CHAPTER 4. Techniques of Circuit Analysis 
P 4.91 (a] 75 V source acting alone: 
Re= 201120 = 100 
i' = 75 = 5 A 
5+ 10 
v' = (5)(10) = 50 V 
+ 
6 A source acting alone: 
51120 = 40 
4+8=12n 
121112 = 60 
... 
50 
+ 
v" 
-
6A 
+ 
"'200 
... 
80 
Hence our circuit reduces to: 
6A 
....----.·+·•----. 
60 
+ v" a 
It follows that 
v~ = 6(6) = 36 V 
and 
80 
120 
!:: 120 
If 4 ( ) v = -- -36 = -12 v 
4+8 
v = v' + v" = 50 - 12 = 38 V 
v2 
[b] p = 20 = 72.2 w 
P 4. 92 70-V source acting alone: 
40 
+ 
v' = 70- 4i~ 
70 = 20i~ + v~ 
., 70 - v~ 
i =---
a 20 
v' v' 70 - v' 11 v' 
·I b b f 3 5 
ib = 2 + 10 - 20 = 20 vb + 10 - · 
v' = v~ + 2i~ 
v~ = v' - 2i~ 
·f 11 ( I 2 ·f) VI 
ib = 20 v - ib + 10 - 3.5 or 
·f 13 f 70 
'lb= -v - -
42 42 
Problems 4-97 
v' = 70 - 4 ( 13 v' - 70) or 
42 42 
v' = 3220 = 1610 V = 34.255 V 
94 47 
50-V source acting alone: 
4-98 CHAPTER 4. Techniques of Circuit Analysis 
v" = -4i~ 
v" = v~ + 2i~ 
v" = -50 + lOi~ 
.,, v" + 50 
id= 10 
v" v" + 50 
i" = i+---
s 2 10 
40 
~ sov 
.,, 11/: vf: v" v" + 50 11 v" + 50 
ib = 20 + i~ = 20 + ; + 10 = 20 v~ + 10 
v~ = v" - 2i~ 
.,, _ 11 ( ,, _ 2 .,,) v" + 50 
ib - 20 v ib + 10 
Thus, v" = _ 4 ( 13 v" + 100) 
42 42 
or 
·II 13 II 100 
zb =-v +-
42 42 
or v" = - 24~ V = -4.255 V 
Hence, v = v' + v" = 1610 _ 200 = 1410 = 30 V 
47 47 47 
P 4.93 45 V source actim: alone: 
SQ lOQ 
io1 = 45/40 = 1.125 A 
10 V source actirnr alone: 
sQ 10Q 
40Q 610v 
lSQ 30Q 
10 30 
is = 40 + 40/3 = 160 A 
. 30 40 
io2 = - 160 · 60 = -0.125 A 
8 A current source acting alone: 
SQ lOQ 
8A 
'------!'·~!-----' 
io3 = (8)(30) = 6 A 
40 
Problems 4-99 
SQ 
8A 
'------<··~!----~ 
io = iol + io2 + io3 = 1.125 - 0.125 + 6 = 7 A 
4-100 CHAPTER 4. Techniques of Circuit Analysis 
P 4.94 Voltage source acting alone: 
SkQ 
35V: 
+ 
20kQ v01 
V 0 1 V 0 1 - 35 (35 - Vol) _ O -+ -5 -
20 5 5 
.·. Vol= 33.6 V 
Current source acting alone: 
Si<P 
.--~~~~>-~~~--, 
SkQ 
+ 
.•. Vo2 = -5.6 V 
Vo = V 0 1 + Vo2 = 33.6 - 5.6 = 28 V 
P 4.95 Voltage source acting alone: 
40Q 
+ 
60Q 
20Q 
80Q 
. 180 180 
ioi = 90 + 40 + 1001125 = 2 + 3 = 5 A 
. ( 80) Val = (3)(20) lOQ = 48 V 
Current source acting alone: 
40Q 
+ 
60V 25Q 
~~~~~-w..~~~~~~~ 
i4 
300 1 7.5A 2 200 
+ 
Solving, V2 = -184 V = Va2; 
. V3 2 A 
i40 = - = -
40 
. _ 7.5(60) _ 5 A 
i30 - 90 -
V3 = -80 V 
io'2 = -i30 - i4o = -5 + 2 = -3 A 
. ·. Vo = Val + Vo2 = 48 - 184 = -136 V 
ia = iol + ia2 = 5 - 3 = 2 A 
Problems 4-101 
4-102 CHAPTER 4. Techniques of Circuit Analysis 
P 4.96 4.5 A source: 
1.80 
4.5A + 12Q 140 100 150 
i' 
-4 r-:(6) · .,= .o =-15A 
. • 'lo 18 . 
20 A source: 
1.80 
120 t 20A 4.20 
if T 
0 
i" = 4·2(20) = 4.67 A 
0 18 
50 V source: 
1.80 100 1.80 
120 140 150 J 120 4.20 +SA 
i~' = 4 · 2~; 5) = -1.167 A 
i 0 = i~ + i~ + i~' = -1.5 + 4.67 - 1.167 = 2 A 
Problems 4-103 
P 4.97 [a] By hypothesis i~ + i~ = 1.5 mA. 
lDmA 
..--~-t~-----.., 
2kQ 
lOkQ 18kQ 
i' ' ' 
i"' = 10 (2) = 1 mA· 
0 (20) ' 
: . i0 = 1.5 + 1 = 2.5 mA 
[b} With all three sources in the circuit write a single node voltage equation. 
P 4.98 [a) 
Vb + Vb - 20 _ 5 _ lQ = O 
18 2 
:. Vb= 45 V 
Vb 
i 0 = - = 2.5 mA 
18 
+ 
60V 200 
Voe = VTh = 75 V; . 60 3 A 'tL = 20 = ; 
Therefore 
15 
RTh = - = 50 
3 
75 - 60 15 
'tL=---=-
RTh RTh 
4-104 CHAPTER 4. Techniques of Circuit Analysis 
P 4.99 [a] xrQ (L-x)rQ v,r : RQ b., 
'----ffw..-O __ _.__{_L~-xM)r~ 
V - V1 V V - V2 O --+-+ = 
2xr R 2r(L - x) 
v [2!r + ~ + 2r(L1- x)l = 2~r + 2r(;2-x) 
v1RL + xR( v2 - v1) 
v=-------
RL + 2rLx-2rx2 
[b] Let D = RL + 2rLx - 2rx2 
dv (RL + 2rLx - 2rx2)R(v2 - v1) - [v1RL + xR(v2 - v1)]2r(L - 2x) 
-
dx D 2 
dv 0 h . -d = w en numerator is zero. 
x 
The numerator simplifies to 
2 2L - v1 RL( v2 - v1) - 2rv1L2 0 x + x+ = 
(v2 - v1) 2r(v2 - v1) 
Solving for the roots of the quadratic yields 
x = L {-v1 ± V1V2 - !!:_(v2 - v1)2} 
V2 -V1 2rL 
[c] x = L {-v1 ± V1V2 - !!:_(v1 - v2)2} 
V2 - V1 2rL 
V2 = 1200 V, V1 = 1000 V, L = 16 km 
r = 5 x 10-5 0/m; R= 3.90 
L 16,000 
V2 - V1 - 1200 - 1000 = 80; 
R ( )2 3.9( -200)2 . 5 
2rL v1 - v2 = (10 x 10-5)(16 x 1Q3) = 0.975 x 10 
x = 80{ -1000 ± Ji.2 x 106 - 0.0975 x 106} 
= 80{-1000±1050} = 80(50) = 4000 m 
[d] 
Vmin -
-
v1RL + R(v2 - v1)x 
RL + 2rLx -2rx2 
Problems 
(1000)(3.9)(16 x 103 ) + 3.9(200)(4000) 
4-105 
(3.9)(16,000) + 10 x 10-5(16,000)( 4000) - 10 x 10-5 (16 x 106) 
- 975 v 
P 4.100 [a] In studying the circuit in Fig. P4.100 we note it contains six meshes and 
six essential nodes. Further study shows that by replacing the parallel 
resistors with their equivalent values the circuit reduces to four meshes 
and four essential nodes as shown in the following diagram. 
The node Voltage approach will require solving three node Voltage 
equations along with equations involving v~ and if3. 
The mesh-current approach will require writing one supermesh equation 
plus three constraint equations involving the three current sources. Thus 
at the outset we know the supermesh equation can be reduced to a single 
unknown current. Since we are interested in the power developed by the 
16 V source, we will retain the mesh current ib and eliminate the mesh 
currents ia, ic And id. 
The supermesh is denoted by the dashed line in the following figure. 
r-------, 
I - I 
, - 16V I 
1vv ~o \ 201 
1+ ; J I r - - - ;-v;:-=.. - 1b I 20 
I 1 
I 
I 
I 
I 
I 
}Q 
4 
) 
a 
L_ _ - - - - - __ _J.9._ - - - J 
[bJ Summing the voltages around the supermesh yields 
-8vy + ~ia + ~ib -16 + 2ib + 2(ic - id)+lie= 0 
Note that Vy= 2ib/3; make that substitution and multiply the equation 
by 12: 
-96 (~ib) + 9ia +Sib - 192 + 24ib + 24(ic - id)+ 12ic = 0 
or 
4--106 
p 4.101 
CHAPTER 4. Techniques of Circuit Analysis 
Now note: 
and 
. . . 
'lx ='le - 'ld 
so 
ia = 2(ie - ia) 3ia = 2ie 
Now use the following constraints: 
Vx . . d 3. 3 = 'le - 'lb an Vx = 4ia 
Therefore 
ia = 4ic - 4ib 
Finally, 
In standard form: 
9ia - 32ib + 36ie - 24ia = 192 
Oia + Oib + 2ie - 3ia = 0 
1 ia + 4ib - 4ie + Oia = 0 
lia + Oib - lie + Oia = 2 
Solving, 
ia = 33.6 V; ib = 23.2 V; ie = 31.6 V; ia = 21.067 V 
P16V = -16ib = -16(23.2) = -371.2 W 
Therefore, the 16 V source deliveres 371.2 W of power. 
8Q 
5\b_ 90 
v 2.,.._ __ -<+ ->---~--'VV'v-----i 
+ 
6Q 
p 4.102 
_v4_-_v_1 + V4 - V2 + V4 - 2 = 0 
2 3 4 
Also, v2 = 5v.6. = 5(v2 - v4) 
Solving, v 1 = -20 V; v2 = -15 V; V4 = -12 V 
Also, Vx = V2 - vi = 5 V Solving, v5 = 8 V 
Solving, 
0 .20 o.20 
uavG 27 ) 
ia c 
0 .30 0 .30 
HOV~ ) 36 ) 
~ ld 
0 .2Q 0 .20 
110 = 27.5ia - 0.3ib - 27ic 
110 = -0.3ia + 36.5ib - 36id 
0 = -27ia + 54.5ic - 0.3id - 27ie 
0 = -36ib - 0.3ic + 72.5id - 36ie 
0 = -27ic - 36id + Slie 
+ 
vl: 
+ 
v 
270 
) 
e 
360 
ia = 19.19 A; ib = 17.36 A; ic = 15.275 A; 
id = 14.39 A; ie = 11.49 A 
So, 
V1 = 27(15.275 - 11.49) = 102.2 V; 
V2 = 36(14.49 - 11.49) = 104.4 V; 
V3 = 18(11.49) = 206.8 V. 
+ 
v 
Problems 4-107 
180 
4-108 CHAPTER 4. Techniques of Circuit Analysis 
p 4.103 lQ 
lQ 
) 
lb 
lQ 
) lQ 
) 2 40\fi : c lQ 
i) 
a 
lQ 
lQ 
) 
1 
lQ 
e 
lQ 
240 = 4ia, - lib - lie - lid - lie 
0 = - lia + Sib - lie + Did+ Oie 
0 = -lia - lib+ 5ie - lid+ Oie 
0 = - lia + Oib - lie + 5id - lie 
0 = -lia + Oib + Oie - lid + 8ie 
A calculator solution yields 
ia = 77.5 A; id = 22.5 A; 
ib = 12.5 A; 
ie = 22.5 A 
. ·. i = ic - id = 0 A 
CHECK: 
ie = 12.5 A; 
SQ 
2Q 
~ 
2Q 
SQ 
LPabs = 1(77.5 - 12.5)2 + 1(77.5 - 22.5)2 + 1(77.5 - 22.5)2 + 1(77.5 - 12.5)2 
+1(12.5)2 + 5(12.5)2 + 1(12.5 - 22.5)2 + 2(22.5)2 + 1(22.5 -· 22.5)2 
+2(22.5)2 + 1(22.5 - 12.5)2 + 5(12.5)2 + 1(12.5)2 
- 4225 + 3025 + 3025 + 4225 + 156.25 + 781.25 
+ 100 + 1012.5 + 1012.5 + 100 + 781.25 + 156.25 = 18,000 w 
240ia = 240(77.5) = 18,000 W (CHECKS) 
p 4.104 dv1 = -R1[R2(R3 + R4) + R3R4] 
dl91 (R1 + R2)(R3 + R4) + R3R4 
dv1 RiR3R4 
dl92 = (R1 + R2)(R3 + R4) + R3R4 
dv2 -R1R3R4 -+---------
dlgl (R1 + R2)(R3 + R4) + R3R4 
dv2 R3R4(R1 + R2) - = ---------
dJ.q2 (R1 + R2)(R3 + R4) + R3R4 
P 4.105 From the solution to Problem 4.104 we have 
dv1 = -25[5(125) + 3750] = _ 175 V /A= _ 14.5833 V /A 
dlgl 30(125) + 3750 12 
and 
dv2 = -(25)(50)(75) = _ 12.5 V/A 
dlgl 30(125) + 3750 
By hypothesis, tJ..191 = 11 - 12 = -1 A 
175 175 
:. tJ..vi=(- 12 )(-1)= 12 =14.5833V 
Thus, v1=25+14.5833 = 39.5833 V 
Also, 
tJ..v2 = (-12.5)(-1) = 12.5 V 
Thus, V2 = 90 + 12.5 = 102.5 V 
The PSpice solution is 
V1 = 39.5830 V 
and 
V2 = 102.5000 V 
These values are in agreement with our predicted values. 
Problems 4-109 
4-110 CHAPTER 4. Techniques of Circuit Analysis 
P 4.106 From the solution to Problem 4.104 we have 
dv1 = (25)(50)(75) = 12.5 V /A 
dlg2 30(125) + 3750 
and 
dv2 = (50)(75)(30) = 15 V /A 
dlg2 30(125) + 3750 
By hypothesis, fl.Ig2 = 17 - 16 = 1 A 
: . fl.v1 = (12.5)(1) = 12.5 V 
Thus, v1 = 25 + 12.5 = 37.5 V 
Also, 
fl.v2 = (15) (1) = 15 V 
Thus, v2 = 90 + 15 = 105 V 
The PSpice solution is 
V1 = 37.5 V 
and 
These values are in agreement with our predicted values. 
P 4.107 From the solutions to Problems 4.104 - 4.106 we have 
dv1 __ 175 V/A· 
dfql - 12 ' 
dv2 
dl = -12.5 V /A; gl 
By hypothesis, 
fl.Iu1=11-12 = -1 A 
fl.lu2 = 17 - 16 = 1 A 
Therefore, 
dv1 
dluz = 12.5 V/A 
dv2 = 15 V/A 
dlg2 
175 
fl.v1 = 12 + 12.5 = 27.0833 V 
Av2 = 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 4.108 By hypothesis, 
ARt ~ 27.5- 25 = 2.50 
AR2 = 4.5-5 = -0.50 
AR3 = 55 - 50 = 5 0 
AR4 = 67.5- 75 = -7.50 
So 
Problems 4-111 
Avt = 0.5833(2.5) - 5.417(-0.5) + 0.45(5) + 0.2(-7.5) = 4.9168 V 
. • . Vt = 25 + 4.9168 = 29.9168 V 
Av2 = 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 
Vt = 29.6710 V 
and 
V2 = 88.5260 V 
Note our predicted values are within a fraction of a volt of the actual values. 
___ ___....,_5 
The Operational Amplifier 
Assessment Problems 
AP 5.1 [a] This is an inverting amplifier, so 
V 0 ~ (-Rt/ ~)vs= (-80/16)v8 , so 
0.4 2.0 3.5 -0.6 -1.6 -2.4 
v0 ( V) -2.0 -10.0 -15.0 3.0 8.0 10.0 
Two of the 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 = -5v8 , Vs = 3 V 
Use the positive power supply value to determine the smallest input 
voltage: 
V 8 = -2 V 
Therefore . - 2 $ V 8 $ 3 V 
AP 5.2 From Assessment Problem 5.1 
V 0 = (-R1/Ri)V8 = (-Rx/16,000)v8 = (-Rx/16,000)(-0.640) 
= 0.64Rx/16,000 = 4 X 10-5 Rx 
U.ae the negative power supply value to determine one limit on the value of Rx: 
so Rx= -15/4 x 10-5 = -375k0 
5-1 
5-2 CHAPTER 5. The Operational Amplifier 
Since we cannot have negative resistor values, the lower limit for Rx is 0. Now 
use the positive power supply value to determine the upper limit on the value 
of Rx: 
4 x 10-5 Rx = 10 so Rx= 10/4 x 10-5 = 250kf2 
Therefore, 
0 <Rx::; 250kf2 
AP 5.3 [a] This is an inverting summing amplifier so 
Va= (-Rt/ Ra)Va +(-Rt/ Rb)vb = -(250/5)va - (250/25)vb = -50va - lOvb 
Substituting the values for Va and Vb: 
Va....:. -50(0.1) - 10(0.25) = -5 - 2.5 = -7.5 V 
(b] Substitute the value for vb into the equation for Va from part (a) and use 
the negative power supply value: 
Va = -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 Va from part (a) and use 
the negative power supply value: 
Va= -50(0.10) ~ lOvb = -5 - lOvb = -10 V; 
Therefore lOvb = 5, so vb = 0.5 V 
[d] The effect of reversing polarity is to change the sign on the vb term in 
each equation from negative to positive. 
Repeat part (a): 
Va= -50va + lOvb = -5 + 2.5 = -2.5 V 
Repeat part (b): 
Va= -50va + 2.5 = -10 V; 50Va = 12.5, Va = 0.25 V 
Repeat part (c), using the value of the positive power supply: 
Va= -5 + lOvb = 15 V; lOvb = 20; Vb= 2.0 V 
AP 5.4 [a] Write a node voltage equation at vn; remember that for an ideal op amp, 
the current into the op amp at the inputs is zero: 
Vn Vn - Va 
4500 + 63 000 = 0 
' 
Problems 5-3 
Solve for v0 in terms of Vn by multiplying both sides by 63,000 and 
collecting terms: 
l4vn + Vn - V0 = 0 so 
Now use voltage division to calculate Vp. We can use voltage division 
because the op amp is ideal, so no current flows into the non-inverting 
input terminal and the 400 m V divides between the 15 kO resistor and 
the Rx resistor: 
Rx ( 
Vp = 15,000 +Rx 0.400) 
Now substitute the value Rx = 60 kO: 
6o,ooo (0 400) - 0 32 v 
Vp = 15,000 + 60,000 . - . 
Finally, remember that for an ideal op amp, Vn = vP, so substitute the 
value of vP into the equation for v0 
V 0 = l5vn = 15vp = 15(0.32) = 4.8 V 
[b] Substitute the expression for Vp into the equation for v0 and set the 
resulting equation equal to the positive power supply value: 
( 0.4Rx ) 
Vo = 15 15,000 + Rx = 5 
l5(0.4Rx) = 5(15,000 + Rx) so Rx = 75 kO 
AP 5.5 [a] Since this is a difference amplifier, we can use the expression for the 
output voltage in terms of the input voltages and the resistor values 
given in Eq. 5.22: 
20(60) 50 
Vo = 10(24) Vb - 10'/Ja 
Simplify this expression and subsitute in the value for vb: 
V 0 = 5( Vb - Va) = 20 - 5va 
Set this expression for v0 to the positive power supply value: 
20 - 5va = 10 V SO Va = 2 V 
Now set the expression for v0 to the negative power supply value: 
20 - 5va = -10 V SO Va= 6 V 
Therefore 2 :::; Va :::; 6 V 
5-4 CHAPTER 5. The Operational Amplifier 
[b) Begin as before by substituting the appropriate values into Eq. 5.22: 
8(60) 
Vo = 10(l2) 'Vb - 5va = 4vb - 5va 
Now substitute the value for vb: 
V 0 = 4(4) - 5va = 16 - 5va 
Set this expression for v0 to the positive power supply value: 
16 - 5va = 10 V SO Va = 1.2 V 
Now set the expression for v0 to the negative power supply value: 
16 - 5va = -10 V SO Va = 5.2 V 
Therefore 1.2 :S Va :::; 5.2 V 
AP 5.6 [a] Replace the op amp with the more realistic model of the op amp from Fig. 
5.15: 
100kQ 
5kQ 5kQ 
F- • + + v Vo ··1 500k0 n 
-.;. 
Write the node voltage equation at the left hand node: 
Vn Vn - Vg Vn - V0 
500,000 + 5000. + 100,000 = 0 
Multiply both sides by 500,000 and simplify: 
Vn + 100vn - 100v9 + 5vn - 5vo = 0 so 21.2vn - v0 = 20v9 
Write the node voltage equation at the right hand node: 
V 0 - 300,000( -vn) Vo - Vn _ O 
5000 + 100,000 -
Multiply through by 100,000 and simplify: 
20v0 + 6 X 106vn + V 0 - Vn = 0 SO 6 X 106Vn + 21v0 = 0 
Use Cramer's method to solve for v0 : 
21.2 -1 
.6. = = 6,000,445.2 
6 x 106 21 
21.2 20v9 6 N 0 = = -120 X 10 Vg 
6 x 106 0 
Vo 
so - = -19.9985 
Vg 
[b] Use Cramer's method again to solve for vn: 
20v9 -1 
Ni= = 420v9 
0 21 
f\11 -5 
Vn = ~ = 6.9995 X 10 Vg 
Vg = 1 V, Vn = 69.995 µ V 
Problems 5--5 
[c] The resistance seen at the input to the op amp is the ratio of the input 
voltage to the input current, so calculate the input current as a function 
of the input voltage: 
. Vg - Vn Vg - 6.9995 x 10-5vg 
i - - --------
g - 5000 - 5000 
Solve for the ratio of v9 to i9 to get the input resistance: 
Vg 5000 _ r. 
Rg = ig = 1 - 6.9995 x 10-5 - 5000.35 ~L 
[d] This is a simple inverting amplifier configuration, so the voltage gain is 
the ratio of the feedback resistance to the input resistance: 
Vp = _ 100,000 = _20 
Vg 5000 
Since this is now an ideal op amp, the voltage difference between the two 
input terminals is zero; since Vp = 0, Vn = 0 
Since there is no current into the inputs of an ideal op amp, the 
resistance seen by the input voltage source is the input resistance: 
R9 = 50000 
5~6 CHAPTER 5. The Operational Amplifier 
Problems 
P 5.1 [a] The five terminals of the op amp are identified as follows: 
p 5.2 
inverting 
input 
[b] The input resistance of an ideal op amp is infinite, which constrains the 
value of the input currents to 0. Thus, in = 0 A. 
[c) The open-loop voltage gain of an ideal op amp is infinite, which constrains 
the difference between the voltage at the two input terminals to 0. Thus, 
(vp - vn) = 0. 
[d] Write a node voltage equation at Vn: 
Vn -1 Vn - V0 
2000 + 8000 = 0 
But Vp = 0 and Vn = Vp = 0. Thus, 
-1 Vo 
2000 - 8000 = 0 SO Vo = -4 V 
Vb-V Vb-V a o_o 
_2_0_ + 160 - ' therefore v0 = 9vb - 8va 
(a] Va = 1.5 V, vb= 0 V, V0 = -12 V 
(b) Va = 3.0 V, Vb= 0 V, V 0 = -18 V (sat) 
[c] Va= 1.0 V, Vb= 2 V, Vo= 10 V 
[d) 1Ja = 4.0 V, Vb= 2 V, V0 =-14V 
[e) Va = 6.0 V, Vb= 8 V, V 0 = 18 V (sat) 
[f] If vb= 4.5 V, V0 = 40.5 - 8Va = ±18 
. . . 2.8125 ::::: Va ::::: 7.3125 V 
P 5.3 V 0 = (1)(9) = 9 V; ii5kn = 1
9
5 = 0.6 mA; 
i6kn = ~ = l.5mA; igkn = ~ = lmA 
9 
. ·. i 0 = -0.6 - 1.5 - 1 = -3.l mA 
p 5.4 
p 5.5 
p 5.6 
Problems 5-7 
Since the current into the inverting input terminal of an ideal op-amp is zero, 
the voltage across the 3.3 Mn resistor is (3.3 x 106)(2.5 x 10-6) or 8.25 V. 
Therefore the voltmeter reads 8.25 V. 
. 120 -6 
[a] Za = 6 X 10 = 20p,A 
Va= -20 X 103ia = -400mV 
Va Va Va - Vo 
[b] 60,000 + 20,000 + 240,000 = 0 
. •• V 0 = 17va = -6.8 V 
[c] ia = 20 p,A 
[d] io = 8~~~0 + ;~O-O~~ = 111.67 µ A 
' ' 
3000 (3) - 1 v - v 
Vp = 3000 + 6000 - - n 
Vn - 5 Vn - V 0 _ O 
10,000 + 5000 -
(1 - 5) + 2(1 - v0 ) = 0 
V 0 = -1.0 V 
Vo 1 6 
iL = 4000 = - 4000 = -250 X 10-
iL = -250µA 
P 5. 7 [a] First, note that Vn = Vp = 3 V 
Let v01 equal the voltage output of the op-amp. Then 
3 - Vg 3 - V 0 1 _ Q 
5000 + 15,000 - ' ·. · Vol = 12 - 3vg 
Also note that v0 1 - 3 = v0 , :. V 0 = 9- 3v9 
15 Vo(V) 
10 
5 
Vg(V) 
0 +-~-.,-~-,-~--=-......,,...~~~--..-~~,...-~-.----_, 
-5 6 7 8 
-1 0 
-15 
-20 
5-8 CHAPTER 5. The Operational Amplifier 
[b) Yes, the circuit designer is correct! 
P 5.8 [a] The circuit shown is a non-inverting amplifier. 
[b) We assume the op amp to be ideal, so Vn = Vp = 750 mV. Write a KCL 
equation at Vn: 
0.75 0.75 - V 0 
20,000 + 100,000 = 0 
Solving, 
V 0 = 4.5 V. 
P 5.9 [a] The gain of an inverting amplifier is the ratio of the feedback resistor to 
the input resistor. If the gain of the inverting amplifier is to be 4, the 
feedback resistor must be 4 times as large as the input resistor. There are 
many possible designs that use only 10 kO resistors. We present two 
here. Use a single 10 kn resistor as the input resistor, and use four 10 kO 
resistors in series as the feedback resistor to give a total of 40 kn. 
lOkO lOkQ 
10k0 
10k0 lOkQ 
o--'W'v-------1 
+ 
Alternately, Use a single 10 kO resistor as the feedback resistor and use 
four 10 kO resistors in parallel as the input resistor to give a total of 2.5 
kn. 
lOkO 
10k0 
10k0 
lOkO 
+ lOkQ 0 + 
Vs Vo 
~ • 
[b) To amplify a 2.5 V signal without saturating the op amp, the power 
supply voltages must be greater than or equal to the product of the input 
voltage and the amplifier gain. Thus, the power supplies should have a 
magnitude of (2.5)(4) = 10 V. 
P 5.10 [a] Let VL:;.. be the voltage from the potentiometer contact to ground. Then 
0-V 0-Vf::;.. 
4000
9 + 20,000 = 0 
Vf::;.. = -5( 40 X 10-3) = -0.2 V 
Problems 5-9 
Vt:,. 6 Vt:,. - Vo O -+ Vt:.+ 1 = 
0: -0: 
( 1 1 ) Vo Vt:,. -+6+-- =--
0: 1-a 1-o: 
.'. v0 =-0.2 [1+6(1-a)+ (l:o:)l 
When a= 0.25, v0 = -0.2(1+4.5 + 3) = -1.7 V 
When o: = 0.8, v0 = -0.2(1 + 1.2 + 0.25) = -0.49 V 
:. -1.7 V:Sv0 :S-0.49 V 
[bJ -0.2 [1+6(1 - o:) + (l: a)] = -12 
o: + 60:(1 - a)+ (1 - o:) = 60a 
o: + 60: - 6a2 + 1 - o: = 600: 
:. 60:2 + 540: - 1=0 so 0: '.::::'. 0.0185 
P 5.11 (a] Replace the combination of v9 , 3.2 kO, and the 4.8 kf2 resistors with its 
Thevenin equivalent. 
3.2k0 
~---Vl/'v-----· l 1.92k0 
0.3 vi~ 0.5 vi~ 
l l 
Then = -[30 + a-170] (0.30) 
Vo 1.92 
At saturation v0 = -10 V; therefore 
-(30+ul70)(o3)=-10 or u=0.2 
1.92 . ' 
Thus for 0 ::::; u < 0.20 the operational amplifier will not saturate. 
5-10 CHAPTER 5. The Operational Amplifier 
[b] When rJ = 0.12, Vo= -(30 + 20.4) (0.30) = -7.875 V 
1.92 
Al Vo + Vo +. O 
SO 180 50.4 'lo = 
. V 0 V 0 7.875 7.875 
'lo = - 180 - 50.4 = 180 + 50.4 mA = 200 µA 
P 5.12 [a] This circuit is an example of an inverting summing amplifier. 
180 180 180 
[b] Vo= - 20 Va - 30 Vb - 60 Ve= -4.5 - 9 + 7.5 = -6 V 
[c] V 0 = -13.5 - 3ve = ±9 
Ve= -7.5 V when v0 = 9 V; 
Ve = 1.5 V when v0 = -9 V 
-7.5 V:::;;ve::s;l.5 V 
P 5.13 [a] Write a KCL equation at the inverting input to the op amp: 
~-~ ~-~ ~-~ ~ ~-~ 
55,000 + 66,000 + 220,000 + 550,000 + 330,000 = 0 
v0 = 14.lvd - 6va - 5vb - l.5vc = 141 - 96 - 60 + 9 = -6 V 
{b] v0 = 141 - 96 - 5vb + 9 = 54 - 5ib 
54 - 5vb = -12 SO Vb = 13.2 V 
54 - 5Vb = 12 SO Vb = 8.4 V 
: . 8.4 V :::;; Vb :::;; 13.2 V 
~-~ ~-~ ~-~ ~ ~-~-0 
p 5·14 [a] 55,000 + 66,000 + 220,000 + 550,000 + Rf -
660 660 . 
12vd - 12va + lOvd - lOvb + 3vd - 3ve + l.2vd + Rr vd = Rr Vo (Rr m kO) 
660 660 
26.2vd + -vd - 12va - lOvb - 3ve = -R Vo 
Rr r 
262 + 6600 - 192 - 120 + 18 = 660 v 
Rr Rr o 
6600 - 32Rr = 660v0 so 32Rr = 6600 - 660v0 
v0 = ±12 but .Rr > 0 
32Rr = 6600 - 660( -12) so Rr = 453. 75 kO 
Problems 5-11 
[b) V 0 = -12 V; A KCL equation at the output node gives 
-12 -12 - 10 
io + 33,000 + 453,750 = O 
:. i0 = 412.12µ A 
P 5.15 We want the following expression for the output voltage: 
This is an inverting summing amplifier, so each input voltage is amplified by a 
gain that is the ratio of the feedback resistance to the resistance in the 
forward path for the input voltage: 
Solve for each input resistance valueto yield the desired gain: 
Ra= 60,000/3 = 20k0 Re= 60,000/4 = 15k0 
Rb = 60,000/5 = 12 kn Rd = 60,000/2 = 30 kn 
The final circuit is shown here: 
Va 
20k0 
60k0 
vb 
12k0 
Ve 
15k0 + 
vct 
30k0 v 
0 
.;,-
p 5.16 V 0 = - [ 4~0 (0.2) + 5~~0 (0.15) + 20~00 (0.4)] 
-6 = -0.1x10-3Rr; Rr = 60kn; o :::; Rr :::; 60 kn 
P 5.17 [a} This circuit is an example of the non-inverting amplifier. 
[b] Use voltage division to calculate vp: 
75,000 3vs 
Vp = 25,000 + 75,000 Vs = 4 
Write a KCL equation at Vn = Vp = 3v8 / 4: 
3vs/4 3vs/4 - V 0 _ O 
8000 + 32,000 -
5-12 CHAPTER 5. The Operational Amplifier 
p 5.18 
Solving, 
V 0 = 12v8 /4 + 3vs/4 = 3.75vs 
[c} 3.75v8 = 15 so Vs= 4 V 
3.75v8 = -9 SO Vs= -2.4 V 
Thus, -2.4 V s; Vs s; 4 V. 
45 
[a} Vp = Vn = 75 Vg = 0.6vg 
0.6vg 0.6vg - V0 _ O· 
15+ 48 - ' 
V 0 = 2.52vg = 2.52(3), V0 = 7.56 V 
[b] V 0 = 2.52vg = ±10 
Vg = ±3.97 V, -3.97 s; Vg s; 3.97 V 
[ 0.6vg 0.6vg - v0 c) 15+ Rr =0 
( 0.6Rr ) ~ +0.6 Vg =Vo= ±10 
. ·. 3Rr + 45 = ±150; 3Rr = 150 - 45; Rr = 35k!1 
P 5.19 [a} This circuit is an example of a non-inverting summing amplifier. 
[b) Write a KCL equation at Vp and solve for Vp in terms of vs: 
Vp-Vs Vp+4 
12,000 + 48,000 = 0 
so 
Now write a KCL equation at Vn and solve for v0 : 
Vn Vn - Vo 
10,000 + 40,000 = 0 so 
Since we assume the op amp is ideal, Vn = Vp. Thus, 
V0 = 5( 4vs/5 - 4/5) = 4vs - 4 
[cJ 4v8 - 4 = 10 so V 8 = 3.5 V 
4v5 -4=-10 so Vs=-1.5 V 
Thus, -1.5 V s; Vs s; 3.5 V. 
P 5.20 [a) This is a non-inverting summing amplifier. 
Problems 5-13 
[ ] 
Vp - Va Vp - Vb 
b 80 x 103 + 64 x 103 = 0 
Vn Vn - Vo 
18,000 + 72,000 = 0 
:. v 0 = 5vn = 5vp = (20/9)va + (25/9)vb = 4.44 V 
[c] Vp = Vn = ~ = 0.889 V 
. Va - Vp 
ia = 80 x 103 = -4.86 µA 
Vb-Vp 
ib = = 4.86µA 
64 x 103 
[d) (20/9) for Va 
(25/9) for Vb 
p 5.21 [a] Vp - Va+ Vp - Vb+ Vp - Ve+ Vp = 0 
Ra Rb Re Rg 
. RbRcRg RaRcRg RaRbRg 
. . Vp = D Va + D 1-'b + D Ve 
where D = RbRcR9 + RaRcR9 + RaRbR9 + RaRbRc 
( 1 1 ) V
0 
Vn Rs+ Rr = Rr 
.'. V 0 = ( 1 + ~:) Vn = kvn 
where k = ( 1 + i) 
or 
5~14 CHAPTER 5. The Operational Amplifier 
:. ~: = ~ = 2 ~ = ~ = 0.75 ~: = ~ = 1.5 
Since Ra= lkO Rb= 2k0 Re= l.5k0 
:. D = [(2)(1.5)(3) + (1)(1.5)(3) + (1)(2)(3) + (1)(2)(1.5)] x 109 = 22.5 x 109 
k(3)(2)(1.5) x 109 = 6 
22.5 x 109 
k = 135 x 109 = 15 
9 x 109 
Rr 
15=1+-
R~ 
Rr = 14 
Rs 
Rr = (14)(15,000) = 210k0 
[b] V 0 = 6(0.5) + 3(2.5) + 4(1) = 14 V 
14.5 
Vn = Vp = - = 0.967 V 
15 
0.5 - 0.967 = -466.67 A 
ia = 1000 µ 
. = 2.5 - 0.967 = 766.67 A 
Zb 2000 µ 
1 - 0.967 = 22.22 A 
ie = 1500 µ 
. = 0·967 = 322.22 A 
Zg 3000 µ 
. = Vn = 0.967 = 64.44 A 
Z.s 15,000 15,000 µ 
p 5.22 [a] Vp - Va + Vp - Vb + Vp - Ve = 0 
Ra Rb Re 
RbRe Ra Re RaRb 
• •• 'Up = --yy-va + -rs-vb + --yy-vc 
where D = Rbllc + RaRc + RaRb 
Vn Vn-Vo O --+ = 20,000 Rr 
( 20~0+1) Vn =Vo 
Re 
Let 20,000 + 1 = k 
kRaRe = l 
D 
kRaRb = 2 
D 
: . Rb = 4Ra = 4 kO 
Re= 2Ra = 2k0 
D = (4)(2) + (1)(2) + (1)(4) = 14 x 106 
k = _}_!}__ = (4)(14) x 106 = 7 
RbRe (4)(2) x 106 
Rr 
20000+ 1 = 7' 
' 
Rr = 120k0 
(b] V 0 = 4(0.75) + 1.0 + 2(1.5) = 7 V 
Vn = v0 /7 = 1 V = Vp 
. Va - Vp 0. 75 - 1 = _ 250 A 
'ta= 1000 - 1000 µ 
Vb-V 1-1 
ib = 4000P = 4000 = OµA 
. Ve - Vp 1.5 - 1 
ie = 2000 - 2000 = 250 µA 
Problems 5-15 
P 5.23 [a] Assume Va is acting alone. Replacing vb with a short circuit yields Vp = 0, 
therefore Vn = 0 and we have 
0-v 0- v' 
R a + R o + in = O, 
a b 
Therefore 
5~ 16 CHAPTER 5. The Operational Amplifier 
Assume Vb is acting alone. Replace Va with a short circuit. Now 
vbR<l 
Vp = Vn = Re + Rd 
II 
Vn Vn - Vo + . - 0 -+ 'tn- ' 
Ra Rb 
therefore 
Rb Rb Rb 
Eq. (5.22) reduces to V0 = Ra Vb - Ra Va= Ra (vb - Va)· 
P 5.24 Use voltage division to find vp: 
25,000 
Vp = 25,000 + 15,000 (S) = 5 V 
Write a KCL equation at Vn and solve it for v0 : 
Vn - Va Vn - Vo 0 ---+ = 
10,000 R1 
so ( R1 ) R1 10,000 + l Vn - 10,000 Va = Vo 
Since the op amp is ideal, Vn = Vp = 5V, so 
( Rt ) R1 
Vo = 2000 + 5 - 10,000 Va 
To satisfy the equation, 
( Ri +5) = 15 
2000 
Thus, R1 = 20 kn. 
and R1 =2 
10,000 
P 5.25 [a] lOkO 375k0 v -~-'\1\11,.--, ..----'\/\II,---., 
a 
30k0 
+ 
v 60k0 p 
Vp Vp - Ve Vp - Vd 
60,000 + 20,000 + 30,000 = 0 
Vn - Va Vn - Vb Vn - Vo _ O 
10,000 + 15,000 + 375,000 -
. ·. V0 - 63.5vn - 37.5va - 25v1 
+ 
v 
0 
- 63.5[(1/2)ve + (1/3)vd] - 37.5va - 25vb 
Problems 5-17 
- 63.5(0.l + 0.2) - 37.5(0.4) - 25(0.8) = -15.95 v 
[b] V 0 = 63.5(0.3) - 37.5(0.4) - 25vb 
±20 = 4.05 - 25v1 
Vb= -0.638 V and Vb= 0.962 V 
-638 ~vb~ 962mV 
~(Ra+ Rb) Rb 33(100) 
P 5.26 [a] V 0 = Ra(Re +Rd) Vb - Ra Va= 20(80) (0.90) - 4(0.45) 
V 0 = 1.8563 - 1.8 = 56.25 ID V 
(b) Vn = Vp = (0.9~~(33) = 371.25mV 
. = (450 - 371.25)10-3 = 3 937r; A 
'la 20 X 103 . Q /J, 
v 450 x 10-3 
R = ~ = = 114 3kf2 
a ia 3.9375 X 10-6 . 
[c] Rinb =Re+ Rd= 80kf2 
5-18 
p 5.27 
CHAPTER 5. The Operational Amplifier 
Rf 
4000 = 7·5; 
Rf 
4000+ 1 = 8"5 
Rf= 30k0 
8.5 (Ra~ Rb)= 7.5 
8.5Rb = 7.5Rb + 7.5Ra 
8.5Ra = 170 kn 
Ra= 20kn 
Rb= 170 - 20 = 150kn 
Rr = 40000 
Problems 5-19 
By hypothesis: Rb/ Ra = 5; Re + Ra = 600 kn; Ra(Ra +Rb) = 2 
Ra(Rc +Ra) 
: . Ra (Ra + 5Ra) = 2 
Ra 600,000 
so Ra = 200 kn; Re = 400 kn 
Also, when v0 = 0 we have 
. ( Ra) . . Vn 1 + Rb = Vai 
. Va - (5/6)va 1 Va 
'ta= = --· 
Ra 6Ra' 
Ra= 3kn; Rb= 15kn 
v0 - (1 + Rr) av - Rf v Rg g Ri g aRg p 5.30 [a] Vp = ( )'Vg aRg+ Rg-aRg 
Vn = Vp =avg - (avg - vg)4 +avg 
Vn - Vg Vn - V0 O + = 
Ri Rf 
- [(a - 1)4 + a]vg 
Rr 
(vn - vg) Ri + Vn - Vo= 0 - (5a - 4)vg 
- ( 5a - 4) ( 2) = lOa - 8 
a Vo a Vo a Vo 
0.0 -8V 0.4 -4 v 0.8 ov 
0.1 -7V 0.5 -3V 0.9 lV 
0.2 -6V 0.6 -2V 1.0 2V 
0.3 -5V 0.7 -1 v 
2 
Ot---t---+~-+--i-~+--+---+-=....,..:::::'--+-__, 
~ -2 
§! -4 
-6 
-8 
alpha 
5~20 CHAPTER 5. The Operational Amplifier 
p 5.31 
p 5.32 
p 5.33 
[b] Rearranging the equation for v0 from (a) gives 
Vo = ( ~~ + 1) Vga + - ( ~~) Vg 
Therefore, 
slope = ( ~: + 1) v9 ; intercept = - ( ~:) v9 
[c] Using the equations from (b), 
-6 = (~~ + 1) v 9 ; 4 = - (~~) Vg 
Solving, 
v - -2 V· g - ' Rt =2 Ri 
20,00 
Vp = lQO 000 (-4) = -0.8 V = Vn 
' 
-0.8 + 4 -0.8 - V 0 
2000 + Rr = O 
• •. V 0 = 0.0016Rt - 0.8 
V 0 = 20 V; Rf = 13 kf! 
v0 = -10 V; Rr = -5.75kf! 
But Rr 2: 0, :. Rr = 13k0 
[ ] A = 95(100 + 5) + 100(5 + 95) = 19 975 
a dm 2(5)(5 + 95) · 
[b] A = (5)(95) - 5(100) = -0.05 
cm (5)(5 + 95) 
[c] CMRR = 119·975 1 = 399.5 
0.05 
A _ (33)(47) - (47)Rx 
cm - 33(47 + R,,J 
A _ 47(33 + 47) + 47(47 +Rx) 
dm - 2(33)(47 +Rx) 
Adm Rx+ 127 
-
Acm 2(33 - Rx) 
Rx+ 127 
2(33 _Rx) = ±750 for the limits on the value of Rx 
If we use +750 Rx= 32.89kn 
If we use - 750 Rx = 33.11 kn 
32.89 kn :::; Rx s 33.11 kn 
P 5.34 [a] 
R 
+ 
v a 
Va Va - Vn Va - Vo O 
Ra + R + R = 
V [_!_ + !] _ Vn = V 0 
aRa R RR 
Va(2+ ~)-vn=V0 
Va - V 0 = -2v9 (1) 
2va +Va (:a) - Va - Vg = V 0 
.". Va ( 1 + ~) - V 0 = Vg (2) 
R 
Problems 5-21 
R 
5--22 CHAPTER 5. The Operational Amplifier 
Now combining equations ( 1) and ( 2) yields 
R 
-Va-= -3v9 
Ra 
Ra 
or Va= 3v9 R 
Hence 
. Va 3v9 'ta=-=-
Ra R 
(b] At saturation V0 =±Vee 
Q.E.D. 
.". Va=± Vee - 2v9 (3) 
and 
.". Va ( 1 + ~) = ± Vee+ Vg ( 4) 
Dividing Eq ( 4) by Eq (3) gives 
l+j!:_= ±Vcc+Vg 
Ra ±Vcc-2Vg 
!!:_ = ±Vee+ Vg _ l = 3v9 
Ra ±Vee - 2v9 ±Vee - 2vg 
or Ra = ( ± V cc - 2vg) R 
3v9 
Q.E.D. 
P 5.35 [a] Assume the op-amp is operating within its linear range, then 
iL = 2_ = 2mA 
1.5 
For RL = 2.5 kn V0 = (2.5 + 1.5)(2) = 8 V 
Now since v0 < 9 V our assumption of linear operation is correct, 
therefore 
iL = 2mA 
[b] 9 = 2(1.5 + RL); RL = 3 kO 
[c] As long as the op-amp is operating in its linear region iL is independent of 
RL. From (b) we found the op-amp is operating in its linear region as 
long as RL :s; 3 kn. Therefore when RL = 6.5 kn the op-amp is saturated. 
We can estimate the value of iL by assuming ip =in« iL. Then 
iL = 9/(1.5 + 6.5) = l.125mA. To justify neglecting thecurrent into the 
op-amp assume the drop across the 47 kn resistor is negligible, and the 
input resistance to the op-amp is at least 500 kO. Then 
ip =in= (3 -1.5)/500 x 10-3 = 3µA. But 3µA « l.125mA, hence our 
assumption is reasonable. 
[d] 
I .ti 
0.5 
0 
0 2 
~ !'--... 
4 
RL(kohm) 
Problems 5-23 
~ t----._ 
P 5.36 [a] Let v01 = output voltage of the amplifier on the left. Let v02 = output 
voltage of the amplifier on the right. Then 
p 5.37 
-90 
Vol = 15(-0.5) = 3 V; 
-120 
V 0 2 = 3{)(0.4) = -1.6 V 
. Vo2 - Vol 
ia = lOOO = -4.6 mA 
[h] ia = 0 when v 0 1 = Vo2 so from (a) v 0 2 = 3 V 
Thus 
-120(v ) = 3 
30 L 
90 
VL = - 120 = - 750 m V 
(320 x 10-3) 2 
[a] P16kn = (16 x 103) = 6.4µW 
[h] v16 kn = (~:) (320) = 80mV 
(80 x 10-3) 2 
P16kn = (16 x 103) = 0.4µW 
[c] Pa = 6.4 = 16 
Pb 0.4 
[d] Yes, the operational amplifier serves several useful purposes: 
• First, it enables the source to control 16 times as much power 
delivered to the load resistor. When a small amount of power controls 
a larger amount of power, we refer to it as power amplification. 
• Second, it allows the full source voltage to appear across the load 
resistor, no matter what the source resistance. This is the voltage 
follower function of the operational amplifier. 
• Third, it allows the load resistor voltage (and thus its current) to be 
set without drawing any current from the input voltage source. This 
is the current amplification function of the circuit. 
5-24 CHAPTER 5. The Operational Amplifier 
P 5.38 [a] Vp = V 8 , 
(R1 + R2) ( R2) Therefore V 0 = R Vs = 1 + R Va 
1 1 
[b] V0 =Va 
[c] Because v0 =Vs, thus the output voltage follows the signal voltage. 
p 5.39 
14.IV----i 
vol 
OmA 
J,, 3kQ 
-20V i 4 
14. IV 
J,, 13kQ 
10v---_, i3 4.7kQ 
vo2 
1.5kQ J,, 
OmA 
lOV il 
. 14.7- 10 1 A Z1= = ID 
4700 
V 0 2 = 10 + (1500)(0.004) = 16 V 
. 14.7- 16 O A 
z3 = 13,000 = - .lm 
V0 1 = 14.7 + 3000(0.0009) = 17.4 V 
P 5.40 vp = 5·6 v9 = 0. 7v9 = 2.8 sin(57r /3)t V 8.0 
p 5.41 
Vn Vn - Vo 
2000 + 18,000 = 0 
: . v0 = 28 sin(51r /3)t V 
V0 = 0 t ~ 0 
At saturation 
28 sin ( 5;) t = ±14; 
. 51r 
sm 3 t= ±0.5 
51r 1r 5?r 71r ll?r 
3t=5, 6' - ' etc. 6' 6 
t = O.lOs, 0.50s, 0.70s, etc. 
V0 (V) 
30 /-, -/ \. 
20 I ' I \ \ I \ 
10 
0 
1.5 
-10 
I 
-20 ' I ..... / 
-30 
It follows directly from the circuit that v0 = -5v9 
From the plot of v9 we have 'V -0 g - ' t<O 
Vg - 4t 0 ~ t ~ 0.5 
Vg - 4-4t 0.5 ~ t ~ 1.5 
Vg 4t-8 1.5 ~ t ~ 2.5 
Vg - 12-4t 2.5 ~ t ~ 3.5 
Vg - 4t-16 3.5 < t < 4.5, etc. 
\ 
Problems 5--25 
2 
' I ' / _.. 
5-26 CHAPTER 5. The Operational Amplifier 
Therefore 
Vo - -20t 0 < t < 0.5 
Vo - 20t- 20 0.5 ~ t ~ 1.5 
Vo - 40 - 20t 1.5 ~ t ~ 2.5 
Vo - 20t ~ 60 2.5 ~ t ~ 3.5 
Vo 80- 20t 3.5 ~ t ~ 4.5, etc. 
These expressions for v0 are valid as long as the op amp is not saturated. 
Since the peak values of v0 are ±6, the output is clipped at ±6. The plot is 
shown below. 
... 
I \ 
I ' 
5 
... 
I \ 
I \ 
t (s) 
0 -1r-~~~~--~~~~--'r~~~~--#'--~~~---'~~~~---. 
-5 
-10 
\ I 
\I 
\ I 
\I 
\ I ,, 
P 5.42 [a] Replace the op amp with the model from Fig. 5.15: 
100k0 
5k0 5k0 
v F500kQ + 
1 
Write two node voltage equations, one at the left node, the other at the 
right node: 
Vn - Vg Vn - V 0 Vn 
5000 + 100,000 + 500,000 = 0 
V 0 + 3 X 105Vn V 0 - Vn V 0 
5000 + 100,000 + 500 = 0 
p 5.43 
Simplify and place in standard form: 
l06vn - 5v0 = lOOvg 
(6 X 106 - l)vn + 221v0 = 0 
Let Vg = 1 V and solve the two simultaneous equations: 
V 0 = -19.9844 V; Vn = 736.lµV 
Thus the voltage gain is v0 /vg = -19.9844. 
[b] From the solution in part (a), Vn = 736.1 µV. 
} . Vg - Vn Vg - 736.1 X 10-6v9 
[c ig = 5000 = 5000 
Vg 5000 
Rg = ig = 1- 736.1x10-6 = 5003.68 0 
Problems 5-27 
[d] For an ideal op amp, the voltage gain is the ratio between the feedback 
resistor and the input resistor: 
V 0 = _ 100,000 = _20 
Vg 5000 
For an ideal op amp, the difference between the voltages at the input 
terminals is zero, and the input resistance of the op amp is infinite. 
Therefore, 
Vn = Vp = 0 V; Rg = 50000 
[ 1 Vn Vn - Vg Vn - V 0 
a 8000 + 600,000 + 240,000 = O or 78.5vn - 2.5v0 = Vg 
V 0 V 0 - Vn V 0 - 100,000( Vp - Vn) _ O 
30,000 + 240,000 + 5000 -
57v0 - Vn - 48 X l05(vp - Vn) = 0 
(vn - vg)(160) 
Vp = Vg + 600 = (11/15)v9 + (4/15)vn 
57v0 - Vn - 48 X 105[(11/15)v9 - (11/15)vn] = 0 
57v0 + 3,520,000vn = 3,520,000v9 
78.5 -2.5 
~= 
3.52 x 106 57 
= 8,804,474.5 
78.5 
3.52 x 106 3.52 x 106v9 
6 = 272.8 X 10 Vg 
N 
V 0 = ,; = 30.98v9 ; Vo= 30.98 
Vg 
5-28 CHAPTER 5. The Operational Amplifier 
-2.5 
[b) Ni= 
3.52 x 106 vg 57 
= 8,800,057vg 
Vn = 999.5mV 
Vp = (11/15)(1000) + (4/15)(999.5) = 999.87mV 
[c] Vp - Vn = 367.94µ V 
[d) i = (1000 - 999.87)10-3 = 836.22 A 
g 160 x 103 p 
[e] Vg Vg - V 0 _ O . S + 240 - , smce Vn = Vp = Vy 
P 5.44 (a] 
31 Vo= 31 V 0 = Vg, 
Vg 
Vn = 'Vp = 1 V; Vp-Vn = 0 V; 
60kQ 
D.lS~SOOkQ + 
l 
240kQ 
Vn - 0.15 Vn Vn - VTh 
60,000 + 500,000 + 240,000 = 0 
VTh + 5 X l04Vn VTh - Vn _ O 
750 + 240,000 -
Solving, VTh = -0.6 V 
ig =0 A 
7500 
• 
+ 
-
.;,. 
Short-circuit current calculation: 
60kQ 
0.15~500kQ + 
l 
240kQ 
Vn Vn - 0.15 Vn - 0 _ O 
500,000 + 60,000 + 240,000 -
.'. Vn = 0.1095 V 
. Vn 5 X 104 
'lsc = 240,000 - 750 Vn = - 7·3 A 
VTh 
RTh = -.- = 82.2m0 
'lsc 
~Q 
-o .6v6 
I L-------· 
Problems 5--·29 
750Q 
i 
SC 
b 
[b] The output resistance of the inverting amplifier is the same as the 
Thevenin resistance, i.e., 
[c] 
Ro = RTh = 82.2 mO 
~Q 
0.6V cb 
+ 
v 
n 
a 
b 
( 150 ) V 0 = 150_082 ( -0.6) = -0.5997 V 
150Q 
5~30 CHAPTER 5. The Operational Amplifi.er 
240kQ 
.--~~Vl/v.~~~~--. 
750Q 
O,OOOvn 
Vn - 0.15 Vn Vn + 0.5997 _ O 
60,000 + 500,000 + 240,000 -
.". Vn = 54.7445 µV 
0.15 - 54.7445 x 10-6 = 2 4991 A 
Zg = 60 000 . /J, 
' 
R9 = 0·.15 = 60,021.6 n 
Zg 
-240 
p 5.45 [a) VTh = -w(0.15) = -0.6 V 
RTh = 0, since op-amp is ideal 
o.J 
I.____ ____ • b 
(b) Ro=RTh=OO 
[c) R9 = 60k0 since Vn = 0 
P 5.46 [a] 
lSOkQ 
Vn - Vg Vn - V 0 _ Q 
25,000 + 150,000 -
V 0 ( 1 + ~) = -6vg 
-6A 
V 0 = (7 +A) Vg 
[b] = -6(150)(0.5) = -2.866 v 
Vo (7 + 150) 
[c] v0 = -6(0.5) = -3 V 
[d] -2.94 = -6(0.5)A 
7+A 
:. A=343 
P 5.47 From Eq. 5.57, 
Vref ( 1 1 1 ) 110 
R + 6.R = Vn R +AR + R - AR + R1 - R1 
Substituting Eq. 5.59 for Vp = Vn: 
Vref Vref (~+~+if) V 0 
R + 6.R = (R - 6.R) (-1-· + - 1- + _1_) R.r 
R+D..R R-D..R Rt 
+ 
v 
0 
Problems 5-31 
5-32 CHAPTER 5. The Operational Amplifi.er 
Rearranging, 
V 0 ( 1 1 ) 
Rt = Vref R - AR - R + AR 
Thus, 
( 2AR ) Vo = Vref R2 _ AR2 Rj 
P 5.48 [a] Use Eq. 5.61 to solve for R1 ; note that since we are using 13 strain gages, 
A= 0.01: 
Rt= voR _ (5)(120) = 2kn 
2L\.vref (2)(0.01)(15) 
[b] Now solve for A given v0 = 50 mV: 
P 5.49 [a] 
A = . v0 R = (0.05)(120) = lOO X 10_6 
2RJVref 2(2000)(15) 
The change in strain gage resistance that corresponds to a 50 m V change 
in output voltage is thus 
AR= AR= (100 x 10-6)(120) = 12 mr! 
v, l R+.6.R I R w. p ;; 'IM n+ R R 
Let R1 = R+AR 
+ 
v 
0 
Problems 5-33 
v0 _ [R + 2Rt] [ RR.rvm l Vin 
Rt - RRt [RR1 + R1R1 + R1R] - R 
V0 [ R+ 2Rt 1 l 
. • Rt= RR1 +RtR1 +R1R - R Vin 
[R2 + 2RRt - Ri(R +Rt) - RR1]R1 
Vo= R{Ri(R +Rt)+ RRt] Vin 
Now substitute R1 = R +AR and get 
-AR(R +Rt )Rtvin 
v - ------'-----=---
0 - R[(R + AR)(R + R1) + RR1] 
If AR<< R 
(R + R1 )R1(-AR)vin v ~--~~---
0 R2(R+2R1) 
[h] v ~ 47 x 104 (48 x 104)(-95)15 ~ -3.384 v 
0 108(95 x 1Q4) 
-95(48 x 104)(47 x 104)15 
[c] Vo= 104[(1.0095)104(48 X 104) + 47 X 108] = - 3·368 V 
p 5.50 [ ] ,..., (R + R1 )R1(-AR)vin a Vo,..., R2(R + 2R.r) 
(R + R1 )(-AR)Rtvin 
v - ------'-----'---
0 - R[(R + AR)(R + R1) + RRt] 
approx value R[(R + AR)(R + R1) + RRt] 
-
true value R2 ( R + 2R1) 
E _ R[(R + AR)(R +Rt)+ RRt] - R2(R + 2Rt) 
rror - R2(R + 2Rt) 
_AR (R+ R1) 
R (R+2R1) 
AR(R+R1) 
.. 3 error = R(R + 2RJ) x 100 
95(48 x 104) x 100 
[h] 3 error= 104(95 x 104) = 0.4835-34 CHAPTER 5. The Operational Amplifier 
.6.R(48 x 104) 
p 5.51 1 = 104(95 x 104) x 100 
.6.R = 9~~0 = 197.91667 n 
. 197.19667 
3 change m R = 104 x 100 ~ 1.983 
P 5.52 [a] It follows directly from the solution to Problem 5.49 that 
[R2 + 2RR1 - R1 (R +Rt) - RR1]R1vin 
v --------------
o- R[R1(R+R1)+RR.r] 
Now R1 = R - .6.R. Substituting into the expression gives 
(R +Rt )Rt(.6.R)vin 
v - ----'----'------
0 - R[(R - .6.R)(R +Rt)+ RRt] 
Now let !::,.R « R and get 
(R + R1 )R1.6.Rv1n 
v "'"'-~----
orv R2(R+2R1) 
[b] It follows directly from the solution to Problem 5.49 that 
approx value R[ ( R - .6.R) ( R + Rt) + RR1] 
-
true value R2(R + 2Rt) 
E _ (R-.6.R)(R+Rt)+RR1-R(R+2
R1) 
rror - R(R + 2R1) 
-t::..R(R+ Rt) 
-
R(R+ 2R1) 
-.6.R(R + R1) 
3 error = R(R + 2Rt) X 100 
[c] R - AR = 9810 n : . .6.R = 10,000 - 9810 = 190 n 
·. v ~ (48 x 104 )(47 x 104)(190)(15) ~ 6 768 v 
• 
0 108 (95 x 104) . 
[d] 3 error = -190(48 x 104)(100) = -0 96o/c 
104 (95 x 104 ) . 0 
------6 
Inductance, Capacitance, and 
Mutual Inductance 
Assessment Problems 
AP 6.1 [a] i9 = 8e-3oot - 8e-1200t A 
v = L dig = -9.6e-300t + 38.4e-12ootv, 
dt 
v(o+) = -9.6 + 38.4 = 28.8 v 
t > o+ 
[b] v = 0 when 38.4e-1200t = 9.6e-300t or t = (ln4)/900 = l.54ms 
[c] p =vi= 384e-1500t - 76.8e-600t - 307.2e-24o0t W 
[d] dp = 0 when e1800t - 12.5e900t + 16 = 0 
dt 
Let x = e900t 
x = 1.44766, 
x = 11.0523, 
and solve the quadratic x 2 - 12.5x + 16 = 0 
ln 1.45 
t = 900 = 411.05 fl,S 
t = ln ll.05 = 2.67ms 
900 
pis maximum at t = 411.05 µs 
[e] Pmax = 384e-1.5(0.41105) - 76.Se-0.6(0.41105) - 307.2e-2.4(0.41105) = 32.72W 
[f] Wis max when i is max, i is max when di/dt is zero. 
When di/dt = 0, v = 0, therefore t = l.54ms. 
[g] irnax = 8[e-0·3(LM) - e-1.2(1.54)] = 3. 78 A 
Wrnax = (1/2)(4 X 10-3)(3.78)2 = 28.6 rnJ 
6-1 
6-2 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
dv d 
AP 6.2 [a] i = C dt = 24 x 10-6 dt [ e-15,oo0t sin 30,000t] 
= (0. 72 cos 30,000t - 0.36 sin 30,000t]e-15•000t A, 
[h] i (:0 ms) = -31.66 mA, v (:0 ms) = 20.505 V, 
p =vi= -649.23mW 
[c] w = (~) Cv2 = 126.13 µJ 
AP6.3 [a] v= (~)fo~idx+v(o-) 
i(O+) = 0.72 A 
= 0_6 } 10_6 h~ 3cos50,000xdx = 100sin50,000tV 
[b] p(t) = vi = [300 cos 50,000t] sin 50,000t 
= 150 sin 100,000t W, P(max) = 150 W 
[c] W(max) = (~) Cv!ax = 0.30(100)2 = 3000 µJ = 3 mJ 
AP 6.4 [a] Leq = 60(240) = 48 mH 
300 
[b] i(o+) = 3 + -5 = -2A 
[c] i = 1~5 r ( -0.03e-5x) dx - 2 = 0.125e-5t - 2.125 A 6 lo+ 
[d] i 1 = 50 r (-0.03e-5x) dx + 3 = 0.1e-5t + 2.9 A 
3 lo+ 
25 lt i2 = - ( -0.03e-5x) dx - 5 = 0.025e-5t - 5.025 A 
6 o+ 
AP 6.5 V1 = 0.5 x 106 rt 240 x 10-6e-10x dx - 10 = -12e-lOt + 2 v .lo+ 
V2 = 0.125 X 106 rt 240 X lQ-6e-lOx dx - 5 = -3e-l0t - 2 V .lo+ 
v1(oo) = 2V, 
AP 6.6 [a] Summing the voltages around mesh 1 yields 
dii d(i2 + ig) . . . . 4dt + 8 dt + 20(zi - z2) + 5(zi + z9 ) = O 
or 
di1 . di2 . ( . dig) 4-d + 25zi + 8-d - 20z2 = - 5z9 + 8-· t t dt 
Summing the voltages around mesh 2 yields 
16d(i2 + ig) + 8dii + 20(i2 - ii)+ 780i2 = 0 
dt dt 
or 
ilii . ili2 . ~g 8dt - 20zi + l6dt + 800z2 = -16dt 
[b] From the solutions given in part (b) 
Problems 6-3 
i 1 (0) = -0.4 - 11.6 + 12 = O; i2(0) = -0.01 - 0.99 + 1 = 0 
These values agree with zero initial energy in the circuit. At infinity, 
ii(oo) = -0.4A; i 2 (oo) = -0.0lA 
When t = oo the circuit reduces to 
,,__~V\~.~~---~-'\1111,.~--i 
SQ 20Q -
1.96A + 7.8V 
+ 
( ) ( 7.8 7.8 ) A ii 00 = - 20 + 780 = -0.4 ; 
From the solutions for ii and i2 we have 
~: = 46.40e-4t - 60e-5t 
~: = 3.96e-4t - 5e-5t 
Also, dig = 7.84e-4t 
dt 
Thus 
4 dii = l85.60e-4t - 240e-5t 
dt 
7800 
i2( oo) = -;~~ = -0.0lA 
6-4 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
25i1 = -10 - 290e-4t + 300e-5t 
8~: = 31.68e-4t - 40e-5t 
20i2 = -0.20 - l9.8oe-4t + 2oe-5t 
5i9 = 9.8 - 9.8e-4t 
8~: = 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 7 -[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 7 -(9.8 + 52.92e-4t) 
-9.8 + oe-5t + (237.08 - 290)e-4t ? -9.8 - 52.92e-4t 
-9.8 - 52.92e-4t = -9.8 - 52.92e-4t (OK) 
Also, 
8~: = 371.20e-4t - 480e-5t 
20i1 = -8 - 232e-4t + 240e-5t 
d' 
16~ = 63.36e-4t - 80e-5t 
dt 
800i2 = -8 - 792e-4t + 800e-5t 
di 
16 d: = 125.44e-4t 
Test: 
371.20e-4t - 480e-5t + 8 + 232e-4t - 240e-5t + 63.36e-4t - 80e-5t 
-8 - 792e-4t + 800e-5t 7 -125.44e-4t 
(8 - 8) + (800 - 480 - 240 - 80)e-5t 
+(371.20 + 232 + 63.36 - 792)e-4t 7 -125.44e-4t 
(800 - 800)e-5t + (666.56 - 792)e-4t 7 -125.44e-4t 
-125.44e-4t = -125.44e-4t (OK) 
Problems 
P 6.1 [a] z 0 
z - 16tA 
i 0.8 - 16tA 
?, - 0 
t<O 
0 :'.St :'.S 25ms 
25 :'.S t :'.S 50 ms 
50ms < t 
di 
[b] v = L dt = 375 x 10-3 (16) = 6V 0 < t S 25rns 
v = 375 x 10-3 (-16) = -6V 25 St S 50rns 
v - 0 t<O 
v - 6V 0 < t < 25ms 
v - -6V 25 < t < 50ms 
v 0 50rns < t 
p= vz 
p - 0 t<O 
p - 96tW 0 < t < 25rns 
p 96t-4.8W 25 < t < 50rns 
p - 0 50rns < t 
1 . 
w = -Lz2 
2 
w - 0 t<O 
w 48t2 J 
w 48t2 - 4.8t + 0.12 J 
0 < t < 25ms 
25 < t < 50rns 
w - 0 50rns < t 
P 6.2 [a] 0 :'.S t :'.S 1 ms : 
i rt 106 rt 
i = L lo Vs dx + i(O) = 300 lo 6 x 10-3 dx + 0 
= 20x 1: = 20tA 
Problems 6-5 
6-6 CHAPTER 6. Inductance, Capacitance, and Mutual. Inductance 
lins :s; t :s; 2 ms : 
106 lt i = - (12 x 10-3 -6x)dx + 20 x 10-3 300 10-3 
: . i = 40t - 10,000t2 - 10 x 10-3 A 
2ms:s;t:s;oo: 
106 lt i = - (0) dx + 30 x 10-3 = 30mA 300. 2xlQ-3 
[b] 30 
25 
~20 
4.: .s 15 
·- 10 
:5 
0 
,,. '"""' 
I/ 
,) 
/ 
7 
7 
0 0.5 1.:5 2 2.:5 3 
t (ms) 
p 6.3 0 st< 00 
103 lnt e-3x It iL = - 30 x 10-3e-3x dx + 2.5 = 7.5-3- + 2.5 4 0 - 0 
= 5 - 2.5e-3t A, 
> .s 
__J 
> 
30 
25 
20 
15 
10 
5 
4 
5 
0 
0 
g 3 
:! 2 
0.2 0.4 0.6 0.8 
t(s) 
0o 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 
t(s) 
p 6.4 di [a] v = L dt 
di 
dt = 20[e-5t - 5te-5t] = 20e-5t(l - 5t) 
v = (100 x 10-6)(20)e-5t(l - 5t) 
= 2e-5t(l - 5t) mV, t > 0 
[b] p =vi= 0.04te-10t(1 - 5t) 
p(lOOms) = 0.04(0.l)e-1 (1 - 0.5) = 735.76 µW 
[c] absorbing 
[d] i(lOOms) = 20(0.l)e-0·5 = 2e-0·5 
1 1 
w = 2Li2 = 2 (100 x 10-
6)(2e-0·5) 2 = 73.58 µJ 
Problems 6-7 
[e] The energy is a maximum where the current is a maximum: 
diL - = 0 when 1 - 5t = 0 or t = 0.2 s dt 
imax = 20(0.2)e-1 = 4e-1 A 
1 
Wmax = 2(100 X 10-6)(4e-1) 2 = 108.27 µJ 
P 6.5 [a] 0 :s; t :s; 2 s : 
v = -25t 
i = 2- rt -25xdx + 0 = -10x2 Jt 
2.5 Jo 2 o 
i = -5t2 A 
2s~t~6s: 
v = -100+ 25t 
i(2) = -20A 
:. i = 1 ft -2 (25x - 100) dx - 20 .5 2 
f t ft 10 
2 
x dx - 40 
2 
dx - 20 
- 5(t2 - 4) - 40(t - 2) - 20 
- 5t2 - 40t + 40 A 
6-8 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
6ssts10s: 
v = 200- 25t 
i(6) = 5(36) - 240 + 40 = -20 A 
i = 1 ft -2 (200 - 25x) dx - 20 .5 6 
ft ;·t 80 dx - 10 x dx - 20 
6 . 6 
80(t - 6) -10(t2 - 36)/2 - 20 = 80t - 5t2 - 320A 
10 s s t s 12 s : 
v = 25t- 300 
i(lO) = 800 - 500 - 320 = -20 A 
v=O 
'l = J_ft (25x - 300) dx - 20 
2.5 10 
_ 10/t x d;i: - 120/t dx - 20 
10 10 
- 5(t2 - 100) - 120(t - 10) - 20 
5t2 - 120t + 680 A 
i(12) = 5(12)2 - 120(12) + 680 = -40 A 
i = - Odx-40 1 ft 
2.5 12 
= -40A 
[b] For 0 :St :S 2s, v = -25tV; i = -5t2 A 
v = 0 when t = 0 so i = 0 A 
t > 12s: 
For 2 :St :S 6s, v = -100 + 25tV; i = 5t2 - 40t + 40A 
v=O when t=4s so i=5(4)2 -40(4)+40=-40A 
For 6 :St :S 10s, v = 200- 25tV; i = -5t2 + 80t - 320A 
v = 0 when t = 8 s so i = -5(8)2 + 80(8) - 320 = 0 A 
For 10 :St :S 12s, v = 25t - 300V; i = 5t2 - 120t + 680A 
v = 0 when t = 12 s so i = 5(12)2 - 120(12) + 680 = -40 A 
For t ~ 12 s, v = O; i = -40A 
p 6.6 
Problems 6-9 
[c] 
12 14 
-10 
-20 
-30 
-40 
-50 
di 
[a] VL = L dt = (56cos140t + 92sin140t]e-20t m V 
dvL 20t . ·. - = [11,760cos 140t - 9680sin 140t]e- mV /s dt 
dvL 
- =0 when dt 
11,760 
tan 140t = 9680 = 1.21 
.'. t = 6.30ms 
Also 140t = 0.8821+7r etc. 
Because of the decaying exponential v L will be maximum thefirst time 
the derivative is zero. 
[b] vL(max) = (56cos0.8821+92sin0.8821)e-0·12602 = 93.997mV 
Note: When 
0.8821+1I' t= . 
' 140 ' VL = -60mV 
lOOO!t . P 6.7 [a] i = 5o 
0
250sm1000xdx-5 
- 5000 j~ sin 1 OOOx dx - 5 
5000 [-cos lOOOx It - 5 
1000 0 
5(1 - cos lOOOt) - 5 
i - -5 cos lOOOt A 
6-10 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
(b] p - vi = ( 250 sin lOOOt) ( -5 cos lOOOt) 
-1250 sin 1 OOOt cos 1 OOOt 
p -625 sin 2000t W 
w - 1 L·2 - 't 
2 
1 
- 2(50 x 10-3)25cos2 1000t 
- 625 cos2 lOOOt mJ 
w [312.5 + 312.5 cos 2000t) mJ. 
300 
200 
100 
2 0 
:> 0.51!: 
t (ms) 
1.51!: 2n 
-100 
-200 
-300 
6 
~ I (ms) 
·- 1n 
-4 
-6 
800 
600 
400 
[ 200 0 t (ms) 
Cl -20 2n 
-400 
-600 
-800 
700 
600 
500 
'S' 400 ,§. 
~ 300 
200 
100 
0 
0 0.5n 1n 1.5 
[c] Absorbing power: 
0.for ~ t ~ 7rmS 
l.57r ~ t ~ 27rms 
Delivering power: 
0 ~ t ~ 0.57rms 
11" ~ t ~ 1.51!" ms 
P 6.8 [a] i(O) = A1 + A2 = 1 
di = -2000A1e-2000t - 8000A2e-soo0t dt 
v = -30A1e-2000t - 120A2e-soo0t V 
v(O) = -30A1 - 120A2 = 60 
Solving, Ai = 2 and A2 = -1 
Thus, 
. _ (2 -2000t -sooot) A z1 - e - e t 2: 0 
v = -6oe-2000t + 12oe-soo0t V, t 2: 0 
[b] p =vi= 300e-10,ooot - 12oe-4oo0t - 120e-16•000t 
p = 0 when 300e6000t - 120e12·000t - 120 = 0 
Let x = e6000t; then 300x - 120x2 - 120 = 0 
Thus x 2 - 2.5x + 1 = 0 so x = 0.5 and x = 2 
Problems 6-11 
If x = e6000t = 0.5, twill be negative. Hence, the solution fort> 0 must 
be x = 2: 
e6000t = 2 so 6000t = ln 2 
ln2 
Thus, t = 6000 = 115.52 µs 
P 6.9 [a] From Problem 6.8 we have 
i = Aie-2ooot + A2e-soo0t A 
v = -30A1e-2000t - 120A2e-sooot V 
i(O) = Ai + A2 = 1 
v(O) = -30A1 - 120A2 = -300 
Solving, 
Thus, 
i = _ 2e-2ooot + 3e-sooot A t 2: 0 
v = 60e-2000t - 360e-soo0ty t 2: 0 
6-12 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
[b] i = 0 when 3e-sooot = 2e-2ooot 
:. e6000t = 1.5 so t = (ln 1.5)/6000 = 67.58 µ,s 
Thus, 
i > 0 for 0 < t ::=; 67.58 µs and i < 0 for 67.58 µs < t < oo 
v = 0 when 60e-2000t = 3600e-sooot 
:. t = (ln6)/6000 = 298.63µs 
Thus, 
v < 0 for 0 ::=; t ::=; 298.63 µ,s and v > 0 for 298.63 µs ::::; t < oo 
Therefore, 
p < 0 for 0 ::=; t::::; 67.58µs and 298.63 µs ::=; t < oo 
(inductor delivers energy) 
p > 0 for 67.58 µs ::=; t < 298.63 µs (inductor stores energy) 
[c] p = vi = 900e-10,ooot - 120e-4000t - 1080e-16,000t W 
/
ti 
... Wstored = pdx + w(O) 
t2 
[ lt2 lt2 lt2] Wstored = 10-3 -90e-lO,OOOx +30e-4000x +67.5e-16,000x + 7.5 X 10-3 
ti ti ti 
- 67.5e-16·000t1 + 7.5 mJ 
where t 1 = 67.58 µs and t 2 = 298.63 µs 
Wstored = 5.11 + 7.5 = 12.61 mJ 
t1 00 
Wextracted = J P dt + J P dt 
0 t2 
= 1:1 [900e-10,000x - 120e-4000x - 1080e-16,000x] dx 
+/oo [900e-10,000x - 120e-4000x - 1080e-16,000x] dx 
tz 
= 10-3 ( _ 90e-10,ooox 1:1 +3oe-4ooox c +67.5e-16,ooox 1:1 ) 
-10-3 (9oe-10,ooox loo +3oe-4000x loo +67.5e-16,000x loo) 
~ ~ ~ 
Problems 6-13 
= 90e-10,000t2 _ 30e-4000t2 _ 67_5e-16,000t2 + 30e-4000t1 
+67.5e-16,000ti - 90e-lO,OOOti - 7.5 mJ 
where ti = 67.58 µs and t2 = 298.63 µs 
Wextra.cted = -12.61 mJ 
Thus, the energy stored equals the energy extracted. 
P 6.10 i = (B1cos5t + B 2 sin 5t)e-t 
i(O) = B 1 = 25A 
di 
dt = (B1cos5t + B2 sin 5t)(-e-t) + e-t(-5B1sin5t + 5B2 cos 5t) 
v = 2ddi = [(10B2 - 2B1) cos 5t - (10B1 + 2B2) sin 5t]e-t t 
v(O) = 100 = lOB2 - 2B1 = lOB2 - 50 . ·. B2 = 150/10 =; 15 A 
Thus, 
i = (25 cos 5t + 15 sin 5t)e-t A, t;:::: 0 
v = (100cos5t- 280sin5t)e-tV, t;:::: 0 
i(0.5) = -6.70A; v(0.5) = -150.23 V 
p(0.5) = (-6.70)(-150.23) = 1007.00W absorbing 
P 6.11 For 0 ::St < 1.2 s: 
iL = - 14 x 10-3 dx + 0 = 0.7 x 10-3t 1 /t 
20 0 
iL(l.2 s) = (0.7 x 10-3)(1.2) = 0.84mA 
Rm= (25)(1000) = 25kD 
Vm(l.2 s) = (0.84 X 10-3)(25 X 103) = 21 V 
6-14 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
P 6.12 p =vi= 40t[e-10t - 10te-20t - e-20t] 
p 6.13 
W = fo00 pdx = fo00 40x[e-10x - 10xe-20x - e-20x] dx = 0.2 J 
This is energy stored in the inductor at t = oo. 
[a] v(20 µs) - 12.5 x 109(20 x 10-6 ) 2 = 5 v (end of first interval) 
v(20 µs) - 106 (20 x 10-6 ) - (12.5)(400) x 10-3 - 10 
5 V (start of second interval) 
v(40µs) - 106 ( 40 x 10-6 ) - (12.5)(1600) x 10-3 - 10 
- 10 V (end of second interval) 
[b] p(lOp.s) = 62.5 x 1012(10-5 ) 3 = 62.5mW, v(lOµs) = l.25V, 
i(lOµs) = 50mA, p(lOµs) =vi= (1.25)(50m) = 62.5mW (checks) 
p(30µs) = 437.50mW, v(30µs) = 8.75V, i(30µs) = 0.05A 
p(30µs) =vi= (8.75)(0.05) = 62.5mW (checks) 
[c] w(lO µs) = 15.625 x 1012 (10 x 10-6) 4 = 0.15625 µJ 
w = 0.5Cv2 = 0.5(0.2 x 10-6)(1.25)2 = 0.15625 µJ 
w(30 µs) = 7.65625 µJ 
w(30 µs) = 0.5(0.2 x 10-6)(8.75)2 = 7.65625 µJ 
P 6.14 ic = C(dv/dt) 
O<t<l: ic = 0.5 x 10-6 (120)t2 = 60t2 11A 
l<t<2: ic = 0.5 x 10-6 (120)(2 - t)2(-1) = -60(2 - t) 2 µA 
60 
ic(µA) 
40 
20 
0 
0.5 
-20 
-40 
-60 
t(s) 
2 
p 6.15 [a) 0:::; t:::; 100/J,S 
C = 0.2µF ~ = 5 x 106 
j ·t v = 5 x 106 
0 
- 0.04dx + 40 
v = -200 x 103t + 40V 0 :::; t :::; 100 p,s 
v(lOO µs) = -20 + 40 = 20 V 
[b] 100 µs :::; t :::; 300 /LS 
v = 5 x 106 r 0.08 dx + 20 = 4 x 105t - 40 + 20 
l1oox10-6 
v = 4 x 105t - 20V 100 :::; t :::; 300 µs 
v(300 µs) = 4 x 105 (300 x 10-6 ) - 20 = 100 V 
[c] 30Qµs:::; t < oo 
v = 5 x 106 ft 0 dx + 100 = 100 
}3oox10-6 
[d] 
v = lOOV, 
-100 
vM 
120 
300 /LS :::; t < 00 
0 100 
dv 
P 6.16 [a] i = C dt = 0, t < 0 
200 300 
dv 
[b] i = C dt = 5e-1000t[cos 3000t + 13 sin 3000tJ mA, t ~ 0 
[c] no, v(o-) = -30 V 
v(o+) = 10 - 40 = -30 v 
[d] yes, i(o-) = 0 A 
i(O+) = 5 mA 
Problems 6-15 
t (us) 
400 500 
6-16 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
[e] v(oo) = lOV 
w = ~Cv2 = ~(0.5 x 10-6)(10)2 = 25 µJ 
50 x 10-3 
P 6.17 [a] i = 10 x 10_6 t = 5 x 103t O~t~lOµs 
i = 50 x 10-3 lO~t~30p,s 
1ox10-6 ·30x10-6 
q = J 5 x 103t dt + j 50 x 10-3 dt 
o 1ox10-6 
t2 ,10x10-6 
- 5 x 103 - +50 x 10-3 (20 x 10-6) 
2 0 
5 x 103 (~)(100 x 10-12) + 1000 x 10-3 x 10-6 
- l.25µC 
[b] i = 200 x 10-3 -5 x 10-3t 30µs ~ t ~ 50µs 
/
50x10-6 
q = 1.25 x 10-6 + [200 x 10-3 - 5 x 103t] dt 
. 30x10-6 
t2 50xl0-6 
- 1.25 x 10-6 + 200 x 10-3 (20 x 10-6) - 5 x 103-1 
2 30x10-e 
[2500 - 900] - 1.25 x 10-6 + 4000 x 10-9 - 5 x 103 2 10-12 
= l.25µC 
Since q = vC, : . v = 1.25/0.25 = 5 V. 
[c] i = -300 x 10-3 + 5 x 10-3t 50µs ~ t < 60µs 
/
60x10-6 
q = 1.25 x 10-6 + [-300 x 10-3 + 5 x 103t] dt 
50xl0-6 
1.25 x 10-6 - 300 x 10-3 (10 x 10-6) 
+5 x 103 [3600 ; 2500] 10-12 
11),c 
1J = 1 x 10-6 = 4 v 
0.25 x 10-6 
c 1 
w = 2 v2 = 2(0.25) x 10-6 (16) = 2 µJ 
P 6.18 [a] v 
250xrn-6 
5 x 106 j 0 100 x 10-3e-1000t dt - 60.6 
e-lOOOt 
1
250x10-6 
500 x 103 -1000 0 -60.6 
= 500(1 - e-0·25) - 60.6 = 50 V 
w ~Cv2 = ~(0.2)(10-6 )(50) 2 =25011J 
[b] v = 500 - 60.6 = 439.40 v 
w = ~(0.2) x 10-6 (439.40) 2 = 19.31mJ = 19,307.24µJ 
P 6.19 [a] w(O) = ~C[v(0)] 2 = ~(0.40) x 10-6 (25)2 = 125 µJ 
[b] v = (Ait + A2)e-i5oot 
v(O) = A2 = 25V 
dv 
- -1500e-i5oot(Ait + A2) + e-i500t(Ai) 
dt 
(-1500Ait - 1500A2 + Ai)e-i5o0t 
dv 
dt (0) = Ai - 1500A2 
i = Cdv i(O) = Cdv(O) 
dt' dt 
dv(O) i(O) 90 x 10-3 
dt - c - 0.40 x 10-6 = 225 x 103 
225 x 103 = Ai - 1500(25) 
Thus, Ai = 2.25 x 105 + 3. 75 x 104 = 262,500 V 
s 
[c] v = (262,500t + 25)e-isoot 
i = c~~ = 0.40 x 10-6 :t (262,500t + 25)e-i5oot 
i = ! [(0.105t + 10 x 10-6)e-i5oot] 
Problems 6-17 
- (0.105t + 10 x 10-6)(-1500)e-1500t + e-isoot(0.105) 
(-157.5t - 15 x 10-3 + 0.105)e-i5oot 
- (0.09 - 157.5t)e-i5o0t A, t ;::::: O 
- (90 - 157,500t)e-1500t mA, t ;::::: 0 
6--18 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
P 6.20 1011(15 + 25) = 8 H 
81112 = 4.8H 
4411(1.2+4.8) = 5.28 H 
21114 = 3.36H 
5.28 + 3.36 = 8.64 H 
P 6.21 61114 = 4.2H 
15.8 + 4.2 = 20 H 
201160 = 15H 
15 + 5 = 20H 
2011so = 16H 
16+ 24 = 40H 
401110 =SH 
Lab = 12 + 8 = 20 H 
P 6.22 (a] I ~i + 
7 .SH~-- v 
~ 
i(t) - -- - l80oe-20xdx -12 1 ;·t 
7.5. 0 
e-20x It 
- 240-~ -12 
-20 0 
- -12(e-20t - 1) - 12 
i(t) -12e-20t A 
[h] 
[c] 
ii (t) 1 /t - - - -l800e-20x dx + 4 
10. 0 
e-20x It 
- 180-- +4 
-20 0 
- -9(e-20t - 1) + 4 
ii(t) - -9e-20t + 13 A 
i2(t) 1 /t - - - - l80oe-20x dx - 16 
30 0 
e-20x It 
- 60-- -16 
-20 0 
-3(e-20t - 1) - 16 
i2(t) - -3e-20t - 13 A 
[d] p =vi= (-l80oe-20t)(-12e-20t) = 21,60oe-40tw 
w = j~pdt = j~ 21,60oe-40t dt 
e-40t i= 
21,600--0 
-4 0 
= 540J 
[e] w = ~(10)(16) + ~(30)(256) = 3920J 
[f] Wtrapped = Winitial - Wdelivered = 3920 - 540 = 3380 J 
[g] Wtrapped = ~(10)(13)2 + ~(30)(13)2 = 3380 J checks 
P 6.23 [a] i 0 (0) = i1 (0) + i2(0) = 5 A 
[h] 
k + 
L.q=lOH~:Oe-"'v 
1 t [ -25xlt 
i 0 - - 10/ 01250e-
25x dx + 5 = -125 e_25 0 
+ 5 
- 5(e-25t - 1) + 5 = 5e-25t A, t ~ 0 
Problems 6-19 
6-20 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
[c] 
Va -
+ 
3.6H 
3.6!£(5e-25t) = -45oe-25t V 
dt 
+ 
-25t 
vb=1250e v 
Ve Va + vb = -450e-25t + 1250e-25t 
- 80oe-25ty 
1 ft Z1 - - - 80oe-25x dx + 10 
8 0 
- 4e-25t - 4 + 10 
i1 - 4e-25t + 6A t~O 
1 ft [d] i2 - -- 800e-25x dx - 5 
32 0 
- e-25t - 1 - 5 
Z2 e-25t - 6A ' t~O 
[e] w(O) = ~(8)(100) + ~(32)(25) + ~(3.6)(25) = 845 J 
1 
[f] Wdel = 2(10)(25) = 125 J 
[g] Wtrapped = 845 - 125 = 720 J 
P 6.24 Vb = 1250e-25t V 
. 5 -25t A Z0 = e 
p = 6250e~5ot W 
w = 6250e-5ox dx = 6250-e- = 125(1- e-50t) W int -50x It 
0 -50 0 
Wtotal = 125 J 
80%Wtotal = 100 J 
Thus, 
125 - 125e-50t = 100; e50t = 5-, t = 32.19ms 
Problems 6-21 
1 1 7 
P 6.25 21 + 28 = 84 :. Ceq = l2µF 
-10 v - 5 v = -15 v 
24+ 12 = 36µF 
- I -
10v
1
~ 21µF 
~ ~ 1+5Vr12µ.F 
:v i28µF I 
1 1 2 
36 + 36 = 36 : . Ceq = 18 µF 
-15 v + 2 v = -13 v 
12 + 20 = 32µF 
18+14 = 32µF 
1 1 2 
32 + 32 = 32 Ceq = l61LF 
8V-13V = -5V 
6-22 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
p 6.26 
1 1 1 5 
C1 = 6.4nF C1 = 8 + 32 = 32; 
l~nF 
C2 = 5.6 + 6.4 = 12nF -15~ 
40~J 12nF 
1 1 1 10 
C3 = 7.2nF C3 = 18 + 12 = 72 ; 
I + C4 = 12.8 + 7.2 = 20 nF 20nFI 25V 
ae 8ftF 
-30~ 
1 1 1 1 1 25VJ 20nF 
C5 = 8 + 20 + 40 = 5; C5 = 5nF + ~v te 
40nF 
Equivalent capacitance is 5 nF with an initial voltage drop of -10 V. 
P 6.27 [a] -2500t 
900e µ 
+ 
109 ft 
Vo - -- 900 X 10-6e-25oox dx + 30 
12. 0 
e-2500x It 
-75,000 -2500 0 +30 
30e-25ooty 
' t 2': 0 
(b] V1 
109 e-2500x It 
- 20 (900 x 10-6) -2500 0 +45 
- 18e-2500t + 27V, t 2': 0 
109 e-2500x It 
[c] V2 - --(900 x 10-6) -15 
30 -2500 0 
- 12e-25oot - 27 V t 2': 0 ' 
[d] p _ vi = (30e-25oot)(goo x 10-6)e-25oot 
- 27 x 10-3e-5oo0t 
w = j~ 27 x 10-3e-5ooot dt 
e-5000t loo - 27x10-3 __ 
-5000 0 
-5.4 x 10-6 (0- 1) = 5.4µJ 
[e] w - ~(20 x 10-9)(45)2 + ~(30 x 10-9)(15)2 
20.25 x 10-6 + 3.375 x 10-6 
- 23.62511,J 
[f] Wtrapped = Winitial - Wdelivered = 23.625 - 5.4 = 18.225 µJ 
[g] Wtrapped - ~(20 X 10-9)(27)2 + ~(30 X 10-9 )(27)2 
- (10 + 15) (27)2 x 10-9 
- 18.225 p,J 
CHECK: 18.225 + 5.4 = 23.625 µJ 
P 6.28 C1 =1+1.5 = 2.5nF 
1 1 1 1 1 
-=-+-+-=-
C2 2.5 12.5 50 2 
:. C2 = 2nF 
vd(O) + Va(O) - Vc(O) = 40 + 15 + 45 = 100 V 
[a] 
+ Black 
vb Box 
109 /t 
vb - -- 50 x 10-6e-25ox dx + 100 
2 . 0 
e-250x It 
- -25,000 -250 0 +100 
100(e-250t - 1) + 100 
- 10oe-250t V 
Problems 6-23 
6-24 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
[b] Va 
109 ft -- 50 X 10-6e-25ox dx + 15 
12.5 0 
[c] 
[d] 
e-250x It 
- -4000 -250 0 +15 
- 16(e-250t - 1) + 15 
- 16e-25ot - 1 V 
Ve 109 It 50 X 10-6e-250x dx - 45 
50. 0 
e-250x It 
- 1000-- -45 
-250 0 
- -4(e-250t - 1) - 45 
- -4e-250t - 41 V 
Vd - - 109 ft 50 X 10-6e-250x dx + 40 
2.5 0 
e-250x r 
-20,000 -250 0 +40 
- 80(e-250t - 1) + 40 
- 8oe-250t - 40 V 
CHECK: Vb - Vd + Va - Ve 
80e-250t - 40 + l6e-250t - 1 + 4e-250t + 41 
10oe-250t V (checks) 
[e] z1 - -10-9~ [8oe-250t - 40] 
dt 
-10-9(-20,oooe-250t) 
- 2oe-2sot µA 
[f] z2 - -1.5 x 10-9 ! [8oe-250t - 40] 
= -1.5 x 10-9(-20,00oe-250t) 
= 3oe-2sot µA 
CHECK: i1 + i2 = 50e~25ot µA = ib 
Problems 6-25 
P 6.29 (a] w(O) - [~(2.5)( 40)2 + ~(12.5)(15)2 + ~(50)(45)2] x 10-9 
- 54,031.25 nJ 
[b] Va( 00) - -lV 
Vc(oo) - -41 v 
va(oo) - -40V 
w(oo) - [~(2.5)(40)2 + !(12.5)(1)2 + !(50)(41)2] x 10-9 
- 44,031.25 nJ 
[c] w = !00 (10oe-250t)(5oe-250t) x 10-6 dt = 10,000 nJ .lo 
CHECK: 54,031.25 - 44,031.25 = 10,000 
[ ] 
07 • 10,000 01 d 10 delivered = 54,031.25 x 100 = 18.5110 
[e] w = 5 x 10-31: e-5oox dx 
- 104(1 - e-5oot) nJ 
104(1 - e-5o0t) = 5000; 
ln2 
Thus, t = 500 = 1.39 ms. 
P 6.30 From Figure 6.17(a) we have 
1 [ 1 1 ] Therefore - = - + - + · · · , 
Ceq C1 C2 
P 6.31 From Fig. 6.18(a) 
e-500t = 0.5 
dv dv dv 
i = Ci- + C2- + · · · = [C1 + C2 + · · ·]-dt dt dt 
Therefore Ceq = C1 + C2 + · · -. Because the capacitors are in parallel, the 
initial voltage on every capacitor must be the same. This initial voltage would 
appear on Ceq· 
6-26 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
p 6.32 
p 6.33 
di 
- 20 x 10-3 ___.!!._ 
dt 
- (20 x 10-3)(50x10-3){e-8000t[-6000sin6000t+12,000cos6000t] 
+ ( -8000e-8000t) [cos 6000t + 2 sin 6000t]} 
e-sooot { 4 cos 6000t - 22 sin 6000t} V 
io(O) = 50mA 
vR(O) = 320(50 X 10-3) = 16 V 
v1(0) = 16 + 4 = 20V 
-106/t 
Ve - -- e-BOx sin 60x dx - 300 
20 0 
VL 
Vo 
-
-
-
-
-
-
5e-80t[80sin60t + 60cos60t] + 300 - 300 
400e-80t sin 60t + 300e-sot cos 60t V 
di0 
5di 
5[-8oe-sot sin 60t + 6oe-80t cos 60t] 
-40oe-sot sin 60t + 30oe-s0t cos 60t V 
(30oe-sot cos 60t - 30oe-BOt cos 60t + 400e-80t sin 60t+ 
400e-Bot sin 60t) 
- 80oe-80t sin 60t V 
P 6.34 [a] 5dig + 40di2 + 90i = 0 
dt dt 2 
40di2 90 . _ 5di9 di+ i2 - - di 
[b] i2 = e-t - 5e-2.25t A 
d' 
:: = -e-t + ll.25e-2·25t A/s 
i9 = lOe-t - lOA 
6-28 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
[c] P<lev - v9 i 9 
960 + 92,480e-4t - 94,400e-5t - 92,480e-9t+ 
93,44oe-10tw 
[d) P<lev( 00) = 960 W 
[e] i 1 (oo) = 4A; i 2 (oo) = lA; ig(oo) = 16A; 
P5n = (16 - 4)2(5) = 720W 
P20n = 32(20) = 180 W 
P6on = 12(60) = 60W 
LPabs = 720+180+ 60 = 960W 
.·. LP<lev = LPabs = 960W 
P 6.37 [a] Rearrange by organizing the equations by diif dt, ii, di2/dt, i2 and transfer 
the ig terms to the right hand side of the equations. We get 
di1 . di2 . . dig 
4di + 25i1 - 8di - 20i2 = 5ig - 8di 
8di1 20i l6di2 so· - 16diu - dt - 1 + di + i 2 - di 
[b] From the given solutions we have 
d' 
~ = -320e-5t + 272e-4t 
dt 
diz = 260e-5t - 204e-4t 
dt 
Thus, 
4 dii = -1280e-5t + l088e-4t 
dt 
25i1 = 100 + 160oe-5t - l 700e-4t 
8 diz = 2080e-5t - 1632e-4t 
dt 
20i2 = 20 - 1040e-5t + 1020e - 4t 
5i9 = 80 - 80e-5t 
8 dig = 640e-5t 
dt 
Problems 6-29 
Thus, 
-1280e-5t + l088e-4t + 100 + l60oe-5t - 170oe-4t - 208oe-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) 
8dii = -2560e-5t + 2176e-4t 
dt 
20i1 = 80 + 1280e-5t - 1360e-4t 
16di2 = 4160e-5t - 3264e-4t 
dt 
80i2 = 80 - 4160e-5t + 4080e-4t 
16dig = 1280e-5t 
dt 
2560e-5t - 2176e-4t - 80 ~ 1280e-5t + 1360e-4t + 4160e-5t - 3264e-4t 
+80 - 4160e-5t + 4080e-4t 7 1280e-5t 
( -80 + 80) + (2560 - 1280 + 4160 - 4160)e-5t 
+(1360 - 2176 - 3264 + 4080)e-4t ? 1280e-5t 
0 + 1280e-5t + oe-4t = 1280e-5t (OK) 
P 6.38 [a] Dot terminal 2; with current entering terminal 2, the flux is right-to-left 
coil 1-2. Assign the current into terminal 4; the flux is left-to-right in coil 
3-4. The flux is in the same direction, due to the topology of the core, so 
dot terminal 4. Hence, 2 and 4 or 1 and 3. 
[b] Dot terminal 1; with current entering terminal 1 the flux is down in coil 
1-2. Assign the current into terminal 4; the flux is right-to-left in coil 3-4. 
Therefore the flux is in the same direction, due to the topology of the 
core, so dot terminal 4. Hence, 1 and 4 or 2 and 3. 
[c] Dot terminal 1; with current entering terminal 1 the flux is up in coil 1-2. 
Assign the current into terminal 4; the flux is left-to-right in coil 3-4. 
Therefore the flux is in the same direction, due to the topologyof the 
core, so dot terminal 4. Hence, 1 and 4 or 2 and 3. 
6-30 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
p 6.39 
p 6.40 
p 6.41 
(d) Dot terminal 2; with current entering terminal 2, the flux is down in coil 
1-2. Assign the current into terminal 4; the flux is down in coil 3-4. 
Therefore, the flux is in the same direction, so dot terminal 4. Hence, 2 
and 4 or 1 and 3. 
When the switch is closed, the induced voltage in the coil connected to the 
source is negative at the dotted terminal. Since the de voltmeter kicks 
up-scale, the induced voltage in the coil connected to the voltmeter is positive 
at the lower terminal. Therefore, dot the upper terminal of the coil connected 
to the voltmeter. 
di di di di di 
[a] Vab =Li dt + L2 dt + M dt + M dt = (L1 + L2 + 2M) dt 
It follows that Lab= (L1 + L2 + 2M) 
di di di di di 
[b) Vab = Li dt - M dt + L2 dt - M dt = (L1 + L2 - 2M) dt 
Therefore Lab = (Li + L2 - 2M) 
[a] v _ L d(i1 - i2) Mdi2 ab - 1 dt + di 
O = L d(i2 - ii) _ Mdi2 Md(i1 - i2) L di2 
1 dt dt + dt + 2 dt 
Collecting coefficients of [diif dt] and [di2/dt], the two mesh-current 
equations become 
di1 di2 
Vab =Li di+ (M - Li) di 
and 
di1 di2 
0 = (M -L1) dt + (L1 +L2 - 2M)di 
Solving for [dii/ dt] gives 
di1 Li + L2 - 2M 
dt = L1L2 - NJ2 Vab 
from which we have 
( L 1L 2 - M
2 
) (di1 ) 
Vab = Li + L2 - 2M di 
L _ L1L2 -M2 ab -
Li +L2 -2M 
[b] If the magnetic polarity of coil 2 is reversed, the sign of M reverses, 
therefore 
L1L2 -M2 
Lab = L1 + L2 + 2M 
Problems 6-31 
( 
M 2 ) (0.1) 2 
p 6.42 [a] L2 = k2 L1 = (0.5)2(0.250) = 160 mH 
Ni = (L; = (250 = 1.25 
N2 yy; Vi60 
[ ] Li 0.250 -6 I b P1 = N'f = (l000) 2 = 0.25 x 10 Wb A 
L 2 0.16 _6 
P2 = Ni = (800)2 = 0.25 x 10 Wb/ A 
L1 400 x 10-6 
P 6.43 Pi= N'f = 2502 = 6.4 nWb/A 
-n = L 2 = 900 x 10-6 = 3 6 Wb/A· ~ r2 Ni 5002 . n , M=kvLiL2 =450µH 
M 450 x 10-6 
Pi2 = P21 = NiN2 = (250)(500) = 3.6 nWb/A 
Pu= Pi -P2i = 6.4- 3.6 = 2.8 nWb/A 
P 6.44 [a] k = M = 19·5 = 0.75 
JLiL2 J676 
[b] Mmax = V676 = 26mH 
L1 N'fP1 (N1 ) 2 
[c] L2 = NiP2 = N2 
:. (~:) 2 = ~~ = 4 
Ni= J4 = 2 
N2 
P 6.45 [a] L1 = Nf P1; 288 x 10-
3 
P1 = 106 = 288 nWb/A 
d</Jn = Pu = 0.5; P21 = 2P11 
d</J2i P21 
. ·. 288 x 10-9 = Pu + P2i = 3Pu 
Pu= 96 nWb/A; P 21 =192 nWb/A 
M = kVL1L2 = (1/3)/(0.288)(0.162) = 72mH 
N
2 
= .NI 72 x 10-3 
NiP21 - (1000)(192 x 10-9 ) = 375 turns 
6~32 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
L2 162 x 10-3 
[b] P2 =Ni = (375)2 = 1152 nWb/A 
[c] P11 = 96 nWb/A (see part (a)] 
[d] </>22 = P22 = P2 - P12 = P2 _ 1 
</>12 P12 P12 P12 
P 6.46 [a] 
P21=P21=192 nWb/A; 
1>22 = 1152 - 1 = 5 
¢12 192 
P2 = 1152 nWb/A 
_!_ = (l +Pu) (l + P22) = (l + Pn) (l + P22) 
k 2 P12 P12 P21 P12 
Therefore 
k2 = P12P21 
(P21 + Pn)(P12 + P22) 
Now note that 
¢1 =</Jn+ ¢21 = P11N1i1 + P21N1i1 = N1i1(P11 + P21) 
and similarly 
¢2 = N2i2(P22 + P12) 
It follows that 
(Pu + P21) = N¢1. 
1i1 
and 
(P22 + P12) = ( ~:2 ) 
Therefore 
k2 = ( ¢12/ N2i2) ( ¢2i/ Ni ii) = ¢12¢21 
(¢i/N1i1)(¢2/N2i2) ¢1¢2 
or 
[b] The fractions ( ¢2i/ ef>1) and ( ¢12/ 1>2) are by definition less than 1.0, 
therefore k < 1. 
P 6.47 [a] W = (0.5)L1ii + (0.5)L2i~ + Mi1i2 
M = o.sj(o.025)(0.1) = 40mH 
w = (0.5)(0.025)(10)2 + (0.5)(0.1)(15)2 + (0.04)(10)(15) = 18.5 J 
Problems 6-33 
(b] w = (0.5)(0.025)(-10)2 + (0.5)(0.1)(-15)2 + (0.04)(-10)(-15) = 18.5 J 
[c] W = (0.5)(0.025)(-10)2 + (0.5)(0.1)(15)2 + (0.04)(-10)(15) = 6.5J 
[d] w = (0.5)(0.025)(10)2 + (0.5)(0.1)(-15)2 + (0.04)(10)(-15) = 6.5J 
P 6.48 [a] M = l.Oj(0.025)(0.1) = 50mH, i1 = lOA 
Therefore 50i~ + 500i2 + 1250 = 0, i~ + l0i2 + 25 = 0 
Therefore i2 = - ( ~O) ± J ( 12°) 2 - 25 = -5 ± v'o 
Therefore i2 = -5A 
[h] No, setting W equal to a negative value will make the quantity under the 
square root sign negative. 
P 6.49 When the button is not pressed we have 
dv d 
C2- = C1-(v -v) dt dt s 
or 
dv C1 dvs 
-
dt (C1 + C2) dt 
Assuming C1 = C2 = C 
dv = O.Sdvs 
dt dt 
or 
v = 0.5v8 (t) + v(O) 
When the button is pressed we have 
6-34 CHAPTER 6. Inductance, Capacitance, and Mutual Inductance 
.l l c2 c3 c3 ~ ~ s -
+ 
c1 
v (t) 
C dv C dv G d(v - V 8 ) _ 0 
1 dt + 3 dt + 2 dt -
dv C2 dvs 
dt C1 + C2 + C3 dt 
Assuming C1 = C2 = C3 = C 
dv 1 dv8 
dt 3 dt 
1 
v = 3vs(t) + v(O) 
Therefore interchanging the fixed capacitor and the button has no effect on 
the change in v(t). 
P 6.50 With no finger touching and equal 10 pF capacitors 
10 
v(t) = 20 ( V 8 (t)) + 0 = 0.5v8 (t) 
With a finger touching 
Let Ce = equivalent capacitance of person touching lamp 
C = (lO)(lOO) = 9.091 F 
e 110 p 
Then C + Ce = 10 + 9.091 = 19.091 pF 
10 
v(t) = 29.091 V 8 = 0.344v8 
b.v(t) = (0.5 - 0.344)vs = 0.156V8 
P 6.51 With no finger on the button the circuit is 
when (2C)dv = o 
dt 
With a finger on the button 
v J )Fixed 
when (3C)~~ = 0 
.·. there is no change in the output voltage of this circuit. 
Problems 6-35 
-----7 
Response of First-Order RL and 
RC Circuits 
Assessment Problems 
AP 7.1 [a] The circuit fort< 0 is shown below. Note that the inductor behaves like a 
short circuit, effectively eliminating the 2 n resistor from the circuit. 
30 60 
+ 
120V ~ v 300 
First combine the 30 n and 6 n resistors in parallel: 
30116 = 50 
Use voltage division to find the voltage drop across the parallel resistors: 
5 
v = 5 + 3 (120) = 75 v 
Now find the current using Ohm's law: 
v 75 
i(o-) = - 6 = - 6 = -12.5A 
[b) w(O) = ~L{~(O) = ~(8 x 10-3)(12.5)2 = 625mJ 
(c) To find the time constant, we need to find the equivalent resistance seen 
by the inductor for t > 0. When the switch opens, only the 2 n resistor 
remains connected to the inductor. Thus, 
L 8 x 10-3 
r=-= =4ms 
R 2 
(d] i(t) = i(O-)et/r = ....., 12.5e-t/o.oo4 = -12.5e-250t A, t 2: 0 
[e] i(5ms) = -12.5e-25o(o.oo5) = -12.5e-L25 = -3.58A 
7-1 
7-2 CHAPTER 7. Response of First-Order RL and RC Circuits 
So w (5rns) = ~Li2(5ms) = ~(8) x 10-3 (3.58)2 = 51.3mJ 
w (dis)= 625 - 51.3 = 573.7mJ 
. . (573.7) 3 d1ss1pated = 625 100 = 91.83 
AP 7.2 [a] First, use the circuit for t < 0 to find the initial current in the inductor: 
+ 
6.4A t V0 100 
Using current division, 
i(O-) = 10 (6.4) = 4A 
10+6 
60 
Now use the circuit fort> 0 to find the equivalent resistance seen by the 
inductor, and use this value to find the time constant: 
60 
l 0 .32H 
i 
40 
L 0.32 Req = 4jj(6+ 10) = 3.20, T = - = - = O.ls 
Req 3.2 
Use the initial inductor current and the time constant to find the current 
in the inductor: 
i(t) = i(o-)e-t/r = 4e-t/o.i = 4e-10t A, t 2: 0 
Use current division to find the current in the 10 n resistor: 
io(t) = 4 + 1~ + 6 (-i) = :o ( -4e-10t) = -0.8e-10t A, t 2: o+ 
Finally, use Ohm's law to find the voltage drop across the 10 n resistor: 
Vo(t) = l0i0 = 10(-Q.8e-l0t) = -8e-l0t V, t > Q+ 
[b] The initial energy stored in the inductor is 
w(O) = ~Li2(0-) = ~(0.32)(4)2 = 2.56J 
Find the energy dissipated in the 4 n resistor by integrating the power 
over all time: 
di 
V4n(t) = L dt = 0.32(-10)(4e-l0t) = -12.8e-10tv, t 2: o+ 
Problems 7-3 
v2 
p4n(t) = ~n = 40.96e-20t W, 
W4n(t) = fo00 40.96e-20tdt = 2.048 J 
Find the percentage of the initial energy in the inductor dissipated in the 
4 n resistor: 
. . (2.048) % d1ss1pated = -- 100 = 80% 
2.56 
AP 7.3 [a] The circuit fort< 0 is shown below. Note that the capacitor behaves like 
an open circuit. 
20k0 J,, 
7 .5mA t 80k0 
+ isok 
v(o-) 50k0 
• 
Find the voltage drop across the open circuit by finding the voltage drop 
across the 50 kn resistor. First use current division to find the current 
through the 50 kn resistor: 
80 x 103 -3 
isok = 80 x 1Q3 + 20 x 103 + 50 x 103 (7.5 x 10 ) = 4 mA 
Use Ohm's law to find the voltage drop: 
v(o-) = (50 x 103)i5ok = (50 x 103)(0.004) = 200V 
[b] To find the time constant, we need to find the equivalent resistance seen 
by the capacitor for t > 0. When the switch opens, only the 50 kn 
resistor remains connected to the capacitor. Thus,T =RC= (50 x 103)(0.4 x 10-6 ) = 20ms 
[c] v(t) = v(o-)e-t/T = 200e-t/o.o2 = 2ooe-50t V, t 2: 0 
[d] w(O) = ~Cv2 = ~(0.4 x 10-6)(200)2 = 8 mJ 
1 1 
[e) w(t) = 2cv2 (t) = 2(0.4 x 10-6)(200e-50t)2 = 8e-100t mJ 
The initial energy is 8 mJ, so when 75% is dissipated, 2 mJ remains: 
8 x 10-3e-100t = 2 x 10-3, e100t = 4, t = (ln4)/100 = 13.86ms 
AP 7.4 [a] This circuit is actually two RC circuits in series, and the requested 
voltage, v0 , is the sum of the voltage drops for the two RC circuits. The 
circuit for t < 0 is shown below: 
7 -4 CHAPTER 7. Response of First-Order RL and RC Circuits 
Find the current in the loop and use it to find the initial voltage drops 
across the two RC circuits: 
. 15 
i = 75,000 = 0.2 mA, 
There are two time constants in the circuit, one for each RC subcircuit. 
75 is the time constant for the 5 µF - 20 kO subcircuit, and 71 is the time 
constant for the 1 µF - 40 kO subcircuit: 
75 = (20 x 103)(5 x 10-6 ) = lOOms; 71 = (40 x 103)(1x10-6 ) = 40ms 
Therefore, 
vs(t) = vs(o-)e-t/715 = 4e-t/O.l = 4e-10t V, t 2: O 
v1(t) = v1(0~)e-t/r1 = se-t/0.04 = 8e-25tv, t;:::: 0 
Finally, 
vo(t) = v1(t) + vs(t) = [8e-25t + 4e-10t] V, t ;::=: 0 
(b] Find the value of the voltage at 60 ms for each subcircuit and use the 
voltage to find the energy at 60 ms: 
v1(60ms) = 8e-25<0·06)'"" l.79V, v5 (60ms) = 4e-I0(0.06) '.:::: 2.20V 
w1(60ms) = 1Cvi(60ms) = ~(1 x 10-6)(1.79)2 '"" 1.59 µJ 
ws(60ms) = !cvg{60ms) = l(5 x 10-6)(2.20)2 ::: 12.05µJ 
w(60ms) = 1.59+12.05 = 13.64µ,J 
Find the initial energy from the initial voltage: 
w(O) = w1 (0) + w2(0) = !(1 x 10-6)(8)2 + ~(5 x 10-6)(4)2 = 72 µJ 
Now calculate the energy dissipated at 60 ms and compare it to the 
initial energy: 
Wdiss = w(O) - w(60ms) = 72 - 13.64 = 58.36 µJ 
% dissipated= (58.36 x 10-6 /72 x 10-6)(100) = 81.05 % 
AP 7.5 [a] Use the circuit at t < 0, shown below, to calculate the initial current in 
the inductor: 
20 
Problems 7-5 
i(O-) = 24/2 = 12A = i(o+) 
Note that i(o-) = i(O+) because the current in an inductor is continuous. 
[h] Use the circuit at t = o+, shown below, to calculate the voltage drop 
across the inductor at o+. Note that this is the same as the voltage drop 
across the 10 n resistor, which has current from two sources - 8 A from 
the current source and 12 A from the initial current through the inductor. 
v(o+) = -10(8 + 12) = -200V 
[c] To calculate the time constant we need the equivalent resistance seen by 
the inductor for t > 0. Only the 10 n resistor is connected to the inductor 
fort> 0. Thus, 
T = L/R = (200 x 10-3/10) = 20ms 
[d] To find i(t), we ;need to find the final value of the current in the inductor. 
When the switch has been in position a for a long time, the circuit 
reduces to the one below: 
101.l 
Note that the inductor behaves as a short circuit and all of the current 
from the 8 A source flows through the short circuit. Thus, 
it= -8A 
Now, 
i(t) =it+ [i(O+) - it]e-t/r = -8 + [12 - (-8)]e-t/o.o2 
= -8 + 2oe-50t A, t > o 
[e] To find v(t), use the relationship between voltage and current for an 
inductor: 
d'(t) 
v(t) = L :t = (200 x 10-3)(-50)(2oe-50t) = -2ooe-50t V, t ~ o+ 
7-6 CHAPTER 7. Response of First-Order RL and RC Circuits 
AP 7.6 (a] + BkO + 40k0 
O -25µ.F ...... v 160k0 
0 
From Example 7.6, 
v0 (t) = -60 + 90e-l00t V 
Write a KVL equation at the top node and use it to find the relationship 
between v0 and VA: 
VA - Vo VA VA+ 75 
8000 + 160,000 + 40,000 = 0 
20v A - 20v0 + VA + 4v A + 300 = 0 
2fro A = 20v0 - 300 
VA= 0.8V0 - 12 
Use the above equation for VA in terms of v0 to find the expression for VA: 
vA(t) = 0.8(-60 + 90e-100t) -12 = -60+ 72e-100tv, 
[b] t ~ o+, since there is no requirement that the voltage be continuous in a 
resistor. 
AP 7.7 [a] Use the circuit shown below, fort< 0, to calculate the initial voltage drop 
across the capacitor: 
60kQ 
+ 
lOm~ t 40kQ vJO-) 25kQ 
i = ( 1~05 xx ~0~3) (10 x 10-3) = 3.2 mA 
vc(o-) = (3.2x10-3)(25x103 ) = 80V so vc(o+) = 80V 
Now use the next circuit, valid for 0::; t::; lOms, to calculate vc(t) for 
that interval: 
60kQ 
• '\/'.II,.~~+-~~~~ 
+ 
25kQ lµF Ve 
Problems 7-7 
For 0 :$ t < lOOms: 
T =RC= (25 x 103)(1x10-6) = 25ms 
vc(t) = vc(O-)etfT = 80e-40ty 0 < t :$ lOms 
[b] Calculate the starting capacitor voltage in the interval t ?:: 10 ms, using 
the capacitor voltage from the previous interval: 
Vc(0.01) = 80e-40(0.0l) = 53.63 V 
Now use the next circuit, valid fort?:: 10 ms, to calculate vc(t) for that 
interval: 
+ + 
25kQ~ 53 • 63V 
lµF ...... Ve ~. lOOkQ 
For t > lOms: 
Req = 25 kn11100 kn = 20 kn 
T = ReqC = (20 X 103 )(1 X 10-6 ) = 0.02 S 
Therefore vc(t) = vc(0.01 +)e-(t-O.Ol)/r = 53.63e-5o(t-O.Ol) V, t?:: 0.01 s 
(c] To calculate the energy dissipated in the 25 kn resistor, integrate the 
power absorbed by the resistor over all time. Use the expression 
p = v2 / R to calculate the power absorbed by the resistor. 
rO.Ol [8oe-40t]2 roo [53.63e-50(t-0.01)]2 
w25 k = lo 25,000 dt + lo.01 25,000 dt = 2·91 m.J 
(d] Repeat the process in part (c), but recognize that the voltage across this 
resistor is non-zero only for the second interval: 
- loo [53.63e-50(t-0.01)]2 _ 
w10okn - 00 000 dt - 0.29mJ . 0.01 1 ' 
We can check our answers by calculating the initial energy stored in the 
capacitor. All of this energy must eventually be dissipated by the 25 kn 
resistor and the 100 kn resistor. 
Check: Wstored = (1/2)(1 X 10-6 )(80)2 = 3.2 mJ 
Wdiss = 2.91 + 0.29 = 3.2 mJ 
7-8 CHAPTER 7. Response of First-Order RL and RC Circuits 
AP 7.8 [aJ Prior to switch a closing at t = 0, there are no sources connected to the 
inductor; thus, i(o-) = 0. 
At the instant A is closed, i(O+) = 0. 
For 0 ~ t ~ls, 
20 0 .80 j,, 
30 i 
The equivalent resistance seen by the 10 V source is 2 + (3110.8). The 
current leaving the 10 V source is 
10 
2 + (3110.8) = 3.8 A 
The final current in the inductor, which is equal to the current in the 
0.8 n resistor is 
3 
IF= 3+0.8(3.8) = 3A 
The resistance seen by the inductor is calculated to find the time 
constant: 
[(2113) + o.8]1]3116 = 1 n L 2 T=-=-=2s 
R 1 
Therefore, 
i = iF + [i(O+) - iF]e-tfr = 3 - 3e-0-5t A, 0 < t < 1 s 
For part (b) we need the value of i(t) at t = 1 s: 
i(l) = 3 - 3e-0·5 = 1.18 A 
[b] Fort> 1 s 
BA 
90 
J-. 30 60 
1 
Use current division to find the final value of the current: 
i = 9 ! 6 ( -8) = -4.8 A 
Problems 7-9 
The equivalent resistance seen by the inductor is used to calculate the 
time constant: 
3]1(9 + 6) = 2.50 L 2 T=- = - =0.8s 
R 2.5 
Therefore, 
i = iF + [i(l +) - iF]e-(t-l)/r 
= -4.8 + 5.98e-i.25<t-i) A, t 2 1 s 
AP 7.9 0 :::=; t ::; 32 ms: 
40kQ 160kQ 
-10~ 
J., 
0.2µF 
.-----jf-----. 
+ 
1 {32xrn-3 1 132x10-3 1 
vo = - RCt .fo -lOdt + 0 = - RCt (-lOt) 0 = - RCt (-320 x 10-
3 ) 
RC1 = (200 x 103)(0.2 x 10-6 ) = 40 x 10-3 
V0 = -25(-320 X 10-3) = 8 V 
t 2 32ms: 
0.2µF 
1 
so -- = 25 
RC1 
~--·-Ji-----. 
90kQ 160kQ 
+ 
1 lt 1 !t 1 Vo= -RC 5dy + 8 =-RC (5y) +8 =-RC 5(t- 32 X 10-3) + 8 
f . 32x10-3 f 32x10-a f 
1 
RC1 =(250x103)(0.2 x 10-6 ) = 50 x 10-3 so RCt = 20 
7-10 CHAPTER 7. Response of First-Order RL and RC Circuits 
V 0 = -20(5)(t - 32 X 10-3) + 8=-lOOt+11.2 
The output will saturate at the negative power supply value: 
-15 = -lOOt + 11.2 t = 262ms 
AP 7.10 {a] Use RC circuit analysis to determine the expression for the voltage at the 
non-inverting input: 
Vp = Vt+ [Va - V1]e-t/r = -2 + (0 + 2)e-t/r 
ljT = 625 
v - -2 + 2e-625t V· p- ' 
Write a KVL equation at the inverting input, and use it to determine v0 : 
Vn Vn - Vo 
10,000 + 40,000 = 0 
.'. Vo= 5vn = 5vp = -10+ 10e-625ty 
The output will saturate at the negative power supply value: 
-10 + 10e-625t = -5; e-625t = 1/2; t = ln2/625 = 1.11 ms 
[b] Use RC circuit analysis to determine the expression for the voltage at the 
non-inverting input: 
Vp =Vt+ [Va - V1Je-t/r = -2 + (1+2)e-625t = -2 + 3e-625t V 
The analysis for v0 is the same as in part (a): 
v0 = 5vp = -10 + 15e-625tV 
The output will saturate at the negative power supply value: 
-10+ 15e-625t = -5; e-625t = 1/3; t = ln3/625 = l.76ms 
Problems 
P 7.1 [a] i(O) = 125/25 = 5 A 
L 4 
[b] T= R= 100 =40rns 
[c] i = 5e-25t A, t > 0 
V1 = -8Qi = -400e-25t V t 2::: 0 
v2 = L dii = 4(-125e-25t) = -5ooe-25t V 
dt 
[dJ Pdiss = i 2 (20) = 25e-50t(20) = 500e-50t W 
Int e-50x It Wdiss = 500e-50x dx = 500--0 = 10 - 10e-50t J 0 -5 0 
Wdiss(12 ms)= 10 - lOe-0·6 = 4.51 J 
w(O) = ~(4)(25) = 50J 
0'1 4.51 ( ) 0'1 
10 dissipated = 50 100 = 9.0210 
P 7.2 (a] t < 0 15k0 
9V • 
-
iJO l 
15kOll15kO = 7.5k0 
ig(O-) = (15 + 7.~) x 103 = 0.4 rnA 
15k0 
ii(o-) = i 2 (0-) = (0.4) x 10-3 i~~j = 0.2 rnA 
[b] ii(o+) = ii(o-) = 0.2rnA 
i2(o+) = -i1(0+) = -o.2rnA 
L 30 x 10-3 
[c] T = R = 30 x lQ3 = 10-6; 
ii(t) = ii(O+)e-t/r 
ii(t) = 0.2e-106t rnA, t > 0 
(when switch is open) 
! = 106 
T 
Problems 7-11 
7-12 CHAPTER 7. Response of First-Order RL and RC Circuits 
when t ~ o+ 
. . ( ) 0 2 -106t A .. i2 t = - . e m , 
[e] The current in a resistor can change instantaneously. The switching 
operation forces i 2 (o-) to equal 0.2mA and i 2(o+) = -0.2mA. 
P 7.3 [a] i 0 (0-) = 0 since the switch is open fort< O. 
(b] Fort= o- the circuit is: 
---7i 
9 
500 1000 
25\11 ~ 3000 
3oon113oon = 15on 
25 - 125 A : · ig = 50 + 150 - m 
iL(o-) = (~~~) i9 = 62.5mA 
[c] For t = o+ the circuit is: 
---7i9 
500 1000 
25Vi ~ 3000 l 
iJO+) 
3oon1110on = 1sn 
25 
. ·. i9 = 50 + 75 = 200 mA 
ia = (!~~) 200 = 150mA 
: . io(o+) = 150 - 62.5 = 87.5 mA 
[d] iL(o+) = iL{O-) = 62.5 mA 
[e] i 0 (00) = ia = 150mA 
2000 
2000 + 
v Jo+) -i 62.5mA 
p 7.4 
Problems 7-13 
[fJ h ( oo) = 0, since the switch short circuits the branch containing the 
200 n resistor and the 50 mH inductor. 
L 50 x 10-3 1 
[g] r = - = = 0.25ms· - = 4000 
R 200 ' r 
:. iL = 0 + (62.5 - O)e-4000t = 62.5e-4oo0t mA, t;:::: 0 
since for t < 0 the curre;nt in the inductor is constant 
[i) Refer to the circuit at t = o+ and note: 
200(0.0625) + VL(o+) = O; :. VL(o+) = -12.5 v 
m VL( 00) = 0, since the current in the inductor is a constant at t = 00. 
[kJ vL(t) = o+ (-12.5-0)e-4000t = -12.5e-4000tv, t;:::: o+ 
[I) io = ia - iL = 150 - 62.5e-4ooot mA, 
[a] ~ = R = lOOe-s0t = 25 n 
i 4e-80t 
1 
[b] r = 80 = 12.5ms 
L 
[c] r = R = 12.5 x 10-3 
L = (12.5)(25) x 10-3 = 312.5 mH 
[d] w(O) = ~L[i(0)] 2 = ~(0.3125)(16) = 2.5 J 
[e] Wdiss = ht 400e-l60:i: dx = 2.5 - 2.5e-l60t 
0.8w(O) = (0.8)(2.5) = 2 J 
2.5 - 2.5e-100t = 2 
Solving, t = 10.06 ms. 
e160t = 5 
P 7.5 w(O) = ~(20 x 10-3)(102) = 1 J 
0.5w(O) = 0.5 J 
. 10 -t/r 
'tR = e 
Pdiss = i~R = lOORe-2t/r 
7-14 CHAPTER 7. Response of First-Order RL and RC Circuits 
p 7.6 
p 7.7 
e-2x/r Ito 
Wdiss = lOOR = -50rR(e-2to/r - 1) = 50L(l - e-2tofr) 
-2/r o 
50L = (50)(20) x 10-3 = 1; t0 = lOµs 
e2to/T = 2; 2t0 = 2t0 R = ln 2 
r L 
R = Lln2 = 20 x 10-3 ln2 = 693.150 
2t0 20 X 10-6 
1 
(a] w(O) = 2LI! 
- 12R -2t/r dt - 12R ! to e-2t/r Ito Wdiss - 0 g e - g -( _-2-/ 7-) 0 
1 1 
= 2J!Rr(l - e-2tofr) = 21;L(l - e-2t0 fr) 
Wdiss = o-w(O) 
. !LJ2(1 _ e-2tofr) = <Y (~L12) . . 2 g 2 g 
1 
€2to/T = --
(1- <Y) 
2t0 = ln [ 1 l · 
'T (1-<Y)' 
R = L ln[l/(1 - a)] 
·· 2t0 
[b] R = (20 x 10-3 ) ln(l/0.5] 
20 x 10-6 
R = 693.150 
(a) iL(O) = 80 = 2A 
40 
R(2to) = ln[l/(1 - a)] 
L 
io(O+) = 80 - 2 = 4 - 2 = 2 A 
20 
( . 80 i 0 oo) = 20 = 4 A 
L 20 
r = - = - x 10-3 = 1 rns R 20 
iL = 2e-1000t A 
i 0 = 4 - iL = 4 - 2e-lOOOt A, 
[c} 4 - 2e-1000t = 3.8 
0.2 = 2e-lOOOt 
elOOOt = 10 
P 7.8 [a] For t < 0 
t = 2.30rns 
150 500 
8 O\ii ~ ~ 500 ~ 
iJO ) itO ) 
i9 =SO= 2A 40 
. 2{50) . 
iL(o-) = (rno) = 1 A= iL(o+) 
For t > 0 
-~ 
< 60Q 20Q 
iL(t) = h(O+)e-t/T A, t ~ 0 
L 0.20 1 
T = R = 5 + 15 = 100 = 0·01 s 
iL(O+) = lA 
iL(t) = e-100t A, t ~ 0 
v0 (t) = -15iL(t) 
Problems 7-15 
30 
+ 
Vo 
600 200 
20 
7-16 CHAPTER 7. Response of First-Order RL and RC Circuits 
p 7.9 p. v~ 11 25 - 20(}tw 2on=20=. e 
/
0.01 
Wdiss = ll.25e-20(}t dt 
. 0 
= 11.25 e-20CJt 10.01 
-200 0 
= 56.25 x 10-3(1- e-2) = 48.64mJ 
1 2 
Wstored = 2(0.2)(1) = 100 rnJ. 
48.64 
% diss = 100 x 100 = 48.64% 
P 7.10 [a] t < 0 
2 .Sm + 16k0 
. (O-) = -2.5(16) = _2 A 
't£ (20) Ill 
t 2: 0 
+ 
120mH OmH ~ 40mH v 1k0 0 
40 x 10-3 
T = = 40 X 10-6 · 103 ' ljT = 25,000 
vo = -1000(-2 x 10-3)e-25•00CJt = 2e-25,ooCJty, 
1 
[h] Wdel = 2(40 X 10-3)(4 X 10-6 ) = 80nJ 
[c] 0.95wdel = 76 nJ 
: . 76 x 10-9 = dt Into 4e-50,00(}t 
0 1000 
+ 
v 1k0 
0 
Problems 7-17 
I
to 
76 X 10-9 = 80 X 10-9e-50,000t 
0 
= 80 X 10-9 (1 - e-50,000t0 ) 
e-50,000to :;:::: 0.05 
50,000t0 = ln 20 so t0 = 59.9 µs 
~ = 5=~9 = 1.498 SO t0 ~ l.5T 
p 7.11 t < 0: 70 
160 .. 
t > 0: 50 
+ 
50 
Re= (20)(5) + 20 = 24fl 
25 
T = ..£ = 96 x 10-3 = 4 ms· 
Re 24 ' 
iL = 2e-250t A 
. 5 . 0 4 -250t A 
'lo= 25 'lL = . e 
50 
50 
200 
~ = 250 
T 
P 7.12 P2on = 20ii, = 20(4)(e-250t)2 = 80e-500t W 
200 
2A 
rX) e-500t 1= 
w200 =lo 80e-5()(:)t dt = 80 _ 500 0 = 160mJ 
w(O) = t(96)(10-3)(4) = 192mJ 
160 
% diss = 192 (100) = 83.33% 
7 -18 CHAPTER 7. Response of First-Order RL and RC Circuits 
P 7.13 [a] v0 (t) = v0 (o+)e-tf7 
v0 (0+)e- 5xl0-'3! 7 = 0.25vo(O+) 
L 5 x 10-3 
r------- R - ln4 
L = 250 x 10- 3 = 180.34 mH 
ln4 
[h] iL(o-) = 60 (~) = lOmA = iL(o+) 
p 7.14 t < 0 
t>O 
Wstored = ~LiL(o+)2 = ~(Rr)(lOO X 10-6) = 25007 µJ. 
iL(t) = 10e-t/T mA 
Pson = i1,(50) = 5000 x 10-6e-2t/T 
5x10-3 
Wdiss = j 
0 
5000 X 10-6e-2t/T dt 
= 5000 X 10-6 t~:::) \~xlO-:~ 
= 2500 X 10- 7 1 - e r 6 [ -lOxI0-3] 
-1ox10-3 21 e r = e- n 4 = 0.0625 
Wdiss = 2500 X 10-67(0.9375) 
o/c d' - 2500 x 10-67(0.9375) 
0 188 - 2500 x 10-67 x 100 
Wdiss = 93. 75% 
25A1 t 100 
Problems 7-19 
+ 
50mH 
Find Thevenin resistance seen by inductor 
VT 1 -. = RTh = - = 0.2 n 
ZT 5 
L 50 x 10-3 
r=-= =250ms· 
R 0.2 ' 
1/r = 4 
+ 
0 .20 50mH 
i 0 = 25e-4t A, t 2:: 0 
P 7.15 [a) t < 0 : 
30 4.50 
--7 9A 
90 
iJO) 
(9)(4.5) = 3 n· 
13.5 ' 
iL(O) = 9 1:_5 = 6A 
7-20 CHAPTER 7. Response of First-Order RL and RC Circuits 
t > 0: 
~ 1000 
ill. 
. ir(200) 2. 
iA. = 300 = 3ir 
2000 
v - 50i . (100)(200) - 50i 2 200. 
r- A.+ir 300 - r3+3ir 
Vr 100 200 
ir =RTh=3+3=lOOn 
1000 
T = L = 200 x 10-3 ~ = 500 
R 100 T 
iL = 6e-5o0t A, t 2:: 0 
[h] vL = 200 x 10-3(-3oooe-500t) = -60oe-500t v, t 2:: o+ 
[c] 
+ 
VL = 50iA + 100iA = 150iA. 
iA. = VL = -4e-500t A 
150 
P 7.16 w(O) = ~(200 x 10-3)(36) = 3.6 J 
i 1000 2000 
P5oiA = -50iA.iL = -50(-4e-500t)(6e-500t) = 12ooe-1000t w 
{oo e-lOOOt loo 
w50iA = lo 12ooe-1000t dt = 1200_1000 0 = 1.2 J 
ot 1.2 ( ) 
10 dissipated = 3_6 100 = 33.333 
P 7.17 [a] t < 0 
40kDll20kn = 13.33kfl 
60kflll30kn = 2okn 
40kQ 
(120 x 10-3)(13.33 x 103) = 1600V 
iJO ) 
~ 
1600 
iL(O-) = 33,333.33 = 48 mA 
t>O 
200rnH 
---748mA 
20k0 60k0 
60kQ 30kQ 
L 0.2 
7 = R = 80,000 = 2·5 µs; 
1 
~ = 400 000 
T ' 
i L ( t) = 48e-40o,ooot mA, 
P6ok = (0.048e-4oo,o00t)2(60,000) = 138.24e-soo,oo0t W 
Problems 7-21 
Wdiss =lot 138.24e-soo,ooox dx = 172.8 x 10-6 [1 - e-8oo,oo0t] J 
w(O) = ~(.2)(48 x 10-3) 2 = 230.4µJ 
0.25w(O) = 57.6 µJ 
172.8(1 - e-8oo,ooot) = 57.6; 
ln 1.5 
t = 800 000 = 0.507 µs 
' 
: . esoo,000t = 1.5 
7-22 CHAPTER 7. Response of First-Order RL and RC Circuits 
p 7.18 
[b] wdiss(total) = 230.4(1 - e-soo,oo0t) µJ 
Wdiss(0.507 µs) = 76.82 µJ 
% = (76.82/230.4)(100) = 33.3% 
[a] t < 0: 10 a 
----:720A 
l 
120 220\11 ~ (50/3) A 
b 
t = o+: 10 a 
----:7220A 
220\11 ~ 
b 
l 
600 (10/3)A 
(50/3) JI.+ 
220 = iab + (50/3) + (10/3), iab = 200A, t = o+ 
[b] At t= oo: ·10 a 
----:72DA 
220 . 120 
iab 
b 
iab = 220/1 = 220 A, t = 00 
10 a 
----:7220A 
l 120 l 600 220\11 ~ 
iab il i2 
b 
[c] i 1 (0) = 50/3, 
2 
71 = 12 x 10-3 = 0.167ms 
15 
72 = 60 x 10-3 = 0.25 ms 
ii(t) = (50/3)e-6000t A, t 2: O 
600 
i2(t)::::: (10/3)e-4000t A, t 2:: 0 
iab = 220 - (50/3)e-6000t - (10/3)e-4o00t A, t 2:: 0 
220 - (50/3)e-6000t - (10/3)e-4000t = 210 
30 =50e-6000t + 1 oe-4000t 
By trial and error 
t = 123.1 µs 
P 7.19 [a] t < 0: 
t = o+: 
t > 0: 
1k0 OmH 
i 
J_, 50mH f 2.5k0 
SmA 
i R = 5et/T mA; 
i R = 5e-5o,ooot mA 
L 
T = - = 20 X lQ-6 
R 
2.5k0 
Problems 7-23 
8mH 
7-24 CHAPTER 7. Response of First-Order RL and RC Circuits 
+ iR~ 
v1 
VR 2.5k0 
+ 
OmH i 8mH + v 0 
i 
0 
VR = (2.5 x 103)(5 x 10-3)e-5o,ooot = 12.5e-5o,ooot V 
v1 = 20 x 10-3(5 x 10-3(-50,000)e-50·000t] = -se-5o,oo0t V 
Vo= -V1 - VR = -7.5e-5o,oo0t v 
103 ft 
[h] io = 48 Jo -7.5e-
5o,ooox dx + 0 = 3.125e-5o,oo0t - 3.125mA 
P 7.20 [a) From the solution to Problem 7.19, 
i R = 5 x 10-3 e-5o,oo0t A 
PR = (25 x 10-6e-ioo,oo0t)(2.5 x 103 ) = 62.5 x 10-3e-ioo,ooot W 
Wctiss = fo00 62.5 x io-3e-Ioo,ooot dt 
e-100,000t loo 
= 62.5 x 10-3 05 = 625 nJ -1 0 
[ ] 1 ·2 ( ) 1 ( -3) ( -3 2 h Wtrapped = 2LeqZR 0 = 2 50 X 10 5 X 10 ) = 625nJ 
CHECK: 
w(O) = ~(20)(25 x 10-6) x 10-3 + ~(80)(25 x 10-6) x 10-3 = 1250nJ 
:. w(O) = Wctiss + Wtrapped 
P 7.21 [a] v1(0-) = v1(0+) = 75 V 
Ceq = 2 x 8/10 = 1.6 p,F 
5k0 
+ 
1.6µ.F ~ 75V 
T = (5)(1.6) x 10-3 = 8ms; 
1 
- = 125 
T 
Problems 7-25 
75 i = - x 10-3e-125t = 15e-125t mA 
5 ' 
-106 it V1 = -- 15 x 10-3e-125x dx + 75 = 60e-125t + 15 V, 
2 . 0 
106 r 
V2 = 8 lo 15 X 10-3e-125x dx + 0 = -15e-125t + 15 V, 
1 
[b] w(O) = "2(2 x 10-6)(5625) = 562511,J 
1 1 
[c] Wtrapped = 2(2 X 10-6)(225) + 2(8 X 10-6)225 = 1125 µJ. 
1 
Wdiss = 2(1.6 X 10-6)(5625) = 4500 µJ. 
Check: Wtrapped + Wdiss = 1125 + 4500 = 5625 µJ; 
P 7.22 [a] R = ~ = 20 kn 
i 
1 1 
[b] - = - = 1000· 
T RC ' 
1 
C = (103)(20 x 103) = 0.05 µF 
1 
[c] T = lOOO = lms 
(d] w(O) = ~(0.05 x 10-6)(104 ) = 250 µJ 
[e] 
. - {to v2 - ro (104}e-2000t 
Wdiss - lo Rdt - lo (20 x 103 ) dt 
e-2000t ]to 
= 0.5 = 250(1 - e-2000t0 ) µJ 
-2000 0 
200 = 250(1 - e-2000t 0 ) 
. .. e-2000to = 0.2; e2000to = 5 
1 
t0 = 2000 In 5; t0 rv 804. 72 /LS 
w(O) = 5625/LJ. 
7-26 CHAPTER 7. Response of First-Order RL and RC Circuits 
P 7.23 [a] t < 0: 
lOkQ 
! ifO-J 5kQ 
ii(o-) = i2(o-) = (:0 x 10-
3 ) = o.2mA 
(b] t > 0: 0 • 4UF 
5kQ lOkQ 
iJD-) iJO-) 
5kQ 
2 
ii(o+) = 10 x 10-3 = 0.2mA 
-2 
i2(o+) = 10 x 10-3 = -0.2 mA 
[c] Capacitor voltage cannot change instantaneously, therefore, 
ii(o-) = ii(o+) = o.2mA 
[ d] Switching can cause an instantaneous change in the current in a resistive 
branch. In this circuit 
i 2(o-) = 0.2 mA and i 2(o+) = -0.2 mA 
(e] Ve= 2e-tfr V, t ~ 0 
T = ReC = 5000(0.4) X 10-6 = 2 X 10-3 
Ve= 2e-500t V, t 2:: 0 
ii = Ve = o.2e-500t mA 
10,000 ' 
[f] i2 = 1~~~0 = -0.2e-500t mA, 
P 7.24 fa] v(O) = (S)(~7J(33) = 118.80V 
Re= (3)(5) = 2kD 
9 
t~O 
t ~ o+ 
T = ReC = (2000)(0.25) X 10-6 = 500 µs; 
v = l18.80e-2000t V t ;:::: 0 
v 
i = -- = 39.6e-2000t mA 
0 3000 
[b] w(O) = ~(0.25)(118.80)2 = 1764.18 µJ 
'l4k = l18.80;-2ooot = 19.8e-2ooot mA 
Problems 7-27 
~ = 2000 
r 
P4k = ((19.8)e-2000t]2(4000) x 10-5 = 1568.16 x 10-3e-4ooot 
e-4000x 1250x10-6 
W4k = 1568.16 X 10-3 _ 4000 O = 392.04(1 - e-l) f1J 
w = 392·04 (1 - e-1 ) x 100 = 14.0507 
IO 1764.18 IO 
P 7.25 [a] t < 0: 
+ 
. (o-) = (25)(8) = 10 A 
'lo (20) m 
vo(o-) = (10)(10) = 100 V 
i2(o-) = 25 - 10 = 15 mA 
v2(0-) = 15(8) = 120V 
t>O 
r =RC= 0.2ms = 200µs; 
2k0 
+ 
1 
- = 5000 
T 
7-28 CHAPTER 7. Response of First-Order RL and RC Circuits 
+ 20V -
0.1/,JY 
i (t) = 20 e-t/r = 10e-5000t mA 
0 2 x 103 ' 
[h] 
+ I ~o --:ir + 
12~Vr .lSµF lO~Vr .30µF =o 
106/t Vo = - 10 X 10-3e-5000x dx + 100 
0.3 0 
105 e-5000x It 
= 3 -5000 0 +lOO 
= -(20/3)e-sooot + (20/3) + 100 
V 0 = [-(20/3)e-SOOOt + (320/3)] V, t 2:: 0 
[c] Wtrapped = (1/2)(0.15) X 10-6 (320/3) 2 + (1/2)(0.3) X 10-6 (320/3)2 
Wtrapped = 2560 µJ. 
Check by combining the capacitors into a single equivalent capacitance of 
0.1 µF with a 20 V initial voltage: 
1 2 1 -6 2 
Wdiss = 2Ceq(V0 ) = 2(0.l X 10 )(20) = 20 µJ 
w(O) = ~(0.15) x 10-6 (120) 2 + ~(0.3 x 10-6)(100)2 = 2580 µJ. 
Wtrapped + Wdiss = w(O) 
2560 + 20 = 2580 OK. 
1.8k0 
P 7.26 (a] t < 0: 
Vo(O) = (60)(10.2) = 51 V 
12 
t > 0: 
+ 
51V ~ (1/6)µ.F 
1 
r = 6(12) x 10-3 = 2ms; 
V 0 = 51e-500t V, t 2:: 0 
+ 
+ 
v 12k0 
!:. = 500 
7 
v2 
p = .....!!... x 10-3 = 216.75 x 10-3e-1000tw 
12 
r2x10-3 
Wdiss =lo 216.75 x 10-3e-lOOOt dt 
= 216.75 x 10-6(1 - e-2) = 187.42 µJ 
[b] w(O) = (~) (~) (51)2 x 10-6 = 216.75µJ 
0.95w(O) = 205.9125 µJ 
68k0 
foto 216.75 X 10-3e-lOOOx dx = 205.9125 X 10-6 
ro lo e-IOOOx dx = 0.95 X 10-3 
Problems 7-29 
:. 1 - e-lOOOto = 0.95; elOOOto = 20; so t 0 = 3 ms 
7-30 CHAPTER 7. Response of First-Order RL and RC Circuits 
p 7.27 t < 0 
+ 
25\i1 ~ 
I • 
+ 
60k0 
vr - 2 X 104io + 60,000ir 
- 20,000( -ir) + 60,000ir = 40,000ir 
+ + 
40k0 v 0 25V ~ 25nF 
T =RC= lms; 
1 
- = 1000 
T 
v0 = 25e-l000t V, t>O 
t ~ o+ 
P 7.28 (a] T =RC= RTh(0.2) x 10-6 = 10-3 ; 
OkO 
VT= 20 x 103 (iT - avA) + 10 x 103ir 
VA = 10 X l03ir 
VT= 30 x 103ir - 20 x 1030:10 x 103iT 
1? = 30 x 103 - 200 x 106 a = 5 x 103 
'lT 
Problems 7-31 
RTh = lOOO = 5k0 
0.2 
. ·. 30 - 200,000a = 5; o: = 125 x 10-6 A/V 
[b] v0 (0) = (0.018)(5000) = 90 V t < 0 
t > 0: 
+ + 
V 0 = 90e-lOOOt V, t 2:: 0 
20k0 
+ 
v 6. 10k0 
VA VA-Vo -6 
10 X 103 + 20,000 - 125 X lO VA = O 
2vA +VA - Vo - 2500 x 10-3VA = 0 
.". VA= 2v0 = 180e~lOOOt V 
7-32 CHAPTER 7. Response of First-Order RL a,nd RC Circuits 
P 7.29 [a] 20k0 
-lOOOt 
22 .5e mA 
~----<~ >---_...._ _ __, 
+ -lOOOt - 90e V + 
+ 
....._ -lOOOt 
0.2µ£' 90e v -lOOOt iQk" 180e V H 
Pds = (-9oe-1000t)(22.5 x 10-3e-1000t) = -2025 x 10-3e-2000t W 
Wds = fo00 Pds dt = -1012.511J . 
. ·. dependent source is delivering 1012.5 µJ 
(180)2e-2000t 
(b] P10k = 10 x 103 
WIQk = fo00 PlOk dt = 1620 /1J 
(90)2e-2000t 
P2ok = 20 x 1Q3 
W20k = fo00 P20k dt = 202.5 µJ 
Wc(O) = ~(0.2) X 10-6 (90)2 = 810 µJ 
L Wdev = 810 + 1012.5 = 1822.5 µJ 
L Wdiss = 202.5 + 1620 = 1822.5 µJ. 
P 7.30 (a] At t = o- the voltage on each capacitor will be 25 x 10-3 x 200 = 5 V, 
positive at the upper terminal. Hence at t ~ o+ we have 
f 50 + 5V....._ 50µ£' 
25mA! t 2000 
+ 
5V....._ 2µ$ 80 
~ 
isd I 
Problems 7-33 
At t = oo, both capacitors will have completely discharged. 
:. isd(oo)=25mA 
[b] isd(t) = 0.025 + ii(t) + i2(t) 
r1 = (5)(2) x 10-6 = 10 µs 
72 = (8)(50 x 10-6) = 400 µs 
. . ( ) -105tA . . i 1 t = e , 
i2(t) = 0.625e-2500t A, t ~ O 
. ·. isd = 25 + lOOOe-lOo,ooot + 625e-2500t mA, 
1 1 
P 7.31 [a] - = 1 + - = 1.25 
Ce 4 
.·. Ce= 0.8µF; v0 (0) = 60 - 10 = 50V 
T = (0.8)(25) X 10-3 = 20ms; 
+ + 
sov~ o .sµr v 25k0 
[b] W 0 = ~ ( 1 X 10-6) (3600) + ~ ( 4 X 10-6) (100) = 2 mJ 
1 
Wdiss = 2(0.8 X 10-6)(2500) = 1 mJ 
3 diss = ~ x 100 = 503 
[ ] . Vo 10-3 2 -50t mA c i 0 = - X = e 
25 
v1 = - 106 jt 2 x 10-3e-5ox dx - 10 = -5oojt e-5ox dx - 10 
4 0 0 
= -500-e- -10 = 10e-50t - 20V t ~ 0 -50x It 
-50 0 
(d) V1 + V2 = V 0 
v2 =Vo - V1 = 50e-50t - 10e-50t + 20 = 40e-50t + 20V t ~ 0 
7-34 CHAPTER 7. Response of First-Order RL and RC Circuits 
p 7.32 
1 1 
[e) Wtrapped = 2(4 X 10-6)(400) + 2(1X10-6 )(400) = lmJ 
Wdiss + Wtrapped = 2 mJ (check) 
[a] The equivalent circuit for t > 0: 
i 
0 
+ + C =40nF 
20V ~C 
eq 
v R 
R =lOkO eq 0 eq eq 
T = 0.4ms; ljT = 2500 
V 0 = 20e-2500t V, t~O 
i 0 = 2e-2500t mA, t ~ o+ 
i = 2e-25o0t ( 15 ) = 0 75e-25o0t mA t > o+ 25k!l 40 . ' 
P25kn = (0.5625 x 10-6e-5000t)(25,000) = 14,062.5 x 10-6e-5000tw 
W25k!l = fo00 14,062.5 X l0-6e-5000t dt = -2.8125 X 10-6(0 - 1) = 2.8125 µJ 
w(O) = ~(0.2 x 10-6)(100) + ~(0.05 x 10-6 )(900) = 32.5 µJ 
01' • ( ) 2.8125 01' 10 d1ss 25 kn = 3 x 100 = 8.6510 2.5 
[b] P625n = 625(2 x 10-3e-2500t)2 = 2.5 x 10-3e-5000t 
W625n = fo00 P625 dt = 0.50 /LJ 
% diss (625r2) = 0·5 x 100 = 1.54% 
32.5 
. 2 -2500t (25) 1 25 -25oot A t > o+ 
i15kn = e 40 = . e m ' 
P15kn = (1.25 x 10-3e-2500t)2(15,000) = 23.4375 x 10-3e-5oo0t W 
W15k0 = fo00 23.4375 X 10-3e-5ooot dt = 4.6875 µJ 
% diss (15kr2) = 14.42% 
[c) L Wdiss = 2.8125 + 0.50 + 4.6875 = 8 µJ 
Wtrapped = w(O) - 2.:.:: Wdiss= 32.5 - 8 = 24.5 µJ 
01' 24.5 07 
10 trapped = 32.5 x 100 = 75.3810 
Check: 8.65 + 1.54 + 14.42 + 75.38 = 99.99 ~ 100% 
P 7.33 After making a Thevenin equivalent we have 
t=O 
~Q 
18ovcb 
LJ 
5kQ 
I 
10 = 180/15 = 12mA 
T = (0.25/20) X 10-3 = 0.125 X 10-4 ; 
Ir = Vs = 180 = 9 mA 
R 20 
_!_ = 80,000 
T 
io = 9 + (12 - 9)e-so,oo0t = 9 + 3e-so,oo0t mA 
Vo = [180 - 12(20]e-so,000t = -6oe-so,ooot V 
P 7.34 t < 0 i (O-) 
1' 
8Q t6A 
t>O 
12Q 5mH SQ 
- v + r~] 32V I L v 0 ~ 48V 
. ( ) _ 32 + 48 _ 4 A 'l£ 00 - -
20 
Problems 7-35 
7-36 CHAPTER 7. Response of First-Order RL and RC Circuits 
L 5 x 10-3 
r = R = 20 = 250 p,s; ~ = 4000 'T 
iL = 4 + (6 - 4)e-4oo0t = 4 + 2e-4o00t A, t ~ 0 
Vo= -Sh+ 48 = -8(4 + 2e-4000t) + 48 = 16- l6e-4oo0ty, 
Check: at t = o+ the circuit is: 
120 6A SQ 
I - vJfl.> + + w.I 
32VQ vJ0+)648V 
I d 
VL(O+) = 32 - 72 + 0 = -40V, 
P 7.35 [a] t < 0 5Q 
+ 
iJO-) 
KVL equation at the top node: 
40 _ vo(o-) vo(o-) vo(o-) 
- - 4 + 20 + 5 
Multiply by 20 and solve: 
-800 = (5 + 1 + 4)v0 ; V0 = -80V 
Problems 7-37 
t>O 
60Q 5Q 
+ 
240\li ~ v 20Q 10mH 
Use voltage division to find the Thevenin voltage: 
20 
VTh = V 0 = 20 + 60 (240) = 60V 
Remove the voltage source and make series and parallel combinations of 
resistors to find the equivalent resistance: 
RTh = 5 + 201160 = 5 + 15 = 20 n 
The simplified circuit is: 
20Q 
60\li ~ 10mH 
L 10 x 10-3 
T = - = = 0.5 ms· 
R 20 ' 
i 0 (00) = 60 = 3A 
20 
! = 2000 
T 
io = io( 00) + [i0 (0+) - i0 ( 00 )]e-t/r 
= 3 + ( -16 - 3)e-2000t = 3 - l9e-2000t A, t 2: 0 
] . ( ) di 0 [b V 0 - 5io + 0.01 dt 
- 15 - 95e-2000t) + 0.01(38,000)(e-2000t) 
- 15 - 95e-2000t + 380e-2000t 
15 + 285e-2000tv, 
P 7.36 (a] Fort< 0, calculate the Thevenin equivalent for the circuit to the left and 
right of the 75 mH inductor. We get 
7-38 CHAPTER 7. Response of First-Order RL and RC Circuits 
8kQ 75mH 15kQ 
120r~5V 
I I 
. - 5 - 120 
i(O )=15k+8k=-5mA 
i(o-) = i(o+) = -5mA 
[b] For t > 0, the circuit reduces to 
75mH 15kQ 
~, 
i 6sv 
Therefore i(oo) = 5/15,000 = 0.333mA 
L 75 x 10-3 
[c] 7 = R = 15,000 = Sµs 
[d] i(t) = i(oo) + [i(O+) - i(oo)]e-tfr 
= 0.333 + [-5 - 0.333]e-2oo,5oot = 0.333 - 5.333e-2oo,5oot mA, t ~ 0 
P 7.37 [a) From Eqs. (7.35) and (7.42) 
i = lfs + (J - lfs) e-(R/L)t 
R 0 R 
v = (Vs - IoR)e-<Rf L)t 
Vs 
:. R = 10; 
Vs - IoR = 200; 
Vs 
10 = -10 + R = OA 
Therefore, Vs = 200V. 
i(oo)=10= 200 so R=200 
R 
R 
L=-=40mH 
500 
Problems 7-39 
[b] i = 10 - 10e-500t; i 2 = 100 - 2ooe-500t + 10oe-1000t 
[b] 
[c] 
[d) 
w = ~Li2 = ~(0.04)[100 - 2ooe-500t + 10oe-1000t] = 2 - 4e-500t + 2e-1000t 
w(oo)=2J 
w(t0 ) = 2 - 4e-500to + 2e-lOOOta = 0.25(2) 
. ·. 1 - 2x + x2 = 0.25 and thus x2 - 2x + 0. 75 = 0 
Solving, x = 1.5 and x = 0.5 but only the second solution is possible 
ln2 
t0 = SOO = 1.386 ms . . . 0.5 = e-500to so 
v0 (oo)=O 
Vo(t) = -lgR2e-[(R1+R2)/L]t V, 
vo(o+)--+ oo, and the duration of v0 ( t) --+ zero 
Vsw = R2io; 
L 
T= 
Ri+R2 
io(o+) =lg; io(oo) =lg R Ri R 
1+ 2 
Therefore i 0 (t) = ~ + [! _ ~] e-[(R1+R2)/L]t Ri+R2 g Ri+R2 
i 0 (t) = Rily + Rzlg e-[(R1+R2)/L]t (R1 +R2) (R1 +R2) 
Therefore Vsw = Ri19 + R2lg e-[(R1 +R2)/L]t (l+R1/ R2) (l+Ri/ R2) ' t ~ o+ 
lvsw(O+)I -+ oo; duration--+ 0 
P 7.39 Opening the inductive circuit causes a very large voltage to be induced across 
the inductor L. This voltage also appears across the switch (part [e] of 
Problem 7.38) causing the switch to arc over. At the same time, the large 
voltage across L damages the meter movement. 
7-40 CHAPTER 7. Response of First-Order RL and RC Circuits 
v 
P 7.40 (a] 
R L 
Vs v 1 ft -- + - + - v dt +I = 0 
R R L o 0 
Differentiating both sides, 
1 dv 1 
--+-v=O 
Rdt L 
dv R 
-+-v=O 
dt L 
dv R 
[b] dt = - L v 
dv R 
-dt= --vdt 
dt L 
In v(t) = - Rt 
Vo L 
so 
R 
dv = --vdt 
L 
v(t) = Vae-(R/L)t = (Vs - Rlo)e-(R/L)t 
P 7.41 Fort< 0 
iJD-) 
lOQ 40Q 
~ + + v<:J>-20Q v ~ 25DV K 
Vx [Vx - 250] [Vx - 250] = O 
20 + 9 50 + 50 
Vx 10 ( Vx - 250) = Q 
20 + 50 
5vx - 5000 + 20vx = O; Vx = 200V 
t>O 
87 .2mH lOQ 40Q 
20Q ~ 
+ v -
<I> 
lOQ l 
~ 250V 
0.9v<P 
87.2mH 
20Q 
lOQ 
+ v -
<I> 
Find Thevenin equivalent with respect to a, b 
lOQ 8Q 
8Q 
Wlrl 
~ 80V 
a~·------''Ml + v<I>-
~ ~8~ 
0 • 9v<I> 
VTh ~ 80 g(VTh - 80) = 0 
18 + 18 
lOQ 8Q 
a&--~~~--~~MA.-~~~~W\r 
+ 
+ v -¢l 
. [· (lOvr)] vr = (ir - 0.9v<t>)l8 = ir - 0.9 l8 18 
Problems 7-41 
7 -42 CHAPTER 7. Response of First-Order RL and RC Circuits 
VT = l8iT - 9vT .·. lOvT = 18ir 
VT -. = RTh = 1.80 
'LT 
87.2mH 
20Q 
i 0 ( oo) = 80/21.8 = 3.67 A 
87.2 3 
T = -2 - x 10-· = 4 ms; 1.8 
1.BQ 
l/r = 250 
~ BOV 
io = 3.67 + (10 - 3.67)e-250t = 3.67 + 6.33e-250t A, t ~ 0 
P 7.42 t < O; lOQ 
5Q 
i 0 (0-) = 5 (0.002) = 0.5 mA 
5+ 15 
t > O; 
-O 002 + Va + Va - Vo = O 
. 5 15 
150 
+ 
vJO+) {- 0.5mA 4Q 
Vo - Va + 5 X 10-4 + Vo - 4ia - 0.001 = 0 
15 4 
t 1mA 
it:.. = Vo - 4it:i.. - 0.001 
4 
Solving, 
We also know that 
v0 (00) = 0 
Find the Thevenin resistance seen by the 2 mH inductor: 
lOQ 
~~~""t--~--'W~-,....-'-t-~~---. 
+ 
Va 15Q+ 4Q 
. VT VT . 
'lT = - +- -4it:i.. 
20 4 
. VT 4 . 'tf:i.. = - - 'lA 
4 
. VT 
'tA= -
20 
iT 1 1 1 2 - = -+- - - = - =0.18 
VT 20 4 5 20 
2 x 10-3 
r = = 0.2ms· 
10 ' 
l/r = 5000 
:. Vo= 0 + (2-0)e-sooot = 2e-soootmV, t ~ o+ 
Problems 7--43 
7-44 CHAPTER 7. Response of First-Order RL and RC Circuits 
P 7.43 [a] Let v be the voltage drop across the parallel branches, positive at the top 
node, then 
- lg + ~g + l1 lot v dx + l2 lot v dx = 0 
~g + (ll + ;2) lot vd.T =lg 
v 1 lt -+- vdx=l9 R9 Le o 
~dv +~=O 
R9 dt Le 
dv R9 -+-v=O 
dt Le 
Therefore v = J9 R9 e-t/7; 
Thus 
. - ~ rt l R -x/T d - lgRg e-x/T It - lgLe (1 - -t/T) 
i 1 -L1 1o 9 9 e x- Li (-1/r) o- Li e 
. L2 
[b] z1(oo) = L L 19 ; 
1+ 2 
p 7.44 t > 0 
. 5 -2otA 'lo= - e J t>O 
Vo = 80io = -400e-20t V, 
+ 
t > o+ 
-40oe-20t = -80; €20t = 5 
1 
t = 20 ln5 = 80.47ms 
800 
1 ·2( ) 1 P 7.45 [a] Wdiss = 2Lei 0 = 2(4)(25) = 50J 
[b] 
ii2H = : 2/: ( -400)e-
20x dx + 5 
= ---- +5 = -e-2ot +-A -100 e-20x It 5 10 
3 -20 0 3 3 
i6H = ~.1: (-400)e-20x dx + 0 
= ---- +o = -e-2ot - -A -200 e-2ox It 10 10 
3 -20 0 3 3 
1 
Wtrapped = 2(18)(100/9) = 100 J 
[c] w(O) = ~(12)(25) = 150 J 
P 7.46 [a] t < 0 
150 
t>O 
+ 
v 480 1' lOA 
4.8 
r = 48 = O.ls; 
iL(oo) = lOA 
Problems 7-45 
1 
- = 10 
T 
iL = 10 + [20 - lO}e-lOt = 10 + 10e-10t A, t;:::: 0 
v0 = 4.8[-100e-10t] = -480e-10t V, 
[b] ii = 1
1
2 lot -480e-lOx dx + 8 = 4e-10t + 4A, 
[c] i 2 = ! {t -480e-rnx dx + 12 = 6e-10t + 6 A, 
8 lo 
7-46 CHAPTER 7. Response of First-Order RL and RC Circuits 
P 7.47 For t < 0, i40mH(O) = 75/5 = 15 A 
Fort> 0, after making a Thevenin equivalent we have 
i<E----
+ 
Vo 
40mH 
115A 
60mH 
i = Vs+ (1 _ Vs) e-t/T 
R 0 R 
.!_ = R = ___!__ x 103 = 40 
T L 100 
4Q 
I 0 =15A; Vs= 100 = 25A R 4 
i = 25 + (15 - 25)e-40t = 25 - 10e-40t A, 
+ 100 v -
t~O 
Vo = 0.04 ~: = 0.04( 400e-40t) = l6e-40t V, t > o+ 
16 
P 7.48 [a] vc(o-) = 20 (30) = 24 V 
( 1 1 )-l Ceq = 30 + 15 = 10 nF 
Fort> 0: 
+ + 
V 0 2QQk0 
r = RC = 200 x 103 x 10 x 10-9 = 2 ms; .!. = 500 
T 
Vo = 24e-500t V, 
Problems 7-4 7 
. V 0 24e-500t _500t 
{b] 'to = 200,000 = 200,000 = 120e p,A 
v = X 120 X 10-6 e-5oox dx + 0 = 16 - l6e-5oot V 1 int 
1 15 x 10-9 0 ' 
t 2:: 0 
P 7.49 [a] The energy delivered to the 200kr2 resistor is equal to the energy stored in 
the equivalent capcitor. From the solution to Problem 7.48 we have 
1 2 1 ( -9)( )2 
W = 2CeqV0 = 2 10 X 10 24 = 2.88µJ 
[b] From the solution to Problem 7.48 we know the voltage on the 15 nF 
capacitor at t = oo is 16 V. THerefore, the voltage across the 30 nF 
capacitor at t = oo is -16 V. It follows that the total energy trapped is 
1 1 
Wtrapped = 2(30 X 10-9)(-16)2 + 2(15 X 10-9)(16)2 = 5.76 µJ 
1 
[c] w(O) = 2 (30 x 10-
9)(242) = 8.64p.J 
Check: Wtrapped + Wdiss = 5.76 + 2.88 = 8.64 = w(O) 
P 7.50 [a] t > 0 
4kQ-=l 8kQ1" 
.-zsnF ~C•f25nF + 5kQ 
,___i_,,._. ___ __.__iu...-__ __,o ~ I -oj 
vo(o-) = vo(O+) = 0 V 
v0 (00) = 40V 
r = (8 x 103)(25) x 10-9 = 0.2 ms 
V 0 = ( 40 - 40e-5000t) V, t > 0 
[b] ic = 25 X lO_gdVo 
dt 
1/r = 5000 
ic = 25 X 10-9 (200,000e-5000t) = 5e-5o00t mA 
v1 = 4(5e-5000t) + 40 - 40e-5oo0t = 40 - 2oe-5oo0t 
V1 -5000t 
i 0 = 20 x 103 = 2 - e mA 
[c] ii(t) = i 0 + ic = 2 + 4e-5000t mA 
7-48 CHAPTER 7. Response of First-Order RL and RC Circuits 
[d] i2(t) = 5 ;
1
103 = 8 - 4e-5000t mA 
[e] ii(o+) = 2 + 4 = 6mA 
Checks: i 1 + i 2 = 10 mA 
10 (1) 
i (o+) = 4 = 5 mA 
c (1 1 1) 
P 7.51 For t < 0 
20mA + 
30mA t 
t>O 
5mA! ,,t,.. 4k0 
3k0 
5+20+4 
4k0 ,,V10mA 12k0 
3k0 
2k0 
+ 
\0(0-) = -3(30) 
= -90V 
2k0 
2k0 
12k0 
2k0 
+ 
v ~ 
0 0 .05µ.F 
~ 0 .05µ.F 
v0 ( 00) = 15 V; T =RC= (5 k)(0.05 µ) = 0.25 ms; 
= 15 - 105e-4oo0t V 
P 7.52 [a] Is = i(O+) = 50 mA; 
IsR = v(oo) = 80 
80 
:. R = 0.05 = 1.6k0 
Va=OV 
1 1 
RC= 2500; C = 2500(1600) = 250 nF 
[b] w(t) = ~(250 x 10-9)[80 - 80e-2500t]2 
= 125 x 10-9 (6400)[1 - e-2500t] 2 
= 800[1 - 2e-25oot + e-5000t] µJ 
Problems 7-49 
_!_ = 4000 
T 
Let x = e-25oot. 
' then 800(1 - 2x + x
2] = 0.64(800) 
: . x 2 - 2x + 0.36 = 0 
ThB two solutions are x = 1.8, x = 0.2 
Only the second solution is valid : . e+25oot = 5 
2500t = In 5 so t = 400 In 5 µs = 643. 787 µs 
P 7.53 [a] vc(o+) = 120 V 
[b] Use voltage division to find the final value of voltage: 
150 
Vc(oo) = 150 + 50 (-200) = -150V 
7-50 CHAPTER 7. Response of First-Order RL and RC Circuits 
[c} Find the Thevenin equivalent with respect to the terminals of the 
capacitor: 
VTh = -150V, RTh = 12.5k + 150kll50k = 5okn, 
Therefore r = ReqC = (50,000)(40 x 10-9) = 2ms 
The simplified circuit for t > 0 is: 
R4 F t=~+ r0 R 
1 
cp 
[d] i(o+) = -150 - 120 = -5.4 mA 
50,000 
[e] Ve = Ve( oo) + [vc(O+) - Ve( oo )]e-t/r 
= -150 + [120- (-150)Je-t/r = -150 + 270e-500tv, t ~ O 
[f] i = C~e = (40 x 10-9)(-500)(270e-500t) = 5.4e-500tmA, t ~ o+ 
P 7.54 [a] Use voltage division to find the initial value of the voltage: 
lOk 
Ve(o+) = V10k = lOk + 15k (-75) = -30V 
[b] Use Ohm's law to find the final value of voltage: 
Vc(oo) = V5k = (5 X 10-3)(5000) = 25V 
[c] Find the Thevenin equivalent with respect to the terminals of the 
capacitor: 
VTh = 25V, RTh = 5k + 20k = 25kr2 
[d] Ve= ve(oo) + [ve(O+) - Vc(oo)]e-t/r 
= 25 + (-30- 25)e-400t = 25 - 55e-400tv, t > O 
We want Ve = 25 - 55e-4o0t = 0: 
Therefore t = ln(55/25) = 1 97 ' 400 · ms 
P 7.55 [a) 
3 .2kQ 
(b] i 0 (00) = 0 
[c] T = RC= (10 x 103)(0.2 x 10-6) = 2 ms 
[d) io = 0 + (6 - O)e-5oot = 6e-5oot mA, t > o+ 
(e) V 0 = 60 - 3.2 X 103i 0 = 60 - 19.2e-SOOt V, t > o+ 
P 7.56 [a] vo(o-) = va(o+) = 48 V 
~+i0 2.5k0 7.5k0 
~ ~~---w.~~ 
:. l12v 
0. 0Bµ;F1.__:_Q ______ __.I 
v0 (00) = -12V; T = 0.8ms; 
V 0 = -12 + (48 - (-12))e-1250t 
vo = -12 + 60e-1250ty, t 2: 0 
[h] i 0 = -0.08 X 10-6 (-75,000e-1250t] 
i 0 = 6e-1250t mA, 
[c] Vg = V 0 - 2.5 X 103i 0 
Vg = -12 + 45e-1250t V 
[d] v9 (0+) = -12 + 45 = 33V 
Checks: 
.£ = 1250 
T 
v9 (0+) = i 0 (0+)7.5 x 103 - 12 = 45 - 12 = 33 V 
v 
iwk = 1ik = -1.2 + 4.5e-125ot mA 
Problems 7-51 
7-52 CHAPTER 7. Response of First-Order RL and RC Circuits 
. Vg 0 4 5 -1250t A 
i3ok = 30k = - . + 1. e m 
-io + i10 + i3o + 1.6 = 0 (ok) 
P 7.57 t < O; 
t = oo: 
iJoo) 
i 20kQ 
lOmA! + 20kQ 
+ 
lOkQ V Joo) 
i 0 (00) = -10 (~~) = -4mA; v0 (00) = i 0 (00)(10) = -40V 
RTh = 10kf21140kn = 8kf2; c = 125nF 
T = (8)(0.125) = 1 ms; ! = 1000 
7 
.". v0 (t) = -40 + lOOe-lOOOt V, 
J 20kQ 
lOmA! -.J... 20kQ 
Jo + lOkQ v 0 
. dvo -lOOOt A 
'le= C dt = -12.5e Ill , t 2:: o+ 
i = Vo = -4 + lOe-lOOOt mA 
10 10 ' 
t 2:: o+ 
. . . (4 2 5 -lOOOt) A 
'lo = 'le + 'llQ = - + . e Ill ' 
ic 
~ 125nF 
P 7.58 Fort> 0 
ib = 400(12) = 100 µA 
48 
RTh = 15k0 
15kQ 
v 0 (00) = -45V; 
r = (15,000)(8)10-6 = 120ms; 
V0 = -45 + 45e-8·33t V, t ~ 0 
Problems 7-53 
1/r == 8.33 
w(t) = ~(8 x 10-6)v~ = 8100(1 - 2e-8.33t + e-16.67t) µ,J 
w(oo) = 8100µJ 
. 8100(1 - 2e-8·33to + e-16·67t 0 ) = 0.90(8100) 
1 - 2x + x 2 = 0.90; 
x2 - 2x + 0.10 = 0 
X1,2 = 1.9487, 0.0513 
e-<25/ 3)to = 0.0513; (25/3)t0 =ln19.4868; t0 = 356.4ms 
7-54 CHAPTER 7. Response of First-Order RL and RC Circuits 
p 7.59 Fort< 0, v0 (0) = 80V 
t > 0: 
---j.ill. 20kQ 
+ 30 x103ill. 
VTh BOkQ • lOOV 
( -100 ) VTh = 30 X 103iD,. + 0.8(100) = 30 X 103 l OO X l 03 + 80 = 50 V 
---j.iT 
+ 
VT 
RTh = -. = 40kf2 
'lT 
80kQ 
t>O ~---W.----~ 
20kQ 
+ I + 40kQ I 
0~VrnF V0 t 50V 
v0 = 50 + (80 - 50)e-tfr 
v0 = 50 + 30e-5000tV 
P 7.60 v0 (0) = 50V; 
RTh = 16kf2 
v0 (00) = 80V 
_!_ = 12,500 
7 
v = 80 + (50 - 80)e-12'500t = 80 - 30e-12,500t V 
_!_ = 5000 
7 
P 7.61 (a] 
f 8R = Ri + ~ 1o: i dx + Vo 
di i 
0 = Rdt + C +0 
di+ _i_ = 0 
dt RC 
di dt 
7 = - RC 
ri(t> dy i r 
}i(O+) y = - RC lo+ dx 
i(t) -t 
ln i(o+) =RC 
:. i(t) = (Is - ~) e-t/RC 
Problems 7-~55 
P 7.62 [a] Let i be the current in the clockwise direction around the circuit. Then 
1 ft 1 ft Vg = iR9 + -C i dx + -0 i dx 1 0 2· 0 
- iR9 + (~1 + ~2 ) J:idx = iR9 + ~eJ:idx 
Now differentiate the equation 
di i di 1 
O = Rg dt + Ce or dt + R9 Ce i = O 
Therefore i = i e-t/R9 Ce = i e-t/r; 
g g 
7--56 CHAPTER 7. Response of First-Order RL and RC Circuits 
v2(t) = VgCi (1 - e-tfT); T = RgCe 
C1 +C2 
C2 C1 
[b] v1(oo) = C G Vy; v2(00) = C C Vy 
1+ 2 i+ 2 
P 7.63 [a] t < 0 
60j 
"M 
l 2.2k0 + (60) (0.6) 
0.3µF I (0. 6+0 .3) 
+ 
0 .6µF (60) (0 .3) 
(0. 6+0. 3) 
t>O 
5k0 
I 
+ + 'l/V'v =7·J 
60V v ~lOOV 
"-~1- _" I 
vo(o-) = vo(o+) = 60V 
v0 (00) = lOOV 
= 40V 
= 20V 
T = (0.2)(5) x 10-3 = 1 ms; 1/r = 1000 
v0 = 100 - 40e-lOOOt V, t 2 0 
[b] i 0 = -C~0 = -0.2 X 10-6(40,000e-lOOOt) 
-106 lot [c] V1 = ~- -8 x 10-3e-iooox dx +40 
0.3 0 
= 66.67- 26.67e-1000tv, t 2 o 
-106 lot [d] v2 = -- -8 x 10-3e-1000x dx + 20 
0.6 0 
= 33.33 - 13.33e-1000t V, t 2 0 
Problems 7-57 
1 -6 2 1 -6 2 
[e] Wtrapped = 2(0.3)10 (66.67) + 2(0.6)10 (33.33) 
= 666.67 + 333.33 = 1000 µJ. 
120 
P 7.64 v0 (0) = 120 (80) = 80V 
v0 (00) = -6(25) = -150V 
.!. = 1000 
7 
v0 = -150 + (80 + 150)e-lOOOt = -150 + 230e-lOOOt V, t ~ 0 
P 7.65 [a] From Example 7.10, 
L = L 1L 2 - M 2 = (8m)(20m) - (10m)2 = 7.5 mH 
eq L 1 + L2 - 2M 8m + 20m - 2(10m) 
Leq (7.5m) 1 
7 = If = 75 = 10,000 
io = 15 - 15 e-10,000t = 0.2 - o.2e-lO,OOOt A t >_ 0 
75 75 
[b) V 0 = 15 - 75i0 = 15 - 75(0.2 - 0.2e-l0,000t) = 15e-lO,OOOt V t ~ o+ 
di1 di2 
[c] V 0 = 0.008-d + 0.01-d t t 
di0 di1 di2 
dt=dt+dt 
di2 = dio _ di1 = 2000e_10,oo0t _ di1 
ili ili ili ili 
15e-10,000t = o.oos~; + 0.01 ( 2oooe-10,000t - ~;) 
d. 
~ = 250oe-10,o00t 
dt 
di1 = 2500e-lO,OOOt dt 
ri1 rt lo dx = 2500 lo e-lO,OOoy dy 
e-10,000y It 
. ._ ii = 2500 = 0.25 - 0.25e-10,oo0t A, 
-10,000 0 
t~O 
7··58 CHAPTER 7. Response of First-Order RL and RC Circuits 
p 7.66 
(d] i2 - i 0 - i1 
- 0.2 - 0.2e-lO,OOOt - 0.25 + 0.25e-lO,OOOt 
- -50 + 5oe-rn,o00t mA, t 2:'.: o 
[e] Vo L di2 + Mdi1 
- 2dt dt 
- 0.02(-500e-10•000t) + 0.01(2500e-10•000t) 
15e-l0,000ty, t > o+ (checks) 
i 1 (0) = 0.25 - 0.25 = O; agrees with initial conditions; 
i 2 (0) = -0.05 + 0.05 = O; agrees with initial conditions; 
The final values of i 0 , i 1, and i 2 can be checked via the conservation of 
Wb-turns: 
i0 (oo)Leq = 0.2 x (7.5m) = 1.5 mWb-turns 
ii(oo)L1 + i2(oo)M = 0.25(8m) - 0.05(10m) = 15 mWb-turns 
i2(oo)L2 + i1(oo)M = -0.05(0.02) + 0.25(0.01) = 15 mWb-turns 
Thus our solutions make sense in terms of known circuit behavior. 
[ J L = (3)(l5) = 2.5H a eq 3 + 15 
Leq 2.5 1 
r = R = 7.5 = 3s 
120 
i 0 (0) = O; i 0 (00) = 7_5 = 16A 
.'. i 0 = 16 - 16e-3t A, t ~ 0 
V 0 = 120 - 7.5i0 = 120e-3t V, 
. - 1 ft 120 -3xd - 40 40 -3tA 
i1 - 3 lo e x - 3 - 3e ' 
. . . 8 8 -3tA t > 0 
i2 = io - i1 = 3 - 3e ' -
t~O 
[h] i 0 (0) = i 1 (0) = i 2 (0) = 0, consistent with initial conditions. 
v0 (0+) = 120 V, consistent with i 0 (0) = 0. 
di1 3t 
Vo = 3dt = 120e- V, 
or 
Problems 7-59 
The voltagesolution is consistent with the current solutions. 
>..1 = 3i1 = 40 - 40e-3t Wb-turns 
.A2 = 15i2 = 40 - 40e-3t Wb-turns 
. ·. A1 = >..2 as it must, since 
d>..1 d.A2 
Vo= dt = dt 
.A.1(00) = >..2(00) = 40 Wb-turns 
>..1(oo) = 3i1(00) = 3(40/3) = 40 Wb-turns 
>..2( oo) = 15i2( oo) = 15(8/3) = 40 Wb-turns 
:. i 1(oo) and i2 (oo) are consistent with >..1(oo) and .:\2(00). 
P 7.67 [aJ From Example 7.10, 
L L 1L2 - M
2 0.125 - 0.0625 0 H - - -5m 
eq - L1+L2+2M - 0.75 + 0.5 -
L 1 1 
r = R = 5000; -:;. = 5000 
:. i 0 (t) = 40- 40e-5oo0t mA, t 2: 0 
[b) V0 = 10 - 250i0 = 10 - 250(0.04 + 0.04e-5ooot = 10e-5oo0t V, 
di1 di2 5000t 
(c) V 0 = 0.5-d - 0.25-d = lOe- V 
t t 
dio = di1 di2 = 2ooe-5000t A/s 
dt dt + dt 
di2 = 2ooe-5000t - di1 
dt dt 
10e-5000t = 0.5~; - 50e-5000t + 0.25~; 
di1 = 80e-5000t dt 
rt1 rt lo dx = lo 80e-5000y dy 
i = e-5oooy = 16 - 16e-5oo0t mA so It 
1 -5000 0 ' 
t~O 
7-60 CHAPTER 7. Response of First-Order RL a.nd RC Circuits 
[d] i2 = io - it = 40 - 40e-5000t - 16 + 16e-5000t 
= 24 - 24e-5o00t mA, t 2: 0 
[e) i 0 (0) = i1 (0) = i2(0) = 0, consistent with zero initial stored energy. 
Vo= Leq ~; = (0.05)(200)e-5000t = 10e-5000t v, t 2: o+ (checks) 
Also, 
di1 di2 5000t 
V0 = 0.5-d - 0.25-d = lOe- V, t 2: 0+ (checks) t t 
di2 di1 -5000t + 
Vo= 0.25-d - 0.25-d = lOe V, t 2: 0 (checks) t t 
v0 (0+) = 10 V, which agrees with i0 (0+) = 0 A 
i 0 (00) = 40mA; i0 (oo)Leq = (0.04)(0.05) = 2 mWb-turns 
ii(oo)L1 +i2(oo)M = (16m)(500) + (24m)(-250) = 2 mWb-turns (ok) 
i2(oo)L2 + ii(oo)M = (24m)(250) + (16m)(-250) = 2 mWb-turns (ok) 
Therefore, the final values of i0 , ii, and i2 are consistent with 
conservation of flux linkage. Hence, the answers make sense in terms of 
known circuit behavior. 
P 7.68 [a] Leq = 4 + 8 - 2(5) = 2 H 
1 - = 25 
T 
i = 4- 4e-25t A 
' t 2: 0 
[b] v1(t) = 4di -5di = - di= -(10oe-25t) = -10oe-25tV t 2: o+ 
dt dt dt ' 
di di di 
[c] v2(t) = 8-d - 5-d = 3-d = 3(10oe-25t) = 300e-25t v, t 2: o+ 
t t t 
[d] i(O) = 4 - 4 = 0, which agrees with initial conditions. 
200 = 50i1 + v1 + v2 = 50(4 - 4e-25t) - 10oe-25t + 300e-25t = 200V 
Therefore, Kirchhoff's voltage law is satisfied for all values oft> 0. 
Thus, the answers make sense in terms of known circuit behavior. 
P 7.69 (a] Leq = 4 + 8 + 2(5) = 22 H 
.!. = 2.273 
T 
i = 4 - 4e-2·273t A, t 2: O 
Problems 7-61 
di di di 
[b] v (t) = 4- + 5- = 9- = 9(9.09e-2·273t) = 81.8le-2·273t v t ;::::: o+ 1 dt dt dt ' 
[c] v (t) = 8di +5di = 13di = 13(9.09e-2·273t) = ll8.l8e-2·273tv t:::::: o+ 2 dt dt dt ' -
[d} i(O) = 0, which agrees with initial conditions. 
200 = 50i1 + v1 + v2 = 50(4 - 4e-2·273t) + 81.8le-2·273t + l18.l8e-2·273t = 200V 
Therefore, Kirchhoff's voltage law is satisfied for all values oft ;::::: 0. 
Thus, the answers make sense in terms of known circuit behavior. 
p 7.70 t < 0: 
0 ::; t ::; 1: 
4Q 
T = 5/0 = 00 
2Q 
iJO-) 
.t lOmH 
1.25A 
iL(t) = l.25e-t/oo = 1.25e-0 = 1.25 
iL(t) = l.25A 
1 ::; t < oo: 
2Q .t lOmH 
1.25 
10 x 10-3 
r = = 5ms· 
2 ' 
1/r = 200 
iL(t) = l.25e-20°Ct-l) A 
7-62 CHAPTER 7. Response of First-Order RL and RC Circuits 
P 7.71 0 ~ t < 3µs: 
+ 
2kQ v0 5nF 
T =RC= (2 x 103)(5 x 10-9 ) = 10 µs; ljT = 100,000 
v0 (0) = OV; v0 (00) = 6V 
V 0 = 6 - 6e- I00,000t V o:::;t:::;3µs 
3µs ~ t < oo: 
2kQ 
Bk QI 
-snF :12v 
t = oo: 
i = 6 - (-l2) = l.8mA 
10 
V 0 ( 00) = 6 - 2i = 2.4 V 
v0 (3 µs) == 6 - 6e-0·3 = 1.555 V 
V 0 = 2.4 + (1.555 - 2.4)e-{t- 3 µS)/r 
RTh = 2kf2ll8H2 = l.6kf2 
T = (1.6)(5) = 8µs; ljT = 125,000 
Vo= 2.4 - 0.845e-125,000(t-3µS) 3 µs :::; t < oo 
P 7. 72 For t < 0: 
SQQ 
lSOQ 
i(O) = 80(150) = 60mA 
200 
0 < t S 250 µs: 
SQQ 
J,, 2SmH 
i 
i(250µs) = 60e-0·5 = 36.39mA 
250 µs St S 650 /J,S: 
SQQ 
2SmH 
R = (50)(75) = 300 
eq 125 
~ = R = 30 x 103 = 1200 
T L 25 
i = 36.39e-1200(t-250x10-6) mA 
650µs < t < oo: 
7SQ 
i(650/J,S) = 36.39e-0.4S = 22.52mA 
Problems 7-63 
J,, 
i ( 0) 
7-64 CHAPTER 7. Response of First-Order RL and RC Circuits 
i = 22.52e-2000(t-650x10-s) mA 
di 
v=L-· 
dt' 
L = 25mH 
~; = 22.52(-2000) x 10-3e-2000(t-650x10-6 ) = _ 45.04e-2000(t-650x10-6 ) 
v = (25 X 10-3)( -45.04)e-2000(t-650x10-s) 
= -l.l3e-2000(t-650x10-6 ) V, t > 650+ µs 
v(lms) = -l.13e-2ooo(35o)xl0-s = -559.12mV 
P 7.73 From the solution to Problem 7.72, the initial energy is 
1 
w(O) = 2(25 rnH)(60 rnA)2 = 45µJ 
For 650 /J,S :::; t < oo: 
w(t) = !(25 mH)(22.52e-2ooo(t-55oxrn-s) mA)2 = (0.04)(45 µJ) 
2 
Solving, 
t = 964. 72 /J,S 
p 7.74 0:::; t:::; 50 µs; 
+ 
• 120kQ ~ 40kQ 
+ 
5 ~ 
9nF 494.6mV vc 
600kQ • .. 200kQ 
Re= 72011240 = 180kf!; T = (~) (180) = lOOµs 
Ve= 494.6e-lO,OOOt mV 
vc(50µs) = 494.6e-0·5 = 300rnV 
50 µs :::;; t < oo: 
-t + 120kQ v1 40kQ 
+ il isw -
9 nF~ 300mV 
-t + 600kQ v2 200kQ 
i2 
Re= 1201140 + 600Jl200=30+150 = 180k0 
T = (~) (180) = lOOµs; .!. = 10,000 
T 
Ve= 300 -10,000(t-50µs) mV 
30 v = _ = 50e-10,000(t-50µs) mV 
1 180 c 
v = 150 J = 250e-10,000(t-50µs)mV 
2 180 c 
i - 1 = 416 7e-10,000(t-50µs) nA 
1 - 120 103 . 
i - 2 = 416.7e-10,000(t-50µs) nA 
2 - 600 103 
isw(lOO µ ) = OA 
P 7. 75 (a] t < 0: 40 
500Vi ~ 
i = 500 = 5A 
g 4+96 
i(o-) = 5( 480) = 4 A = i(o+) 
600 
120Q 
480Q 
Problems 7-65 
7 ~66 CHAPTER 7. Response of First-Order RL and RC Circuits 
[b] 0 ~ t ~ lOOµs: 
i = 4e-t/r 
_!_ = R = 120 + 9611480 = 10 OOO 
T L 20 x 10-3 ' 
i = 4e-10,000t 
i(25µs) = 4e-104C25)xio-6 = 4e-0·25 = 3.12 A 
[c] i(lOOµs) = 4e-1 = 1.47 A 
lOOµs ~ t < oo: 
-7i 120Q 
20mH 
1 R 120 
- = - = - x 103 = 6000 
T L 20 
i = l.47e-6000(t-10ox10-6 ) A 
i(200ps) = l.47e-6ooo(IOO)xl0-s = l.47e-0·6 = 807.59mA 
[d] 0 ~ t ~ 100 /1S: 
i = 4e-lO,OOOt 
v = L di= (20 x 10-3)(4)(-104)e-104t = -80oe-104ty 
dt 
v(loo-µs) = -8ooe-la4(lOOxl0-6 ) = -80oe-1 = -294.30 v 
[e] lOOµs < t < oo: 
i = l.47e-6000(t-1oox10-6 ) 
v = (20 x 10-3)(1.47)(-6000)e-6000(t-1oox10-a) 
= -176.58e-6000(t-1oox10-a) V 
v(10o+µs) = -176.58V 
Problems 7-67 
p 7.76 t < 0: ..---------'Vl/'v • 
4. 7kQ + 
50mA t 2kQ 
vc(o-) = (50)(2000) x 10-3 = lOOV = vc(o+) 
0 :St< 250ms: 
+ 
lOOV 
r = oo; 1/r = O; v0 = lOOe-0 = 100 V 
250ms < t < oo: 
+ 
lOOV='" 0. 16µF ,. 6. 25kQ 
T = (6.25)(0.16)10-3 = 1 ms; 1/r = 1000; Vo = lOOe-lOOO(t-0.25) V 
Summary: 
V 0 = lOOV, 0::; t :S 250ms 
Vo= lOOe-lOOO(t-0.25) V, 250ms ::=; t < oo 
7-68 CHAPTER 7. Response of First-Order RL and RC Circuits 
P 7.77 Note that fort> 0, v0 = (35/40)ve, where Ve is the voltage across the 50 nF 
capacitor. Thus we will find Ve first. 
t<O 
35kQ SkQ r -¥/1,----------.+ 
280¥ ~ v lOkQ c: 
280 
Vc(O) = 50(10) = 56 V 
0 _::; t :S 400 /IS: 
SkQ 
+ 
35kQ SOnF ~ v JO) 
Re= (10)(40) = 8k0 
50 
T = (8 X 103)(50 X 10-9 ) = 400 µs, 
Ve = 56e-2500t V, t 2: 0 
ve(400µs) = 56e-1 = 20.60V 
400 µs :S t :S 1.4 ms: 
SkQ 
T = (40 x 103)(50 x 10-9 ) = 2ms, 
Ve = 20.60e-500(t-400x10-6) V 
lOkQ 
2:. = 2500 
T 
1 - = 500 
T 
1.4ms :::; t < oo: 
5kQ 
35kQ 
T = 400µs, ..!:. = 2500 
T 
+ 
10k0 
vc(L4ms) = 20.60e-5oo(i4oo-4oo)rn-e = 20.60e-0·5 = 12.50V 
vc(L6ms) = 12.50e-25oo(1.6-1.4)rn-3 = 12.50e-0·5 = 7.58 V 
V 0 = (35/40)(7.58) = 6.63V 
P 7.78 w(O) = ~(50 x 10-9)(56)2 = 78.4µJ 
0 $ t ::;; 400 J-tS: 
Ve = 56e-2500t : v2 = 3136e-5ooot 
c 
400x10-6 
w10k = J 0 3136 x 10-4e-5ooot dt 
e-5000t 
1
4oox10-6 
= 3136 x 10-4 -5000 0 
= -6272 x 10-8 (e-2 -1) = 54.23µJ 
1.4 ms ::;; t < oo: 
Problems 7-69 
Ve = 12.50e-2500(t-1.4x10-3) V; v; = 156.13e-5000(t-1.4x10-3) 
W10k = / 00 156.13 X 10-4e-5ooo(t-1.4xrn-3 ) dt 
1.4x10-s 
e-5000(t-1.4x10-3 ) loo 
= 156.13 x 10-4 _____ _ 
-5000 1.4x10-s 
= -311.83 x 10-8 (0 - 1) = 3.12 µJ 
7--70 CHAPTER 7. Response of First-Order RL and RC Circuits 
W10k = 54.23 + 3.12 = 57.35 µJ 
57.35 
% = 78.4 (100) = 73.15% 
To check, find the energy dissipated in the 40 kO resistance: 
0 ::; t < 400 µs: 
v 2 = 3136e-5ooot c 
3136 -3 -5000t 
P4ok = 4o x 10 e , 
. e-5000t 
1
4oox10-6 
W4ok = 784 X 10-4 _ 5000 O 
= -156.8(10-7)(e-2 -1) = 13.56mJ 
400 µs < t -:::;1 ms: 
Ve = 20.60e-500(t-400x10-6); 
10-a 
W4ok = 106.10 X 10-4 { 
}4oox10-a 
v~ = 424.41e-1000(t-4oox10-6) 
e-lOOO(t-400x 10-6) 
1
10-3 
= 106.10 x 10-4------
-1000 400x 10-a 
= -106.10(10-7)(e-0·6 - 1) = 4.79mJ 
1.4ms s; t < oo: 
v~ = 156.13e-5000(t-1.4x10-a) 
Note in this interval the energy dissipated in the 40k resistor will be 1/4th 
that dissipated in the lOk resistor. 
1 
W4ok = 4(3.12) = 0.78 µJ 
W4ok = 13.56 + 6. 71 + 0. 78 = 21.05 µJ 
W4ok + W1Qk = 57.35 + 21.05 = 78.40 µJ 
Problems 7-71 
P 7. 79 [a] 0 s t s 2 µs 
20kQ 
[h] 
0-----'tM·-----o 
+ + 
100V iLi 40mH V0 
iL(O) = O; iL(oo) = 5mA 
L 0.04 
7 = R = 20,000 = Zµs 
h = 5 - 5e-500,000t mA, 0 S t S 2 µs 
V 0 = {0.04)((500,000)(0.Q05)e-500,000tJ = 100e-500,000t V, 
2µsst<oo 
20kQ 
iL(2µs) = 5- 5e-1 ::::::: 3.16mA 
iL( oo) = O; r = 2 µs; 1/r = 500,000 
iL = O + (3.16 _ O)e-500,ooo(t-2µs) mA 
= 3.16e-5oo,ooo(t-2µS) mA, 2 µs St< oo 
Vo= L ~~ = (0.04)(3.16 X 10-3)l-500,000e-5oo,ooo(t-2µs)J 
= (-5)(4)(3.16)e-500,000(t-2 µS) 
= _ 63.21e-5oo,ooo(t-2µS) V, 
V0(V) 
120 
2µs<t<oo 
36.79V 
o+ st< 2µs 
100 
80 
60 
40 
20 
0 -t--~---r-----1------.-------,---~t(~µ....:.,s) 
-20 
-40 
-60 -63.21V 
-80 
7-·72 CHAPTER 7. Response of First-Order RL and RC Circuits 
[c] v0 (4µs) = -23.25V 
+23.25 
i 0 = 20 OOO = l.16mA 
' 
P 7.80 [a] i 0 (0) = O; i 0 ( oo) = 25 mA 
~ = R = SOOO x 103 = 32,000 
T L 250 
i 0 = (25 - 25e-32·000t) mA, 0 < t :::; 50 µs 
v0 = 0.25~; = 2ooe-32·000tv, O:::; t:::; 50µs 
50 µs:::; t < oo: 
i 0 (50µs) = 25 - 25e-l.6 = 19.95mA; 
io = 19.95e-32,ooo(t-5ox10-6) mA 
Vo= (0.25) ddio = -159.62e-32,000(t-50µs) 
t 
V 0 - 0 
io(oo) = 0 
:. t < 0: 
osts50µs: 2ooe-32,oo0t V 
50 µs S t < oo : Vo - -159.62e-32,000(t-50µS) 
[b) vo(50-µs) = 200e-i.6 = 40.38V 
va(so+ /LS) = -159.62 V 
[c) i 0 (50-µs) = i 0 (50+µs) = 19.95mA 
P 7.81 [a] 0 s t s 6 ms: 
ve(O+) = O; Ve( oo) = 40 V; 
RC= 500 x 103(0.02 x 10-6) = lOms; 
Ve = 40 - 40e-lOOt 
Vo = 40 - 40 + 40e-lOOt = 40e-lOOt V, 
6ms St< oo: 
ve(6ms) = 40-40e-0·6 =18.0SV 
Ve(oo)=OV 
r = lOms; 1/r = 100 
Ve= 18.05e-100(t-0.006) V 
Va= -Ve= -18.05e-100(t-0.006) V, 
1/RC = 100 
Osts6ms 
t:;::: 6ms 
[b) 
V0(V) 
50 
40 
30 
20 
10 
0 
-10 2 4 6 
-20 
L 
-30 
P 7.82 [a] t < O; v0 = 0 
0::; t::; lOms: 
r = (50)(0.4) x 10-3 = 20ms; 1/r = 50 
[b] 
V0 = 40 - 40e-50t V, 0 ::; t ::; 10 ms 
v0 (10ms) = 40(1- e-0·5 ) = 15.74 V 
10 ms::; t < 20ms: 
V0 = -40 + 55.74e-5o(t-O.Ol) V 
v0 (20ms) = -40 + 55.74e-0·5 = -6.19V 
20 ms ::;; t < oo: 
Vo= -6.19e-50(t-0.02) V 
0 ,. 
40 ---------
30 
20 
-10 
-20 
-30 
-40 
t (ms) 
[c] t::; 0: v0 = 0 
0::; t < lOms: 
r = 10(0.4 x 10-3 ) = 4 :ms 
Problems 7-73 
t(ms) 
8 10 
7-7 4 CHAPTER 7. Response of First-Order RL and RC Circuits 
0 $ t $ lOms 
v0 (10ms) = 40-40e-2·5 = 36.72V 
lOms $ t $ 20ms: 
v0 = -40 + 76.72e-25o(t-O.Ol) V, lOms $ t $ 20ms 
v0 (20ms) = -40 + 76.72e-2·5 = -33.7V 
20ms $ t $ oo: 
Vo = -33. 7 e-250(t-0.02) V, 20ms::; t $ oo 
40 ---------
30 
20 
-10 
-20 
-30 
-40 
t (ms) 
P 7.83 (a] r = RC = (8000)(100) x 10-9 = 800 /ls; 
i 0 = V 0 = 0 t < 0 
io(O+) = 20 (~) = 15mA, 
. ·. i 0 = 15e-1250t mA 
i6kO = 20 - 15e-1250t mA 
: • V 0 = 120 - 90e-1250t V 
i 0 (00) = 0 
Ve = Vo - 2 X 103i 0 = 120 - 120e-1250t V 
Vc(0.5ms) = 120 -120e-0·625 = 55.77V 
. . ( + ) -55. 77 A 
.. i 0 0.5 ms = 8 = -6.97m 
i 0 (00) = 0 
l/r = 1250 
0 $ t $ 0.5ms 
io = -6.97e-1250(t-500µs) mA, 0.5+ms $ t < oo 
Vo = -6000io = 41.83e-1250(t -500µs) V o.s+ ms $ t < 00 
Problems 7-75 
Summary part (a) 
i 0 = 0 t<O 
io = 15e-1250t mA (o+ St S 0.5- ms) 
io = -6.97e-125o(t-500µs) mA 0.5+ms < t < oo 
V0 = 0 t<O 
v0 = 120 - 90e-1250t V, 0 St S 0.5-ms 
Vo= 41.83e-1250(t-500µs) V, 0.5+ms St< oo 
[h] io(o-) = 0 
io(o+) = 15mA 
io(0.5- ms)= 15e-0·625 = 8.03mA 
io(0.5+ ms)= -6.97mA 
[c] vo(o-) = 0 
vo(o+) = 30V 
vo(0.5- ms) = 120 - 90e-0·625 .:__ 71.83 V 
vo(0.5+ ms) = 41.83 
(d] 
15 
io (mA) 
10 
5 
t (ms) 
-2 -1.5 -1 -0.5 05 ·2 
-5 
-10 
fe1 
VO (V) 80 
70 
60 
50 
40 
30 
20 
10 
t (ms) 
-2 -1.5 -1 -0.5 0 0.5 1.5 2 
7-76 
p 7.84 
CHAPTER 7. Response of First-Order RL and RC Circuits 
+ - Vi;p + 
6kQ 
VT= 2000iT + 6000(iT + 15 x 10-4v<1>) = 8000iT + 9vq, 
= 8000iT + 9( -2000ir) 
VT -. = -10,000 
iT 
25mA SH 
-lOkQ 
t ..._____., 
8 
T = -- x 10-3 = -08ms· -10 . ' 
i = 25el250t mA 
: . 25e1250t x 10-3 = 12; 
ljT = -1250 
ln480 
t = -- = 4.94ms 
1250 
p 7.85 t > 0: 
v 
T 
3 
75x10 iA 
25kQ 
r =RC= -40 x 103 (0.025) x 10-6 = -10-3 
Ve = 25e1000t V; 25e1000t = 50,000 
lOOOt = ln 2000 t = 7.6ms 
P 7.86 [a] 
VT= 2000iu 
iu = ~ (iT + /3iu) = 0. 75ir + 0. 75/3iu 
iu(l - 0.75/3) = 0.15iT 
. 0.75ir 
i - . 
u - 1 - 0. 75/3 ) 
2000i = 1500ir 
u (1 - 0. 75,B) 
VT 1500 
RTh = ir = 1 _ O. 7513 = -3000 
1 - 0. 75,B = -0.5 
[b] Find VTh; 
:. /3 = 2 
2kQ 
VTh - 120 VTh - 2(VTh -120) -
2000 + 6000 2000 - 0 
VTh = 180V 
Problems 7-77 
7-78 CHAPTER 7. Response of First-Order RL and RC Circuits 
-3kQ r 0' 180~ l 
di 
180 = -3000i + 0.3 dt 
300mH 
~~ = 600 + 10,000i = 10,000( i + 0.06) 
di 
i + 0.06 = 10,000 dt 
fi d~ 06 = ft 10,000 dx lo x + . lo 
. ·. i = -60 + 60e1o,ooot mA 
~~ = (60 x 10-3)(10,000)e10•000t = 600e10•000t 
d" 
v = 0.3 d; = 180e10'000t = 36,000; elO,OOOt = 200 
ln200 
. ·. t = lO,OOO = 529.83 µs 
P 7.87 Find the Thevenin equivalent with respect to the terminals of the capacitor. 
RTh calculation: 
+ 
8kQ 
. Vy Vy Vy 
Zy = 8000 + 20 000 - ll.5 20 000 
' ' 
Zy 2.5 + 1 - 11.5 -8 
-= ---
20,000 20,000 
Vy = -20,000 = _ 25000 
ir 8 
4kQ 
i 16kQ 
i.!I. 
Open circuit voltage calculation: 
11.5i + 
8k0 v oc 
Voe Voe - V1 _ ll.fii = O 
8000 + 4000 6. 
V1 - Voe+ V1 ~ 1.6 X lQ-3 = O 
4000 16,000 
. V1 
't6_ = 16,000 
Solving, Voe = -12.4 V 
-25000 
16nF~l2.4V 
1- I 
vc(O) = O; Ve(oo) = -12.4 V 
4kQ 
-t 16k0 
iA 
T =RC= (-2500)(16 x 10-9 ) = -40 x 10-6 ; 
Ve = -12.4 + 12.4e25,000t = 930 
e25,000t = 76; 25,000t = In 76; t = 173.23µs 
P 7.88 [a] 
+ 
T = (25)(2) x 10-3 = 50ms; 1/7 = 20 
Problems 7-79 
+ 
v1 
1 
- = -25,000 
T 
7-·80 CHAPTER 7. Response of First-Order RL and RC Circuits 
p 7.89 
Vc(oo) = 0 
. 80 -20t = 5· . . e ' e20t = 16; 
ln16 
t = W = 138.63ms 
[bJ o+ < t < 138.63 ms: 
i = (2 x 10-6)(-1600e-20t) = -3.2e-20tmA 
t > 138.63+ ms: 
T = (2)(4) x 10-3 = 8ms; 1/T = 125 
Vc(138.63+ ms)= 5 V; Vc(oo) = 80V 
Ve= 80 _ 75e-125(t-0.13863) V, t 2: 138.63 ms 
i = 2 x 10-6(9375)e-125(t-O.I3863) 
= 18.75e-125<t-o.i3s63) mA, t 2:: 138.63+ ms 
[cJ 80 - 75e-125At = 0.85(80) = 68 
80 - 68 = 75e-125At = 12 
e125At = 6.25; b.t = In 6'25 rv 14 66 ms 
125 . 
-250 x 106t 
Vo= R 
R = (-250 x 106)(3 x 10-3) = 50 x 103 = 50kr2 
-15 
p 7_90 0 - 15 _ Cdv0 = O· 
R dt ' 
-15 
dv0 = RC dt 
p 7.91 
-15 -250 x 106t 
Vo = RC t + v0 (0) = R + 5 = -15 
R = 250 x 106 (8 x 10-3) = lOOkO 
20 
[a] CdvP vP - vb _ O· 
dt + R - ' 
therefore 
But Vn = Vp 
therefore 
dvn Vn dvp Vp Vb 
Therefore dt + RC = dt + RC = RC 
Problems 7-81 
Therefore dv0 1 ( ) dt = RC Vb - Va ; 
1 ft 
V 0 = RC lo (Vb - Va) dy 
[b] The output is the integral of the difference between vb and Va and then 
scaled by a factor of 1 /RC. 
1 ft 
[c] Vo= RC lo (Vb - Va) dx 
RC= ( 40) x 103 (25) x 10-9 = 1 ms 
Vb - Va= 50mV 
ft 
V0 = 50 lo dx = 50t; 50tsat = 12; tsat = 240 ms 
p 7.92 - l 5(20) - 6 v 
V2 - (50) -
6+4 d 
50,000 + C dt (6 - Vo) = O 
dv0 10 X 106 
dt 50,000(0.5) = 400 
dv0 = 400dt; V0 = 400t + v0 (0) 
V 0 (0) = 6-16 = -lOV 
:. V 0 = 400t - lOV 
0 = 400t0 -10 
10 
t0 = - = 25ms 
400 
7-82 CHAPTER 7. Response of First-Order RL and RC Circuits 
1 rt 
P 7.93 V0 = RC Jo (Vb - Va) dy + v0 (0) 
RC= (40 x 103)(12.5 x 10-9) = 500 x 10-6 = 0.5ms 
1 
RC= 2000; Vb - Va= 10 - (-5) = 15mV 
v0 (0) = 15- 45 = -30mV 
v0 = (2000)(15) X 10-3t - 30 X 10-3 = (30,000t - 30) m V 
V2 = 10 + (15 - 10)e-2000t mV = [10 + 5e-2000t} mV 
VJ = Vo - Vp = (30,000t ~ 40 - 5e-2000t) m V 
P 7.94 (a] RC = 40(50)x 10-6 = 2 ms; 
[b] 0 :S t :S 50 ms : 
v0 = -500 ht -0.50dx = 250tV 
[c] 50ms :St :S lOOms; 
v0 (0.05) = 250(0.05) = 12.5 V 
1 
RC= 500; V 0 = 0, t < 0 
v0 (t) = -500 lot 0.50dx + 12.5 = -250(t - 0.05) + 12.5 = -250t + 25 V 
0.05 
(d] lOOms < t < oo: 
v0 (0.l) = -25 + 25 = OV 
v0 (t) = OV 
14 
12 
10 
4 
2 
00 20 40 60 80 100 
t (ms) 
Problems 7-83 
P 7.95 Write a KCL equation at the inverting input to the op amp, where the voltage 
is 0: 
Note that this first-order differential equation is in the same form as Eq. 7.50 
if ls= -v9 / Ry,. Therefore, its solution is the same as Eq. 7.51: 
[a) V 0 = 0, t < 0 
[b] R1C1=(4x106 )(50 x 10-9 ) = 0.2; 
-v9 R1 =-(-0.5)(4x106 ) = 50 
Ri 40,000 
Vo= Vo(O) = 0 
Vo = 50 + (0 - 50)e-5t = 50(1 - e-5t) V, 0St:S50ms 
[c] - 1- =5 
R1C1 
-u9 R1 -(0.5)(4 x 106 ) 
Ri = 40,000 = - 50 
Vo = v0 (0.05) = 50(1 - e-0·25) ~ 11.06 V 
V 0 = -50 + [11.06 - (-50)Je-5(t-o.o5) 
= 61.06e-5<t-o.o5) - 50V, 50ms St S lOOms 
1 
[d] -=5 
R1C1 
Vo= Vo(0.10) = 61.06e-0·25 - 50 ~ -2.45 V 
Vo= 0 + (-2.45 - O)e-5(t-O.l) = -2.45e-5(t-O.l) V, lOOms::; t:::; oo 
7-84 CHAPTER 7. Response of First-Order RL and RC Circuits 
14 
12 
10 
8 
-2 
-4 
40 60 
t (ms) 
- wfo fb resistor - - wf fb resistor 
P 7.96 [a] RC= (200 x 103 )(25 x 10-9) = 5 x 10-3; 
0 :::; t :::; 5 µs: 
Vg = 0.6 X 106t 
v0 = -200 lot 0.6 X 106x dx + 0 
x2 It = -12 x 107 - = -6 x 107t 2 
2 0 
80 
1 
-=200 
RC 
v0 (5µs) = -6 X 107 (5 X 10-6)2 = -1.5 X 10-3 y 
5 µs :::; t :'.S 15 µs: 
Vg = 6 - 0.6 X 106t 
V 0 = -200 rt (6- 0.6 X 106x) dx - 1.5 X 10-3 
lsx10-s 
= - [12oox It + 12 x 101 x2 It ] - 1.5 x 10-3 
5x10-6 2 5xlo-s 
= -1200t + 6 x 10-3 + 6 x 107t 2 - 1.5 x 10-3 - 1.5 x 10-3 
= 6 x 107t 2 - 1200t + 3 x 10-3 
v0 (15 µs) = 6 x 107 (15 x 10-6 ) 2 - 1200(15 x 10-6) + 3 x 10-3 
= -1.5 x 10-3 
l511s :'.S t :'.S 20 µs: 
Vg = -12 + 0.6 X l06t 
V 0 = -200 rt (-12 + 0.6 X l06x) dx - 1.5 X 10-3 
}15x10-s 
= - [2400x It -12 x 107 x2 It ] - 1.5 x 10-3 
l1sx10-s 2 l1sx10-s 
= 2400t - 36 x 10-3 - 6 x 107 t2 + 13.5 x 10-3 - 1.5 x 10-3 
= -6 x 107t2 + 2400t - 24 x 10-3 
Problems 7-85 
v0 (20p,s) = -6 X 107(20 X 10-6) 2 + 2400(20 X 10-6) - 24 x 10-3 = 0 
[b] 
0.0 
0 5 10 15 20 
-0.5 
-1.0 
> .s -1.5 
0 
> 
-2.0 
-2.5 
-3.0 
t (us) 
[c) The output voltage will also repeat. This follows from the observation that 
at t = 20 µs the output voltage is zero, hence there is no energy stored in 
the capacitor. This means the circuit is in the same state at t = 20 µs as 
it was at t = 0, thus as v9 repeats itself, so will v0 • 
P 7.97 [a] While T2 has been ON, C2 is charged to Vee, positive on the left terminal. 
At the instant T1 turns ON the capacitor C2 is connected across b2 - e2 , 
thus Vbe2 =-Vee. This negative voltage snaps T2 OFF. Now the polarity 
of the voltage on C2 starts to reverse, that is, the right-hand terminal of 
C2 starts to charge toward +Vee. At the same time, C1 is charging 
toward Vee, positive on the right. At the instant the charge on C2 
reaches zero, Vbe2 is zero, T2 turns ON. This makes Vbel =-Vee and T1 
snaps OFF. Now the capacitors C1 and C2 start to charge with the 
polarities to turn T1 ON and T2 OFF. This switching action repeats itself 
over and over as long as the circuit is energized. At the instant T1 turns 
ON, the voltage controlling the state of T2 is governed by the following 
circuit: 
+ R2 
v cc + 
c2 I :cc vbe2 ·-
It follows that vbe2 =Vee - 2Veee-t/R2e2 _ 
[b] While T2 is OFF and T1 is ON, the output voltage Vce2 is the same as the 
voltage across Ci, thus 
7-86 CHAPTER 7. Response of First-Order RL and RC Circuits 
+ RL 
v - + cc 
I cl I v ce2 ·-
It follows that Vce2 = Vee - V00e-t/ RLCi. 
[c] T2 will be OFF until 7Jbe2 reaches zero. As soon as 7Jbe2 is zero, ib2 will 
become positive and turn T2 ON. 7Jbe2 = 0 when Vee - 2Vcce-t/ R202 = 0, 
or when t = R2C2 In 2. 
[d] When t = R2C2 ln 2, we have 
~, TT TT e-[(R2C2 ln2)/(RLC1)] lT TT e-10 ln2,....., TT • ce2 = V CC - V CC = V CC - V CC = V CC 
(e] Before T1 turns ON, ib1 is zero. At the instant T1 turns ON, we have 
+ 
. Vee+ Vee -t/RLCi 
'lb1 = -- --e 
Ri R1 
[f] At the instant T2 turns back ON, t = R2C2 ln2; therefore 
ibi = Vee + Vee e-10 1n2 ,....., Vee 
Ri R1 Ri 
When T2 turns OFF, ib1 drops to zero instantaneously. 
[g] 
Vce2 
etc. 
[h] 
etc. 
t 
P 7 .98 [a] toFF2 = R2C2 ln 2 = 18 x 103 (2 x 10-9) ln 2 ~ 25 µs 
[b] toN2 = R1C1 ln2 ~ 25µs 
[c] toFFl = R1C1 ln2 ~ 25µs 
(d) toNl = R2C2 ln 2,...., 25 µs 
9 9 
[e) ib1 = 3 + 18 = 3.5mA 
[f] ih1 = 1
9
8 + ~e-2516 ,...., 0.5465 mA 
(g) Vce2 = 9 - 9e-2516 rv 8.86 V 
P 7.99 [a) toFF2 = R2C2 ln 2 = (18 x 103)(2.8 x 10-9) ln 2 ,...., 35 µs 
[b) toN2 = R1C1 ln2 rv 37.4p,s 
(c] toFFl = R1C1 ln2 ~ 37.4µs 
[d) toNl = R2C2 ln 2 = 35 µs 
[e] ib1 = 3.5mA 
[f] ib1 = 1
9
8 + 3e-35!9 rv 0.561 mA 
[g) Vce2 = 9 - 9e-35!9 rv 8.81 V 
Problems 7-87 
Note in this circuit T2 is OFF 35 JJB and ON 37.4 fJB of every cycle, 
whereas T1 is ON 35 µs and OFF 37.4 µs every cycle. 
P 7.100 If R1 = R2 = 50RL = lOOkO, then 
48 x 10-6 
C1 = 100 x 1Q3 ln 2 = 692.49 pF; 
36 x 10-6 
C2 = lOO x 103 ln 2 = 519.37pF 
then 
48 x 10-6 
C1 = 12 x lQ3 In 2 = 5.77nF; 
36 x 10-6 
C2 = 12 x lQ3 ln2 = 4.33nF 
Therefore 692.49 pF < C1 :s; 5. 77 nF and 519.37 pF :s; C2 :s; 4.33 nF 
7-88 CHAPTER 7. Response of First-Order RL and RC Circuits 
P 7.101 (a] T2 is normally ON since its base current ib2 is greater than zero, i.e., 
ib2 = Vce/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 
is charged to Vee, positive at the left terminal. This is a stable state; 
there is nothing to disturb this condition if the circuit is left to itself. 
[b] When S is closed momentarily, Vbe2 is changed to - Vee and T2 snaps 
OFF. The instant T2 turns OFF, vce2 jumps to VccRi/(R1 + RL) and ib1 
jumps to Vec/(R1 + RL), which turns T1 ON. 
[c] As soon as T1 turns ON, the charge on C starts to reverse polarity. Since 
Vbe2 is the same as the voltage across C, it starts to increase from - Vee 
toward +Vee- However, T2 turns ON as soon as Vbe2 = 0. The equation 
for Vbe2 is Vbe2 =Vee - 2Vcce-t/Re. 11be2 = 0 when t =RC ln2, therefore 
T2 stays OFF for RC In 2 seconds. 
P 7.102 [a] Fort< 0, Vce2 = 0. When the switch is momentarily closed, Vce2 jumps to 
( Vee ) 6(5) Vce2 = Ri + RL Ri = 25 = 1.2 V 
T2 remains open for (23,083)(250) x 10-12 ln2,...., 4µs. 
1.4 
1.2 -i----------r 
- 1 ~ 0.8 
B o.6 
> 0.4 
0.2 
0 -+-r-i.........-.,..,.......,....,...,,..,..,....-.....~r-r+-.-r-T"""T"r"r-rTT-r-T........-r-r-r.,_,...,......,.......,.......,........""T"T"~ 
0 2 4 6 8 10 
t(us) 
[b] ib2 = V~c = 259.93 pA, -5 < t::::; Oµs 
0 < t <RC ln2 
- 259.93 + 30oe-0·2xl06 (t-4xio-s) µA, 
600 
500 
- 400 
<( 
2.. 300 
N----i 
.a 200 
100 
Problems 7-89 
RC ln2 < t 
o.....,........,..,._....,_..,..,_..,...,,,.......,..,~,...,....,...,,..........,,...,..,..,~,...,....,...,~,.,....,..,,.,...,...,.., 
0 2 4 6 8 10 
t(us} 
P 7.103 [a} We want the lamp to be in its nonconducting state for no more than 10 s, 
the value of t 0 : 
1-6 
10 = R(lO x 10-6 ) In -- and R = 1.091 Mn 
4-6 
[bJ When the lamp is conducting 
Vr - 20 x 103 (6) - 08 v 
Th - 20 X 103 + 1.091 X lQ6 - O.l 
RTh = 20 kll L091 M = 19,640 n 
So, 
-6 4 - 0.108 (tc - t0 ) = (19,640)(10 X 10 )In l _ O.l08 = 0.289 S 
The flash lasts for 0.289 s. 
P 7.104 [a] At t = 0 we have 
BOOkQ 
+ + 
T = (800)(25) x 10-3 = 20 sec; 
Vc(oo) = 40V; Vc(O) = 5V 
1/r = 0.05 
7-90 CHAPTER 7. Response of First-Order RL and RC Circuits 
40 - 35e-0.05to = 15; : . e0.05to = 1.4 
t 0 = 20 In 1.4 s = 6. 73 s 
At t = t 0 we have 
800k0 
40r25]1>'_1;V 
+ 
vc lOkO 
The Thevenin equivalent with respect to the capacitor is 
(800/81)0 
40 (10) '.:J_ l;V 
BlO vC:> µ J I -
r = (800) (25) x 10-3 = 20 s-
81 81 ' 
40 
Ve( to)= 15 V; Ve( oo) = 81 v 
v (t) = - + 15 - - e-4.05(t-to) v = - + --e-4.05(t-to) 40 ( 40) 40 1175 
c 81 81 81 81 
40 + 1175e-4.05(t-to)= 5 
81 81 
1175 e-4.05(t-to) = 365 
81 81 
4.05(t-to) = 1175 = 3 22 
e 365 · 
1 ,......, 
t - to = 4.05 ln 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 
71 R (k0) 
r1++ 
40vC:> 25µ>' J sv v, 
I - -
r = 25R x 10-3 . 
' 1/r = 40/R 
Ve = 40 - 35e-(40/R)t 
40 - 35e-(40/R)to = 15 
R 
t 0 = 40 ln 1.4, R in kn 
At t =to: 
71 R (k0) r+ 
4ov6 25µF ~ 1sv vc lOkO 
+ 
10 400 
VTh = R + 10 ( 40) = R + 10; 
lOR 
RTh = R+ lOkO 
1 4(R+10) (25)(10R) x 10-3 0.25R T- . - . - R+lO - R+lO' -T R 
400 ( 400 ) _ 4(R+lO) (t-t ) 
Ve= + 15- e R 0 
R+lO R+lO 
or 
400 [ 15R - 250] _ 4(R~10) (t-to) _ 
R+lO+ R+lO e - 5 
( l5R - 250) e- 4(R~10> (t-to) = 5R - 350 
R+lO (R+lO) 
4(R+10) (t-t ) 3R - 50 
eR o=---
R-70 
R (3R-50) 
t-to= 4(R+l0) In R-70 
At 12 flashes per minute t0 + (t - t0 ) = 5 s 
R R (3R-50) 
40 In 1.4 + 4(R + 10) ln R - 70 = 5 
~ 
dominant 
term 
Problems 7-91 
Start the trial-and-error procedure by setting (R/40) In 1.4 = 5, then 
R = 200/(In 1.4) or 594.40k0. If R = 594.40k0 then t - t0 rv 0.29s. 
Second trial set (R/40) lnl.4 = 4.7s or R = 558.74k0. 
With R = 558. 7 4 kO, t - t0 "" 0.30 s 
This procedure converges to R = 559.3k0. 
7-92 CHAPTER 7. Response of First-Order RL and RC Circuits 
P 7.105 [a] t0 =RC In ( ~:: = ~) = (3700)(250 x 10-6 ) In ( =~~~) 
= l.80s 
RCRL l (Vmax - VTh) tc - to= n 
R + RL Vmin - VTh 
RL 1.3 
R + RL 1.3 + 3.7 = 0·25; RC = (3700)(25010-
6 ) = 0.925 s 
v; = 1000(1.3) = 260 V· 
Th 1.3 + 3.7 ' RTh = 3.7kllL3k = 962 n 
tc - t 0 = (0.925)(0.26) ln(640/40) = 0.67 S 
tc = 1.8 + 0.67 = 2.47 S 
60 
flashes/min = 2.47 = 24.32 
fb] 0 < t S to: 
VL = 1000 - 700e-tfri 
r 1 =RC= 0.925s 
to St S tc: 
VL = 260 + 640e-(t-t.,)/r2 
r2 = RThc = 962(250) x 10-6 = 0.2405 s 
0 St S to: i = 1000 - VL = 2-e-t/0.925 A 
3700 37 
i = 1000 - VL = 74 _ 64 e-(t-to)/0.2405 
3700 370 370 
Graphically, i versus t is 
i (A) 
7 
1 -----
37 
A2 I 
--+------
1 
Problems 7-93 
The average value of i will equal the areas (A1 + A2 ) divided by tc. 
Ai +A2 
.·. iavg= ---
ic 
Ai = !__ [to e-t/0.925 dt 
37 lo 
= 6.475 (1- e-In 7) = 0.15 A-s 
37 
Jtc 7 4 _ 64e-(t-to}/0.2405 A2 = to 370 dt 
_ 74 ( _ ) 15.392( -lni6 _ l) 
- 370 tc to + 370 e 
= 17.797 l lB _ 15.392(l - -lni6) 
370 n 370 e 
= 0.09436 A-s 
iavg = (0.15 + 0.09436) (1000) = 99.06 mA 
0.925 ln 7 + 0.2405 ln 16 
[c] Pavg = (1000)(99.06 X 10-3) = 99.06 W 
N f k h I = (99.06)(24)(365) = 867 77 o. o w rs yr 1000 . 
Cost/year = (867.77)(0.05) = 43.39 dollaxs/year 
P 7.106 [a] Replace the circuit attached to the capacitor with its Thevenin equivalent, 
where the equivalent resistance is the paxallel combination of the two 
resistors, and the open-circuit voltage is obtained by voltage division 
across the lamp resistance. The resulting circuit is 
RTh 
VThrl :, 
RRL 
RTh=RllRL= R+RL; 
From this circuit, 
vc(O) = Vmaxi 
Thus, 
vc(t) = VTh + (Vmax - VTh)e-<t-to)/r 
where 
RRLC 
r=---
R+RL 
7-94 CHAPTER 7. Response of First-Order RL and RC Circuits 
{b) Now, set vc(tc) = Vmin and solve for (tc - to): 
VTh + (Vmax - VTh)e-(tc-to)/r = Vmin 
V,. - VTh e-(tc-to)/r = _m_1n __ _ 
Vmax - VTh 
-(fc - to) = ln Vmin - VTh 
T Vmax - VTh 
P 7.107 [a] 0 :::; t S 0.5: 
. _ 21 (30 _ 21) -t/r 
i-60+ 60 60 e 
i = 0.35 + 0.15e-60t/L 
where T = L/R. 
i(0.5) = 0.35 + O.l5e-3o/L = 0.40 
. e30/L = 3· . . ' L = l~03 = 27.31 H 
[b] 0 St< tr, where tr is the time the relay releases: 
i = 0 + (~~ - 0) e-GGt/L = 0.5e-50t/L 
. 0 4 = 0 5e-60tr/L. . . . . / ' e60tr/L = 1.25 
_ 27.31ln1.25 ~ 0 10 fr - GO - . S 
-------8 
Natural and Step Responses of 
RLC Circuits 
Assessment Problems 
1 1 
AP 8.1 (a] (2RC)2 - LC' 
1 
[b] a= 5000 = 2RC' 
therefore C = 500 nF 
therefore C = 1 µF 
s1,2 = -5000 ± 25 x 106 - (l03~bl06 ) = ( -5000 ± j5000) rad/s 
1 
[c] rm= 20,000, 
vLC 
therefore C = 125nF 
S1,2 = [-40 ± J ( 40)2 - 202] 103 , 
s1 = -5.36 krad/s, S2 = -74.64krad/s 
AP 8.2 iL - 1 ft 
50 x 10-3 0 
[-14e-5000x + 26e-20,000x] dx + 30 X 10-3 
{
-14e-5000x It 26e-20,000t It} 
- 20 -5000 0 + -20,000 0 + 30 x 10- 3 
- 56 x 10-3 (e-5ooot - 1) - 26 x 10-3 (e-2o,oo0t -1) + 30 x 10-3 
- [56e-5oo0t - 56 - 26e-2o,ooot + 26 + 30] mA 
_ 56e-5ooot _ 26e-20,ooot mA, t ~ 0 
AP 8.3 From the given values of R, L, and C, s1 = -lOkrad/s and s2 = -40 krad/s. 
8-1 
8-2 CHAPTER 8. Natural and Step Responses of RLC Circuits 
[h] ic(o+) = -(iL(o+) + iR(o+)) = -(-4 + o) = 4A 
[c] cdvcj~+) = ic(o+) = 4, therefore dvcj~+) = ~ = 4 x 108 v /s 
[d] v = [A1e-io,oo0t + A2e-40•000t] V, t > o+ 
dv(o+) = -10 OOOA -40 OOOA 
dt ' 1 ' 2 
Therefor€ Ai + 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,oo0t] V, t ?: O 
1 
AP 8.4 [a] 2RC = 8000, therefore R = 62.5 0 
[b] i (o+) = lOV = 160 A 
R 62.5!1 ID 
ic(O+) = -(iL(O+) + iR(O+)) = -80- 160 - -240mA = Cdv~~+) 
dv(o+) -240m 
Therefore dt = C = -240kV /s 
[c) B1 = v(o+) = lOV, 
Therefore 6000B2 - 8000B1 = -240,000, 
(d) iL = -(iR + ic); iR = v/R; 
dv 
ic=G-
dt 
v = e-sooot[lO cos 6000t - 83° sin 6000t] V 
B2 = (-80/3) V 
1280 
Therefore iR = e-so00t[l60 cos 6000t - - 3- sin 6000t] mA 
ic = e-BOOOt [-240 cos 6000t + 4~0 sin 6000tJ mA 
iL = 10e-soo0t[8cos6000t + 8: sin6000tJ mA, t?: 0 
( 1 )
2 1 106 
AP 8·5 (a) 2RG = LC = 4' 
[b] 0.5CVi = 12.5 x 10-3 , 
[c) 0.5LI~ = 12.5 x 10-3 , 
1 
therefore 2RG = 500, R = 100 0 
therefore Vo = 50 V 
Io= 250mA 
Problems 8-3 
dv(o+) -- D1 - ,...,D2 [d] D2 = v(o+) = 50, dt .,. 
iR(O+) = 1
5i0 = 500 mA 
Therefore ic(O+) = -(500 + 250) = -750mA 
Therefore dv(O+) = -750 x 
10- 3 = -75 OOOV/s 
dt c ' 
Therefore D1 - aD2 = - 75,000; 
1 
a= 2RC = 500, D 1 = -50,000 V /s 
[e] v = [5oe-500t - 50,000te-500t] V 
iR = ~ = [0.5e-5o0t - 500te-500t] A, 
. + Vo 40 
AP 8.6 [a] iR(O ) = R = 500 = 0.08 A 
[h] ic(O+) = I - iR(o+) - iL(o+) = ~ 1 - 0.08 - 0.5 = -1.58 A 
[ ] dh ( o+) = Vo = 40 = 62 5 A/ 
c dt L 0.64 . 8 
1 1 
[d] a= 2RC = 1000; LC = 1,562,500; s1,2 = -1000 ± j750 rad/s 
[ ] . . B' -at t B' -at . t e iL = i1 + 1e coswa + 2e smwa,, i1 =I= -lA 
therefore B~ = 1.5 A 
diL(O+) . I I I ( I ) 
dt = 62.5 = -aB1 + wdB2 , therefore B 2 = 25 12 A 
Therefore iL(t) = -1+e-looot[l.5cos750t + (25/12) sin 750t] A, t 2: 0 
LdiL lOOOt . [f] v(t) = dt = 40e- [cos 750t - (154/3) sm 750t)V t 2: 0 
AP 8.7 (a] i(O+) = 0, since there is no source connected to L fort< 0. 
[b] vc(o+) = vc(o-) = ( 15 ~~k 9 k) (80) = 50 V 
[c] 50 + 80i(o+) + L di~~+) = 100, di~~+) = 10,000 A/s 
[d] a = 8000· - 1- = 100 x 106 • 
' LC ' 
[e] i =if+ e-at[B~ coswdt + B; sinwdt]; 
s1,2 = -8000 ± j6000 rad/s 
i1 = 0, i(O+) = 0 
Therefore 
I di(o+) I f 
B 1 = O; dt = 10,000 = -aB1 + wdB2 
Therefore B; = 1.67 A; i = l.67e-soo0t sin 6000t A, t 2: 0 
8-4 CHAPTER 8. Natural and Step Responses of RLC Circuits 
VJ= lOOV 
therefore 50 = 100 + B~ 
B~ = -50V; 
Therefore B' = ~B' = (8000) (-50) = -66.67V 2 Wd l 6000 
Therefore vc(t) = 100- e-8000t[50cos6000t + 66.67sin6000t] V, t 2: 0 
Problems 
p 8.1 
1 109 
[a] a= 2RC = (10,000)(8) = 12,500 
2 1 109 
WO = LC = (1.25)(8) = 108 
81,2 = -12,500 ± j(l.5625 - 1)108 = -12,500 ± 7500 
81 = -5000 rad/s 
82 = -20,000 rad/s 
[b] overdamped 
[c] wd = j w~ - a 2 
. ·. a 2 = w~ - w~ = 108 - 36 x 106 = 0.64 x 108 
a = 0.8 x 104 = 8000 
. 1 
2RC = 8000; 
109 
: . R = (l6,000)(8) = 7812.5 n 
[d] 8 1 = -8000 + j6000 rad/s; s2 = -8000 - j6000 rad/s 
1 4 1 
[e] a = 10 = 2RC; : . R = 2C(l04) = 6250 n 
P 8.2 [a] -a+ ija2 -w; = -5000 
-a - /a2 - w; = -20,000 
:. -2a = -25,000 
a= 12,500rad/s 
1 106 
2RC - 2R(0.05) = 12'500 
R =soon 
2 J a2 - w~ = 15,000 
4(o:2 - w~) = 225 x 106 
: . w0 = 10,000rad/s 
2 8 1 
WO= 10 =LC 
1 
:. L = wsc = 200mH 
[b) ia = v~) = -6.25e-5ooot + 25e-2o,000t mA, t 2: o+ 
dv(t) 
ic = c--;J,t = l.25e-5000t - 2oe-2o,000t mA, t 2: o+ 
iL = -(iR + ic) = 5e-5oo0t - 5e-2o,oo0t mA, t > o+ 
P 8.3 [a] a = 4000; Wd = 3000 
:. w~ = w~ + o:2 = 9 x 106 + 16 x 106 = 25 x 106 
- 1-= 25 x 106 
LC 
1 
L = (25 x 106)(50 x 10-9) = 0.8H = 800mH 
1 
[b) a= 2RC 
1 109 
:. R = 2o:C = (8000)(50) = 25oon 
[c) v;, = v(O) = 125 V 
Problems 8-5 
8-6 CHAPTER 8. Natural and Step Responses of RLC Circuits 
(d) I 0 = iL(O) = -iR(O) - ic(O) 
iR(O) = V,R0 = 125 x 10-3 = 50mA 
2.5 
dv 
ic(O) = C dt (0) 
~~ = 125{e-4000t[-3000sin3000t-6000cos3000t]-
4000e-4oo0t [cos 3000t - 2 sin 3000t] 
ddv (0) = 125{1(-6000) -4000} = -125 x 104 
t 
C~~ (0) = -125 x 104 (50 x 10-9 ) = -6250 x 10-5 = -62.5 mA 
. ·. I 0 = -50 + 62.5 = 12.5 mA 
dv 
(e) dt - 125e-4000t[5000 sin 3000t - 10,000 cos 3000t] 
- 625 x 103e-4000t(sjn3000t- 2cos3000t] 
dv 
C- = 31,250 x 10-6e-4000t(sin 3000t - 2 cos 3000t) 
dt 
ic(t) = 31.25e-4000t(sin3000t - 2cos 3000t) mA 
iR ( t) = 50e-4oo0t (cos 3000t - 2 sin 3000t) mA 
iL(t) = -iR(t) - ic(t) 
= e-4o00t(l2.5 cos 3000t + 68. 75 sin 3000t) mA, t 2: 0 
CHECK: 
dh 
- { -400oe-4000t[12.5 cos 3000t + 68. 75 sin 3000t} 
dt 
+e-4000t[-37.5 x 103 sin3000t 
+ 206.25 x 103 cos 3000t] x 10-3 
- e-4o00t[156.25 cos 3000t - 312.5 sin 3000t] 
L ~~ - e-4oo0t[125 cos 3000t - 250 sin 3000t] 
- 125e-4000t[cos 3000t - 2 sin 3000t] V 
( 1 )
2 1 
P 8.4 [a] 2RC = LC = ( 4000)2
 
1 
:. C = (16 x 106)(5) = 12.5 nF 
2~0 = 4000 
109 
:. R = (8000)(12.5) = lOkn 
v(O) = D 2 = 25V 
iR(O) = ~~ = 2.5mA 
ic(O) = -2.5 - 5 = ~7.5mA 
dv(O) = D - 4000D = - 7·5 x io-
3 
= -6 x 105 
dt 1 2 12.5 x 10-9 
. ·. D1 = -6 x 105 + 4000(25) = -5 x 105 V /s 
[b] v = -5 x 105te-4o00t + 25e-4ooot 
dv = (20 x 108t - 6 x 105]e-4000t 
dt 
ic = c~~ = 12.5 x 10-9 [20 x 108t - 6 x 105]e-4oo0t 
= (25,000t - 7.5)e-4000t mA, t > 0 
P 8.5 [a) 2a = 200; a= lOOrad/s 
. 2Ja2 -w~ = 120; W0 = 80rad/s 
1 1 
C = 2aR = 200(200) = 25 µF 
1 106 
L = w~C = (80)2(25) = 6.25 H 
ic(O+) = A1 +A2=15mA 
die dh diR 
dt + dt + dt = 0 
dic(O) diL(O) diR(O) 
-
dt dt dt 
Problems 8-7 
8-8 CHAPTER 8. Natural and Step Responses of RLC Circuits 
diL(O) = _Q_ = O A/s 
dt 6.25 
diR(O) 1 dv(O) 1 ic(O) 15 x 10-3 
dt = R ~ = R 0- = (200)(25 x 10-6) = 3 A/s 
dic(O) = -3 A/s 
dt 
160A1 + 40A2 = 3 
4A1 + A2+ = 75 x 10-3 ; 
:. ic = 2oe-160t - 5e-40t mA, 
Ai= 20mA; 
t 2: 0 
[b J By hypothesis 
v = A3e-l60t + A4e-40t, t 2: 0 
v(O) = A3 + A4 = 0 
dv(O) = 15 x 10-3 = 600 V /s 
dt 25 x 10-6 
-l60A3 - 40A4 = 600; A3 = -5V; 
v = -5e-160t + 5e-40t V, t 2: 0 
v 
[c] ia(t) = 200 = -25e-160t + 25e-40t mA, 
[d} iL = -iR - ic 
. 5 -160t 20 -40t mA iL = e - e , 
P 8.6 [a] iR(O) = 2~~0 = 45mA 
iL(O) = -30mA 
t 2: 0 
t 2: o+ 
ic(O) = -iL(O) - iR(O) = 30 - 45 = -15mA 
1 109 
[b) a= 2RC = (4000)(10) = 25,000 
2 1 (103)(109) 8 
WO = LC = (250)(10) = 4 x 10 
A2 = -5mA 
s1,2 - -25,000 ± /6.25 x 108 - 108 (4) = -25,000±15,000 
s1 = -10,000 rad/s; s2 = -40,000 rad/s 
p 8.7 
Problems 8-9 
v(O) = Ai + A2 = 90 
dv -15 x 10-3 
dt (0) = -104A1 - 4A2 x 104 = 10 x 10_9 = -1.5 x 10
6V/s 
-A1 -4A2 = -150 
.·. -3A2 = -60; Ai= 70 
v = 7oe-io,o00t + 2oe-4o,000t V, t 2: o 
(c) ic = cdv 
dt 
- 10 x 10-9 (-70 x 104e-lO,OOOt - 80 x 104e-40,o00t] 
- -1e-io,000t - 8e-4o,oo0t mA 
iR = 35e-lO,OOOt + 10e-40,000t mA 
iL = -ic - iR = -28e-io,OOOt - 2e-4o,oo0t mA, t 2: 0 
1 109 
a = 2RC = (5000)(10) = 2 x 104 
Critical damping: 
a2 = w2 
0 
. ( +) 90 
'lR 0 = 2500 = 36mA 
v(O) = D2 = 90 
dv (O) _ D _ aD _ ic(O) _ -6 x 10-3 _ 5 
dt - 1 2 - c - 10 x 10-9 - -6 x 10 
D1 = aD2 - 6 x 105 = (2 x 104)(90) - 6 x 105 = 120 x 104 
v = (120 x 104t + 90)e-20·000t V, t 2: 0 
8-10 CHAPTER 8. Natural and Step Responses of RLC Circuits 
p 8.8 
1 3 x 109 
-2R-C = (25,000)(10) = 12,000 
1 8 -=4x10 
LC 
s1,2 = -12,000 ± j16,000rad/s 
. ·. response is underdamped 
v(t) = B1e-12,ooot cos 16,000t + B2e-12,oo0t sin 16,000t 
iR(O+) = (l2,:~0/3) = 21.6 mA 
ic(o+) = [-iL(o+) + iR(o+)] = -[-30 + 21.6] = 8.4mA 
dv(O+) = B.4 x 10- 3 = 840 OOOV/s 
dt 10 x 10-9 ' 
dv(O) 
-;ft = -12,00081 + 16,000B2 = 840,000 
or - 3B1 + 4B2 = 210; :. B2=120V 
v(t) = 90e-12,o00t cos 16,000t + 120e-12,oo0t sin 16,000t V, t ~ 0 
P 8.9 a= 2000/2 = 1000 
1 106 
R = 2aC = (2000)(18) = 27·78 n 
v(o+) = -24 v 
-24 
iR(o+) = 27_78 = -864mA 
dv = 2400e-200t + 21 60oe-1800t 
dt ' 
dv~~+) = 2400 + 21,600 = 24,000V /s 
ic(o+) = 18 x 10-6 (24,000) = 432 rnA 
iL(O+) = -(iR(O+) + ic(O+)] = -[-864 + 432} = 432 mA 
2 1 109 6 
P 8.10 [a] w0 = LC = 40 = 25 x 10 
w0 = 5000 rad/s 
1 
2RC = 5000; 
1 
R = 10,000C 
109 
R = 8 x 104 = 12.5 kO 
[h] v(t) = D1te-50o0t + D2e-5000t 
v(O) = -25 V = D2 
dv = (D1t - 25)(-5000e-5000t) + D1e-5oo0t 
dt 
dv (0) = 125 x 103 + D1 = ic(O) 
dt c 
ic(O) = -iR(O) - iL(O) 
-25 
iR(O) = - = -2mA 
12.5 
ic(O) = 2- (-1) = 3mA 
dv 3 x 10-3 
. . dt (0) = 8 x 10-9 = 0.375 x 106 = 3.75 x 105 
1.25 x 105 + D 1 = 3.75 x 105 
D1=2.5 x 105 = 25 x 104V/s 
:. v(t) = (25 x 104t - 25)e-5000t V, t 2: 0 
. dv 
[c] z0 (t) = 0 when dt (t) = 0 
Problems 8-11 
dv = (25 x 104t - 25)(-5000)e-5o00t + e-5oo0t(25 x 104) 
dt 
= (375,000 - 125 x 107t)e-5000t 
dv 
dt = 0 when 125 x 107t 1 = 375,000; t1 = 300µs 
v(300µs) = 50e-L5 = 11.16 V 
8-12 CHAPTER 8. Natural and Step Responses of RLC Circuits 
11.16 
[d] iL(300µs) = -iR(300µs) = -- = 0.89mA 
12.5 
wc(300µs) = 4 x 10-9(11.16)2 = 497.87nJ 
WL(300µs) = (2.5)(0.89)2 x 10-6 = 1991.48nJ 
w(300µs) =We+ WL = 2489.35nJ 
w(O) = 4 x 10-9 (625) + 2.5(10-6 ) = 5000nJ 
3 remaining = 2~~~·i5 (100) = 49.793 
1 
P 8.11 [a] a= 2RC = 1 rad/s 
2 1 
WO= LC= 10 
wd = JlO - 1 = 3 rad/s 
. ·. v = B1 e-t cos 3t + B2e-t sin 3t 
v(O) = B1 = O; 
12 = -aB1 + wdB2 = -1(0) + 3B2 
B2 =4 
v = 4e-tsin3tV, t?: 0 
dv -t . 
(b] dt = 4e (3 COS 3t - Sill 3t) 
dv 3 
-(o+) = - = 12 V/s 
dt 0.25 
dv 
dt = 0 when 3cos3t = sin3t or tan3t = 3 
: . 3t1 = 1.25, ti = 416.35 ms 
3t2 = 1.25 + 7r, t2 = 1463.55ms 
3t3 = 1.25 + 27r, t3 = 2510.74ms 
[c] t3 - ti= 2094.40ms; 
[d) t2 - ti= 1047.20ms; 
271" 271" 
Ta= - = - = 2094.40ms 
Wd 3 
~d = 209i.40 = 1047.20 ms 
[e) v(t1) = 4e-<0·41635) sin 3(0.41635) = 2.50 V 
v(t2 ) = 4e-(1.46355) sin 3(1.46355) = -0.88 V 
v(t3 ) = 4e-<2•51074) sin3(2.51074) = 0.31 V 
[f] 
·1.0 
t (s) 
P 8.12 [a] a = O; Wd = w0 = JlO = 3.16rad/s 
dv 
C-d (0) = -iL(O) = 3 ,t 
v(O) = B1 = O; 
12 = -aB1 + wdB2 = -0 + JlOB2 
:. B2 = 12/JiO = 3.79V 
v = 3.79sin3.16tV1 t ~ 0 
[b] 27rf = 3.16; 
[c] 3.79 V 
f = 3.16 rv 0.50Hz 
271" 
P 8.13 From the form of the solution we have 
Problems 8-13 
4 
We know both v(O) and dv(O+)/dt will be real numbers. To facilitate the 
algebra we let these numbers be K1 and K2, respectively. Then our two 
simultaneous equations are 
8-14 CHAPTER 8. Natural and Step Responses of RLC Circuits 
The characteristic determinate is 
1 1 
~= 
The numerator determinates are 
We see from these expressions that A1 =Ai 
P 8.14 By definition, B1 = A1 + A2 • From the solution to Problem 8.13 we have 
But K1 is v(O), therefore, B1 = v(O), which is identical to Eq. (8.30). 
By definition, B2 = j(A1 - A2 ). From Problem 8.13 we have 
It follows that 
but K - dv(o+) 
2 - dt 
Thus we have 
dv 
-d (o+) = -aB1 + wdB2, ,t 
which is identical to Eq. (8.31). 
p 8.15 
p 8.16 
Problems 8-15 
1 
(a] a = 2RC = 1000J2, therefore overdamped 
S1 = -414.21, S2 = -2414.21 
[ dv~~+)] = ic~+) = 98,000 V /s 
Therefore - 414.21A1 - 2414.21A2 = 98,000 
(b) 
A1 = 49, A2 = -49 
v(t) = 49[e-414.2lt _ e-2414.21t] V, t ~ 0 
v(t) (V) 
30 
25 
20 
15 
10 
5 
0 
0 2 3 4 
t (ms) 
Example 8.4: Vmax ,...., 7 4.1 V at 1.4 ms 
Example 8.5: Vmax ,...., 36.1 V at 1.0 ms 
Problem 8.15: Vmax ,...., 28.2 V at 0.9 ms 
+ 
150k0 l 
ioti 
v _ 104 ir(150 x 103) (150)(60)106 . 
T - 210 X 103 + 210 X 103 'tT 
5 
60k0 
~VT = 1500 X 103 9000 X 103 = 10,500 X l03 = 50 kO 
'tT 210 + 210 210 
6 7 
8-16 CHAPTER 8. Natural and Step Responses of RLC Circuits 
p 8.17 
75 
V. = -(6) = 45V· 
0 10 ' 
Io= 0 
ic(O) = -iR(O) - iL(O) = - 50~~00 = -0.9 mA 
ic(O) = -0.9 x 106 = -720 x 103 
c 1.25 
2 1 109 
WO= LC = (8)(1.25) = 108; W0 = 10
4 rad/s 
1 109 
a= 2RC = (2)(50)(1.25)x 103 = 3000 rad/s 
wd = J(loo - 64) x 106 = 6000 rad/s 
v0 (0) = B1 = 45V 
~0 (0) = 6000B2 - 8000B1 = -720 x 103 
. ·. 6000B2 = 8000( 45) - 720 x 103 ; B2 = -60V 
v0 = 45e-SOOOt cos 6000t - 60e-sooot sin 6000t V, t 2: 0 
150k0 
104 ir(150 x 10
3) (150)(60)106 . 
V - + 'lT 
T - 210 X 103 210 X 103 
l 60k0 
iq:, 
_vr = 1500 x 1.03 9000 x 103 = 10,500 x 103 = 50 kn 
ir 210 + 210 210 
75 
Vo= 10 (6) = 45V; Io= 0 
p 8.18 
ic(O) = -in(O) - k(O) = - 50~~00 = -0.9mA 
ic(O) -0.9m 9 0 3 --= =-OxlO c 10-9 
2 1 1 8 
WO = LC = (10)(10-9 ) = 10 ; 
Wo = 10,000 rad/s 
1 1 
a= 2RC = (2)(50,000)(10-9 ) = lO,OOO rad/s 
so the response is critically damped 
v0 (0) = D2 = 45V 
dvo 3 
-(0) = D1 - aD2 = -900 x 10 
dt 
Problems 8-17 
:. D 1 =-900x103 + (10,000)(45); :. D1 = -450,000V/s 
vo = -450,000te-10'ooot + 45e-rn,oo0tv, t 2: 0 
150k0 
4 ir(150 x 103) (150)(60)10
6 . 
Vr = 10 210 X lQ3 + 210 X 1Q3 'lT 
l 60k0 
i.p 
3 . 3 
_vr = 1500 x 10 9000 x 10 = 10,500 10a = 50 kn 
ir 210 + 210 210 x 
75 
Vo= 10 (6) = 45V; ! 0 = 0 
45 
ic(O) = -iR(O) - iL(O) = - 50,000 = -0.9 mA 
8-18 CHAPTER 8. Natural and Step Responses of RLC Circuits 
ic(O) -0.9 = -1125 x 103 -c = 800 x 10-12 
2 1 1 
WO= LC = (12.5)(800 x lQ-12) = 108; Wo = 10,000 rad/s 
1 1 
a= 2RC = (2)(50,000)(800 x 10-12) = 12'500 rad/s 
a2 > w2 
0 so the response is overdamped 
s1,2 = -a± j a2 - w; = -12,500 ± J (12,500)2 - 108 = -12,500 ± 7500 
: . s1,2 = -5000 r/s, -20,000 r/s 
Ai + A2 = Vo = 45 and - 5000A1 - 20,000A2 - -1125 x 103 
:. Ai= -15, A2 = 60 
Vo = -15e-5000t + 60e-20,ooot V, t :?: 0 
p 8.19 t < 0: Va= 60V, I 0 = 45mA 
+ 
l 60V 162. SnF lH 
45mA 
t > 0: 
+ 
~ 62.5nF 1.6k0 Vo 1H 
in(O) = 1~~0 = 37.5 mA; iL(O) = 45mA 
Problems 8-19 
ic(O) = -37.5 - 45 = -82.5mA 
1 109 
a= 2RC = 3200(62.5) = 5000 rad/s 
2 1 109 6 
w =-=-=l6xl0 
0 LC 62.5 
S1,2 = -5000 ± v'25 X 106 - 16 X 106 = -5000 ± 3000 
s1 = -2000 rad/s; s2 = -8000 rad/s 
. A -2000t +A -aoo0t .. Vo= 1e 2e 
dvo (0) = -2000A - 8000A = - 82·5 x 10- 3 = -1320 103 
dt 1 2 62.5 x 10-9 x 
Solving, Ai= -140V, A2 = 200V 
: . v0 = -140e-2000t + 200e-SOOOt V, t ;:::: 0 
p 8.20 w2 - ~1- - 16 x 106 = 25 x 106 
0 - LC - 0.64 
1 16 x 106 
a = 2RC = 4000 = 4000 rad/s; 
wd = J (25 - 16) x 1Q6 = 3000 rad/s 
S1,2 = -4000 ± j3000rad/s 
v0 (0) = B1 = 60V 
iR(O) = 2~~0 = 30mA 
iL(O) = 45mA 
ic(O) = -iR(O) - iL(O) = -75 mA 
a2 = 16 x 103 
8~20 CHAPTER 8. Natural and Step Responses of RLC Circuits 
p 8.21 
ic(O) = (-75 x 10-3 )(16 x 106) = -12 x 105 
c 
dvo ( ) 5 - 0 = -4000B1 + 3000B2 = -12 x 10 
dt 
. ·. 3B2 = 4B1 - 1200 = 240 - 1200 = -960; B2 = -320 V 
v0 ( t) = 60e-4000t cos 3000t - 320e-4ooot sin 3000t V, t 2: 0 
2 1 - 16 x 106 = 108. 
WO - LC 0.16 ' 
= _1_ = 16 x 106 = 104 
a 2RC 1600 
. ·. a 2 = w; (critical damping) 
iR(O) = S: = 75mA 
iL(O) = 45mA 
ic(O) = -120mA 
dv0 ( ) dt 0 = - l0,000D2 + Di 
ic~O) = (-120 x 10-3)(16 x 106 ) = -1920 x 103 
v0 (t) = (60 - 132 x 104t)e-io,000tv, t > O 
P 8.22 [a] v = L ( ~~) = 16[e-2o,ooot - e-80,000t] V, t ;:::: O 
v [b] iR = R = 40(e-20,ooot - e-80,000tJ mA, t 2: 0+ 
[c] ic =I - iL - iR = {-8e-20•000t + 32e-80·000t] mA, 
Problems 8-21 
P 8.23 [a] v = L ( ~~) = 40e-32•000tsin24,000tV, t > 0 
[b] ic(t) =I - iR - iL = 24 x 10-3 - 6~5 - iL 
= [24e-32•000t cos 24,000t - 32e-32•000t sin 24,000t] mA, 
P 8.24 v = L ( ~~) = 960,000te-40•000t V, t .2 0 
2 1 106 4 
p 8.25 WO = LC = (20)(5) = 10 ; w0 = 100 rad/s 
1 106 104 
a = 2RC = (1600)(5) = 8o = 125 rad/s 
81,2 = -125 ± /(125)2 - 104 =-125±15 
81 = -50 rad/s; 82 = -200 rad/s 
ft= 15mA 
:. -30= 15+A~ +A;; A~ + A; = -45 x 10-3 
diL = -50A' - 200A' = 60 = 3 
dt 1 2 20 
Solving, A~= -40mA; A;= -5mA 
iL = 15 - 40e-50t - 5e-200t mA, t .2 0 
1 106 
P 8.26 a = 2RC = (2S00)(5) = 80; a? = 6400 
w2 = 104 • 
0 ' 
wd = V104 - 6400 = 60 rad/s 
81,2 = -a± jwd = -80 ± j60 rad/s 
iL = 15 + B~ e-80t cos 60t + B~e-sot sin 60t 
-30 = 15+ B~ :. B~ = -45mA 
diL ( ) t t -d 0 = -80B1 +60B2 =3 
t 
:. B~ = -lOmA 
iL = 15 - 45e-80t cos 60t - lOe-sot sin 60t mA, t ;:=:: 0 
8-22 CHAPTER 8. Natural and Step Responses of RLC Circuits 
1 106 
p 8·27 a= 2RC = (2000)(5) = lOO 
a 2 = 104 = w2 0 critical damping 
iL(O) = -30 = 15 + n;; :. n; = -45mA 
~~ (o) = -10on; + D~ = 3000 x 10-3 
. ·. D~ = 3000 x 10-3 + 100( -45 x 10-3) = -1500 x 10-3 
iL = 15 - 1500te-10ot - 45e-100t mA, t ~ 0 
1 106 
p 8·28 a= 2RC = (1600)(6.25) = lOO; a2 = 104 
1 106 
w; = LC = (25)(6.25) = 6400 
812 = -100 ± J104 - 6400 = -100 ± 60 
' 
8 1 = -40 rad/s; 82 = -160 rad/s 
. dv0 ( ) / / . . dt 0 = 0 = -40A1 - 160A2 
Solving, A~= 40V, 
t > o+ 
Problems 8-23 
30 
It= 800 = 37.5mA; io(O) = 0 
0 = 37.5 x 10-3 +A~ +A;, A~ +A; = -37.5 x 10-3 
dio (0) = 30 = -40A' - 160A' 
dt 25 1 2 
Solving, A~= -40mA; A;= 2.5mA 
io = 37.5 - 40e-40t + 2.5e-160t mA, t ~ 0 
[b] dio = [1600e-40t - 400e-160t] x 10-3 
dt 
L ~; = 25(1.6)e-4ot - 25(0.4)e-160t 
:. V 0 = 40e-40t - 10e-l60t V, t ~ 0 
P 8.30 For t > 0 
37 .5mA! t 8000 
6 .25,uF ~ 
1 
a= -- = 100· 
2RC ' 
1 
LC= 6400 
81,2 = -100 ± 60 
81 = -40 rad/ s; 8 2 = -160rad/s 
ic(o+) = 37.5mA 
dvo(O+) ic(o+) 
dt - 6.25 x 10-6 = 6000V /s 
8-24 CHAPTER 8. Natural and Step Responses of RLC Circuits 
dvo(O+) = -40A' - 160A' 
dt 1 2 
-40A~ - 16oA; = 6000 
A~+ 4A; = -150 
:. A~= 50V; 
v0 .....: 50e-40t - 50e~l60t V, t ~ 0 
P 8.31 [a) From the solution to Prob. 8.30 8 1 = -40rad/s and 82 = -160rad/s, 
therefore 
i 0 = 11 + A~e-40t + A;e-160t 
11=37.5mA; 
:. 0 = 37.5+A~ +A;; 
It follows that 
-40A~ - l60A; = 0 
A~= -50mA; A;= 12.5mA 
. ·. io = 37.5 - 50e-40t + 12.5e-160t mA, t ~ O 
[b] dio 2 -40t 2 -160t dt = e - e 
di 
Va = £-5!.. = 25[2e-40t - 2e-160t] 
dt 
v0 = 50e-4ot - 50e-160t V, t ~ 0 
P 8.32 iL(O-) = iL(O+) = 30 mA 
Fort> 0 
30ml\i t 4000 
+ 
1 
a = 2RC = 1000 rad/ s; 
2 1 4 
w = - =64x 10 
0 LC 
s1 = -400rad/s s2 = -1600rad/s 
v0 (00) = 0 =Vt 
ic(o+) = -30 + 30 + o = o 
dvo = 0 
dt 
dvo (0) = -400A' - l600A' dt 1 2 
A~ + 400A~ = O; 
A~= O; 
v0 = 0 fort~ 0 
Note: v0 (0) = O; v0 (00) = O; 
dv0 (0) = O 
dt 
Problems 8-25 
Hence the 30 mA current circulates between the current source and the ideal 
inductor in the equivalent circuit. In the original circuit the 12 V source 
sustains a current of 30mA in the inductor. This is an example of a circuit 
going directly into steady state when the switch is closed. There is no 
transient period, or interval. 
p 8.33 t < 0: 
-- +_1000 -v0 (0 ) - v0 (0 ) - 1250 (25) - 20 V 
8-26 CHAPTER 8. Natural and Step Responses of RLC Circuits 
t>O 
+ 
lOOmAI t 2000 
10µ.E' ~ 20V 
20 
-100 + 0.2 + ic(O+) + 0 = O; 
1 106 
2RC (400)(10) = 250rad/s 
1 106 w; = LC = lO(l.6) = 62,500 
a:2 = w~ critically damped 
V1 =0 
dvo(O) = -250D; + D~ = 0 
dt 
v 0 (0+) = 20 = n; 
D~ = 25oD; = 5000 V /s 
.". V0 = 5000te-25ot + 20e-250t V, t 2:: Q+ 
[b] iL = 11 + n;te-25ot + D~e-25ot 
+ 
v 
0 
11 = lOOmA; diL(o+) = 20 = 12.5 A/s 
dt 1.6 
:. 0 = 100 + D~; D~ = -lOOmA; 
-250D~ + n; = 12.5; n; = -12.5 A/s 
. ·. h = 100 - 12,500te-25ot - 100e-250t mA t 2:: 0 
p 8.34 [a] WL = ('° pdt = ! 09 Voh dt .lo .lo 
Vo = 5000te-250t + 20e-250t V 
iL = 0.1 - 12.5te-250t - o.1e-250t A 
Problems 8-27 
p = 2e-25ot + 500te-250t - 750te-5oot - 62,500t2e-5oot - 2e-500t W 
WL = J00 e-250t dt + 250J00 te-250t dt - 375J00 te-5oot_ 
2 0 0 0 
31,250 J~ t2e-5oot dt - J~ e-5oot dt 
e-250t loo 250 -250t loo 
= -250 o + (250)2e (-250t - 1) o -
375 e-5oot(-500t - l)loo 
(500)2 0 
311250 e-500t(5002t2 + lOOOt + 2)1 00 
(-500)3 0 
(~=r 
All the upper limits evaluate to zero hence 
WL 1 250 375 (31,250)(2) 
2 = 250 + 62,500 25 x 1Q4 (5)31Q6 
WL=8+8-3-l-4=8mJ 
1 
500 
Note this value corresponds to the final energy stored in the inductor, i.e. 
1 ( 2 wL(oo) = 2(1.6) 0.1) = 8mJ. 
[b] v = 5000te-250t + 20e-250t V 
iR = _..:!:_ = 25te-250t + 0 1e-250t A 200 , . 
PR= viR = 2e-500t[62,500t2 + 500t + lJ 
WR = fo00 PR dt 
......!!: = 62,500 t2e-500t dt + 500 te-500t dt + e-5o0t dtw Joo Joo Joo 
2 0 0 0 
500e-5oot loo e-500t loo 
25 x 104 (-500t- l) 0 + (-500) 0 
Since all the upper limits evaluate to zero we have 
WR 62,500(2) 500 1 
2 = 125 x 106 + 25 x 104 + 500 
8-28 CHAPTER 8. Natural and Step Responses of RLC Circuits 
WR=2+4+4= lOmJ 
[c] 100 = iR + ic + iL (mA) 
iR + iL = 25,000te-250t + 1ooe-250t + 100 - 12,500te-250t - 10oe-250t mA 
= 100 + 12,500te-250t mA 
ic = 100 - ( iR. + iL) = -12,500te-250t mA = -12.5te-250t A 
Pc= vie= [5000te-250t + 2oe-250t](-12.5te-250t] 
= -250[250t2e-500t + te-500t] 
~ = 250 t'0 t2e-5o0t dt + f''0 te-5oot dt 
-250 lo lo 
we 25oe-500t 1= e-500t 1= 
-250 = -125 x 1Q6 [25 x l04t2 + lOOOt + 2] o + 25 x 1Q4 (-500t - 1) o 
Since all upper limits evaluate to zero we have 
w = -250(250)(2) - 250(1) = -1000 x 10-6 - 10 x 10-4 = -2 J 
c 125 x 106 25 x 104 m. 
Note this 2 mJ corresponds to the initial energy stored in the capacitor, 
i.e., 
wc(O) = ~(10 x 10-6)(20)2 = 2mJ. 
Thus we ( oo) = 0 mJ which agrees with the final value of v = 0. 
(d] i 8 = lOOmA 
pB(del) = lOOvmW 
= 0.1[5000te-250t + 2oe-250t] 
= 2e-250t + 500te-25ot W 
Ws = 1= e-250t dt + 1= 250te-250t dt 
2 0 0 
e-2sot 1= 250e-2s0t 1= 
= -250 0 + 62,500 (-250t - l) 0 
1 1 
= 250 + 250 
2(2) 4 
Ws = 250 = 250 = 16 mJ 
[e] WL = 8 mJ (absorbed) 
WR= lOmJ (absorbed) 
we= 2mJ (delivered) 
Ws = 16mJ (delivered) 
L Wdel = Wabs = 18 rnJ. 
p 8.35 t < 0: 
t > 0: 
iL = 3/150 = 20 mA 
120mAI-.), 3000 
30011150 = 100 n 
lOOmAI-.), 1000 
iL(O) = 20mA, iL(oo) = -lOOmA 
~ 0 .5µ.F 
~ 0 .5µ.F 
2 1 109 6 
WO = LC = (31.25)(0.5) = 64 x 10 ; Wo = 8000 rad/s 
1 106 
a= 2RC = (200)(0.5) = l04; a
2 = 100 x 106 
a 2 - w; = (100 - 64)106 = 36 x 106 
81,2 = -10,000 ± 6000 
81 = -4000 rad/s; 8 2 = -16,000 rad/s 
iL(oo) =It= -lOOmA 
iL(O) =A~+ A~+ It= 20rnA 
Problems 8-29 
1500--i20ml\ 
8-30 CHAPTER 8. Natural and Step Responses of RLC Circuits 
A~+ A~ - 100 = 20 so A~+ A~= 120mA 
diL ( ) 1 dt 0 = 0 = -4000A1 - 16,000A2 
Solving, A~= 160mA, A~= -40mA 
iL = -100 + l60e-4000t - 40e-16,oo0t mA, t;:::: O 
P 8.36 vc(O+) = 1(240) = 120V 
iL(oo) = 2: 0 x 10-3 = 48mA 
1 106 
a= 2RC = 2(2500)(5) = 40 
1 106 
w2 =-=-=2500 
0 LC 400 
a 2 = 1600· ' 
a2 < w2· 
Ol underdamped 
s1,2 = -40 ±jJ2500- 1600 = -40 ±j30 rad/s 
iL - If + B~ e-at cos wdt + B~e-at sin wdt 
- 48 + B~ e-40t cos 30t + B~e-40t sin 30t 
B~ = 60 - 48 = 12 mA 
diL ( ) 1 1 120 3 dt 0 = 30B2 - 40B1 = 80 = 1.5 = 1500 x 10-
30B~ = 40(12) x 10-3 + 1500 x 10-3 ; B; = 66mA 
iL = 48 + 12e-40t cos30t + 66e-40tsin30tmA, t > 0 
P 8.37 [a] 2a = 5000; a= 2500rad/s 
.j a 2 - w2 = 1500· 
0 ' 
R 
a=~= 2500· 
2L ' 
1 
w2 = - = 4 x 106 · 
0 LC ' 
R = 25,0000 
w2 = 4 x 106 · 
0 ' 
Wa = 2000rad/s 
R= 5000L 
109 
L = 4 x 106 (50) = 5H 
[b] i(O) = 0 
L did"(O) = Vc(O); 
t 
~(50) x 10-9v~(O) = 90 x 10-6 
:. v~(O) = 3600; Vc(O) = 60V 
di(O) = 60 = 12 A/s 
dt 5 
[c] i(t) = Aie-1ooot + A2e-4ooot 
i(O) = Ai + A2 = 0 
di(O) d:t = -1000A1 - 4000A2 = 12 
Solving, 
:. Ai= 4mA; A2 = -4mA 
i(t) = 4e-1000t - 4e-4000t mA t 2:: 0 
[d] di(t) = -4e-1000t + 16e-4ooot 
dt 
di 
- = 0 when 16e-40o0t = 4e-lOOOt 
dt 
or e3000t = 4 
ln4 
t = 3000µs = 462.10 µs 
[e] imax = 4e-0.4621 - 4e-L8484 = l.89mA 
di 
[f] VL(t) = 5 dt = (-20e-l000t + 80e-4000t) V, t 2:: o+ 
P 8.38 a= 800rad/s; wd = 600rad/s 
w2 - a 2 = 36 x 104 · 
0 ' 
R 
a= - =800· 
2L ' 
1 
- = 100x104 · 
LC ' 
:. R=3.20 
w~ = 100 X 104; w0 = lOOOrad/s 
R = 1600£ 
106 
L = (100 x 104)(500) = 2 mH 
at t = o+ 
Problems 8-31 
8~32 CHAPTER 8. Natural and Step Responses of RLC Circuits 
3.20 2mH 
---)x0 + vJO+) - + 
500µ.F ~ 12V 
di(o+) = -12 = -6000 A/s 
dt 0.002 
di(o+) 
. ·. dt = 600B2 - 800B1 = -6000 
. ·. 600B2 = 800B1 - 6000; 
: . i = -lOe-soot sin 600t A, t 2: 0 
P 8.39 From Prob. 8.38 we know Ve will be of the form 
From Prob. 8.38 we have 
Vc(O) = 12V = B3 
and 
dvc(O) = ic(O) = O 
dt c 
dvc(O) = 600B - 800B 
dt 4 3 
. ·. 600B4 = 800B3 + O; 
vc( t) = 12e-SOOt cos 600t + 16e-s00t sin 600t V t > 0 
P 8.40 [a] t < 0: 
120 
i =--=15mA· 
0 8000 ' 
Vo= (5000)(0.015) = 75 V 
t >0: 
R 5000 
a= 2L = 2(l) = 2500 rad/s 
2 1 109 6 4 
WO = LC - (1)(250) = 4 x 10 = 400 x 10 
a2 -w~ = 625 x 104 -400 x 104 = 225 x 104 
:. 812=-2500±1500 
' 
s1 = -1000 rad/s s2 = -4000 rad/s 
io(O) = A1 + A2 = 15 x 10-3 
di0 ( ) di 0 = -1000A1 - 4000A2 = 0 
Solving, A1 = 20mA; 
io(t) = 20e-lOOOt - 5e-4000t rnA, 
[b] Vo(t) = Aie-lOOOt + A2e-4000t 
A2 = -5mA 
t:;:::: o+ 
dv0 -15 X 10-3 
dt (0) = -lOOOA1 - 4000A2 = 250 x 10_9 
Solving, Ai= 80V; 
Check: 
000 . 1dio 5 'l 0 + di= Vo 
5000i0 = lOOe- lOOOt - 25e-4o00t 
dio = -20e-1000t + 20e-40oot 
dt 
t:;:::: o+ 
5000i + dio = 80e-lOOOt - 5e-4000t V 
0 dt 
(checks) 
Problems 8-33 
8-34 CHAPTER 8. Natural and Step Responses of RLC Circuits 
1 109 108 
p 8.41 [a] w~ = LC = (0.25)(160) = 4 = 25 x 106 
R 
a= - = w0 = 5000 rad/s 
2L 
:. R = (5000)(2)£ = 25000 
[h] i(O) = iL(O) = 24 mA 
vL(O) = 90 - (0.024)(2500) = 30V 
di(O) = ~ = 120 A/ 
dt 0.25 s 
[c] Ve= Dite-5ooot + D2e-5ooot 
vc(O) = D2 = 90V 
dvc (0) =Di - 5000D2 = ic(O) = -iL(O) 
& c c 
24 x 10-3 
Di - 450,000 = - 160 x 10_9 = -150,000 
:. Di= 300,000 V /s 
vc = 300,000te-5000t + 90e-5ooot V, 
P 8.42 [a] Fort> 0: 
+ 
Since i(O-) = i(O+) = 0 
va(o+) = 300V 
(b] Va= 200i + 5 X 104 lot idx + 300 
dva 200 di 5 104. -= -+ x i 
dt dt 
5H 
dva(o+) = 2oodi(o+) + 5 x 104i(o+) = 2oodi(o+) 
dt dt dt 
-L di(O+) = 300 
dt 
di~~+) = -0.2(300) = -60A/s 
:. dva(o+) = -12,000V /s 
dt 
R 800 
[c] a= 2L = 10 = 80rad/s 
2 1 106 
WO= LC= (5)(20) = 104 
s1,2 = -80 ± V6400 - 104 == -80 ± j60 rad/s 
Underdamped: 
Va = B1 e-BOt cos 60t + B2e-s0t sin 60t 
Problems 8-35 
Va(O) = B1 = 300V 
dva(O) 
dt = -80B1 + 60B2 = -12,000; . ·. B2 = 200 V 
Va = 300e-BOt COS 60t + 200e-SOt sin 60t V, t 2: o+ 
2 1 1 6 
WO = LC = (0.100)(200 x lQ-9) = 50 x 10 
R 200 
a = 2L = 2(0.100) = lOOO; 
0 2 <w2 
0 
underdamped 
s1,2 = -1000 ± j7000 rad/s 
i(O) = B1 = 280 mA 
di 
dt (0) = 7000B2 - 1000B1 = 0 
1 
:. B2 = 7B1 = 40mA 
i = 280e-1000t cos 7000t + 40e-looot sin 7000t mA, 
8-36 CHAPTER 8. Natural and Step Responses of RLC Circuits 
p 8.44 t < 0: 
300 
+ 
100 100 
80 i(O 
i(O) = 100 = 100 = 5 A 
4+8+8 20 
v0 (0) = 100 - 5(4) -10(5) (~~) = 70V 
t > 0: 
2H 
80 
R 20 
a=-=-=5 
2L 4 ' 
w2 = _1_ = 100 = 50 
0 LC 2 
120 
a 2 = 25 
w~ > a 2 underdamped 
B -at t B -at . t v 0 = 1e coswd + 2e smwd ,; 
+ + 
70V 
v0 (0) = B1 = 70V 
Cdvo(O) = -5 
dt ' 
dvo = - 5 x 103 = -500V /s 
dt 10 
~0 (0) = -5B1 + 5B2 = -500 
5B2 = -500 + 5B1 =-500+350; B2 = -150/5 = -30V 
: . v 0 = 70e-5t cos 5t - 30e-5t sin 5t V, t ? 0 
Problems 8-37 
R 
P 8.45 a= 2L = 5000rad/s 
1 109 
w2 = - = - = 50 x 106 
0 LC 20 
s1,2 = -5000 ± V25 x 106 - 50 x 106 = -5000 ± j5000rad/s 
v0 = VJ + B~ e-sooot cos 5000t + B~e-5oo0t sin 5000t 
v0 (00) = 40V; B~ = -40V 
dvo(O) = 0 = 5000B~ - 5000B~ 
dt 
: . B~ = B~ = -40 V 
v0 = 40 - 40e-so00t cos 5000t - 40e-5oo0t sin 5000t V, t ;::: 0 
R 
P 8.46 a= 2L = 5000rad/s 
2 1 1 6 
WO= LC= (0.4)(100 x 10-9) = 25 x 10 
w0 = 5000 rad/s 
The response is therefore critically damped 
v0 (0) = 0 = Vi + D~ 
v0 (00) = 40V; D~ = -40V 
dv0 (0) _ O _ D' _ D' 
- - 1 a 2 dt 
:. D~ = (5000)(-40) = -200,000 V /s 
Vo= 40 - 200,000te-5000t - 40e-sooot V, t ;::: 0 
8--38 CHAPTER 8. Natural and Step Responses of RLC Circuits 
R 
P 8.47 a = 2£ = 5000 rad/s 
2 1 1 6 
WO = LC = (0.4)(156.25 x 10-9) = 16 x lO w0 = 4000 rad/s 
The response is therefore overdamped 
s 1,2 = -5000 ± V50002 - 40002 = -5000 ± 3000 = -2000rad/s, - 8000rad/s, 
Vo(O) = 0 = VJ +A~ +A~ 
v0 (00) = 40V; :. A~+~= -40V 
. ·. A~ = -53.33 V, A~ = 13.33 V 
v0 = 40 - 53.33e-2000t + 13.33e-sooot V, t ~ 0 
P 8.48 [a] Let i be the current in the direction of the voltage drop v0 (t). Then by 
hypothesis 
. . + B' -at t B' -at . t i = 'tf le COSWd + 2e SlllWd' 
it= i{oo) = 0, 
L di(O) = O 
dt ' 
i(O) =Vy = B~ 
R 
d"(O) 
therefore ~t = 0 
di[( B' B') ( B' B') . ] -at dt = Wd 2 - a 1 cos wdt - a 2 + wd 1 sm wdt e 
Therefore wdB~ - aB~ = O; B~= ~B~ = ~ Vg 
Wd Wd R 
Therefore 
dv0 Wd - = 0 when tanwdt = -
dt a 
Therefore wdt = tan-1(wd/a) (smallest t) 
t = _.!._ tan-1 (wd) 
Wd a 
P 8.49 [a] From Problem 8.48 we have 
-Vg -at • t 
Vo= RC e SlllWd 'wd 
R 120 
a= 2L = O.Ol = 12,000rad/s 
1 1012 
w~ = Lc = 250o = 400 x 106 
Wd = /w~ - a 2 = 16krad/s 
-(-600)109 
(120)(500)(16) x 103 = 625 
. ·. v0 = 625e-12'000t sin 16,000t V 
Problems 8--39 
8--40 CHAPTER 8. Natural and Step Responses of RLC Circuits 
[b] From Problem 8.48 
1 _1 (wd) 1 _1 (16,000) 
td = wd tan ~ = 16,000 tan 12,000 
td = 57.96µs 
[c) Vmax = 625e-0·012<57·96) sin[(0.016)(57.96)] = 249.42 V 
[dJ R = 12 O; o: = 1200rad/s 
wd = 19,963.97 rad/s 
v0 = 5009.02e-1200t sin 19,963.97t V, t 2:: 0 
td = 75.67 µs 
Vmax = 4565.96 V 
P 8.50 ic(O) = O; v0 (0) = 200V 
R 4 
o: = 2£ = 2(0.04) = 50rad/s 
1 103 
w~ = LC = 0.4 = 2500 
0:2 = w2· 
Ol 
critical damping 
Vt= lOOV 
v0 (0) = 100 + D~ = 200; 
dvo (0) = -50D~ + D~ = 0 
dt 
D~ = 50D~ ~ 5000 V /s 
D~ = lOOV 
'11o = 100 + 5000te-50t + lOOe-5ot V, t 2:: 0 
P 8.51 [a] t < 0: 4k0 12k0 
+ 
28\n ; 
I • 
. - 48 
z0 (0 ) = 16,000 = 3 mA 
vc(o-) = 20 - (12,000)(0.003) = -16 v 
t = o+: 
+ 
v c 
12kflll24kfl = 8kfl 
24k0 
12k0 + 
+ 
vJO ) 
~ 20V 
I 
:. vo(o+) = (0.003)(8000) - 16 = 24- 16 = 8V 
and vL(o+) = 20- 8 = 12V 
[b] v0 (t) = 8000i0 +VO 
dvo (t) = 8000di0 + dvc 
dt dt dt 
dvo (o+) = 8000 dio (o+) dvc (o+) 
dt dt + dt 
vL(o+) = L ddio (o+) 
t 
dio (o+) = vL(o+) = E = 60 A/s 
dt L 0.2 
cd~c (o+) = io(o+) 
3 dvc( + 3 X 10-
dt 0 ) = 8 x 10-9 = 375,000 
dv 
dt0 (o+) = 8000(60) + 375,000 = 855,000 V /s 
Problems 8-41 
8-42 CHAPTER 8. Natural and Step Responses of RLC Circuits 
2 1 109 6 
le) w0 = LC = 1.6 = 625 X 10 ; w0 = 25,000 rad/s 
R 8000 
a = 2L = 0.4 = 20,000 rad/s; a
2 = 400 x 106 
a2 < w2 
0 
underdamped 
s 1,2 = -20,000 ± j15,000 rad/s 
v0 (t) = V1 + B~e-2o,o00t cos 15,000t + B~e-20,000t sin 15,000t 
Vt= Vo(oo) = 20V 
B~ = -12V 
-20,000B~ + 15,000B~ = 855,000 
Solving, 
v0 (t) = 20 - 12e-2o,oo0t cos 15,000t + 41e-2o,ooot sin 15,000t V, 
p 8.52 t < 0: 
2000 
160\'1 ; 10000 
-160 
iL(O) = 1600 = -lOOmA 
vc(O) = lOOOiL(O) = -lOOV 
t > 0: 
..--~~V\f'v~~~v.r.~~~,-YYY-,~--, 
j 6000 4000 100mH I + 
60vQ 
1 
400nF t 
R 1000 
a = 2L = 200 x 103 = 5000 rad/s 
w2 = _1_ = (109)(103) = 10s = 25 x 106 
0 LC (100)(400) 4 
Problems 8-43 
w0 = 5000 rad/s critical damping 
vc(O) = -lOOV; Vt= -60V 
:. -100 = -60 + D~; 
Cd;~ (0) = iL(O) = -100 x 10-3 
dvc (0) = -lOO x io-3 = -250 000 V /s 
dt 400 x 10-9 ' 
. ·. D~ = 5000( -40) - 250,000 = -450,000 
vc(t) = -60- 450,000te-5000t - 40e-5000tv, t;:::: 0 
P 8.53 [a] Ve= Vt+ [B~ coswdt + B~ sinwdt] e-at 
dvc [( B' B') ( B' B') . ] -at -d = wd 2 - a 1 cos wdt - a 2 + wd 1 sm wdt e 
t 
Since the initial stored energy is zero, 
vc(O+) = 0 and dvc(O+) = 0 
dt 
It follows that B~ = - Vt and B' - aBi 2-
Wd 
When these values are substituted into the expression for [dvc/ dt], we get 
dv (w2 ) Therefore dtc = w: Ve-at sinwdt 
[ ] dvc b dt = 0 when 
where n = 1, 2, 3, ... 
'Tl/Tr 
Therefore t = -
Wd 
8-44 CHAPTER 8. Natural and Step Responses of RLC Circuits 
p 8.55 
p 8.56 
n1r 
[c] When tn = -, coswdtn = cosn1r = (-l)n 
Wd 
and sin wdt = sin n7r = 0 
Therefore 'IJc(tn) = V[l - (-lre-an7r/wd] 
[d] It follows from [c) that 
V(t1) = V + Ve-(a7r/wd) and Vc(t3) = V + Ve-(3a7r/wd) 
v (t1) - V e-(011r/wd) 
Therefore c = = e<2a7r/wd) 
Vc(t3) - V e-(3a7r/wd) 
31r x 2x 
Td = t3 - ti = - - - = - ms 
12 12 12 
= 12,000 1 [ 13.505] = 5000· 
a 2x n 0.985 ' 
2x 
wd = Td = 12,000rad/s 
w~ = w~ + o.2 = 144 x 106 + 25 x 106 = 169 x 106 
1 
L = (l69)(0.2) = 29.6mH; R = 2aL
 = 295.86 n 
At t = 0 the voltage across each capacitor is zero. It follows that since the 
operational amplifiers are ideal, the current in the 500 kfl is zero. Therefore 
there cannot be an instantaneous change in the current in the 1 µF capacitor. 
Since the capacitor current equals C(dv0 /dt), the derivative must be zero. 
d2v 
[a) From Example 8.13 dt2° = 2 
dg(t) dv0 
therefore ~ = 2, g(t) = dt 
g(t) - g(O) = 2t; g(t) = 2t + g(O); (O) = dv0 (0) g dt 
11 .. lE' "ic 
~----11~ ---=:? 
500kQ 
• + -=-.::;"':"" 
5V lR 
1=70A 
+ 
ov 
. 5 3 . 0 dv0 (0) iR = - x 10- = 10 µA = ic = -
500 dt 
dv0 (0) = -10 X 10-6 = -lO = (O) 
dt 1x10-6 g 
dvo = 2t-10 
dt 
dv0 = 2tdt - lOdt 
V 0 - v0 (0) = t2 - lOt; v0 (0) = 8 V 
V 0 = t 2 - lOt + 8, 0 :S t :S tsat 
[bJ t2 - lOt + 8 = -9 
t 2 - lOt + 17 = 0 
trv2.17s 
P 8.57 Part (1) - Example 8.14, with R1 and R2 removed: 
[a] Ra = 100 kO; Ci= O.lµF; 
Problems ~45 
1 1 
RaC1 = lOO RbC2 = 40 
therefore 
dv0 (0) 2 
[b] Since v0 (0) = 0 = dt , our solution is v0 = 500t 
The second op-amp will saturate when 
V0 = 6 V, Or tsat = J6/500 '""'0.1095 S 
dvo1 1 
(c] - = ---Vg = -25 
dt RaC1 
[d] Since v0 1(0) = 0, v0 1 = -25tV 
At t = 0.1095s, V 0 1 rv -2.74 V 
Therefore the second amplifier saturates before the first amplifier 
saturates. Our expressions are valid for 0 :St :S 0.1095 s. Once the second 
op-amp saturates, our linear model is no longer valid. 
Part (2) --- Example 8.14 with v01 (0) = -2 V and v0 (0) = 4 V: 
[a] Initial conditions will not change the differential equation; hence the 
equation is the same as Example 8.14. 
8-46 CHAPTER 8. Natural and Step Responses of RLC Circuits 
[b] v0 = 5 + A~e-lOt + A;e-20t (from Example 8.14) 
v0 (0) = 4 = 5 +A~ +A; 
lµF iJO+} 
1ooko ;141,ao)mA 
------1~ 
• 
25kQ 
w,~--+ 
2V iR(2/25)mA ov 
~+ ~-
_!_ + ic(o+) - ~ = o 
100 25 
i (o+) = _!_mA = Cdvo(O+) 
c 100 dt 
dv0 (0+) = 0.04 x 10-3 = 40 V /s 
dt 10-6 
dvo = - lOA' e-10t - 20A' e-20t 
dt 1 2 
dvo (o+) = -10A~ - 20A; = 40 
dt 
- 4V + 
Therefore -A~ - 2A~ = 4 and A~ + A~ = -1 
Thus, A~= 2 and A~= -3 
v0 = 5 + 2e-l0t - 3e-20t V 
[c] Same as Example 8.14: 
dvo1 dt + 20v0 1 = -25 
[d] From Example 8.14: 
V0 1(00) = -l.25V; 
Therefore 
v1(0) = -2V (given) 
v0 1 = -1.25 + (-2 + l.25)e-20t = -1.25 - 0. 75e-20t V 
] d2v0 1 P 8.58 [a - = v 
dt2 R 1 C1R2C2 g 
1 10-6 
R1C1R2C2 - (50)(20)(2)(4) x 10-6 x 10-6 = 125 
d2 Vo 
dt2 = l25v9 
0::; t < 0.2-: 
v9 = 400mV 
d2vo = 50 
dt2 
() dv0 Let gt = dt, 
l g(t) lot dx = 50 dy 
g(O) 0 
then dg = 50 or dg = 50 dt 
dt 
g(t) - g(O) = 50t, g(O) = d~o (0) = O 
() dv0 gt = dt = 50t 
dv0 = 50tdt 
Problems 8-47 
1vo(t) lot dx = 50 xdx; 
Vo(O) 0 
Vo(t) - Vo(O) = 25t2 , v0 (0) = 0 
v0 (t) = 25t2 V, Q < t < 0.2-
dv0 1 1 - = ---Vg = -l0v9 = -4 
dt RiC1 
dv0 1 = -4dt 
1Vo1(t) lot dx = -4 dy 
Vo1(0) 0 
Va1(t)-Vo1(0) = -4t, 
V0 1(t) = -4tV, 
0.2+ < t S: fsat: 
d2v0 
dt2 = -12.5, () 
dv0 
let gt = dt 
d~~t) = -12.5; dg(t) = -12.5dt 
1g(t) it dx = -12.5 dy 
g(0.2+) . 0.2 . 
9(t) - 9(0.2+) = -12.s(t - 0.2) = -12.5t + 2.5 
8-48 CHAPTER 8. Natural and Step Responses of RLC Circuits 
( 2+) = dv0 (0.2+) go. dt 
0 dv0 (0.2+) = 0- V0 1(0.2+) 
dt 20 x 103 
Vo1(0.2+) = vo(o.2-) = -4(0.2) = -0.80V 
. 0 dvo1 (0.2+) = o.so = 0 A 
. . dt 20 x 103 4 µ 
dv0 1 (o.2+) = 40 x 10-6 = 10 V /s 
dt 4 X lQ-G 
dv0 
g(t) = -12.5t + 2.5+10=-12.5t+12.5 = dt 
dv0 = -12.5t dt + 12.5 dt 
ro(t) dx = rt -12.5y dy + rt 12.5 dy 
lv0 (0.2+) lo.2+ Jo.2+ 
Vo(t) - vo(o.2+) = -6.25y2 \t +12.5y It 
0.2 0.2 
v0 (t) = v0 (0.2+) - 6.25t2 + 0.25 + 12.St - 2.5 
vo(o.2+) = vo(o.2-) = 1 V 
:. v0 (t) = -6.25t2 + 12.5t - 1.25 V, 0.2+ ::; t::; tsat 
d~;l = -10(-0.1) = 1, 0.2+ < t::; tsat 
dv0 1 = dt; l V01(t) lot dx= dy 
• Vo1(0.2+) 0.2+ 
.". V0 1(t)=t-lV, 
Summary: 
0::; t ::; 0.2-s: Vol = -4t V, V 0 = 25t2 V 
0.2+s < t ::; tsat : V0 1 = t - 1 V1 V 0 = -6.25t2 + 12.5t - 1.25 V 
[b) -10 = -6.25t;at + 12.5tsat - 1.25 
. ·. 6.25t~t - 12.Stsat - 8. 75 = 0 
t;at - 2tsat - 1.4 = 0 
tsat = 1 ± J2 + 1.4 = 1 ± 1.844 
:. tsat = 2.844sec 
V0 1(tsat) = 1.844 - 1=0.844 V 
P 8.59 7 1 = (0.25 x 106 )(2 x 10-6) = 0.50s 
1 
- =2; 
71 
72 = (0.25 X 106)(4X 10-6) = 1 s; 
d2v0 dv0 
-d 2 +3-d +2v0 = 50 t t 
s2 + 3s + 2 = 0 
( s + 1) ( s + 2) = O; 
v - Tr +A' e-t +A' e-2t. 
0 - Vf 1 2 ' 
:. A~= -50, A~= 25V 
vo(t) = 25 - 50e-t + 25e-2t V, 
50 
V1=- =25V 
2 
dv0 (0) = 0 =-A~ - 2A; 
dt 
0 :::; t :::; 0.2 s 
Problems 8~49 
dvol dt + 2vol = -4; :. v01=-2+2e-2tv, 0:::; t:::; 0.2s 
v0 (0.2) = 25 - 50e-0·2 + 25e-0.4 = 0.8215 V 
v0 1(0.2) = -2 + 2e-0.4 = -0.6594 V 
At t = 0.2s 
4µf 
20kQ l i:'L 
+ -•+---w. 
-0. 6594V 0. 8215v 
. = 0 + 0.6594 = 32 97 A ic 20 x 1Q3 . µ 
8-50 CHAPTER 8. Natural and Step Responses of RLC Circuits 
dv0 
C dt = 32.97 µA; d~0 = 32~97 = 8_24 V /s 
0.2s $ t < oo: 
d2vo dv0 
d 2 +3-d +2 = -12.5 t t 
V0 ( 00) = -6.25 
· ·• Vo = -6.25 +A~ e-(t-0.2) + A~e-2<t-0.2) 
0.8215 = -6.25 + A~ + A~ 
dvo ( ) , 1 dt 0.2 = 8.24 = -A1 - 2A2 
:. A~+ A~= 7.07; -A~ - 2A~ = 8.24 
A~= 22.38; A~= -15.31 
:. Vo= -6.25 + 22.38e-(t-0.2) - 15.31e-2(t-0.2) V, 0.2 $ t < 00 
dvo1 
dt+2vo1=1 
Vo1 = 0.5 + (-0.6594 - l)e-2<t-0.2) = 0.5 - 1.66e-2<t-0.2) V, 
P 8.60 {a] 
R v R a 
•+ ¥Al 
v 
g re + 
20 dva Va - v9 Va = O 
dt + R + R 
( ) dva Va _ Vg . 1 Therefore dt + RC - 2RC, 
c c 
0.5R 
+ 
0.2 $ t < 00 
dvb Va 
(2) Therefore dt +RC= O, 
2vb CdVb Cd(vb - Vo) = Q 
R + dt + dt 
dVb 
v =-RC-
a dt 
dvb Vb 1 dv0 
( 3) Therefore dt + RC = 2 dt 
Problems 8-51 
From (2) we have dva - RCd
2vb and v = -RCdvb 
dt - - dt2 a dt 
When these are substituted into (1) we get 
( ) RCd
2vb dvb _ v9 
4 - dt2 - dt - 2RC 
Now differentiate (3) to get 
(5) d2vb _1_ dvb = ~ d
2vo 
dt2 + RC dt 2 dt2 
But from ( 4) we have 
(6) d2vb _2_ dvb = 
dtz +RC dt 
Now substitute (6) into (5) 
d2v0 Vg 
[b] When Ri C1 = RzC2 = RC : dt2 = Rzcz 
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 8.61 (a] J(t) - inertial force+ frictional force+ spring force 
_ M[d2x/dt2] + D[dx/dt] + Kx 
[b] d2x = j_ _ (D) (dx)- (K) x 
dt2 M M dt M 
d2x 
Given VA = dt2 , then 
1 rt (Jlx) 1 dx 
VB= - R1C1 lo dy2 dy = - R1C1 dt 
1 1·t 1 
vc=--c VBdy= RCGX R2 2 o Ri 2 1 2 
8~52 CHAPTER 8. Natural and Step Responses of RLC Circuits 
[-Rs] VF= R1 f(t), 
Therefore 
Therefore M = R7 
Rs' 
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 8.62 [a] Given that the current response is underdamped we know i will be of the 
form 
i = 11 + [B~ coswdt + B~ sinwdt]e-<Y.t 
where 
and 
The capacitor will force the final value of i to be zero, therefore 11 = 0. 
By hypothesis i(O+) = Vdc/ R therefore B~ =~cl R. 
At t = o+ the voltage across the primary winding is zero hence 
di(o+)/dt = o. 
From our equation for i we have 
di [( f f) ( f I) · 1 -at dt = wdB2 - aB1 cos wdt - wdB1 + aB2 sm wdt e · 
Hence 
di(o+) , , 
d = wdB2 - aB1 = 0 t 
Thus 
B' _ ~B' _ a:Vac 
2 - wa 1 - waR 
It follows directly that 
. Vac[ a . ] a.t 
i = R coswat + wa smwat e-
[h] Since waB~ - a:B~ = 0 it follows that 
di ( B' B' ) -at . dt = - wa 1 + a 2 e sm wat 
But and 
Therefore 
B' B' _ Wa Vdc a 2Vac _ Vac [w~ + a:2 ] 
wa 1 + a 2 - R + R - R wa wa 
2 2 2 1 
But wd+a =w0 =LC 
Hence 
B' B' Vdc wa 1 + a 2 = waRLC 
Now since 
di 
V1 = L dt we get 
v1 = -L Vdc e-at sinwat = - Vac e-a.t sinwdt 
waRLC waRC 
. di 
[c] v = Va - iR - L-e c dt 
iR = Vdc ( coswdt + :d sinwat) e-at 
Problems 8-53 
8-54 CHAPTER 8. Natural and Step Responses of RLC Circuits 
1 -1 (Wd) tmax = - tan -
wd o: 
Note that because tanO is periodic, i.e., tanO = tan(O ± n7r), where n is an 
integer, there are an infinite number of solutions fort where dvsp/dt = 0, that 
lS 
tan-1(wd/a) ± mr 
t=-------
Wd 
Because of e-at in the expression for Vsp and knowing t 2: 0 we know Vsp will 
be maximum when t has its smallest positive value. Hence 
P 8.64 [a) Ve= Vdc[l - e-at coswdt +Ke-at sinwdt] 
~c = Vdc :t {1 + e-at(K sinwdt - coswdt)] 
= Vdc{(-ae-at)(K sinwdt- coswdt)+ 
e-at [wdK cos wdt + Wd sin wdt]} 
= Vdce-at[(wd - aK) sinwdt +(a+ wdK) coswdt] 
or tanwat = [a; waK] 
a -wa 
wat ± n1f = tarC1 [-a-K+_w_a_K_] 
a -wa 
R 4 x 103 
a= 2£ = 6 = 666.67rad/s 
Wa= 
109 
- 2 - (666.67)
2 = 28,859.81 rad/s 
1. 
K = ~ (-1- - a) = 21.63 
wa RC 
Problems 8-55 
The smallest positive value oft occurs when n = 1, therefore 
tcmax = 55.23 µs 
[h] Vc(tcmax) = 12(1 - e-O:tcmax coswdtcmax + Ke-O:tcmo:x sinwatcmax] 
= 262.42V 
[c] From the text example the voltage across the spark plug reaches its 
maximum value in 53.63 µ,s. If the spark plug does not fire the capacitor 
voltage peaks in 55.23 µs. \Vhen 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 8.65 [a) w = ~L[i(o+)] 2 = ~(5)(16) x 10-3 = 40mJ 
R 3 x 103 
[h) a= 2£ = 10 = 300rad/s 
Wd= 
1Q9 
- 2- - (300)
2 = 28,282.68 rad/s 
1. 5 
1 106 4 x 106 
Re= 0.75 = 3 
1 -1 (Wd) tmax = - tan - = 55.16 µs Wd a 
12(50)( 4 x 106) -o:t . 
Vsp (tmax) = 12 - 3(28,282_68) e max SlllWdtmax = -27,808.04 V 
8-56 CHAPTER 8. Natural and Step Responses of RLC Circuits 
[c] Ve (tma:x) = 12[1 - e-O!tma.x coswdtmax + Ke-atmax sinwdtmax) 
K = ~ [-1- - a] = 47.13 
Wd RC 
Ve (tma:x) = 568.15 V 
------9 
Sinusoidal Steady State Analysis 
Assessment Problems 
AP 9.1 [a] V = 170/-40° V 
[b] 10 sin(lOOOt + 20°) = 10 cos(lOOOt - 70°) 
. . I= 10/-70° A 
[c] I= 5L36.87° + 10/-53.13° 
= 4 + j3 + 6 - j8 = 10 - j5 = ll.18L-26.57° A 
[d] sin(20,0007rt + 30°) = cos(20,0007rt - 60°) 
Thus, 
v = 300/45° - lOOL-60° = 212.13 + j212.13 - (50 - j86.60) 
= 162.13 + j298. 73 = 339.90L6L51° m v 
AP 9.2 [a] v = 18.6cos(wt - 54°) V 
[b] I= 20/45° - 50L - 30° = 14.14 + jl4.14 - 43.3 + j25 
= -29.16 + j39.14 = 48.81/126.68° 
Therefore i = 48.81 cos( wt+ 126.68°) mA 
[c] v = 20 + i80 - 30L15° = 20 + j80 - 28.98 - j7. 76 
= -8.98 + j72.24 = 12. 79L97.08° 
v = 72.79cos(wt + 97.08°) V 
AP 9.3 [a] wL = (104)(20 x 10-3) = 200 0 
[b) ZL = jwL = j200 n 
9-1 
9-2 CHAPTER 9. Sinusoidal Steady State Analysis 
[c} V L = IZL = (10/30°)(200L90°) x 10-3 = 2/120° V 
(dJ VL = 2 cos(lO,OOOt + 120°) V 
-1 -1 
AP 9.4 [a] Xe= wC = 4000(5 x 10-6) = -500 
[b) Ze =}Xe= -j500 
(c) I= ~ = 30~ = 0.6/115° A 
Ze 50L-90° --
(d) i = 0.6 cos( 4000t + 115°) A 
AP 9.5 11 = 100L25° = 90.63 + j42.26 
12 = 100L145° = -81.92 + J57.36 
13 = lOOL-95° = -8.71 - j99.62 
l4 = -(11 + 12 + l3) = (0 + jO) A, therefore i 4 = 0 A 
. 125L-60° 125 o 
AP 9.6 [aJ I= IZl/Bz = TZf /(-60 - Oz) 
But -60 - Oz= -105° 
Z = 90+ jl60+ jXe 
:. Bz = 45° 
1 
Xe= -70!1; Xe= - wC = -70 
1 
C = (70)(5000) = 2'86 µF 
[bJ I= Vs= 125L=filr = 0.982L-105°A-
Z (90 + j90) ' 
AP 9.7 [a] 200 
III= 0.982A 
x _ _.. 
--~\1\1\.~----<~ • y 
50 25µ.F 
5mH 
w = 2000rad/s 
-1 
wL = 100, - = -200 
wC 
. . 20(j10) . 
Zxy = 2011110 + 5 + J20 = (20 + jlO) + 5 - J20 
= 4 + JB + 5 - j20 = (9 - J12) n 
Problems 9-3 
[b] wL = 40!1, 
-1 
-=-5!1 
wC 
z = 5 - )0 5 + 20llJ"40 = 5 - J0 5 + [(20)(j40)l xy 20 + j40 
= 5 - j5 + rn + JB = (21 + J3) n 
(c] Zx = [ 20(jwL) l + (5 - j106) 
Y 20+ jwL 25w 
20w2 L2 ·400wL ·106 + J +5 J - 4_0_0_+_w_2_L_2 400 + w2 L2 - 25w 
The impedance will be purely resistive when the j terms cancel, 'i.e., 
400wL 106 
-
400 + w2 L 2 25w 
Solving for w yields w = 4000 rad/ s. 
20w2L2 
[d] Zxy = 400 + w2 L2 + 5 = 10 + 5 = 15 n 
AP 9.8 The frequency 4000 rad/s was found to give Zxy = 15 !l in Assessment 
Problem 9.7. Thus, 
V = 150/0°, Is = V = 150f!L = lO[Oo A 
Zxy 15 -
Using current division, 
1£ = 20 !0J20 (lo) = 5 - J5 = 1.01 f-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!1, 40!1, and 10!1 resistors with its 
equivalent wye, the circuit becomes 
!.;.. 14Q l 
136'.Ji ~ 
c_ 
20Q 
4Q 
9-4 CHAPTER 9. Sinusoidal Steady State Analysis 
AP 9.10 
The circuit is further simplified by combining the parallel branches,(20 + j40)ll(5 - j15) = (12 - j16) n 
Therefore I= 136L!r_ = 4/28.07° A 
14 + 12 - j16 + 4 
V1 = 240L53.13° = 144 + j192 v 
V2 = 96/-90° = -j96V 
jwL . j(4000)(15 x 10-3 ) = j600 
1 .6xl06 . 
jwC = -l (4000)(25) = - 16on 
Perform a source transformation: 
_V_1 = 144 + j192 = 3.2 _ .2.4 A 
j60 j60 1 
V2 .96 . 
- = -1- = '-14.8A 
20 20 
Combine the parallel impedances: 
1 1 1 1 j5 1 
y = j60 + 30 + -j60 + 20 = j60 = 12 
1 z = - = 12n y 
+ 
V 0 = 12(3.2 + j2.4) = 38.4 + j28.8 V = 48L36.87° V 
v0 = 48cos(4000t + 36.87°) V 
Problems 9-5 
AP 9.11 Use the lower node as the reference node. Let Y 1 =node voltage across the 
200 resistor and YTh =node voltage across the capacitor. Writing the node 
voltage equations gives us 
y 1 - 2L4 o y 1 - lOlx = 0 and y -j 10 ( OI ) 20 ___Q_ + j 10 Th = 10 - j 10 1 x 
We also have 
Solving th~se equations for YTh gives YTh = l0j45°V. To find the Thevenin 
impedance, we remove the independent current source and apply a test 
voltage source at the terminals a, b. Thus 
j100 
200 
It follows from the circuit that 
lOix = (20 + jlO)Ix 
Therefore 
Yr Yr 
Ix= 0 and Ir= -jlO + 10 
100 
Yr 
ZTh = Ir ) therefore ZTh = (5 - j5) n 
..... -j100 
AP 9.12 The phasor domain circuit is as shown in the following diagram: 
+ 
50 v ..... -j(20/9)0 j50 
T 
9-6 CHAPTER 9. Sinusoidal Steady State Analysis 
The node voltage equation is 
V V v v - lOOL-90° 
-lO + 5 + -j(20/9) + j5 + 20-- = 0 
Therefore V = 10 - j30 = 31.62/-71.57° 
Therefore v = 31.62 cos(50,000t - 71.57°) V 
AP 9.13 Let Iai lb, and le be the three clockwise mesh currents going from left to 
right. Summing the voltages around meshes a and b gives 
33.8 = (1 + j2)1a + (3 - j5)(Ia - h) 
and 
But 
therefore 
le= -0.75[-j5(1a - lb)]. 
Solving for I = Ia = 29 + j2 = 29.07 L3.95° A. 
AP 9.14 [a] M = 0.4v'0.0625 = 0.1 H, wM =son 
Z22 = 40 + j800(0.125) + 360 + j800(0.25) = ( 400 + j300) n 
Therefore IZ22I =soon, Z22 = ( 400 - j300) n 
( 80 )
2 zT = 500 ( 400 - j3oo) = (10.24 - j7.68) n 
[b] I - 245.20 - l". L I': 0 A 
1 - 184+100 + j400 + ZT - 0·00 - 03·13 
ii = 0.5cos(800t - 53.13°) A 
[c] 12 = (j~:) 11 = 500~:~.870 (0.5/ - 53.13°) = 0.08/0° A 
i2 = 80 cos soot mA 
Problems 9-7 
AP 9.15 
11 = Vs _____ 2_5_x_l_0_3/_0_0 __ _ 
Z 1 + 2s2 Z2 1500 + j6000 + (25)2(4 - j14.4) 
= 4 + j3 = 5/36.87° A 
V 1 =Vs - Z1l1 = 25,000/0° - (4 + j3)(1500 + j6000) 
= 37,000- j28,500 
V2 = - 2~ V1 = -1480 + j1140 = 1868.15/142.39°V 
12 = V 2 = 1868.15/142.39° = 1251216_870 A 
Z2 4 - j14.4 
9-8 CHAPTER 9. Sinusoidal Steady State Analysis 
Problems 
P 9.1 [a] w = 2Jrf = 2407rrad/s, 
[h] T = 1/ f = 8.33ms 
w f = - = 120Hz 
211" 
p 9.2 
p 9.3 
[c] Vm = lOOV 
[d] v(O) = 100 cos( 45°) = 70. 71 V 
45°(211') 1l" [e) ¢ = 45°; ¢ = = - = 0 7854 rad 360° 4 . 
[f] V = 0 when 2407rt + 45° = 90°. Now resolve the units: 
45° 1l" 
(24071' rad/s)t = 57.30 /rad = 4 rad, 
(g] ( dv / dt) = (-100)24071' sin(2407rt + 45°) 
(dv/dt) = 0 when 240Jrt + 45° = 180° 
135° 371' 
2407rt = 57,30 /rad= 4 rad or 
Therefore t = 3.125 ms 
t = l.042ms 
60cos(wt+30) 
time (s) 
[a] Left as¢> becomes more positive 
[bJ Right 
BOcos(wt-30) 
60cos(wt-60) 
T 1250 250 [a] 2 = - 6- + 6 = 250p.s; T = 500µs 
1 106 f = - = - = 2000Hz T 500 
[b] v = Vm sin( wt+ 0) 
w = 2xf = 4000x rad/s 
( -250 ) 4000x - 6 - x 10-
6 + 0 = O; 
v = Vm sin[40007rt + 30°] 
75 = Vmsin30°; Vm = 150V 
0 =~rad= 30° 
6 
v = 150 sin[40001rt + 30°) = 150 cos[4000xt - 60°) V 
P 9.4 [aJ By hypothesis 
i = lOcos(wt + 0) 
~; = -lOw sin( wt+ 0) 
. ·. lOw = 20,000x; 
w 
[b] f = 2x = 1000 Hz; 
w = 2000x rad/ s 
1 
T = f = 1 ms = 1000 µs 
150 3 
1000 = 20' () = -90 - 2-(360) = -144° 20 
: . i = 10 cos(2000xt - 144°) A 
P 9.5 [a] 170V 
[b) 2x f = 120x; f = 60Hz 
[c) w = 1207r = 376.99 rad/s 
-7l" -7l" 
[d] O(rad) = 180 (60) = T = -1.05 rad 
[e] 0 = -60° 
. 1 1 
[f] T = f = 60 = 16.67 ms 
7l" 
[g) 120xt - 3 = O; 
1 
:. t = 360 = 2.78ms 
[hJ v = 110 cos [ 1207l" (t + 0·::5 ) - i] 
= 170 cos[120xt + (157r /18) - (Jr /3)) 
= 170cos[120xt + (Jr /2)) 
= -170 sin 120xt V 
Problems 9~9 
9-10 CHAPTER 9. Sinusoidal Steady State Analysis 
[i] 1207r(t - t 0 ) - (7r/3) = 1207rt - (7r/2) 
7f 
1207rt = -· 
0 6' 
25 
t =-ms 
0 18 
[j] 1201l"(t - t 0 ) - (7r /3) = 1207rt 
25 
t 0 = 9ms 
f to+T p 9.6 u = to v,;, cos2 (wt + ¢) dt 
f to+T 1 1 = v,;, -2 + - cos(2wt + 2¢) dt 
• to 2 
= ;, { .JL0 +T dt + J::+T cos(2wt + 2¢) dt} 
= ~~ { T + 2~ [sin(2wt + 2¢) l!:+T)} 
= 17;} { T + 2~ [sin(2wt0 + 47r + 2¢) - sin(2wt0 + 2¢)]} 
= V~ (~) + :w (0) = Vr?i (~) 
P 9.7 Vm = J2Vr·ms = J2(120) = 169.71 V 
p 9.8 Vrms = 1 lT/2 27f - V~sin2 -T tdt T. o 
{T/2 (21!" ) V:2 {T/2 ( 47r ) V:2T Jo v;,sin2 rt dt = ;n Jo 1 - cos rt dt = 1~ 
f1VIT Vm 
Therefore Vrms = VT 4 = 2 
P 9.9 [a] The numerical values of the terms in Eq. 9.8 are 
Vm = 100, R/ L = 533.33, wL=30 
.j R2 + w2 £2 = 50 
¢ = 60°, () = tan-1 30/40, () = 36.87° 
i = [-1.84e-533·33t + 2cos(400t + 23.13°)) A, t ~ 0 
[b] Transient component = - l.84e-533·33t A 
Steady-state component = 2 cos(400t + 23.13°) A 
[cJ By direct substitution into Eq 9.9, i(l.875 ms)= 133.61 mA 
Problems 9-11 
[dJ 2 A, 400rad/s, 23.13° 
[e] The current lags the voltage by 36.87°. 
P 9.10 [a] From Eq. 9.9 we have 
Ldi = VmRcos(</>-8)e-(R/L)t_ wLVmsin(wt+<f>-8) 
dt y' R2 + w2 £2 y' R2 + w2 £2 
Ri = -VmR cos(</> - 8)e-(R/L)t + VmR cos( wt+</> - 8) 
y'R2 +w2£2 y'R2 +w2£2 
L di Ri = V. [R cos( wt+</> - B) - wL sin(wt + </> - 8)] 
dt + m y'R2 + w2£2 
But 
R 
y' = cos 8 and R2 +w2L2 
wL . 8 ----;:=:=== = Sln y'R2 +w2L2 
Therefore the right-hand side reduces to 
Vm cos(wt + </>) 
At t = 0, Eq. 9.9 reduces to 
'(O) -Vm cos(</> - 8) v;rt cos(</> - fJ) 
i = + =0 y'R2 -w2£2 JR2 + w2£2 
[b] i88 = y' Vm 2 2 cos( wt+</> - 8) R2 +w L 
Therefore 
L di88 -wLVm . ( ,1,. 8) - = sm wt + '+' -dt y' R2 + w2 £2 
and 
VmR ( ) Riss = J 2 2 cos wt+ </> - () R +w2L 
Ldiss R' =V. [Rcos(wt+</>-8)-wLsin(wt+</>-O)l 
dt + iss m y' R2 + w2 £2 
= Vm cos(wt + </>) 
P 9.11 [a] Y = 100/45° + 500/ - 60° = 483.86/ - 48.48° 
y = 483.86 cos(300t - 48.48°) 
[b) y = 250/30° - 150/50° = 120.51/4.8° 
y = 120.51cos(377t+4.8°) 
9-12 CHAPTER 9. Sinusoidal Steady State Analysis 
[cJ Y = 60L60° - i20/ - 215° + 100/90° = 152.88L32.94° 
y = 152.88 cos(lOOt + 32.94°) 
[d] Y = 100L40° + 100Lrno0 + 100L - 80° = o 
y=O 
P 9.12 [a] 50Hz 
[bJ Bv = 0° 
I= 340&_ = 340 L- goo= 8.5L - 900: 
jwL wL , 
[cJ 340 = 8.5; wL = 40CT 
wL 
[d] L = ~ = 400mH=127.32mH 
1007r 1r 
(e] ZL = jwL = j400 
P 9.13 [a] w = 21ff = 8011" x 103 = 251.33krad/s = 251,327.41rad/s 
[h] I= 2·5 ;1:~~&_ = jwC(2.5 x 10-3)L0° = 2.5 x 10-3wCL90° 
[c] 125.66 x 10-6 = 2.5 x 10-3 wC 
1 2.5 x 10-3 
wC - 125.66 x 10_6 = 19.89 CT, . . Xe = -19.89 CT 
1 1 
[d] C = 19.89(w) - (19.89)(80?r x 103 ) 
C = 0.2 x 10-6 = 0.2µF 
[e] Zc = j (~~) = -j19.89 CT 
P 9.14 [a] V 9 = 150/20°; 19 = 30L - 52° 
. z Vu / on . . =I= 5 12 ~{,, 
g 
[b] i 9 lags v9 by 72°: 
27r j = 80007r; f = 4000 Hz; T = 1// = 250µs 
i 9 lags v9 by ;6~ (250) = 50 µs 
P 9.15 [a] jwL = j(5 x 104)(40 x 10-6) = j2 0. 
1 106 
jwC = -j 5 x 104 = -j20!1; 19 = 20L-20° A 
10 
200 ~ -j200 
[bJ V0 = 20/-20° Ze 
Z 1 v 1 .1 1 
e = Ye ; .le = 20 + J 20 + 1 + j2 
Ye = 0.05 + j0.05 + 0.20 - j0.40 = 0.25 - j0.35 s 
Ze = 0.25 ~ j0.35 = 2.32L54.46° n 
Vo = (20L-20° )(2.32/54.46°) = 46.4/34.46° V 
[c] v0 = 46.4cos(5 x 104t + 34.46°) V 
P 9.16 (a] 
500L60° o 
[h] I= 400 + j700 - j400 = 1L23.13 A 
[c] i = 1cos(8000t+23.13°) A 
P 9.17 [a] Z1 = Ri - j 0
1 
w 1 
Problems 9-13 
j20 
z2 = R2/jwC2 _ R2 _ R2 - jwR§C2 
R2 + (1/jwC2) 1 + jwR2C2 1 + w2R~Ci 
9-14 CHAPTER 9. Sinusoidal Steady State Analysis 
500 
[b] Ri = 1 + (64 x 108)(25 x 104)(625 x 10-18) = 250 O 
2 
Ci = (64 x 108)(25 x 104)(25 x lQ-9) = 50 nF 
1 . 
P 9.18 [a] Y2 = R
2 
+ JwC2 
1 jwC1 Yi= -----
R1 + (1/jwC1) 1 + jwR1 C1 
Therefore Yi= Y2 when 
w2R1C'f_ + jwC1 
1 +w2RrC'f 
[b] R = 1 + (4 x 108)(4 x 106)(2500 x 10-18) = 2500 = 2 5kD 
2 (4 x 108)(2 x 103)(2500x 10-18) . 
C2 = 50 x 10-9 = lOnF 
5 
P 9.19 [a] Z1 = Ri + jwL1 
z2 = R2(jwL2) _ w2L~R2 + jwL2R~ 
R2 + jwL2 R~ + w2L~ 
[b] R = (4 x 108)(6.25)(5 x 104) = 2_5 x 104 
1 25 x 108 + (4 x 108)(6.25) 
:. R1 = 25k0 
Li = (25 x 108)2.5 = 1.25 H 
50 x 108 
1 j 
P 9.20 [a] Y2 = - - -
R2 wL2 
1 Yi=----
R1 + jwL1 
R1 - jwL1 
Rt +w2Lf 
Therefore 12 =Yi when 
and 
R2 +w2L2 
L l 1 
2 = 2£ w 1 
[b] R = 25 x 106 + 108(0.25) = 10 103 
2 5xl~ x 
:. R2 = lOkrl 
50 x 106 
L2 = 108(0.5) = 1 H 
p 9·21 [a] y = 4 ~ j3 + 16: j12 + -j~OO 
= 0.16 + j0.12 + 0.04- j0.03 + j0.01 
= 0.2 + jO.l = 223.6L26.57° mS 
(b] G = 200mS 
(c] B = lOOmS 
Problems 9-15 
[d] I= 50/0° A, v = _!_ = 50 = 223.61L-26.57° v Y o.223L26.57° 
I = ~ = 223.6L-26.570 = 2.24L63.430 A 
c Zc lOOL-90° 
ic = 2.24cos(wt + 63.43°) A, Im= 2.24A 
. (4000)(109 /jw625) 
P 9·22 [a] Zab=J5w+ 4000+(109/j625w) 
4 x 1012 
= j5w + 25x1()5jw+1Q9 
4 x 107 
= j5w + 104 + j25w 
4 x 1011 100 x 107w 
= i 5w + lQ8 + 625w2 - j 108 + 625w2 
109 
·. · 5 = 108 + 625w2 
5 x 108 + 3125w2 = 109 
w = 4 x 102 = 400rad/s 
[b] z (400) = ·2000+ (4000)(-j4000) = 2krl 
ab J 4000 - j4000 
P 9.23 Z1 =10 - j40rl 
z = (5 - jl0)(10 + j30) = 10 _ .100 
2 15 + j20 J 
9-16 CHAPTER 9. Sinusoidal Steady State Analysis 
z = 2o(j20) = rn ·10 n 
3 20 + j20 + J 
:. Zab = Z1 + Z2 + Z3 = 30 - j40!1=50L-53.13° n 
P 9.24 First find the admittance of the parallel branches 
Yp = 6 ~ j 2 + 4 +1j 12 + ~ + j~O = 0.375- j0.1258 
Z = I.. = 1 = 2 4 ·o 8 n 
p Yp 0.375 - j0.125 . + J . 
Zab = -j12.8 + 2.4 + j0.8 + 13.6 = 16 - j12 !1 
Yab = ~b = 16 ~ j12 = 0.04 + j0.03 s 
= 40 + j30mS = 50/36.87° mS 
. 1000 
P 9.25 z = 400 + j(5)(4o) - 1 (5)(o.4) = sooL - 36.87° n 
I = 750L!L. x 10- 3 = 1.5L36.87° mA 
0 5ooL - 36.87° 
i 0 (t) = 1.5 cos(5000t + 36.87°) mA 
P 9.26 Vg = 50/ - 45° V; lg= lOOL - 8.13° mA 
z = Vg = 500L- 36.87° n = 400 - j3oon 
lg 
Z = 400 + j ( 0.04w - 2·5 : 106 ) 
0.04w - 2.5 x 106 = -300 
w 
w2 + 7500w - 62.5 x 106 = 0 
. w = -3750 ± v (3750)2 + 62.5 x 106 = -3750 ± 8750 
w>O, :. w = 5000rad/s 
p 9.27 ZL = j(5000)(48 x 10-3 ) = j2400 
Zc = (5000)(2~g x 10-6) = -j800 
Construct the phasor domain equivalent circuit: 
+ 
Vo 2400 800 
0.21!_ 0 t - J; 
.___I _r80____,0 ~ j2400 
Using current division: 
(80 + j240) . 
I= 240- j80 + 80+ j240(0.2) = O.l +JO.IA 
V 0 = 2401 = 24 + j24 = 33.94L45° 
v0 = 33.94cos(5000t + 45°) V 
1 - 109 - ·4ooon P 9·28 jwC (31.25)(8000) - -J 
jwL = j8000(500)10-3 = j4000 n 
V 9 = 64L0°V 
-j40000 
r-=;~q 
64~0 vcb : 20000 j 40000 
z = (2ooo)(i4000) = 1600 ·800 n 
e 2000 + j4000 + J 
Zr = 1600 + j800 - j4000 = 1600 - j3200 0 
Problems 9-17 
9-18 CHAPTER 9. Sinusoidal Steady State Analysis 
64/0° . 
lg= l600-Ja200 =8+1l6mA 
V 0 = Zelg = (1600 + j800)(0.0()8 + j0.016) = j32 = 32[!30° V 
Vo= 32 cos(8000t + 90°) V 
P 9.29 [a] 
250 
r 1200 
v 
o;r Va 
i ;{.,c ~-j800 
j400 40dmA 
+ 
..J, ( 40+ j80 )mA 
Va= (120 + j40)(0.04LQ.'.J = 4.8 + jl.6 V 
I = 4·8 + jl.6 = 20 ·2omA 
b 160- j80 + J 
le= 40/0° + (20 + j20) + (40 + j80) mA = 100 + jlOOmA 
Vg = 25Ic +Va= 25(0.100 + j0.100) + 4.8 + jl.6 = 7.3 + j4.l v 
(b] ib = 28.28 cos(800t + 45°) mA 
ic = 141.42 cos(800t + 45°) mA 
vg = 8.37 cos(800t + 29.32°) V 
[ ] 1 - 109 - - "10 n p 9·30 a J-.w-C j8 x 105(125) - J H 
jwL = j8 x 105 (25 x 10-6 ) = j20 n 
z = ( - 110) (20) = 4 - .8 n 
e 20 - jlO J 
lg= 5[0° 
Vg = 19 Ze = 5(4- j8) = 20 - j40V 
40 -jBO 120 
6~ 
I~--
+ v 
j200 g Vo 
v = (20 - J40)(j2o) = 44 - J'8 = 44.72 1 - 10.30° v 
0 (16 + j12) L 
Vo= 44.72 cos(8 X 105t - 10.30°) V 
f=4xl05 
7r 
1 1f 
T = f = 4 x 10s = 2.57r µs 
10.30 
360 (2.57r) = 224.82ns 
v0 lags i 9 by 224.82 ns 
P 9.31 18 = 15[0° rnA 
1 - 106 ~soon 
jwC j0.05(8000) = -J 
jwL = j8000(1.25) = jl0,000 0 
After two source transformations we have 
-j2.5k0 
15k0 ~ 30k0 
jlOkO 
15 knlJ30 kO = 10 kO 
10-3 1 1 
Yo = w + -j2500 + j104 = 10-4(1 + j3) 
104 
Za = .3 = (1 - j3) kO l+J 
Vo= IgZo = (10)(1- j3) = 10- j30 = 31.62L- 71.57°V 
v0 = 31.62 cos(8000t - 71.57°) V 
Problems 9-19 
+ 
9-20 CHAPTER 9. Sinusoidal Steady State Analysis 
p 9.32 + 
Ict 
vd 
~ -j5 0 
~Ib j50 
60 /J}.°V + - + v 
+ I b z 
v v c z 
60 
Va= j21a = j2(-j5) = 10L0° V 
Ve 
le= 6- j8 
50L!t 5/r,::3 13° 3 ·4 A 10L - 53.13° = 0 · = + J 
+ 
j2 0 
Va 
lra 
+ lrc 
-j8 0 
lb= le -Ia= 3 + j4- (-j5) = 3 + j9A = 9.49L71.57° A 
Vb= lb(j5) = (3 + j9)(j5):::::: -45 + j15 V 
Vd + Vz = 60L0°; vd = 60- 5- J15 = 55 - j15 v 
lz =Id - lb= 3 + jll - 3 - j9 = j2 A 
Z = ~z = 5 +.;15 = 7.5- j2.50 
z J 
P 9.33 V 2 is the voltage across the -jlOO impedance. 
(40 + j30) - (100 - j50) 40 + j30 (40 + j30) - V2 _ O 
20 + j5 + z -
Problems 9-21 
:. V2 = 40 + j30 + (3 - j4)Z 
V 2 - (40 + j30) V2 _ (20 .30) V2 - (100 - j50) = O z + -jlO +J + 3+jl 
Substituting the expression for V2 found at the start and simplifying yields 
z = 12 + jl60 
P 9.34 Simplify the top triangle using series and parallel combinations: 
(1 + Jl)ll(l - Jl) = 1 n 
Convert the lower left delta to a wye: 
z = (jl)(l) = ·1 n 
1 1 + jl - jl J 
z = (-Jl)(l) = - ·1 n 
2 1 + jl - jl J 
z = (Jl)(-Jl) = 1 n 
3 1 + jl - jl 
Convert the lower right delta to a wye: 
z = (-Jl)(l) = - ·1n 
4 1 + jl - jl J 
Zs= (-jl)(jl) = 1 n 
1 + jl - jl 
z = (Jl)(l) = ·1 n 
6 1 + jl - jl J 
jlQ 
9-22 CHAPTER 9. Sinusoidal Steady State Analysis 
The resulting circuit is shown below: 
10 
• b 
Simplify the middle portion of the circuit by making series and parallel 
combinations: 
(1 + jl - jl)ll(l + 1) = 1112 = 2/3 o 
Zab = -jl + 2/3 + jl = 2/30 
P 9.35 (a] Yp = 10 : j 2w + j4 x 10-3w 
10 - j2w ·4 x 10-3w 
100 + 4w2 + J 
10 j2w . 3 
100+4w2 +J4 x 10- w ----100+4w2 
Yp is real when 
_ 3 2w 
4 x 10 w = 100+4w2 
or w2 = 100; w = 10 rad/s; f = 5/7r = l.59Hz 
10 
[h] Yp(lOrad/s) = 500 = 20mS 
103 
Zp(lOrad/s) = 20 = 500 
Z(lOrad/s) = 50 + 150 = 2000 
Vg 10L0° I 0 
Io= 200 A= 200 = 50&_.mA 
i 0 = 50 cos lOt mA 
Problems 9-23 
P 9.36 [a] z = 4000 - j 109 + 104(j2w) 
g 25w 104 + j2w 
= 4000 - . 109 2 x 1Q4jw(104 - j2w) 
.7 25w + 108 + 4w2 
109 4 x 104w2 2 x 108w 
= 4000 - j 25w + 1Q8 + 4w2 + j 108 + 4w2 
109 0.2 x 109w -----
25w 108 + 4w2 
108 + 4w2 = 5w2 
w2 = 108. 
l w = 10,000rad/s 
[b] When w = 10,000rad/s 
P 9.37 [a] 
4 x 104(104) 2 
Zg = 4000 + 108 + 4(l04)2 = 12,000 n 
. 45L0° 0 
.. lg= 12 OOO = 3.75LQ_mA 
' 
109 
Z1 = 4000 - j 2 04 = 4000 - j4000D 5xl . 
V 0 = 45L0° - (3.75 x 10-3)(4000 - j4000) = 45 - (15 - jl5) 
= 30 + jl5 = 33.54/26.57° v 
v0 = 33.54cos(10,000t + 26.57°) V 
1 3 Yi = 5000 = 0.2 x 10- s 
1 
Y2=----
1200+ j0.2w 
1200 . 0.2w 
= 1.44 x 106 + 0.04w2 - 1 1.44 x 106 + 0.04w2 
Y3 = jw50 x 10-9 
Yr = Yi + Y2 + Y3 
For ig and v0 to be in phase the j component of Yr must be zero; thus, 
50 ic-9 0.2w 
w x ) = 1.44 x 106 + 0.04w2 
9-24 CHAPTER 9. Sinusoidal Steady State Analysis 
or 
0.2 x 109 
0.04w2 + 1.44 x 106 = = 4 x 106 
50 
:. 0.04w2 = 2.56 x 106 :. w = 8000rad/s = 8ktad/s 
[b] y; - 0 2 x 10-3 1200 - 0 1 - 3 s 
T - • + 1.44 X 106 + 0.04(64) X 106 - ·5 X O 
:. Zr= 20000 
Vo= (2.5 X 10-3 L0°)(2000) = 5L0° 
V 0 = 5 COS 8000t V 
R 
p 9·38 [a] Zp = R + {~fywC) - 1 +;RC 
12,500 12,500 
-
1 + j(1000)(12,500)C 1 + jl2.5 x 106C 
12,500(1 - jl2.5 x 106C) 
-
1 + 156.25 x 1012C2 
12,500 . 156.25 x 109C 
= 1 + 156.25 x 1012c2 - 1 1 + 156.25 x 1012c 2 
jwL = j1000(5) = j5000 
156.25 x 109C 
5000 = 1 + 156.25 x 1012c 2 
781.25 x 1015C2 - 156.25 x 109C + 5000 = o 
c 2 - 20 x 10-sc + 64 x 10-16 = o 
C1,2 = 10 x 10-8 ± vlOO x 10-16 - 64 x 10-16 
C1 =10 x 10-8 + 6 x 10-8 = 16 x 10-8 = 0.16µF 
C2 = 10 x 10-8 - 6 x 10-s = 4 x 10-8 = 0.04µF 
12,500 
[b] Re = 1 + 156.25 x 1012C2 
When C = 160nF Re= 25000; 
I = 250L!r_ = 0.1 10° A· i9 = 100 cos lOOOt rnA g 2500 L'_ ' 
When C = 40 nF Re = 10,000 O; 
I = 25oi!r_ = 0.025[0° A· i9 = 25 cos lOOOt rnA 
g 10 000 - ' 
' 
Problems 9-25 
109 
P 9.39 [a] Z1 = 1600 - j 104(62_5) = 1600 - j1600 n 
z _ 4000(j104L) _ 4 x 105£ 2 + j16 x 104£ 
1 - 4000 + jl04L- 16 + 100£2 
4 x 105£ 2 16 x 104£ 
Zr = Z1 + Z2 = 1600 + 16 + 100L2 - jl600 + j 16 + 100L2 
Zr is resistive when 
16 x 104£ 
16 + 100£2 = 1600 
£ 2 - L + 0.16 = 0 
or 
Solving, L 1 = 0.8 H and L 2 = 0.2 H. 
[h] When L = 0.8 H: 
z = 1600 4 x 105(0·54) = 4800 n 
r + 16 + 64 
lg= 96Loo x 10-3 = 20 ioo mA 
4.8 [_ 
ig = 20 cos 10,000t mA 
When L = 0.2 H: 
z = 1600 + 4 x 105(o.o4) = 2400 n 
r 16+4 
ig = 40cos 10,000tmA 
P 9.40 Step 1 to Step 2: 
75/0° = -J·4.167 = 4.167 / - 90° A 
j18 
Step 2 to Step 3: 
(j18)ll24 = (jl8)(24) = 8.64 + Jll.52 n 
24 + j18 
Step 3 to Step 4: 
( 4.167 / - 90°)(8.64 + jll.52) = 60/ - 36.87° v 
9~26 CHAPTER 9. Sinusoidal Steady State Analysis 
Step 1 -j220 j 400 
7 sl!!'_v [ '---~2-4_0_ a 
Step 3 
.----.----a 
8 .640 
P 9.41 Step 1 to Step 2: 
(l6L0°)(25) = 400L0° V 
Step 2 to Step 3: 
25 + 15 + J30 = (40 + J30) n 
Step 3 to Step 4: 
400LQ:'.. = 8/ - 36.87° A 
(40 + j30) 
Step 4 to Step 5: 
Step 2 
Step 4 8 .640 
(- "50)(40 + '30) 
(40+j30ll-J50= J . ~ =50-j25f2 
40 + 330 - ]50 
step 1 
~--+--" y y ., '-'t------. a 
j300 
250 -j500 
150 
Step 3 Step 4 
a 
-j220 
240 
j400 
b 
j 11.520 
... a 
~-__..,... __ _,_ __ .... a 
400Lb_0 v j300 
-j5o0 
400 
step s ~-_,_ ___ a 
8/-36. 87°A 500 
~ -j 250 
L------------b 
Problems 9-27 
P 9.42 [a] jwL = j(5000)(50) x 10-3 = j2500 
1 1 
jwC = -j (5000)(400 x 10-9 ) = -j5000 
j2500 
2500 
22.36L'i_6.565° :i: 
v 
... -j5000 
j2500 
Using voltage division, 
(250+j25o)11 ( -j5oo) . o o 
Vab = j250 + (250 + j250)Jl(-j500) (23.36/26.565 ) = 20&_ 
VTh = Vab = 20/0°V 
[b] Remove the voltage source and combine impedances in parallel to find 
ZTh = Zab: 
1 1 1 . 
Yab = j250 + 250 + j250 + -j500 = 2 - J4 mS 
ZTh = Zab = ,; = 100 + j200f2 
Lab 
[c] 
1000 j2000 
.-----"w.~~--"rvYY'~~---•a 
20/s/v l 
l.______ __ •b 
9-28 CHAPTER 9. Sinusoidal Steady State Analysis 
p 9.43 
j120 
120~ 
a 
~0v 
120 ~ 
~ 
:Cb 
(27 + j12)Ia - 3Ib = -87 L0° 
-31a + (27 - j12)1b = 87 L0° 
Solving, 
a 
120 
120 
-jl20 
Ia= -2.4167 + jl.21; lb = 2.4167 + jl.21 
VTh = 12Ia + (12 - j12)1b = 14.5/0° V 
Short Circuit Test: 
j120 
120~ 
. a 
120 
~ 
SC 
(27 + j12)1a - 3Jb - 12lsc = -87 
-31a + (27 - j12)1b - (12 - j12)1sc = 87 
-121a - (12 - j12)Jb + (24 - j12)Isc = 0 
Solving, 
Problems 9-29 
z = VTh = 14.5&_ = 14.50 
Th lsc 1L0° 
Alternate calculation for ZTh: 
~-------aa 
j120 
120 120 
120 
~ -j120 
2:: z = 12 + 3 + 12 - j12 = 21 - ji2 
36 12 
Zi = 27- j12 - 9-j4 
36 - j36 12 - jl2 
z2 = 27 - j12 - 9-j4 
23 = 12(12 - j12) = 48 - j48 
27-j12 9-j4 
'---------..aa 
j120 
L--------b 
12 12(14 + j5) 
Za = 12 + j12 + g _ j 4 - g _ j 4 
12 - j12 12(10 - j5) 
zb = 12 + g - j4 - g - j4 
9-30 CHAPTER 9. Sinusoidal Steady State Analysis 
p 9.44 
z II z = 165 - j2o 
a b 18 - j8 
z z llZ = 48 - j48 165 - j20 = 14.50 
3 + a b 9 - j4 + 18 - j8 
:IN f 
:IN f 
+ 
100~20° 
mV 
+ 
1~0 mA 
jlDDO 
~ -j 1000 
IN = 0-3~ 600 + ( -3/210°) rnA, ZN in kn 
o.1~ + lL3oo = o.3~~ 500 + (-3/2100) 
o.3/ - 60° - o.1L120° = 1L300 + 312100 
ZN - --
z = o.3L - 500 - o.1L1200 = o.2L900 = ·o.2kn 
N lL30° + 3/210° - J 
I = O.l~ + 1/30° = 1.5/30° mA 
N 0.2L90° - -
.--------<r-----e a 
i.sb_o t j2000 
mA 
L-..._ ___ .__ ____ b 
Problems 9-31 
P 9.45 JWL = jl.6 X 106 (25 x 10-6) = j40 [2 
1 10-6 x 109 
jwC- jl.6(25) =-J25 n 
j40 0 
+ >r 
T 
~ -j 25 0 25 0 
VT = j40Ir + 151.o. + 251.o. 
Ir(-j25) -jlr 
I~= 25 - j25 = 1 - jl 
V = ·401 +40(-jlr) 
T J T 1 - jl 
~; = Zab = j40 + 20(-j)(l + j) = 20 + j20 !1 = 28.28/45° n 
p 9.46 _1_ = (10-3)(109) = 12k!1 
wC1 25(10/3) 
_1_ = (10-3)(109) = 24k!1 
wC2 25(5/3) 
-7'I.T lk0 19I:T 
----"---11~---'lll/lr---1'---C ~ 
+ -j 12k0 
I o.12sk!.l 
2D'I.;Y 1 
VT= (1 - j12)1r + 20Ir(0.125) 
ZTh = ~; = 3.5-j12k!1 
-j24k0 
147k0 
I 
9-32 CHAPTER 9. Sinusoidal Steady State Analysis 
P 9.47 Short circuit current 
p 9.48 
113 = 2113 
-j2 
20~t -j20 
SC 
The Norton impedance is the same as the Thevenin impedance. Find it using 
a test source 
If-
T 
+ 
v 
T 
z, =Yr= (-2-j2)113 = -2-j2 =0.4- T 2 0 
1·h Ir [(1 - j2)/1]1,a 1 - j2 J 
6000 j1500 -j1500 
ii I 
+ ;o 400 
V 1 - 75 0.02V1(40) Y1 _ O 
150(4 + jl) - 40- j150 + 40- j150 -
p 9.49 
v - 75(4 - jl5) 
1 - 16 - j12 
V _ 40V1 4 
Th - 40 - jl50 - 4 - j15 
75(4 - j15) 
16 - j12 
75 /' 6 8 0 v 4 - j3 = 15 ,{ . 7 
I = 75 =~A 
SC 600 8 
Z VTh I 0 Th = -1- = 120L36.87 = 96 + j72 Q 
SC 
960 j720 
~a 
Jisb6.87°v 
~I ~~~~~~~-·b 
-j2500 
Solving, 
V2 = -1- j0.75V = l.25/216.87°V 
_ . _ - 25 /0° _ h 0 
lsc - -14> - - -2o/O mA 
1000 -
i.25/216.87° I o . n 
ZTh = -25 x lQ-3/0o = 50 36.87 n = 40 + J30H 
Problems 9-33 
9-34 CHAPTER 9. Sinusoidal Steady State Analysis 
ZN= ZTh = 50/36.87° = 40 + j300 
j300 
,__ __ ...___ __ .,. b 
P 9.50 [a] jlOQ 
1 Vr Vr - aVr/10 
T = 10 + jlQ 
~ = ]:_ + (1 - a/10) = (10 - a)+ jlO 
VT 10 jlO jlOO 
. ZTh = Vr = 1000 + jlOO(lO - a) 
Ir (10 - a)2 + 100 
ZTh is real when a= 10. 
[h.l 1000 n ZTh = 100 = 10~~ 
[ c] ZTh = 5 + j 5 
1000 
(10 - a)2 + 100 = 5; (10 - a)2 = 100 
:. 10- a= ±10; a= 10 =F 10 
a=O; a= 20 
But the j term can only equal the real term with a = 0. Thus, a = 0. 
{d] ZTh will be inductive when a< 10. 
p 9.51 j400 
+ 600 + 
Solving for V 1 yields 
V 1 =30-j40V 
v - v 1 ( ·20) - ( j ) v 
0 
- 60 + j20 J - 3 + j 1 
V 0 = 15 + j5 V = 15.81/18.43° V 
P 9.52 jwL = j(5000)(0.4 x 10-3) = j2 n 
1 106 
jwC = -j (5000)(50) = -j40 
V 91=10L53.13° = 6 + j8V 
Vg2 = 8L-90° = -j8V 
j20 
Problems 9-35 
-j4 0 
( 6+j 8) V ~..---_.ry'lrr..... __ _._6_0 _ _;:;ib -j 8V 
V 0 - 6 - j8 Vo Vo+ (-j8) _ O 
j2 + 6 + -j4 -
Solving, 
v0 (t) = 12cos5000tV 
9-36 CHAPTER 9. Sinusoidal Steady State Analysis 
P 9.53 jwL = j104 (1.2 x 10-3) = j12 n 
- 1- = -jl06 = - ·2on 
jwC 5 x 104 J 
Va= 100[- 90° = -jlOOV 
Vb= 500[0° = 500V 
-j200 
j120 200 
-j 10 QI} : : soov 
~a Ji 
Solving, 
v 1 = 160[53.13° v = 96 + j128 v 
-jlOO - 96 - j128 -jlOO - 500 
Ia = j12 + -j20 
= -14- j17 = 22.02[- 129.47° A 
ia = 22.02 cos(lO,OOOt - 129.47°) A 
500 - 96 - j128 500 + jlOO 
lb= 20 + -j20 
= 15.2 + jl8.6 = 24.02[50.74° A 
ib = 24.02cos(10,000t + 50.74°) A 
Problems 9-37 
p 9.54 
j30 
V1 -j30 Vo 50 
+ Vg -j5V 
5/0°A t j2 0 
.. -j5V -
(5 + j6)Vo + lOV1 = 30 
_ 5 V1 - Vo V1 + j5 _ O + ·3 + ·3 --J J 
Vg = V 1 - V 0 = 9 - j5 - jlO = 9 - j15 = 17.49/ - 59.04° V 
p 9.55 
j400 
+ 
V0 250 ~ -j500 
Vo Vo 321 = 0 
25 + -j50 +. 0 
(2 + j)V0 = -160010 
V 0 = (-640 + j320)10 
I _ V1 - (Vo/4) 
0 - j40 
:. V 1 = (-160 + j120)I0 
9-38 CHAPTER 9. Sinusoidal Steady State Analysis 
v 
17 = 2~ +lo = ( -8 + j6)I0 + I0 = ( - 7 + j6)I0 
:. Ia= (-/: j 6) = -1.4- jl.2 A= 1.84/ - 139.40°
 A 
V 0 = (-640 + j320)10 = 1280 + j320 = 1319.39[14.04° V 
Solving, 
Vo = 72 + j96 = 120[53.13° V 
P 9.57 Va = 60L0° V; vb= 90L90°V 
jwL = j(4 x 104)(125 x 10-6 ) = j50 
. ·106 
-y = -J = -j200 
wC 40,000(1.25) 
200 -j200 
jSO~ 1j90V 
60 = (20 + j5)1a - j5lb 
j90 = -j5la - j15Ib 
Solving, 
Ia = 2.25 - j2.25 A; lb= -6.75 + j0.75A 
Io= Ia - lb= 9 - j3 = 9.49L - 18.43° A 
i0 (t) = 9.49cos(40,000t-18.43°)A 
Problems 9-39 
P 9.58 From the solution to Problem 9.52 the phasor-domain circuit is 
10[53.13° = (6 + j2)11 - 612 
8[-90° = -611 + (6 - j4)12 
Solving, 
v0 (t) = 12 COS 5Q00t V 
p 9.59 j30 
---------" '------~ 
-j30 ~ 5(] 
+-----i 
• -j5V 
j31a + 5(1a - lb) - j3(1a - 5) = 0 
j2(Ib - 5) + 5(Ib - Ia) - j5 = 0 
Solving, 
Ia= -j3; 19 = -j3 = 3/ - 90° A 
9-40 CHAPTER 9. Sinusoidal Steady State Analysis 
0 
p 9.60 2 0 A 
.---~~~~··~!~~~~ 
IOOL0° = (5 + j5)11 - 512 - j513 
50/0° = -511 + (5 - j5)12 + j513 
Solving, 
11 = 58 - j20A; 12 = 58 + jlOA; 13 = 28 + jOA 
Ia = 13 + 2 = 30 + jO A 
lb = 11 - 13 = 58 - j20 - 28 = 30 - j20 A 
le= 12 - l3 = 58 + jlO- 28 = 30 + jlOA 
Id= 11 - 12 = 58 - j20- 58 - jlO = -j30A 
P 9.61 jwL :::::: j5000(14 x 10-3) = j70 0 
1 
- -J =-·4000 
jv.1C (5000)(0.5 x 10-6 ) J 
~ 
j70Q -j400Q500 
lqi 
12/siv : r1a rib 1600 
- 590Iqi 
72l0° = (50 + j70)1a - 501b + 590(-lb) 
+ 
Vo 
p 9.62 
0 = -50Ia - 590(-Ib) + (210- j400)Ib 
Solving, 
lb= (50 - j50) mA 
Vo= 160Ib = 8 - j8 = ll.31L - 45° 
Vo = 11.31 cos(5000t - 45°) V 
106 
Zo = 600-j (5000)(o.25) = 600- j80on 
Problems 9-41 
zT = 300 + ]2000 + 600- jBOo = 900 + j12oon = 1500L53.13° n 
V = V Z 0 = (75L0°)(1000L - 53.13°) = 50L _ 106.260 V 
0 9 Zr 1500/53.13° -
Vo= 50cos(500ot - 106.26°) V 
1 106 
P 9.63 - = -j- = -jlOOO 
jwC 104 
jwL = j(500)(1) = j500 n 
Let Z1 = 50 - jlOO D; 
19 = 125L0° mA 
Z2 = 250 + j5000 
125~(50 - jlOO) 
(300 + j400) 
= -12.5 - j25 mA = 27.95L - 116.57° mA 
i 0 = 27.95cos(500t -116.57°)mA 
P 9.64 V 9 = 1.2L0° V; 
1 106 
-=-=-jlOkO 
jwC jlOO 
Let Va= voltage across 1 µF capacitor, positive at upper terminal 
Then: 
Va-1.2~ Va Va_O· 
10 + -jlO + 10 - ' 
0 - Va 0 - Vo _ o· 
10 + 200 - ' 
:. Va= (0.48 - j0.24) V 
.". V 0 = -9.6 + j4.8 = 10.73L153.43° V 
Vo = 10. 73 cos(lOOt + 153.43°) V 
9-42 CHAPTER 9. Sinusoidal Stea.dy Sta.te Analysis 
P 9.65 (a] 2DDkQ 
lOkQ lOkQ 
Va -1.2L0° ·w V Va _ 
10 000 + J Co a + 10 000 - O 
' ' 
v - 1.2 
a - 2 + jl04wC0 
V 0 = -20Va (see solution to Prob. 9.73) 
-24 24~ 
Vo= 2 + "106C - 2 + J'106Co .J 0 
. ·. denominator angle = 60° 
tan60° = J3 
106Co = J3 
2 
2v'3 
or C0 = 106 = 2J3 µF = 3.46 µF 
[b] v = 24L!..filr = 6/120° v 
0 2+j2v'3 -
Vo = 6 cos(lOOt + 120°) V 
P 9.66 (a] Vg = 2L0° V 
VP= 18~0 Vg = l.6LOO; 
1.6 1.6 - Vo _ O 
160 + Zp -
Z _ (200)(1/jwC) 
P - 200 + (1/jwC) 
j~C - jlO~~~.l) = -j105 = -j100kf2 
-lOV 
47kQ 
p 9.67 
z = 200(-jlOO) = 40 - ·8 kn 
p 200 - j 100 J 0 
Vo= 1.6 + ~O = 2 - j0.8 = 2.15L - 21.80° 
v0 = 2.15cos(105t - 21.80°) V 
[b] VP= 0.8VmL0°; 
0.8Vm 0.8Vm - V 0 O 
16() + 40 - j80 = 
Problems 9-43 
40- j80 . Vo= 0.8Vm + 160 Vm(0.8) = 0.8Vm(l.25 - J0.5) 
. · j0.8Vm(l.25 - j0.5)j ~ 5 
Vm ~ 4.64V 
1 1012 ·10kn 
jwC1 
- jl06(100) = -J 
1 1012 
jwC2 
- j(106)(so) = -j20kn 
:. (-2 + j7)Va - jV0 = j40 
0 - Va 0 - V 0 _ O· 
10 + -jlO - ' 
:. (7 + j)V0 = j40 
40kQ 
-jlOkQ 
6V 
+ 
9-44 CHAPTER 9. Sinusoidal Steady State Analysis 
Vo = /!0j = 0.8 + j5.6 = 5.657 /81.87° V 
Vo(t) = 5.657 cos(106t + 81.87°) V 
1 -j109 . 
p 9·68 (a] jwC - (2 x 105)(12.5) = -J400!1 
V 0 Vn Vn 
-j400 = 200 + -j400 
V 0 = Vn - j2Vn = (1- j2)Vn 
V _ V 9 (1/jwC0 ) _ V 9 
P - 500 + (1/jwC0 ) - 1 + j(500)(2 X 105)C0 
V 9 = 10L0°V 
10/0° 
Vp= l+jWsCo =Vn 
. v = (1 - j2)10L!r_ 
• • 0 1 + j108Co 
!Vol= J5(10) = 10 
j1+1016c; 
Solving, 
C0 = 20nF 
[b] v = lO(l - j 2) = 10L- 126.87° 
0 1 + j2 
Vo = 10 cos(2 X 105t - 126.87°) V 
P 9.69 (a] 
OA 
+-------.---------~ !-----------...... + 
vab v 
0 
Problems 9--45 
Because the op-amps are ideal Iin = I 0 , thus 
Z _ Vab _ Vab. I _ V ab - V 0 
ab - Iin - L' o - Z 
Vo2 =-(~:)Vol= -KV01 = -KVab 
Vo= Vo2 = -KVab 
Io= Vab - (-KVab) = (1 +K)Vab 
z z 
Vab Z 
Zab = (1 + K)Vab z = (1 + K) 
1 
[b] Z = jwC; 
1 
Zab = jwC(l + K); 
P 9. 70 [a] Superposition must be used because the frequencies of the two sources are 
different. 
(bJ For w = 80,000 rad/s: 
200 -jlOU 
V' - 5 V' V' 
0
20 + jlo + jlo = 0 
V' (_!_ _1_ __1_) _ ~ 
0 20 + jlO + -jlO - 20 
:. v~ = 5[0°V 
V' 
I~= jlO = -j0.5 = 500[- 90° mA 
For w = 320,000 rad/s: 
200 
+ p:"·-j2.~ 
v" j 400 62.sfd'v 
20llj40 = 16 + j8 n 
9-46 CHAPTER 9. Sinusoidal Steady State Analysis 
16+ '8 
V" = J (2.5L0°) = 2.643L7.59° V 16 + j8 - j2.5 - --
V " ·1" L 0 A • • 0 = j 40 = 66.08 - 82.4 m 
Thus, 
i 0 (t) = (500sin80,000t + 66.08cos(320,000t - 82.4°)] mA, t 2:: 0 
P 9.71 [a} Superposition must be used because the frequencies of the two sources are 
different. 
[b) For w = 2000 rad/ s: 
20/-36 I 8 7°V ~ 
10ll-j5=2-j40 
For w = 5000 rad/s: 
j50 
so 
+ 
j20 -j 50 
'----..----1~·-~ 
+ 
100 Va1 
V 1 = 2 - j 4 (20 I - 36.87° = 31.62 I - 55.3° V 
o 2 - j4 + j2 L L 
-j20 
vo2 100 
1-i1oh6. 2f{'v 
J5IJ10 = 2 + j4n 
v = 2 + J4 (10 116.26°) = 15.81 134.69° v o2 2 + j 4 _ j 2 L L 
Thus, 
v0 (t) = [31.62 cos(2000t - 55.3°) + 15.81cos(5000t+34.69°)] V, t 2:: 0 
P 9.72 [a] jwL1 = jwL2 = j(l0,000)(1x10-3 ) = jlOO 
jwM = j(l0,000)(0.5 x 10-3 ) = j5 n 
200 = (5 + jlO)lg + j51L 
0 = j51g + (15 + jlO)IL 
Solving, 
lg= 10- j15A; 
i 9 = 18.03 cos(lO,OOOt - 56.31°) A 
iL = 5 cos(lO,OOOt - 180°) A 
[h] k = M = 0.5 = 0.5 
JL1L2 JI 
(c] When t = 5011" µs, 
jlO~ 15Q 
~ I)_ 
• 
10,000t = (10,000)(5011") x 10-6 = 0.511" = 7r/2rad = 90° 
i 9 (501l" µs) = 18.03 cos(90 - 56.31°) = 15 A 
i£(501l"J1.s) = 5cos(90+ 180°) = OA 
Problems 9-47 
w = ~L1iI + ~L2i~ + Mi1i2 = ~(1 x 10-3)(15)2 + 0 + 0 = 112.5 mJ 
When t = 10011" µs, 
10,000t = 1l" rad = 180° 
i9 (1001l"µs) = -lOA 
iL(l001l"µs) = 5 A 
w = ~(1x10-3)(10) 2 + ~(1 x 10-3)(5)2 + 0.5 x 10-3(-10)(5) = 37.5mJ 
P 9.73 [a] jwL1 = j(50)(5) = j250!1 
jwL2 = j(50)(20) = jlOOOD 
1 109 
jwC - j(50 x 103)(40) = -jSOOD 
9-48 CHAPTER 9. Sinusoidal Steady State Analysis 
· Z22 = 75 + 300 + jlOOO- j500 = 375 + j5000 
z;2 = 375 ~ j5oon 
M = kVL1L2 = lOk x 10-3 
wM = (50)(10k) = 500k 
z. = [ 56~:] 2 (375 - j500) = k2(240- j320) !1 
Zin= 120 + j250 + 240k2 - j320k2 
IZinl = [(120 + 240k2) 2 + (250 - 320k2)2}~ 
d!Zinl = ~[(120 + 240k2) 2 + (250 - 320k2)2J-! x 
dk 2 
[2(120 + 240k2)480k + 2(250 - 320k2)(-640k)] 
djZinl O h 
~=wen 
960k(120 + 240k2) - 1280k(250 - 320k2 ) = 0 
: . k2 = 0.32; : . k = JD.32 = 0.5657 
[b] Zin (min) = 120 + 240(0.32) + j(250 - 0.32(320)] 
= 196.8 + j147.6 = 246L36.87° n 
369L0° 
11 (max) = 246L36.870 = 1.5/ - 36.87° A 
: . i1 (peak) = 1.5 A 
Note~ You can test that the k value obtained from setting dlZinl/dt = 0 
leads to a minimum by noting 0 < k :$ 1. If k = 1, 
Zin= 360- j70 = 366.74L- 11° n 
Thus, 
!Zinlk=l > IZinlk=v1f.32 
If k = 0, 
Zin= 120 + j250 = 277.31[64.36° n 
Thus, 
p 9.74 ZTh = 30 + j200 + (50/25)2(15 - j20) = 90 + j1200 
225L0° . o 
vTh = 15 + fio (150) = 450L36.87 v 
j120Q 
P 9.75 jwL1 = j(25 x 103)(3.2 x 10-3) = j80 n 
jwL2 = j(25 x 103)(12.8 x 10-3 ) = j320 fl 
1 109 
jwC - j(25 x 103)(250) = -Jl50n 
jwM = j(25 x l03)kj (3.2)(12.8) x 10-3 = jl60k fl 
Z22 = 40 + j320 - jl60 = 40 + jl60 !1 
z;2 = 40- j160!1 
Z,. = [140 ~~~601 ]' (40- jl60) = 37.647k2 - j150.588k2 
Problems 9--49 
Zab = 10 + j80 + 37.647k2 - jl50.588k2 = (10 + 37.647k2) + j(80 - 150.588k2) 
Zab is resistive when 
80 - 150.588k2 = 0 or k2 = 0.53125 
:. Zab = 10 + (37.647)(0.53125) = 30!1 
P 9.76 [a] jwL2 = j(500)103(500)10-6 = j250 fl 
1 109 
jwC = j(500 x 103)(20) = -jlOOO 
Z22 = 150 + 50 + j250 - jlOO = 200 + j150 fl 
z;2 = 200 - JI5on 
wM = (500 x 103)(100 x 10-6) = 50!1 
Zr= ( 25~0) 2 [200 - j150] = 8 - j6 fl 
9-50 CHAPTER 9. Sinusoidal Steady State Analysis 
p 9.77 
(b] Zab = Ri + jwL1 + 8 - j6 
jwL1 = j(500 x la3)(80 x 10-6 ) = j40!1 
Zab = 20 + j340 
--· 
1 ' 
25' 
b-~ 
ideal 
z = 2500 V 3/25 = 2500V3 
ab 2512 625 l3 
ideal 
= 4ZL = 4(200 + jl50) = (800 + j600) !1 
P 9.78 In Eq. 9.69 replace w2 M 2 with k2w2 L 1L2 and then write Xab as 
X L k2w2 L1L2(wL2 + wLL) 
ab= w 1 - R~2 + (wL2 +wLL)2 
= wLi { l _ k2wL2(wL2 + wLL) } 
R~2 + (wL2 + wLL)2 
For Xab to be negative requires 
or 
Problems 9-51 
which reduces to 
But k s;; 1 hence it is impossible to satisfy the inequality. Therefore Xab can 
never be negative if XL is an inductive reactance. 
P 9.79 [a] 
Vi Ni 
V2 = N2' 
Vi+ V2 = ZLl1 = (~~ + 1) V2 
Zab = liZL 
(Ni/N2 + 1)(1 + Ni/N2)l1 
ZL 
:. Zab = [l + (Ni/N2)]2 Q.E.D. 
(b] Assume dot on the N2 coil is moved to the lower terminal. Then 
Ni Ni V1 = --V2 and 12 = --11 
N2 N2 
As before 
V2 
Zab = 11 + 12 and V1 + V2 = ZLl1 
. V2 ZLI1 
. . Zab = (1 - Ni/N2)I1 - [1 - (Ni/N2)J2I1 
ZL 
Zab = [l _ (Ni/ N2 )]2 Q.E.D. 
9-52 CHAPTER 9. Sinusoidal Steady State Analysis 
P 9.80 [a] 
( N1 )
2 
Zab = 1 + N
2 
ZL Q.E.D. 
[b) Assume dot on N 2 is moved to the lower terminal, then 
V1 -V2 -Ni --- V1=--V2 
Ni N2 ' N2 
As in part [a] 
-Ni 
12 = N
2 
11 
and Zab -- V1 + V2 V2 = (12 + l1)ZL 
11Zab = (1- Ni/N2)V2 = (1- Ni/N2)(l - NifN2)ZLii 
11 11 
Zab = [1- (Ni/N2)J2 ZL Q.E.D. 
p 9.81 
Problems 9-53 
240 240 
[a] I = 24 + j 32 = (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 
240 
le= j7.5 A, Ze = j 7.5 = -j32 n 
Vs= 240 + (0.1 + j0.8)10 = 241 + j8 = 241.13{1.90° V 
[c] Let le denote the magnitude of the current in the capacitor branch. Then 
I= (10 - j7.5 + jlc) = 10 + j(le - 7.5) A 
Vs = 240{a = 240 + (0.1 + j0.8)[10 + j(le - 7.5)] 
= (247 - 0.81c) + j(7.25 + O.lJc) 
It follows that 
240coso: = (247 - 0.8lc) and 240sino: = (7.25 + O.lic) 
Now square each term and then add to generate the quadratic equation 
1; - 605. 77 le + 5325.48 = O; 
Therefore 
le = 302.88 ± 293.96 
le= 8.92A (smallest value) and Ze = 240/j8.92 = -j26.900. 
P 9.82 The phasor domain equivalent circuit is 
+ 
~~ + :!m. 
2 Rl -jXC 
vmLQ_o + Vo + 
+ v 
R2 J:R R .....Ill 2 x 
I= Vm 
Rx-iXc 
As Rx V¥ies from 0 to oo, the amplitude of v0 remains constant and its phase 
angle increases from G° to -180°, as shown in the following phasor diagram: 
9~54 CHAPTER 9. Sinusoidal Steady State Analysis 
P 9.83 [a] 
[b) 
+ + 
/ 
I 
:r 
,,.,,.-~--IR 
+ 
440LQ.0 v 220 
I = 440 440 = 20 - ·20 A 
t 22 + j22 J 
j22Q 
Vt= (0.2 + jl.6)(20 - j20) = 36 + j28 = 45.61L37.87° V(rms) 
Vs= 440L0° +Vt= 476 + j28 = 476.82L3.37° V 
440 440 440 . [c] lt=-+-+--=20+JOA 22 j22 -j22 
Vt= (0.2 + jl.6)(20 + jO) = 4 + j32 = 32.25L82.87° 
Vs= 440 +Vt= 444 + j32 = 445.15L4.12° 
p 9.84 
Problems 9-55 
- 120 240 _ . _ , . r;O [a] 11 - 24 + 8.4 + j 6.3 - 23.29 - J13.71 - 27.02L-30.o A 
I = 120 _ 120 = 5 100 A 
2 12 24 L__ 
120 240 . 0 
l3 = 12 + 8.4 + J6_3 = 28.29 - 113.71 = 31.44L-25.87 A 
I = 120 = 5 ioo A· 
4 24 0:!_ ' Is= 
120 = 10L0° A 12 -
I 240 . L 0 A 6 = 8.4 + j6.3 = 18.29 - J 13. 71 = 22.86 -36.87 
[b] When fuse A is interrupted, 
11 = 0 
12 = 10 + 5 = 15 A 
l3 = 15A 
l4 = -5A 
l5 = lOA 
16 = 5A 
[c] The clock and television set were fed from the uninterrupted side of the 
circuit, that is, the 12 n load includes the clock and the TV set. 
[d] No, the motor current drops to 5 A, well below its normal running value of 
22.86A. 
(e] After fuse A opens, the current in fuse B is only 15 A. 
P 9.85 [aJ The circuit is redrawn, with mesh currents identified: 
~ 
lQ 
• J + 12ofo_0 v 
13 .2fil0 kV ~ ~- - ~--a-
~ ~) Q :Ib IL 
The mesh current equations are: 
120 L0° = 231a - 21b - 201c 
120 L0° = -21a + 431b - 40Ic 
Solving, 
2DQ:J 
:IC 
40Q 
Ia= 24/0° A 4 = 21.96L0° A le= l9.40L0° A 
lOQ 
9-56 CHAPTER 9. Sinusoidal Steady State Analysis 
p 9.86 
The branch currents are: 
11 = la = 24L0° A 
12 =Ia - lb= 2.04[0° A 
l3 = lb = 2l.96L0° A 
l4 = le = 19.40[0° A 
l5 = Ia - le = 4. 6 [0° A 
16 =lb - le= 2.55[0° A 
[b] Let N1 be the number of turns on the primary winding; because the 
secondary winding is center-tapped, let 2N2 be the total turns on the 
secondary. From Fig. 9.58, 
[a] 
13,200 240 N2 1 or -=-
The ampere turn balance requires 
Nilp = N2l1 + N213 
Therefore, 
Ip = ~: (l1 + l3) = l~O (24 + 21.96) = 0.42/0° A 
:I 
p lQ 
• • + 
12:~0vJ 
13 .2~0 kV - 2Q a 
~ 
• + 
J 120~0v 
b 
The three mesh current equations are 
120/0° = 231a - 21b - 20Ie 
120[0° = -21a + 23Jb - 20Ic 
0 = -20Ia - 20Jb + 50Ic 
Solving, 
20Q 
J 
c 
20Q 
Ia= 24/0° A; le = 19.2[0° A 
lOQ 
Problems 9-57 
N2 N2 
(b] Ip = Ni (I1 + I3) = Ni (Ia+ lb 
1 = 110 (24 + 24) = 0.436 A 
[ c] When the two loads are equal, more current is drawn from the primary. 
P 9.87 (a] 
[c] 
R 
J 
3 
R 
R 
125 = (R + 0.05 + j0.05)11 - (0.03 + j0.03)12 - Rl3 
125 = -(0.03 + j0.03)11 +(R+0.05 + j0.05)12 - Rl3 
Subtracting the above two equations gives 
0 = (R + 0.08 + j0.08)11 - (R + 0.08 + j0.08)12 
Since 11 = 12 (from part [a]) V1 = V2 
jO. 02Q 
250 = (660.04 + j0.04)Ia - 660Ib 
0 = -6601a + 670Ib 
lOQ 
9-58 CHAPTER 9. Sinusoidal Steady State Analysis 
[d] 
Solving, 
Ia = 25.275945L - 0.231714° = 25.275738 - j0.10222 A 
lb = 24.898692L - 0.231713° = 24.898488 - j0.100694 A 
11 =Ia - lb = 0.37725 - j0.001526 A 
V1 = 6011 = 22.635 - j0.09156 = 22.635185L - 0.231764° v 
v 2 = 600I1 = 226.35 - j0.9156 = 226.35185L - 0.231764° v 
125 = (60.05 + j0.05)11 - (0.03 + j0.03)12 - 6013 
125 = -(0.03 + j0.03)11 + (600.05 + j0.05)12 - 60013 
0 = -6011 - 60012 + 67013 
Solving, 
11 = 26.97 L - 0.24° = 26.97 - j0.113A 
12 = 25.10/ - 0.24° = 25.10 - j0.104A 
13 = 24.90/ - 0.24° = 24.90 - j0.104A 
V1 = 60(11 - 13) = 124.4L - 0.27° v 
V2 = 600(12 - 13) = 124.6L - 0.20° v 
lOQ 
[e] Because an open neutral can result in severely unbalanced voltages across 
the 125 V loads. 
P 9.88 [a] Let N1 = primary winding turns and 2N2 = secondary winding turns. 
Then 
14,000 250 N2 1 
Ni - 2N2; N1 = 112 = a 
In part c), 
Ip= 2ala 
I _ 2N2Ia _ -1.I 
P - N1 - 56 a 
1 . 
= 56 (25.28 - J0.10) 
Ip = 451.4 - jl.8 mA 
In part d), 
IpN1 = I1N2 + I2N2 
N2 
Ip= Ni (I1 + I2 ) 
= 1~2 (26.97 - j0.11+25.10 - j0.10) 
= 1~2 (52.07 - j0.22) 
Ip = 464.9 - jl.9 mA 
Problems 9~59 
[b] Yes, because the neutral conductor carries non-zero current whenever the 
load is not balanced. 
-----10 
Sinusoidal Steady State Power 
Calculations 
Assessment Problems 
AP 10.1 [a] V = 100/ - 45° V, I= 20/15° A 
Therefore 
1 
P = 2(100)(20) cos[-45 - (15)] = 500W, A-+ B 
Q = lOOOsin-60° = -866.03 VAR, B----+ A 
(b] V = 100/ - 45°, I= 20/165° 
P = lOOOcos(-210°) = -866.03W, B-+ A 
Q = lOOOsin(-210°) = 500VAR, A-+ B 
[c) V = lOOL - 45°, I= 20/ - 105° 
P = 1000 cos(60°) = 500 W, A -+ B 
Q = 1000sin(60°) = 866.03VAR, A-+ B 
[d] V = 100/0°, I= 20/120° 
P = 1000 cos( -120°) = -500 W, B -+ A 
Q = lOOOsin(-120°) = -866.03VAR, B--+ A 
AP 10.2 
pf = cos(Bv - Bi) = cos[15 - (75)] = cos(-60°) = 0.5 leading 
rf = sin(Bv - Bi)= sin(-60°) = -0.866 
10-1 
10-2 CHAPTER 10. Sinusoidal Steady State Power Calculations 
AP 10.3 
From Ex. 9.4 L = Iµ = O.l8 A 
eff J3 J3 
AP 10.4 [a] Z = (39 + j26)1i(-j52) = 48 - j20 = 52[ - 22.62° n 
250[0° 
Therefore Ie = 48 _ j 20 +J_ + j 4 = 4.85[18.08° A(rms) 
VL = Zie = (52/- 22.62°)(4.85[18.08°) = 252.20[ - 4.54° V(rms) 
IL= 39 :~26 = 5.38[- 38.23° A(rms) 
[b) SL = v Lii = (252.20/ - 4.54°)(5.38/ + 38.23°) = 1357 /33.69° 
= (1129.09 + j752.73) VA 
PL= 1129.09W; QL = 752.73VAR 
[c] Pe = 11£1 21 = ( 4.85)2 • 1 = 23.52 W; Qt = !Itl24 = 94.09 VAR 
[d) 89 ( delivering) = 250I; = (1152.62 - j376.36) VA 
Therefore the source is delivering 1152.62 W and absorbing 376.36 
magnetizing VAR. 
[ ] Q = IVLl2 = (252.20)2 = -1223.18 VAR e cap -52 -52 
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 == 2501* = (40,000 - j30,000) 
Therefore I*= 160 - j120 = 200L - 36.87° A(rms) 
I= 200[36.87° A(rms) 
z = v = 250 = 1.25/ - 36.87° = (1 - ·o. 75) n I 200L36.87° J 
Therefore R = 1 n, Xe= -0.750 
AP 10.6 
Parallel circuit derivation 
p = (250)2 . 
R ' 
(250)2 
therefore R = 40,000 = 1.5625 n 
Q = (250)
2 . 
Xe' 
therefore X = (250) 2 = -2.083D 
c -30,000 
S1 = 15,000(0.6) + j15,000(0.8) = 9000 + j12,000 VA 
S2 = 6000(0.8) - j6000(0.6) = 4800 - j3600 VA 
Sr = S1 + S2 = 13,800 + j8400 VA 
Sr= 2001*; therefore I*= 69 + j42 I= 69-j42A 
Problems 10-3 
Vs= 200 + jI = 200 + j69 + 42 = 242 + j69 = 251.64/15.91° V(rms) 
AP 10. 7 fa) The phasor domain equivalent circuit and the Thevenin equivalent are 
shown below: 
Phasor domain equivalent circuit: 
jlBQ 
a 
20Q -j40Q~ 
4Q 
'-----___... ___ _.__ __ -'\1\1\; ____ b 
Thevenin equivalent: 
20Q jlOQ 
~ 47 .43i=_18 .43°V 
~'------•b 
v = 3 - )800 = 48 - ·24 = 53.67L - 26.57° v 
Th 20 - j40 J 
_:800 
zTh = 4 + j18 + 20 ~ j 40 = 20 + ,jlO = 22.36L26.57° n 
For maximum power transfer' ZL = (20 - jlO) n 
10-4 CHAPTER 10. Sinusoidal Steady State Power Calculations 
AP 10.8 
[b] I = 53.67 I - 26.570 = l.34L - 26.570A 
40 
( 1.34)
2 
Therefore P = v'2 20 = 17.96 W 
[cJ RL = IZThl = 22.360 
[d] I = 53.67 L - 26.570 = 1.23/ - 39.850 A 
42.36 + jlO 
( 123)
2 
Therefore P = ·~ (22.36) = 17W 
34Q jSOQ 
• 
b jlOOQ--:t 
• 
Mesh current equations: 
Solving, 
P = ~(3.5)2(100) = 612.50W 
lOOQ 
jSOOQ 
j400Q ~ jlOOOQ 
AP 10.9 [a] 
• • 
~ 
1 
375Q ~ 400Q 
2 
248 = j400l1 - j500I2 + 375(11 - 12) 
0 = 375(12 - 11) + jlOOOl2 - j500I1 + 40012 
Solving, 
11 = 0.80 - j0.62 A; 12 = 0.4 - j0.3 = 0.5,I' - 36.87° 
:. P = 1(0.25)(400) = 50W 
[b) 11 - 12 = 0.4 - j0.32 A 
1 2 
?375 = 2111 - 121 (375) = 49.20W 
1 
[cJ P9 = 2 (248)(0.8) = 99.20W 
L Pabs = 50 + 49.2 = 99.20 W (checks) 
AP 10.10 [a] VTh = 210V; V2 = iV1; 
Short circuit equations: 
840 = 8011 - 2012 + V1 
0 = 20(12 - 11) - v 2 
210 
:. 12 = 14A; RTh = 14 = 150 
( 210)
2 
(b] Pmax = 3Q 15 = 735W 
AP 10.11 [a] VTh = -4(146/0°) = -584L0° V(nns) 
Short circuit equations: 
146/0° = 8011 - 2012 + V1 
0 = 20(12 - 11) - V2 
Problems 10-5 
.·. 12 = -146/365 = -0.40A; 
-584 
RTh = -- = 14600 
-0.4 
( -584)
2 
[b] p = 2920 1460 = 58.40 w 
lQ-6 CHAPTER 10. Sinusoidal Steady State Power Calculations 
Problems 
P 10.1 p = P + Pcos2wt- Qsin2wt; dp . dt = -2w P sm 2wt - 2wQ cos 2wt 
dp 
-- = 0 when 
dt 
- 2w P sin 2wt = 2wQ cos 2wt 
p 
p 
cos2wt = y'P2 + Q2 ; sin 2wt = - v' p2~ Q2 
or tan2wt = - Q 
p 
Let(}= tan-1(-Q/ P)~ then pis maximum when 2wt =(}and pis minimum 
when 2wt = (0 + 7r). 
Th £ _ p . p. p _ Q( -Q) _ p I p2 Q2 . ere ore Pmax - + v' pz + Q2 v' p 2 + Q2 - + V + 
and Pmin = P - P · v' p2p+ Q2 - Q · v' p2~ Q2 = P - V P 2 + Q2 
P 10.2 [a] P = ~(340)(20) cos(60-15) = 3400cos45° = 2404.16W (abs) 
Q = 3400sin45° = 2404.16 VAR (abs) 
1 
[b] P = 2(16)(75)cos(-15-60) = 600cos(-75°) = 155.29W (abs) 
Q = 600sin(-75°) = -579.56 VAR (del) 
1 
[c] P = 2(625)(4)cos(40-150) = 1250cos(-110°) = -427.53W (del) 
Q = 1250sin(-110°) = -1174.62 VAR (del) 
1 
[d] P = 2(180)(10) cos(l30 - 20) = 900cos(l10°) = -307.82 W (del) 
Q = 900sin(ll0°) = 845.72 VAR (abs) 
Problems 10-7 
P 10.3 [a] coffee maker = 1200 W radio = 71 W 
television = 145 W portable heater = 1322 W 
EP=2738W 
2738 
Therefore Ieff. = 120 = 22.82 A 
Yes, the breaker will trip. 
1538 [b] LP = 2738 - 1200 = 1538 W; Jeff. = 120 = 12.82 A 
Yes, the breaker will not trip if the current is reduced to 12.82 A. 
P 10.4 lg = 30l0° mA; 
1 106 . 
jwC - j(25 x 103)(40) =-JU) 
jwL = j(25 x 103)(40) x 10-6 = jl n 
20 50 
/_o ITn J'ln 30~ mA© -JlH u 
Z1 = -jlll(5 + jl) = 0.2 - jl n 
Zeq = 2 + Z1 = 2.2 - jl 0 
( 30 )
2 
Pg= llrmsl2Re{Zeq} = y12 x 10-3 (2.2) = 990µW 
1 109 
p l0.5 wC = (5000)(80) = 25000 
z = -j2500(7500) = 750- '22500 
f 7500 - j2500 J 
zi = 15oon 
Zr 750 - j2250 = 0.5 _ .1.5 · · zi - 1500 1 
10-8 CHAPTER 10. Sinusoidal Steady State Power Calculations 
Y 0 = (-0.5 + jl.5)(4) = -2 + j6 = 6.32L108.43° V 
P = ! V~ = ! (4)(10) = 20 x 10-3 = 20mW 
2R 21000 
1 P 10.6 jwL = jl0,000(10-3) = jlO 0; - 106 = -j400 
jwC jl0,000(2.5) 
jlOO + lO:C!l 
'--V-0 ~~-:~--j_4_0~0~~~~~_,' 200 
v [-1- 1 + j0.25] = 15 
0 -j40 + 20+ jlO 
Y 0 = 300-jlOOV 
Ia= Yo = 2.5 + j7.5A 
-j40 
I 0 = l5L0° - Ia = 15 - 2.5 - j7.5 = 12.5 - j7.5 = 14.58/ - 30.9° A 
P 10.7 [a] line loss = 50,000 - 40,000 = lOkW 
line loss = jl9 j220 :. j19 j2 = 500 
jl9 j = vi500A 
1Igl2 RL = 40,000 
Thus, 
20Q -jX,,-Q 80Q 
2500.{Q_0 ~t-----"J'IV---, j 60Q 
V (r:ms) 
Problems 10--9 
IZI = j (100)2 + (60 - Xt) 2 II I= 2500 
9 j10,ooo + (60 - Xt)2 
625 x 104 
:. 10,000 + (60 - Xt) 2 = 500 = 12,500 
Solving, ( 60 - Xt) = ±50. 
Thus, Xt = 10 n 
[bJ If X1 =ion: 
or 
2500 . 
19 = lOO + j 50 = 20 - J 10 A 
Xt =non 
89 = -25oOI; = -50- j25kVA 
Thus, the voltage source is delivering 50 KW and 25 magnetizing K vars. 
Q-iIO = II9 l2Xt = 500(-10) = -5000VAR 
Therefore the line reactance is generating 5 magnetizing kvars. 
Qj60 = 1Igl2 XL= 500(60) = 30,000 VAR 
Therefore the load reactance is absorbing 30 magnetizing kva.rs. 
L Qgen = 25,000kVAR = L Qabs 
If Xt =non: 
I _ 2500 _ . A 
g - 100 - j50 - 20 + J 10 
89 = -2500I; = -50 + j25 kV A 
Thus, the voltage source is delivering 50 kW and absorbing 25 
magnetizing kvars. 
Q-j110=119 12 (-110) = 500(-110) = -55kVAR 
Therefore the line reactance is generating 55 magnetizing kvars. The load 
continues to absorb 30 magnetizing kvars. 
LQgen = 55kVAR = LQabs 
10-10 CHAPTER 10. Sinusoidal Steady State Power Calculations 
P 10.8 [a] P = ! (go)2 = 3W 
2 1350 
1 (90)2 
Q = 2 (1012.5) = 4 VAR 
Pma:x = P + J p2 + Q2 = 3 + V(3)2 + (4)2 = 8 W(del) 
[b] Pmin = 3 - 5 = -2W(abs) 
[c] P = 4 W from (a) 
[d) Q = 4 VAR from (a) 
[e] absorb, because Q > 0 
[f) pf = cos( Ov - Oi) 
I= 3
90
0 + .1i02 = 0.0667 - j0.08889 = 111.llL- 53.13° mA 1 5 J 1 .5 
. ·. pf = cos(O + 53.13°) = 0.6 lagging 
[g) rf = sin(53.l3°) = 0.8 
P 10.9 [a] From the solution to Problem 9.56 we have: 
Vo= 72 + j96 = 120L53.13° V 
89 = -~V01; = -~(72 + j96)(15) = -540- j720VA 
Therefore, the independent current source is delivering 540 W and 720 
magnetizing vars. 
I1 = ~0 = 15L53.13° A 
Psn = ~(15)2 (8) = 900W 
Therefore, the 8 n resistor is absorbing 900 W. 
v 
I~= --!!---0 = -9.6 + j7.2 = 12L143.l3° A -11 
Problems 10-11 
Qcap = ~(12)2(-10) = -720VAR 
Therefore, the -jlO n capacitor is delivering 720 magnetizing vars. 
2.5Ic.. = -24 + j18 V 
I _ Vo - 2.5Ic.. 72 + j96 + 24 - j18 
2 - j5 - j5 
= 15.6- jl9.2A = 24.72L - 50.91° A 
Therefore, the j5 0 inductor is absorbing 1530 magnetizing vars. 
S2.5J6 = !(2.51c..)I; = !(-24 + j18)(15.6 + j19.2) 
= -360- j90VA 
Thus the dependent source is delivering 360 Wand 90 magnetizing vars. 
[b] L Pgen = 360 + 540 = 900 W = L Pabs 
[c] L Qgen = 720 + 90 + 720 = 1530 VAR= L Qabs 
P 10.10 [a] From the solution to Problem 9.57 we have 
200 
Ia= 2.25 - j2.25A; lb= -6.75 + j0.75A; Io= 9 - j3A 
S6ov = -~(60)1: = -30(2.25 + j2.25) = -67.5 - j67.5 VA 
Thus, the 60 V source is developing 67.5 W and 67.5 magnetizing vars. 
S90v = -!(j90)1~ = -j45(-6.75 - j0.75) 
= -33. 75 + j303. 75 VA 
Thus, the 90 V source is delivering 33. 75 W and absorbing 303. 75 
magnetizing vars. 
P20n = ~IIal2 (20) = 101.25 W 
10-12 CHAPTER 10. Sinusoidal Steady State Power Calculations 
Thus the 20 n resistor is absorbing 101.25 W. 
1 2 Q-j2on = 2llbl (-20) = -461.25 VAR 
Thus the -j200 capacitor is developing 461.25 magnetizing vars. 
Qi5f1. = ~llol 2 (5) = 225 VAR 
Thus the j5 0 inductor is absorbing 225 magnetizing vars. 
[b] L Pdev = 67.5 + 33. 75 = 101.25 W = L Pabs 
(cJ L Qdev = 67.5 + 461.25 = 528.75 VAR 
L Qabs = 225 + 303. 75 = 528. 75 VAR= L Qdev 
v2 
P 10.11 Wdc = ~cT; 
1 i·t,,+T Vdc = T v; dt = Vrms = Veff 
• to 
P 10.12 [a] leff = 60/110 rv 0.545A; (b] Jeff= (60 + 80)/110 rv l.273A 
P 10.13 [a] Area under one cycle of v;: 
A= (400)(4)(20 x 10-6) + 10,000(2)(20 x 10-6 ) 
= 21,600(20 x 10-6 ) 
Mean value of v;: 
M.V. = A = 21,600(20 x 10-6) = 3600 
120 x 10-6 120 x 10-6 
:. Vrms = J360o = 60V(rms) 
[b] p = V:.~1s = 3600 = 300 W 
R 12 
Problems 10-13 
30 
P 10.14 i(t) = 40 x 103t = 750t 0:::; t:::; 40ms 
i(t) = M - ~~ x 103t 40ms:::; t:::; 50ms 
i(t) = 0 when t = 50ms 
. ·. M = 3000(50 x 10-3 ) = 150 
i(t) = 150 - 3000t 40 ms :::; t :::; 50 ms 
; 1000 { l0.04 1·0.05 } . ·. lrms = \, - 0 (750)2t2 dt + (150 - 3000t)2 dt v 5 . 0 . 0.04 
,0.04 t3 I 0.04 
.lo (750)2t2 dt = (750)23 0 = 12 
(150 - 3000t)2 = 22,500 - 9 x 105t + 9 x 106t2 
{0.05 
Jn 22,500 dt = 225 
0.04 
{0.05 10.05 Jn 9 x 105t dt = 45 X 104t2 = 405 
Q04 Q04 
{0.05 10.05 
9 x 106 Jn t2 dt = 3 x 106t3 = 183 
0.04 0.04 
:. Irrns = j20{12 + (225 - 405+183)} = v'300 = 17.32 A 
P 10.15 P = 12 R TillS : R = 24 x 103 = 80 n . 300 
P 10.16 lg= 30L0° mA 
jwL = j(lOO)(lO) = jlOOOn; 
1 106 -
jwC - j(100)(2) = -J50oon 
4k0 
+ 
30/0° mA t jlkO 
I = 30~(.jlOOO) = 3.75 "2/135° A 
0 4000 - j4000 v .t. ' m 
10-14 CHAPTER 10. Sinusoidal Steady State Power Calculations 
P = IIol;rns(4000) = (3.75)2(4000) = 56.25mW 
Q = llol;ms(-5000) = -70.3125mVAR 
S = P+ jQ = 56.25 - j70.3125mVA 
ISi = 90.044mVA 
1 106 
P 10.17 [a] jwC = jl05 = -jlOO 
jwL = j105 (50 x 10-6 ) = j5 n 
SJ2 
-J·100 ·so J 
F
1I1-g---------J'""Y'n_,....,,_~~) ....._I_2 ___, 
so& 
V ( ['ffiS) 
z = -jlO + (5)(j~) + 7.5 = 10 - j7.5 n 
5+J5 
5010° . 
19 = - = 3.2 + J2.4 A 10- j7.5 
S9 = _!y91; = -25(3.2 - j2.4) = -80 + j60VA 2 
P = 80W(abs); Q = 60VAR(del) 
ISi = ISgl = lOOVA 
[b] 11 = Ig(jS) = !(3.2 + j2.4)(1 + jl) = 0.4 + j2.8 A 
5+j5 2 
P50 = ~11112 (5) = 20W 
P1.5n = ~j19 j 2 (7.5) = 60W 
L Pdiss = 20 + 60 = 80 W = L Pdev 
7. s 0 
] 19 5 1 [c Ijs = 5 + j 5 = 2(3.2 + j2.4)(1- jl) = 2.8 - j0.4A 
P 10.18 [a] 
1 2 
Qjsn = 2lljsl (5) = 20VAR(abs) 
Q-jwn = ~II9 j 2 (-10) = -80VAR(dev) 
LQabs = 20 + 60 = 80VAR = LQ<lev 
500 
---7 Ig 
+ 
0 
340&: -jlOOO ..... 
Vo 
V ( rms) J,, 
I1 
Va Va-340 Va 
-jlOO+ 50 +80+j60=0 
: . V 0 = 238 - j34 V 
lg = 340 - 25308 + j34 = 2.04 + j0.68 A 
S9 = V 9I; = (340)(2.04 - j0.68) 
= 693.6- j231.2VA 
800 
[h] Source is delivering 693.6 W. 
[c] Source is absorbing 231.2 magnetizing VAR. 
[d] 11 = -~~O = 0.34 + j2.38A 
S1 = Voli = (238 - j34)(0.34 - j2.38) 
= O-j578VA 
I V 0 _ 238 - j34 = 1.7 _ .1.7 A 
2 = 80 + j60 80 + j60 J 
82 = Vol2 = (238 - j34)(1.7 + jl.7) 
= 462.4 + j346.8 VA 
) 
Sson = II9 12(50) + jO = (2.15) 2(50) = 231.2 W 
I2 
Problems 10-15 
j60 0 
10-16 CHAPTER 10. Sinusoidal Steady State Power Calculations 
(e] L Pdel = 693.6 W 
I: Pdiss = 462.4 + 23i.2 = 693.6 w 
: · L Pdel = L Pdiss = 693.6 W 
[f] L Qabs = 231.2 + 346.8 = 578 VAR 
L Qdev = 578 VAR 
:. L mag VAR dev = L mag VAR abs = 578 
P 10.19 (a] Let VL = VmL0°: 
+ , 0 
240~ 
v (rms) 
lQ j8Q 
SL = 250(0.6 + j0.8) = 150 + j200 v A 
I* - 150 .200. 
£ - TT +) v; ' 
Vrn m 
( 150 .200) . 240LO = Vm + Vm - J Vm (1 + J8) 
240VmLe = V~ + (150- j200)(1 + j8) = V~ + 1750 + jlOOO 
240Vmcos0=V~+1750; 240Vmsin0=1000 
(240)2V~ = (V~ + 1750)2 + 10002 
57,600V~ = V~ + 3500V~ + (3.0625 + 1) x 106 
or 
V~ - 54,lOOV,;_ + 4,062,500 = 0 
Solving, 
v~ = 21,050 ± 26,974.8; 
If Vm = 232.43 V: 
Vm = 232.43V and Vm = 8.67V 
. 1000 
sm (} = (232.43) (240) = 0.0179; 
If Vm = 8.67 V: 
1000 
sin()= (8.67)(240) = 0.4805; 
(} = 1.03° 
0 = 28.72° 
[b] 
~240~0v 
-~~ 0 
\ > 232.43LQ. v 
r.) 1.08/-53.13°A 
0 
28.84/-53.13 v 
(R=lQ) 
(X=8Q) 
p 10.20 ST = 52,800 - j 56~8°0 (0.6) = 52,800 - j39,600 v A 
S1 = 40,000(0.96 + j0.28) = 38,400 + jll,200 VA 
82 = ST - 81 = 14AOO - j50,800 = 52,801.52/ - 7 4.17° v A 
rf = sin(-74.17°) = -0.9621 
pf = cos(-74.17°) = 0.2727 leading 
P 10.21 [aJ Z1 = 12 + j(27r)(60)(15 x 10-3) = 13.27 /25.23° fl 
pf = cos(25.23°) = 0.9 lagging 
rf = sin(25.23°) = 0.43 
Z2 = 80 - 27r(60)(l{ x 10_6 ) = 184.08/ - 64.24° fl 
pf = cos(-64.24°) = 0.43 leading 
rf = sin( -64.24°) = -0.9 
Zs= 400+Zp 
z = jwL(l/jwC) 
P jwL + 1/jwC 
jwL 
l-w2LC 
Problems 10-17 
10-18 CHAPTER 10. Sinusoidal Steady State Power Calculations 
j(1207r)(20) . 
- 1 - (120n-)2(20)(5 x 10~6) = -J570;57 n 
Z3 = 400 - )570.67 = 696.90L - 54.97° n 
pf = cos(-54.97°) = 0.57 leading 
rf = sin( -54.97°) = -0.82 
[b] y = Yi + 12 + Y3 
1 y; - . 
1 - 13.27 L25.23°' 
1 y;- . 
2 - 184.08L - 64.24°' 
Y = 71.35 - j26.05mS 
z = _!_ = 13.16/20.06° n y 
pf = cos(20.06°) = 0.94 lagging 
rf = sin(20.06°) = 0.343 
~ - 1 
3 - 696.90L - 54.97° 
P 10.22 [a] S1 = 18 + j24kVA; S2 = 36- j48kVA; S3 = 18 + jOkVA 
24001* = (72 - j24) x 103; : . I = 30 + j 10 A 
z = 302!0~10 = 12 - J24n = 75.89L- 18.43° n 
[b] pf = cos( -18.43°) = 0.9487 leading 
P 10.23 [a] From the solution to Problem 10.22 we have 
IL= 30 + jlOA(rms) 
Vs= 2400L0° + (30 + jl0)(0.2 + jl.6) = 2390 + j50 
= 2390.52/1.20° V(rms) 
[b] lh I = JI00o 
Pp_= (1000)(0.2) = 200W Qe = (1000)(1.6) = 1600VAR 
[c) Ps = 72,000 + 200 = 72.2kW Qs = -24,000+1600 = -22.4kVAR 
72 
{d] TJ = 72.2 (100) = 99.72% 
Problems 10-19 
p 10.24 
jlOO 
+ 
j4800 
24ooul 
v {rms} 
2400Ii = 24,000 + j18,000 
Ii = 10 + j7.5; :. 11 =10- j7.5A(rms) 
2400I; = 48,000 - j30,000 
1; = 20 - jl2.5; :. 12 = 20 + j12.5A(rms) 
I = 2400&_ = 40 ·oA· 3 60 +J ' 
240010° . 
l4 = j 480 = 0 - J5A 
V 9 = 2400 + (70)(j10) = 2400 + j700 = 2500L16.26° V(rms) 
P 10.25 [a] 81 = 24,960 + j47,040 VA 
s = IVLl2 = (480)2 = 23 040- ·23 040VA 2 z2 5 + j5 ' J ' 
Si + S2 = 48,000 + j24,000 VA 
4801~ = 48,000 + j24,000; IL= 100 - j50A(rms) 
Vg = VL + IL(0.02 + j0.20) = 480 + (100 - j50)(0.02 + j0.20) 
= 492 + j19 = 492.37 L2.21° Vrms 
IV9 j = 492.37Vrms 
1 1 
[b] T = f = 60 = 16.67ms 
2.21° t 
--= 
360° 16.67 ms' 
t = 102.39 µs 
10-20 CHAPTER 10. Sinusoidal Steady State Power Calculations 
(c) VL lags Vg by 2.21° or 102.31µs 
P 10.26 [a] 
v 
~: 
0.050 
+ 
125h,_0 
( r:ms) 
0 .050 
I 5000 - j2000 40 .16 A ( ) 1 = = -J rms 125 
I2 = 3750 - jl500 = 30- jl2A (rms) 
125 
I - 8000 + jO - 32 ·o A ( ) 3 - 250 - + J rms 
: . lg 1 = 72 - j 16 A ( rms) 
In= 11 - 12 = 10- j4A (rms) 
Ig2 = 62 - j12A 
+ 
50~0 
( r:ms) 
Vg1 =0.05Ig1 +125 + jO + 0.141n = 130- jl.36V(rms) 
Vg2 = -0.14In + 125 + jO + 0.05Ig2 = 126.7 - j0.04 V(rms) 
Sg1 = [(130- jl.36)(72 + j16)J = [9381.76 + j1982.08}VA 
Sg2 = [(126.7 - j0.04)(62 + j12)J = [7855.88 + jl517.92] VA 
Note: Both sources are delivering average power and magnetizing VAR to 
the circuit. 
[b] Po.05 = Jlg112(0.05) = 272W 
Po.15 = JinJ2(0.14) = 16.24 W 
Po.05 = IIg212(0.05) = 199.4 W 
L: Pdis = 212 + 16.24 + 199.4 + 5000 + 3150 + sooo = 11,237.64 w 
P 10.27 [a] 
L Pdev = 9381. 76 + 7855.88 = 17,237.64 W = L Pdis 
L Qabs = 2000 + 1500 -' 3500 VAR 
L Qdel = 1982.08 + 1517.92 = 3500 VAR = L Qabs 
120I~ = 1800 + j600; 
1201; = 1200 - j900; 
120 j480 
11 = 15 - j5 A{rms) 
12 = 10 + j7.5A{rms) 
240 240 
13 = 12 + j 48 = 20 - j5A(rms) 
Ig1=11 + l3 = 35 - jlOA 
Sg1 =120(35 + jlO) = 4200+ jl200VA 
Problems 1(}-21 
Thus the V gl source is delivering 4200 W and 1200 magnetizing vars. 
lg2 = l2+l3 = 30+j2.5A(rms) 
Sg2 = 120(30 - j2.5) = 3600 - j300 VA 
Thus the V g2 source is delivering 3600 W and absorbing 300 magnetizing 
vars. 
[b] L Pgen = 4200 + 3600 = 7800W 
(240)2 L Pabs = 1800 + 1200 + 12 = 7800 W = L Pgen 
L Qdel = 1200 + 900 = 2100VAR 
(240)2 L Qabs = 300 + 600 + 48 = 2100 VAR = L Q del 
10-22 CHAPTER 10. Sinusoidal Steady State Power Calculations 
P 10.28 S1=1200+1196 + 516 + jO = 2912 + jOVA 
p 10.29 
2912 . . 11 = 120 +JO= 24.27 +JOA 
S2 = 600 + 279 + 88 + 512 + jO = 1479 + jOVA 
1479 . . 
12 = 120 +JO= 12.33 +JOA 
83 = 4474 + 12,200 + jO = 16,674 + jOVA 
· 1 16'674 ·o 69 48 ·o A . . 3 = 240 + J = . + J 
191=11 + l3 = 93.75 + jOA 
Breakers will not trip since both feeder currents are less than 100 A. 
v 
fl 
lQ j8Q 
+ 
2sooLQ_0 v 250kVA 
0.96 lag 
IL= 240,000- j70,000 = 96 _ j 2BA(rms) 
2500 
It== 96- j28 + jlc = 96 + j(Ic - 28) 
Vs= 2500 + (1 + j8)(96 + j(Ic - 28)] 
= (2820 - 81c) + j(740 +le) 
1Vsl 2 = (2820 - 8Ic)2 + (740 + Ic) 2 = (2500)2 
. ·. 6516 - 43,640Ic + 2,250,000 = 0 
p 10.30 
Problems 10-23 
43,640 ± J ( 43,640)2 - 4(65)(2,250,000) 
le = 2(65) 
= 335.69 ± 279.42 = 56.27 A(rms)* 
*Select the smaller value of le to minimize the magnitude of lp. 
2500 
Xe = --- = -44.43 
56.27 
1 
C = (44.43)(12011") = 59·7 µF 
7200~ 0 [a] I= 140 + j 480 = 14.4/ - 73.74 A(rms) 
P = (14.4)2(2) = 414.72W 
1 138 - j460 
[b] YL = 138 + j460 - 230,644 
. C . 460 Xe= -230,644 = _ 501.400 . . -JW = -J 230 644 460 
' 
[ ] z = 230'644 = 1611 33 n C L 138 . 
[d] I = 7200 = 4.30/ - 0.68° A 
1673.33 + j20 
p = ( 4.30)2(2) = 37.02 w 
37.02 
[e] 3 = 414_ 72 (100) = 8.933 
Thus the power loss after the capacitor is added is 8.933 of the power 
loss before the capacitor is added. 
P 10.31 [a] SL= 24 + j7kVA 
1251~ = (24 + j7) x 103 ; I~= 192 + j56 A(rms) 
:. IL= 192 - j56A(rms) 
Vs= 125 + (192 - j56)(0.006 + j0.048) = 128.84 + j8.88 
= 129.15/3.94° V(rms) 
IVsl = 129.15 V(rms) 
(b] Pt= 1Itl2 (0.006) = (200)2(0.006) = 240W 
10-24 CHAPTER 10. Sinusoidal Steady State Power Calculations 
[c] (125)2 = -7000· 
Xe ' Xe= -2.230 
1 
- wC = -2.23; 1 C =