821
Answers to Selected Problems
AppendixHChapter 11.1 110 giga-watt hours
1.3 a) 111.6 seconds
b) 2480 bytes
1.6 0.10 mm
1.9
1.12 a) from A to B
b) from B to A
c) from B to A
d) from A to B
1.17 a)
b)
c)
1.24 a)
b) (delivered)
c)
d) (extracted)
e) , , ,
1.26
Chapter 22.2
2.3 Not valid, due to and current sources inthe rightmost branch
2.6 Not valid, since the voltage drop between the topand bottom nodes is different due to different volt-age sources in the left and right branches
2.8 Not valid, since the voltage drop between the topand bottom nodes is different due to different volt-ages in the left and right branches
2.10 resistor
2.11 resistor4 kÆ
8 kÆ
5 A4 A
1700 W
770 mW
0 mJ4 mJ4 mJ0 mJ
5.196 mW
t = 2.366 s
5.196 mW
t = 0.634 s
21.67 mJ
1.24 mJ
3.1 mW
4800 W
2400 W
2000 W
600 W
6 sin 4000t mC
2.14 a) A current source in parallel with a resistor
b)
2.17 a)
80 W
20 Æ5 A
10
80
40506070
3020
00 75 150 225 300 400 500
iS (mA)
vS (V)
b) A source in series with a resistor
c)
d)
e)
f) A linear model cannot predict non-linearbehavior
2.18 a)
b)
c)
d) , ,
e)
2.19 a) ,
b)
c) Power developed and dissipated is
2.24 a) , , , ,
b)
c) Power developed and dissipated is
2.28 a)
b) Power developed and dissipated is 741 mW
4.5 V
83.33 W
83.33 W
0 W16.67 W11.11 W33.33 W22.22 W
768 W
192 V
1.6 A2.4 A
125 W
20 W80 W25 W
40 V
0.5 A
2 A
500 mA
375 mA
125 mA
200 Æ75 V
89251_26_AppH 4/5/07 2:22 PM Page 821
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
2.34 , so a warning sign should be postedand precautions taken
2.35
2.36 a) ; ;
b) ; ;
c) All values are much greater than a few minutes
2.37 a)
b) No, will cause a shock
2.38
Chapter 33.1 a) and , and ; simplified
circuit is
b) and , and ; simplifiedcircuit is
240 �
340 �
480 �
10 V�
�
200 Æ140 Æ300 Æ180 Æ
6 k�10 k� 12 k�
11 k�
2 mA
7 kÆ5 kÆ8 kÆ3 kÆ
3000 V
12 V>800 Æ = 15 mA
40 V
ttrunk = 70,422.54 s
tleg = 7071.13 starm = 1414.23 s
Ptrunk = 7.40 W
Pleg = 29.59 WParm = 59.17 W
RA
RA
RT
RLRL
V�
�
i = 385 mA c) , and , and ; simplifiedcircuit is
3.2 a) and , and ; simplifiedcircuit is
b) and , and ; simplified circuit is
c) and , , and ;simplified circuit is
3.5 a)
b)
c)
3.6 a)
b)
c)
3.13 a)
b) ,
c) ,
3.15 a) ,
b)
3.21 , , , 41.2 kÆ8240 Æ1030 Æ257.5 Æ
1>8 W
15 kÆ60 kÆ
12,408 Æ17,672 Æ
1.32 W1.88 W
66 V
125 kÆ
14 Æ
6 Æ
50 Æ
500 Æ
6 kÆ
25 k�
75 k�
25 k�
0.5 V�
�
150 kÆ50 kÆ75 kÆ300 kÆ100 kÆ
4.2857 �
8 V
20 � 6 �
40 � 4 ���
18 Æ9 Æ5 Æ30 Æ
18 �7.5 �
12 �
200 mA
21 Æ28 Æ20 Æ12 Æ
50 V�
�75 �150 �
30 Æ45 Æ60 Æ50 Æ40 Æ
822 Answers to Selected Problems
89251_26_AppH 4/5/07 2:23 PM Page 822
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Answers to Selected Problems 823
3.22 a)
b)
c)
d)
3.23 a)
b)
c)
3.30 a)
b)
c) Yes
3.33 a)
b)
c)
3.48 a)
b)
c) ,
d) ,
3.52 ,
3.53 a) -connected — — become Y-connected— — ; equivalent resistance is
b) Y-connected — — become -connected— — ; equivalent resistance
is
c) Convert the delta connection — — to its equivalent wye. Convert the wye connection
— — to its equivalent delta.
3.54
3.71 , , ,
, , ,
, ,
3.72
3.73 a) , , ,
, , ,
, ,
b) , or ;
, or ;
, or ;
= PdissPdel = 997.5 W
150 W>mib2Rb = 7.5 Wib = 33.75 A
150 W>mi22R2 = 187.5 Wi2 = 16.23 A
150 W>mi12R1 = 187.5 Wi1 = 17.52 A
Rd = 0.0244 ÆRc = 0.0066 ÆRb = 0.0066 Æ
Ra = 0.0244 ÆR5 = 0.6106 ÆR4 = 0.7122 Æ
R3 = 0.768 ÆR2 = 0.7122 ÆR1 = 0.6106 Æ
= PdelPdiss = 624 W
Rd = 0.0259 ÆRc = 0.0068 ÆRb = 0.0068 Æ
Ra = 0.0259 ÆR5 = 1.0372 ÆR4 = 1.1435 Æ
R3 = 1.2 ÆR2 = 1.1435 ÆR1 = 1.0372 Æ
99 Æ
R6R4R3
R6R5R4
50 Æ45 Æ270 Æ90 Æ
¢R5R4R2
50 Æ6 Æ18 Æ3 ÆR4R3R2¢
24 V48 V
90 mW900 Æ
180 mW800 Æ
25 mA
900 Æ
50 Æ
4950 Æ
99,950 Æ
1>2500
im =
(25>12)
50 + (25>12) (imeas) = 1
25 imeas
20 mA
13.33 mA
3.2 V
1.33 V
3.2 V
12 V
1.2 mA Chapter 4
4.1 a) 5
b) 3
c) , ,
d) 2
e) ,
4.2 a) 11
b) 10
c) 11
d) 10
e) 5
f) 5
g) 7
4.3 a) 2
b) 5
c) 7
d) 1, 4, 7
4.4 a) 10
b) 4
c) 4
d) Avoid the three meshes with current sources
4.6
4.9 ,
4.10 a) , , , ,
b)
4.19
4.20 a)
b)
4.21
4.26 ,
4.27
4.31 a)
b)
4.32 a)
b) 1140 W, absorbed
1140 W, developed
-1.72 A, 1.08 A, 2.8 A
9.8 A, -0.2 A, -10 A
26 V
1.2 W200 V
3.2 V
165 W
165 W
375 W
582 W
-1 A3 A2 A2 A4 A
90 V25 V
10 V
R3i3 + R5i5 - R4i4 = 0
R1i1 + R3i3 - R2i2 = 0
i5 - i2 - i3 = 0
- i1 + i4 + i3 = 0- ig + i1 + i2 = 0
89251_26_AppH 4/5/07 2:23 PM Page 823
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
824 Answers to Selected Problems
4.37
4.38
4.41 a)
b)
c) checks
4.42 a)
b)
c)
4.47
4.50 a)
b)
4.54 a) Mesh current method requires fewer equations
b)
c) No—you can solve for the voltage from themesh currents
d)
4.56 a) The constraint equations are easier to formu-late if the node voltage method is used
b)
4.59 a)
b)
4.62 a)
b)
4.63 ,
4.66 (down), 10
4.67 a)
b) %
4.71 , 43.2
4.77 ,
4.79 ,
4.80 a)
b)
4.91 a)
b)
4.92
4.105 ,
4.106 ,
4.107 , v2 = 117.5 Vv1 = 52.083 V
v2 = 105 Vv1 = 37.5 V
v2 = 102.5 Vv1 = 39.583 V
30 V
72.2 W
38 V
48 W
12 Æ
150 Æ16.67 Æ
8 Æ0 V
kÆ-86.40 V
-5.67
45.28 V
kÆ8 mA
16 Æ48 V
1 A
1 A
3 mA
3 mA
180 W
200 mW
4 mW
1319.685 W
5.7 A, 4.6 A, 0.97 A, -1.1 A, 3.63 A
99 W
2.912 mW
200 mW
-5.2 mA
6847.36 W
2643.36 W
2700 W
98 W Chapter 55.1 a)
b) The input resistance;
c) The open-loop voltage gain;
d)
5.2 a)
b) (saturates)
c)
d)
e) (saturates)
f)
5.3
5.8 a) Non-inverting amplifier
b)
5.9 a) Many possible designs; one uses a single input resistor and four series-connected resistors in the feedback path
b)
5.12 a) Inverting summing amplifier
b)
c)
5.13 a)
b)
5.15
20 k�
Vcc
Vo�Vcc
Va
12 k�Vb
15 k�Vc
30 k�
60 k�
Vd �
��
�
8.4 V … vb … 13.2 V
-6 V
-7.5 V … vc … 1.5 V
-6 V
;10 V
10 kÆ
10 kÆ
4.5 V
-3.1 mA
2.8125 V … va … 7.3125 V
18 V
-14 V
10 V
-18 V
-12 V
vo = -4 V
(vp - vn) = 0
in = 0
Positivepower supply
Negativepower supply
Output
Non-invertinginput
Invertinginput �
�
89251_26_AppH 4/5/07 2:24 PM Page 824
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Answers to Selected Problems 825
5.17 a) Non-inverting amplifier
b)
c)
5.18 a)
b)
c)
5.24
5.25 a)
b)
5.26 a)
b)
c)
5.32 a) 19.975
b)
c) 399.5
5.33
5.42 a)
b)
c)
d) , ,
5.43 a) 30.98
b) ,
c)
d)
e) 31, , ,
5.48 a)
b)
Chapter 66.1 a)
50 ms 6 ti = 0
25 … t … 50 msi = 0.8 - 16t A
0 … t … 25 msi = 16t A
t 6 0i = 0
12 mÆ
2 kÆ
0 A0 V1 V
836.22 pA
367.94 mV
999.87 mV999.5 mV
5000 Æ0 V-20
5003.68 Æ
736.1 mV
-19.9844
32.89 kÆ … Rx … 33.11 kÆ
-0.05
80 kÆ
114.3 kÆ
56.25 mV
-638 mV … vb … 962 mV
-15.95 V
20 kÆ
35 kÆ
-3.97 V … vg … 3.97 V
7.56 V
-2.4 V … vs … 4 V
3.75vs
6.14
t(s)0.5 1 21.5
60
40
20
00
�20
�40
�60
ic(�A)
t(s)
5
4
3
2
1
0.1 0.2 0.30
0 0.4 0.5 10.6 0.7 0.8 0.9
iL(A)
t(s)
30
25
20
15
10
5
0.20
0 0.4 10.6 0.8
vL(mV)
b)
6.3
50 ms 6 tw = 0
25 6 t 6 50 msw = 48t 2
- 4.8t + 0.12 J
0 6 t 6 25 msw = 48t 2 J
t 6 0w = 0
50 ms 6 tp = 0
25 6 t 6 50 msp = 96t - 4.8 W
0 6 t 6 25 msp = 96t W
t 6 0p = 0
50 ms 6 tv = 0
25 6 t 6 50 msv = -6 V
0 6 t 6 25 msv = 6 V
t 6 0v = 0
89251_26_AppH 4/5/07 2:24 PM Page 825
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
826 Answers to Selected Problems
6.15 a)
b)
c)
d)
100 V
4 * 105t - 20 V
-200 * 103t + 40 V 6.43 ,
6.47 a)
b)
c)
d)
6.48 a)
b) No
6.49
6.51 a)
b)
Chapter 77.1 a) 5 A
b) 40 ms
c) A, V,
d)
7.2 a) 0.2 mA, 0.2 mA
b) 0.2 mA,
c)
d)
e) The current in a resistor can changeinstantaneously
7.3 a) 0 A
b) 62.5 mA
c) 87.5 mA
d) 62.5 mA
e) 150 mA
f) 0 A
g)
h) 0 V
i)
j) 0 V
k)
l)
7.21 a) , ,
b)
c) , 4500 mJ1125 mJ
5625 mJ
-15e-125t+ 15 V60e-125t
+ 15 V15e-125tmA
150 - 62.5e-4000t mA
-12.5e-4000t V
-12.5 V
62.5e-4000t mA
-0.2e-106t mA
0.2e-106t mA
-0.2 mA
9.02%
-500e-25t V-400e-25t5e-25t
3C dv
dt= 0
2C dv
dt= 0
v =
13
vs(t) + v(0)
-5 A
6.5 J
6.5 J
18.5 J
18.5 J
3.6 nWb>A2.8 nWb>A
t(�s)5004003002001000�100
120
100
80
60
40
20
0
v(V)
6.21 20 H
6.22 a)
b)
c)
d) 540 J
e) 3920 J
f) 3380 J
g) (checks)
6.26 5 nF, initial voltage is
6.27 a)
b)
c)
d)
e)
f)
g)
6.34 a)
b)
c)
d) , which is consistent with the circuit’sbehavior
6.42 a) , 1.25
b) , 0.25 * 10-6 Wb>A0.25 * 10-6 Wb>A160 mH
-48.75 V
-105e-t+ 56.26e-2.25t V
50e-t
40[-e-t+ 11.25e-2.25t] + 90[e-t
- 5e-2.25t] =
40di2
dt+ 90i2 = -5
dig
dt
18.225 mJ
12
(20 * 10-9 )(27)2
+
12
(30 * 10-9 )(27)2
=
18.225 m J
23.625 m J
5.4 m J
12e-2500t- 27 V
18e-2500t+ 27 V
30e-2500t V
-10 V
12
(10)(13)2+
12
(30)(13)2= 3380 J
-3e-20t- 13 A
-9e-20t+ 13 A
-12e-20t A
89251_26_AppH 4/5/07 2:25 PM Page 826
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Answers to Selected Problems 827
7.24 a)
b)
7.33 mA,
7.34 a) A,
b) V, 0 V
7.35 a)
b)
7.50 a)
b)
c)
d)
e) 6 mA
7.51
7.53 a) 120 V
b)
c) 2 ms
d)
e)
f)
7.54 a)
b) 25 V
c) 2.5 ms
d) 1.97 ms
7.65 a)
b)
c)
d)
e) Yes
7.67 a)
b)
c)
d)
e) Yes
7.72
7.76 V, ms;
V, ms
7.86 a) 2
b) s
7.87 s173.23 m
529.83 m
… t 6 q250vo = 100e-1000(t - 0.25)
0 … t … 250vo = 100
-559.12 mV
24 - 24e-5000t mA
16 - 16e-5000t mA
10e-5000t V
40 - 40e-5000t mA
-50 + 50e-10,000t mA
0.25 - 0.25e-10,000t A
15e-10,000t V
0.2 - 0.2e-10,000t A
-30 V
5.4e-500t mA
-150 + 270e-500t V
-5.4 mA
-150 V
15 - 105e-4000t V
8 - 4e-5000t mA
2 + 4e-5000t mA
2 - e-5000t mA
40 - 40e-5000t V
15 + 285e-2000t V
3 - 19e-2000t A
-40
16 - 16e-4000t V4 + 2e-4000t
-60e-80,000t V9 + 3e-80,000t
14.05%
39.6e-2000t mA 7.92 25 ms
7.93 mV, mV,
7.103 a) 1.091 M
b) 0.29 s
7.104 a)
b) 559.3
7.105 a)
b) 99.06 mA
c) per year
Chapter 88.1 a) ,
b) overdamped
c)
d) ,
e)
8.2 a) , 200 mH, ,
b) ,,
8.3 a) 800 mH
b)
c) 125 V
d) 12.5 mA
e)
8.4 a) , 12.5 nF, , 25 V
b)
8.5 a) , , 6.25 H, F, 20 mA,
b)
c)
d)
8.19
8.20
8.21
8.25
8.26
8.27
8.45 40 - 40e-5000t cos 5000t - 40e-5000t sin 5000t V
15 - 1500te-100t- 45e-100t mA
15 - 45e-80t cos 60t - 10e-80t sin 60t mA
15 - 40e-50t- 5e-200t mA
(60 - 132 * 104t)e-10,000t V
60e-4000t cos 3000t - 320e4000t sin 3000t V
-140e-2000t+ 200e-8000t V
5e-160t- 20e-40t mA
-25e-160t+ 25e-40t mA
-5e-160t+ 5e-40t V
-5 mA25 m80 rad>s100 rad>s
(25,000t - 7.5)e-4000t mA
-5 * 105 V>s10 kÆ
e-4000t(12.5 cos 3000t + 68.75 sin 3000t) mA
2500 Æ
1.25e-5000t- 20e-20,000t mA
5e-5000t- 5e-20,000t mA
-6.25e-5000t+ 25e-20,000t mA
10,000 rad>s12,500 rad>s800 Æ
6250 Æ
-8000 - j6000 rad>s-8000 + j6000 rad>s7812.5 Æ
-20,000 rad>s-5000 rad>s
$43.39
24.3 flashes>min
kÆ
8.55 flashes>min
Æ
30,000t - 40 - 5e-2000t mV10 + 5e-2000t30,000t - 30
89251_26_AppH 4/5/07 2:26 PM Page 827
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
828 Answers to Selected Problems
8.46
8.47
8.58 a) s: V, V; 0.2 s :,
b) 2.844 s
8.59 : , ;s: ,
8.64 a) s
b) 262.42 V
c) s,
8.65 a) 40 mJ
b)
c) 568.15 V
Chapter 99.1 a) 120 Hz
b) 8.33 ms
c) 100 V
d) 70.71 V
e) , 0.7854 rad
f) 1.042 ms
g) 3.125 ms
9.5 a) 170 V
b) 60 Hz
c)
d) rad
e)
f) 16.67 ms
g) 2.78 ms
h)
i)
j)
9.8
9.9 a)
b) A,
c) 133.61 mA
d) 2 A, ,
e) 36.87°
23.13°400 rad>s
2 cos(400t + 23.13°) A-1.84e-533.33t
-1.84e-533.33t+ 2 cos(400t + 23.13°) A
Vm
2
25>9 ms
25>18 ms
-170 sin 120pt V
-60°
-1.05
377 rad>s
45°
-27,808.04 V
v(tmax) = 262.15 Vtmax = 53.63 m
55.23 m
0.5 - 1.66e-2(t - 0.2) V-6.25 + 22.38e-(t - 0.2)
- 15.31e-2(t - 0.2) Vt Ú 0.2-2 + 2e-2t V25 - 50e-t
+ 25e-2t V0 … t … 0.2 s
t - 1 V-6.25t 2
+ 12.5t - 1.25 V… t … tsat-4t25t
20 … t … 0.2
40 - 53.33e-2000t+ 13.33e-8000t V
40 - 200,000te-5000t- 40e-5000t V 9.12 a) 50 Hz
b)
c)
d) 127.32 mH
e)
9.13 a)
b)
c)
d) F
e)
9.14 a)
b) s
9.15 a)
b)
c)
9.16 a)
b)
c)
9.21 a)
b) 200 mS
c) 100 mS
d) 2.24 A
9.26
9.27
9.28
9.34
9.40 ,
9.41 , 50 - j25 Æ8l -36.87° A
8.64 + j11.52 Æ60l -36.87° V
2>3 Æ
32 cos (8000t + 90°) V
33.94 cos (5000t + 45°) V
5000 rad>s
223.6l26.57° mS
1 cos(8000t + 23.13°) A
1l23.13° A
60� V500 �j400 �
400 � j700 �
I
�
�
46.4 cos(5 * 104t + 34.46°) V
46.4l34.46° V
�20� A20 j2 ��j20 �
1 �
20 �
50 m
5l72° Æ
-j19.89 Æ
0.2 m
-19.89 Æ
90°
251,327.41 rad>sj40 Æ
40 Æ
-90°
89251_26_AppH 4/5/07 2:27 PM Page 828
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Answers to Selected Problems 829
9.47 20 A,
9.51
9.56
9.58
9.61
9.72 a) A,
b) 0.5
c) 112.5 mJ, 37.5 mJ
9.73 a) 0.5657
b) 1.5 A
9.77
9.81 a)
b) ,
c)
9.82
9.85 a) , ,, ,
,
b)
9.86 a) 0 A
b) 0.436 A
c) When the two loads are equal, more current isdrawn from the primary
Chapter 1010.2 a) 2404.16 W (abs), 2404.16 VAR (abs)
b) 155.29 W (abs), VAR (del)
c) W (del), VAR (del)
d) W (del), 845.72 VAR (abs)
10.3 a) Yes
b) Yes
10.13 a) 60 V(rms)
b) 300 W
-307.82
-1174.62-427.53
-579.56
0.42l0° A
I6 = 2.55l0° AI5 = 4.6l0° AI4 = 19.40l0° AI3 = 21.96l0° A
I2 = 2.04l0° AI1 = 24l0° A
Rx� ��Vm/2
Vo � (Vm/2) � IRx
Vo�IRx
IRx
I
Rx� 0
Vm/2 Vm
-j26.90 Æ
241 + j8 = 241.13l1.90° V-j32 Æ
247 + j7.25 = 247.11l1.68° V
800 + j600 Æ
5 cos (10,000t - 180°) A18.03 cos(10,000t - 56.31°)
11.31 cos(5000t - 45°) V
12 cos 5000t V
72 + j96 = 120l53.13° V
15.81l18.43° V
0.4 - j1.2 Æ 10.17 a)
b) Balances (80 W)
c) Balances (80 VAR)
10.21 a) 0.9 lagging, 0.43; 0.43 leading, ;0.57 leading,
b) 0.94 lagging, 0.343
10.22 a)
b) 0.9487 leading
10.41 a)
b) 16.875 mW
10.44 a)
b) 280 W
c)
10.45 a) 3240 W
b) 6480 W
10.49 a) 9 W
b) , 8 mH
c) 17.31 W, yes
d) 18.75 W
e) , 9 mH
f) Yes
10.50 a) 8 mH, , 18.26 W
b) Yes
c) Yes
10.58 160 W
10.59 a) 8
b) 250 W
10.66 a)
b)
c) Yes
10.67 a)
b) 1125 W
10.68 , 24 Æ36 Æ
= PM
2
PM - PL
PH =
V2(R1 + R2)R1R2
=
V2V2
PL
¢V2
PL-
V2
PM≤ ¢ V2
PM≤
PL =
V2
R1 + R2 PM =
V2
R2
28.8 Æ
28.8 Æ
31.62 Æ
30 Æ
20 Æ
9.72%
140l0° V(rms)
30 + j10 kÆ
72 - j24 = 75.89l -18.43° Æ
-0.82-0.9
-80 W(del), 60 VAR, 100 VAR
89251_26_AppH 4/5/07 2:28 PM Page 829
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
830 Answers to Selected Problems
Chapter 1111.1 a) abc
b) acb
11.2 a) Balanced, positive phase sequence
b) Balanced, negative phase sequence
c) Balanced, positive phase sequence
d) Balanced, negative phase sequence
e) Unbalanced, due to unequal amplitudes
f) Unbalanced, due to unequal phase angleseparation
11.8 ,,
11.9 a) 32.84 A(rms)
b) V(rms)
11.10 a) , ,
b) , ,
c) ,,
d) ,,
11.14 a)
b)
c)
11.15
11.16
11.24
11.25 a)
b) 519.62 V(rms)
11.41 197.29 W, 476.63 W
1833.46l22° VA
6120l36.61° VA
10.82l71.34° A(rms)
159.5l29.34° V(rms)
35.3l116.63° V(rms)
0.23l156.87° A(rms)
�
�39 � j33 �
1 � j3 �
�
�
VAN150�20
V (rms)
IaA
VCA = 8071.28l -154.37° VVBC = 8071.28l85.63° VVAB = 8071.28l -34.37° V
VCN = 4659.96l -124.37° VVBN = 4659.96l115.63° VVAN = 4659.96l -4.37° V
Vca = 8313.84l -150° VVbc = 8313.84l90° VVab = 8313.84l -30° V
IcC = 24l -136.26° AIbB = 24l103.74° AIaA = 24l -16.26° A
12,845.94
vCA = 13,799.25 cos (vt - 150°) VvBC = 13,799.25 cos (vt + 90°) VvAB = 13,799.25 cos (vt - 30°) V
11.42 a)
Thus,
b)
11.52 a)
b)
11.53 a)
b)
11.54 , so the voltage is below theacceptable level of 13 kV. Thus, when the load atthe substation drops to zero, the capacitor bankmust be switched off.
11.55 kW, kW
Chapter 1212.1 a)
b)
c)
12.2 a)
b)
12.5 a) 1.0
b) 0
c)
12.6 a) 52
b) 6.25
q
6)u(t - 6) - 5(t - 9)u(t - 9)3)u(t + 3) + 5(t - 3)u(t - 3) + 5(t -5(t + 9)u(t + 9) - 5(t + 6)u(t + 6) - 5(t +
(5t - 50)u(t - 10)(2.5t + 50)u(t + 20) - 2.5tu(t) +
(30 - 3t)t[u(t) - u(t - 10)]
a50 sin p
2 tbu(t) - a50 sin
p
2 tbu(t - 4)
8) - u(t - 12)]30t)[u(t) - u(t - 8)] + (-360 + 30t)[u(t -(120 + 30t)[(u(t + 4) - u(t)] + (120 -
PL(after) = 40.83PL(before) = 81.66
ƒ Vab ƒ = 12,548.8 V
50.14 mF
16.71 mF
1.2MW
1.2 MW
1.2 MVAR
1.70 MVA
45�
(a)
-4172.79 VAR
13(W2 - W1) = 13VLIL sin u = QT
30°)] = 2VLIL sin u sin 30° = VLIL sin u
W2 - W1 = VLIL[ cos (u - 30°) - cos (u +
89251_26_AppH 4/5/07 2:29 PM Page 830
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Answers to Selected Problems 831
12.7
12.13 a)
b)
c)
d)
e)
12.17 a)
b)
c) check
12.18 a)
b)
c) 2
d) check
12.24 a)
b)
c)
l5e-tt cosh t6 =
s 2
+ 2s + 2
s 2(s + 2)2
l5t sin bt6 =
2bs
(s 2
+ b2)2
l5t 56 =
120
s 6
= (-1)nl5t nf(t)6
So d
nF(s)ds
n = (-1)n
L
q
0-
t nf(t)e-st
dt
d 3F(s)
ds 3 =
L
q
0-
- t3f(t)e-st dt
d 2F(s)
ds 2 =
L
q
0-
t2f(t)e-st dt
So l5tf(t)6 = -
dF(s)ds
= -L
q
0-
tf(t)e-st dt
dF(s)ds
=
d
ds BL
q
0-
f(t)e-st dtR
-v2
s 2
+ v2
sv
s 2
+ v2
1
s 3
1s(s + a)
sinh u + s[cosh u]
(s 2
- 1)
1
s 2
v cos u + s sin u
s 2
+ v2
v
s 2
+ v2
1
(s + a)2
3>812.25 a)
b)
12.26
12.40 a)
b)
c)
d)
12.41 a)
b)
c)
d)
e)
12.42 c)
12.47 a)
b)
c)
d)
Chapter 13
13.4 a)
b) Zeros at and
; pole at 0- 1000 - j3000 rad>s-1000 + j3000 rad>s
5[s 2
+ 2000s + 107]s
f(0+) = 56, f(q) = 8
f(0+) = 11, f(q) = 0
f(0+) = 8, f(q) = 10
f(0+) = 18, f(q) = 0
d¿(t) + 5d(t) + 50e- 20tu(t)
36.87)]u(t)[50te- t cos(2t - 16.26°) + 20e- t cos(2t +
[(2t - 1.5t 2
+ 1)e- 2t ]u(t)
[10t - 5 + 10e- 2t cos (t + 53.13°)]u(t)
[10 - 40te- 2t+ 20e- 2t]u(t)
[40t - 8 + 16e- 10t]u(t)
[8 + 50e- 7t cos (24t + 16.26°)]u(t)
[5e- 2t+ 10e- 6t cos (8t - 53.13°)]u(t)
[10 + 5e- 2t- 8e- 3t
+ e- 5t]u(t)
[3e- t+ 6e- 2t
+ 9e- 3t]u(t)
200s 2
(s 2
+ 40s + 64)(s 2
+ 100)
l5t sin bt6 =
b
s 2
+ b2
= lb f(t)tr
=
L
q
0-
f(t)B -e-st
- tR dt
=
L
q
0-
f(t)B e-tu
- t2 qsR dt
=
L
q
0-
f(t)L
q
se-ut
du dt
=
L
q
0-
BL
q
sf(t)e-ut
duR dt
L
q
sF(u)du =
L
q
sBL
q
0-
f(t)e-
ut dtR du
89251_26_AppH 4/5/07 2:30 PM Page 831
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
832 Answers to Selected Problems
13.5 a)
b) Zero at 0; poles at and
13.9 a)
b)
c)
13.10 a)
b)
c)
d)
e)
13.15 a)
b) Compare solution at and tocircuit at and t = qt = 0
t = qt = 0
[5 - 6e- 5t+ 4e- 20t]u(t) V
- [25,000t + 10]e- 5000tu(t) mA
[5 * 105te- 5000t+ 100e- 5000t]u(t) V
-0.01(s + 7500)
(s + 5000)2
100(s + 104)
s 2
+ 104s + 25 * 106
10 k� 108/s � 0.01/s A106 A 4s � Vo
IL
�
�
(-8e- 500t+ 8e- 2000t)u(t) V
-12,000
s 2
+ 2500s + 106
2000 �
15 10�3
s0.8 106
s0.8s �
�
�
Vo
2000 �
0.8 106
s12 10�3 V
0.8s �
�
�
�
�Vo
�
�
A
-4000 rad>s -1000 rad>s25 * 106s
s 2
+ 5000s + 4 * 106
13.16 a)
b)
c)
13.26 a)
b)
c) Compare solutions at and tocircuit at and
13.27 a)
b) Initial values: 15 A, 12 A; final values: 4 A, 4 A
c)
13.35
13.36 a)
b) Initial value is 0, final value is 650 mA
c)
13.37 a)
b)
13.42 a)
b)
13.49 a) , no zeros, pole at
b) , zero at 0, pole at
c) , zero at 0, pole at -3 * 106 rad>ss
s + 3 * 106
-50 rad>ss
s + 50
-50 rad>s50s + 50
(56 - 108e- 2000t+ 52e- 3000t)u(t) V
6 * 104(s + 4000) + 96 * 106
s(s + 2000)(s + 3000)
(-51e- 200t+ 51e- 850t)u(t) mA
(51e- 200t- 51e- 850t)u(t) mA
(650 - 425e- 200t- 225e- 850t
)u(t) mA
276.25(s + 400)s(s + 200)(s + 850)
[5 - 5000te- 1000t- 5e- 1000t]u(t) mA
(4 + 27e- 2t- 19e- 3t)u(t) A
(4 - 27e- 2t+ 38e- 3t)u(t) A,
12s 2
+ 63s + 24s(s + 2)(s + 3)
15s 2
+ 15s + 24s(s + 2)(s + 3)
,
t = qt = 0t = qt = 0
(75 + 5e- 10,000t- 80e- 40,000t)u(t) mA
(256e- 40,000t- 4e- 10,000t)u(t) V
3.35e- 1000t cos (3000t - 26.57°)u(t) A
3(s + 2500)
s 2
+ 2000s + 107
0.02s �
40 �
0.06 V
Io
120/s V
270/s V�
��
�
� �
106
5s�
89251_26_AppH 4/5/07 2:31 PM Page 832
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Answers to Selected Problems 833
d)
t(ms)
500
400
300
200
100
�200
�1002.5
00 5 12.5 15 17.5 207.5 10
vo(V)
c) Voltage spikes in Problem 13.89 but not here
Chapter 1414.1 a)
b)
c) V,V,
14.2 a)
b)
14.12 a)
b)
14.13 a)
b)
14.21 a)
b)
c) 7.5
159.15 kHz
1 Mrad>s60 kÆ
4 kÆ
917.03 Hz
9.95 kÆ
1640.85 Hz
392.70 Æ
vo(3vc) = 15.81 cos(18,000t - 71.57°) Vvo(0.3vc) = 47.89 cos(1800t - 16.70°)vo(vc) = 35.36 cos(6000t - 45°)
0.3162l -71.57°0.9578l -16.70°, H( j3vc) =
H( jvc) = 0.7071l -45°, H( j0.3vc) =
954.93 Hz
13.90 a)
b)
-20.58e-1475pt+ 172.62 cos(120pt - 83.15°) V
t(ms)
200
100150
�150
50
�50
�200
�100
2.50
0 5 12.5 15 17.5 207.5 10
vo(V)
d) , no zeros, pole at
e) , no zeros, pole at
13.57
13.58
13.74
13.75 a)
b)
13.76 a)
b)
c)
13.83 a)
b)
c)
d)
e)
f)
g)
13.84 a)
b)
c)
d)
e)
f)
g)
13.89 a) ;
b)
c) V0 = 122.06l6.85° V(rms)
v0(0+) = 424.26 V
6.85°) V
v0 = 252.89e- 1475pt+ 172.62 cos (120pt +
+
30012s + 1475p
V0 =
1440p(122.9212s - 3000p12
(s + 1475p)(s 2
+ 14,400p2)
iL(0-) = iL(0+) = 35.36 A
i2(0-) = i2(0+) = 0 A
[-1.6 * 10- 3d(t)] - [7200e-2 * 106tu(t)] V
-0.6 e-2 * 106tu(t) A
0.6e-2 * 106tu(t) A
-0.6 A
0.2 A
0.6 A
0.8 A
20 V
4 V
16 V
32d(t) mA
0 V
20 V
80 V
11.68 cos(2000t + 30.96°) V
(35e- 2000t- 25e- 4000t)u(t) V
s(s + 9000)(s + 2000)(s + 4000)
4 cos(8000t - 161.57°) V
-16 * 104s
(s + 8000)(s + 16,000)
50 cos(8000t + 36.87°) V
(1 - e)e- t V
(e - 1)e- t V
-125 rad>s100s + 125
- 3 * 106 rad>s3 * 106
s + 3 * 106
89251_26_AppH 4/5/07 2:32 PM Page 833
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
834 Answers to Selected Problems
d)
e)
f)
g)
h)
14.22 a) ,
b) ,
c)
14.33 a)
b)
c) 16
d)
e)
f)
g)
h)
14.34 a) ,
b) ,
c)
14.43 a) ,
b) ,
c)
14.44 H, F,
14.45 63.7 times as large as the DTMF tones
Chapter 1515.4 a) ,
b)
�
� �
�
vo
�
�
vi
3.18 k�
5 nF
0.57 k�
kÆR2 = 3.18kÆR1 = 0.57
0.344|Vpeak|C = 0.057 mL = 0.225
|V1209Hz| = 0.344|Vpeak|
|V770Hz| = |V852Hz| = 0.948|Vpeak|
|V697Hz| = |V941Hz| = 0.707|Vpeak|
0.10 mF0.39 H
6.25 kHz
53.22 kHz46.97 kHz
101.32 mH254.65 Æ
79.58 kHz
1.31 Mrad>s8.25 Mrad>s1.234 MHz
7.75 Mrad>s
1.27 MHz
8 Mrad>s795.77 Hz
8.37 kHz7.57 kHz
16 mH8 kÆ
21.22 kHz
170.12 kHz
1068.89 krad>s148.90 kHz
935.55 krad>s 15.7 a) ,
b)
15.13 a) 1 H, 1 F,
b) 0.9 H, 0.11 nF, 3.6
c)
15.14 a)
b)
c) 5 k , 2 H, 0.2 nF
d)
e)
15.30 Hz, Hz,,
15.31 , ; if then
15.33 a) 4
b) dB-48.16
kÆRf = 4kÆRi = 1RH = 20.7 ÆRL = 784.6 Æ
RH = 821.64 ÆRL = 21.64 ÆfC2 = 49,037.85fC1 = 1291.4
2500s
s 2
+ 2500s + 25 * 108
2 H �
�
vo
�
�
vi
0.2 nF
5 k�
Æ
(1>Q)s
s 2
+ (1>Q)s + 1
1>L = 1>Q F
0.9 H
�
�
vo
�
�
vi
0.11 nF
3.6 k�
kÆ
0.04 Æ
vo
�
�
vi
�
�
�
�
23.29 k�
5.85 k�680 pF
kÆR2 = 23.29kÆR1 = 5.85
89251_26_AppH 4/5/07 2:33 PM Page 834
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Answers to Selected Problems 835
15.36 a) First stage: 208.05 nF, 30.44 nF; second stage:86.12 nF, 73.53 nF
b)
15.37 a) ,
b)
15.58 a),
b)
c)
15.59 Choose , . Then,, dB,
dB
15.60 Choose k , then ,nFC1 = 7.96
kÆR2 = 400ÆR1 = 100
|H( j>R2C1)| = 17.04|H( jv)|max = 20.01C = 39 nF
R2 = 100 kÆR1 = 11.1 kÆ
s 2
+ 64 * 106p2
s 2
+ 533.33ps + 64 * 106p2
�
��
��
�
265 �132.5 �
265 �
261 �
4.4 �
150 nF 150 nF
300 nF
vi
�
�
va
2C = 300 nF4.4 Æ, C = 150 nFR = 265 Æ, sR = 261 Æ, (1 - s)R =
���
�
900 �
1.81 k�vi
�
�va
25 nF 25 nF
1800.63 Æ900.32 Æ
��
�� �
�
�
�
2 k�2 �2 k�
208 nF
2 k�
v1
va
86 nF
30 nF 74 nF
Chapter 1616.1 a) ,
b) Hz, kHz
c) ,
d) for all k; for k even;
for k odd
V for all k;
for all k
e)
16.2 a)
b)
c)
16.3
16.10 a) 100 Hz
b) no
c) yes
d) yes
e) yes
f) , for all k, for k even,
for k oddbk =
80
p2k2 sin kp
4
bk = 0ak = 0av = 0
600p
aq
n = 2,4,6 cos(np>2) cos(nvot)
(n2- 1)
V
300p
+ 50 cos vot -
2Vm
p aq
n = 2,4,6,Á
1
1 - n2 cos nvot V
Vm
p+
Vm
2 sin vot +
2Vm
p B1 + 2 a
q
n = 1
1
1 - 4n2 cos nvotR V
4Vm
p a
q
n = 1,3,5,Á 1n
sin nvot V
50p
aq
n = 1 1n
¢sin np
4+ sin
np
2≤ cos nvot V
vb(t) = 18.75 +
200p
aq
n = 1,3,5,Á 1n
¢2 - cos np
3≤ sin nvot V
va(t) =
bkb = 0
akb =
50kpb sin
kp
4+ sin
kp
2r
bka =
200pkB2 - cos
pk
3R V
bka = 0aka = 0
avb = 18.75 Vava = 0
fob = 125foa = 11,111.11
vob = 785,398.16 rad>svoa = 69,813.17 rad>s
89251_26_AppH 4/5/07 2:34 PM Page 835
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
836 Answers to Selected Problems
16.11 a)
b) no
c) yes
d) no
16.18 a) where
b) 888.92 mV
16.27
16.28 a)
b) The 5th harmonic at is eliminated bythe bandreject filter whose center frequency is
16.32 a) 287.06 W
b) 300 W
c)
16.33 41.52 mW
16.36 a) 74.5356 V(rms)
b) 74.5306 V(rms)
16.37 a) 77.9578 V(rms)
b)
c) 46.1880 V(rms),
16.44 ,
,
16.45 a) 4000 W
b) 7.72 A
c) -10.57%
n = ;1, ;2, Á
Cn =
Im
n2p2 [2 cos(np>2) + np sin(np>2) - 2]
C0 =
Im
4
-0.0156%
-2.55%
-4.31%
50 krad>s50 krad>s
118.74 cos(70,000t - 171.70°) V
278.78 cos(30,000t + 174.64°) +
839.82 cos(10,000t - 1.19°) +
123.69°) + 17.83 cos(10,000t - 68.20°) V
214.66 cos (2000t - 26.57°) + 44.38 cos (6000t +
Akl -uk =
0.756k
¢ 2pk
- j1≤ mV, for odd k
aq
n = 1,3,5An cos(nvot - un )
p
6 rad>s 16.48 a)
b)
16.49 a)
00
100
200
300
400424
270
13590 67.5 54 45
500
4vo 5vo 6vo1vo 2vo 3vo
Ak
00
90
4vo 5vo 6vo1vo 2vo 3vo
�uk(deg)
�1
�1
�3�5�7
2
4
6
8
10�Cn�
n7531
�3�5�7
�135
135
�90
9045
�45
1 3 5 7n
uk�
00
1 2 3k
Ak (V)
4 5 6 7 8
2468
101214161820
0 1 2 3k4 5 6 7 8
0
45
90
135uk�
89251_26_AppH 4/5/07 2:34 PM Page 836
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
Answers to Selected Problems 837
b)
Chapter 17
17.2 a)
b)
c)
�15
0.20.10.0
0.30.40.5
1.00.90.80.70.6
�10 0v (rad/s)
�5 5 10 15
�F(v)�
At
2
2A
v2t b2 cos
vt
2+ vt sin
vt
2- 2 r
00
100
200
300
400
500
2 4�2�4�6
00
90
�90
21 43 5�1�3�5 �2�4�6
Cn
�un(deg)
17.3 a)
b) 0
c)
17.4 a)
b)
c)
d)
e)
17.19 a)
b)
17.22 a)
b) Yes, check the initial conditions and final values
17.28 a)
b) 5 V
c) 5 V
d)
e) Yes
17.32 166.67 cos(2500t + 90°) mA
(12.5e-t- 7.5e-5t)u(t) V
5e5tu(- t) + (12.5e-t- 7.5e-5t)u(t) V
20sgn(t) - 40e-50tu(t) V
F(v) : p[d(v - vo) + d(v + vo)]
t
2• sin[(v + vo)t>2]
(v + vo)(t>2)+
t
2• sin[(v - vo)t>2]
(v - vo)(t>2)
e-jvto
v - vo
a2+ (v - vo)2 +
v + vo
a2+ (v + vo)2
a
a2+ (v - vo)2 +
a
a2+ (v + vo)2
j48av(a2
- v2)
(a2+ v2)4
2(a2- v2)
(a2+ v2)2
t (s)
f(t)
10
�1.0
�0.4�0.2
0.00.2
0.60.81.0
50�5�10
�0.6�0.8
0.4
A
pvot 2[vot cos(vot>2) - 2 sin(vot>2)]
89251_26_AppH 4/5/07 2:35 PM Page 837
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.
838 Answers to Selected Problems
17.39 a)
b)
c)
d) 720 J
e) 28 J
v
00
1
2
3
4
5
2 4 6 8 10�2�4�6�8�10
Vo(v)
v
00
5
10
15
20
25
2 4 6 8 10�2�4�6�8�10
Vg(v)
15e-5t- 5e-25t
]u(t) + 10e5tu(- t) V f)
g)
Chapter1818.2 ; ; ;
18.4 ; ; ;
18.5 ; ; ;
18.11 ; ; ;
18.12 a) ; ;;
b) ; ;;
18.30 15.625
18.32 a)
b) 11.20 mW
c)
18.37 7.5 W
18.38 3.88 V
2.88 mW
28l180° V(rms)
a22 = -0.025a21 = -5 * 10-7 Sa12 = -25 Æa11 = -4 * 10-4
a22 = -0.025a21 = -5 * 10-7 Sa12 = -25 Æa11 = -4 * 10-4
h22 = 20 * 10-6 S
h21 = 40h12 = 1 * 10-4h11 = 1000 Æ
b22 = 50b21 = 20 mSb12 = 2 MÆb11 = 500
z22 = 22 Æz21 = 18 Æz12 = 18 Æz11 = 20 Æ
z22 = 80 Æz21 = 20 Æz12 = 20 Æz11 = 25 Æ
96.93%
95.95%
89251_26_AppH 4/5/07 2:35 PM Page 838
Electric Circuits, Eighth Edition, by James A. Nilsson and Susan A. Riedel.ISBN-13: 978-0-13-198922-1. © 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.
This material is protected by Copyright and written permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise.