Solution Manual + Answer Key

Solution Manual for Circuit Analysis Theory and Practice 5th

Edition by Allan H. Robbins

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Book Title: Circuit Analysis Theory and Practice

Edition: 5th Edition

Author: Allan H. Robbins

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Instructor’s Solutions Manual
to Accompany
Circuit Analysis: Theory and Practice, 5e

Allan H. Robbins
Wilhelm C. Miller

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A u s t r a l i a C a n a d a M e x i c o S i n g a p o r e S p a i n U n i t e d K i n g d o m U n i t e d S t a t e s
Instructor’s Solutions Manual to accompany: © 2013. 2007 Delmar, Cengage Learning
Circuit Analysis: Theory and Practice, 5th edition
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1 2 3 4 5 14 13 12
Contents

Con ten ts

Answers to Odd-Numbered Questions

Chapter 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
2 Voltage and Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
3 Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
4 Ohm’s Law, Power, and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
5 Series Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
6 Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
7 Series-Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
8 Methods of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
9 Network Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28
10 Capacitors and Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 34
11 Capacitor Charging, Discharging, and Simple Waveshaping Circuits . . . . . . . . . . . . . . . 38
12 Magnetism and Magnetic Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
13 Inductance and Inductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
14 Inductive Transients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49
15 ac Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54
16 R, L, and C Elements and the Impedance Concept . . . . . . . . . . . . . . . . . . . . . . . . . 59
17 Power in ac Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62
18 AC Series-Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 65
19 Methods of ac Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
20 ac Network Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
21 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 87
22 Filters and the Bode Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92
23 Transformers and Coupled Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 100
24 Three-Phase Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105
25 Nonsinusoidal Waveforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110
iv Contents

Answers to Even-Numbered Questions

Chapter 1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 117
2 Voltage and Current . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119
3 Resistance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121
4 Ohm’s Law, Power, and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124
5 Series Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 127
6 Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131
7 Series-Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 135
8 Methods of Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 139
9 Network Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144
10 Capacitors and Capacitance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 150
11 Capacitor Charging, Discharging, and Simple Waveshaping Circuits . . . . . . . . . . . . . . 155
12 Magnetism and Magnetic Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 161
13 Inductance and Inductors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 165
14 Inductive Transients . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 168
15 ac Fundamentals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 173
16 R, L, and C Elements and the Impedance Concept . . . . . . . . . . . . . . . . . . . . . . . . 180
17 Power in ac Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
18 ac Series-Parallel Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 186
19 Methods of ac Analysis . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194
20 ac Network Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 200
21 Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 210
22 Filters and the Bode Plot . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 216
23 Transformers and Coupled Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 226
24 Three-Phase Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 230
25 Nonsinusoidal Waveforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 235

Lab Manual Answers

Lab 1 Voltage and Current Measurement and Ohm’s Law . . . . . . . . . . . . . . . . . . . . . . . . 243
Lab 2 Series dc Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251
Lab 3 Parallel dc Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 263
Lab 4 Series-Parallel dc Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 269
Lab 5 Potentiometers and Rheostats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 277
Lab 6 Superposition Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 283
Lab 7 Thévenin’s and Norton’s Theorems (dc) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293
Contents v

Lab 8 Capacitors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 299

Lab 9 Capacitor Charging and Discharging . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 367
Lab 10 Inductors in dc Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 319
Lab 11 The Oscilloscope (Part 1) Familiarization and Basic Measurements . . . . . . . . . . . . . . . 333
Lab 12 Basic ac Measurements: Period, Frequency, and Voltage (The Oscilloscope Part 2) . . . . . . . 341
Lab 13 ac Voltage and Current (The Oscilloscope Part 3, Additional Measurement Techniques) . . . 347
Lab 14 Capacitive Reactance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 357
Lab 15 Inductive Reactance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 365
Lab 16 Power in ac Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 375
Lab 17 Series ac Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 383
Lab 18 Parallel ac Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391
Lab 19 Series-Parallel ac Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 397
Lab 20 Thévenin’s and Norton’s Theorems (ac) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 481
Lab 21 Series Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 415
Lab 22 Parallel Resonance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 425
Lab 23 RC and RL Low-Pass Filter Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433
Lab 24 RC and RL High-Pass Filter Circuits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 445
Lab 25 Bandpass Filter . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 455
Lab 26 Voltage, Current, and Power in Balanced Three-Phase Systems . . . . . . . . . . . . . . . . . 461
Lab 27 The Iron-Core Transformer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 471
Lab 28 Mutual Inductance and Loosely Coupled Circuits . . . . . . . . . . . . . . . . . . . . . . . . 483
Chapter

1
Introduction

1.3 Converting Units

1. a. 27 min × (60 s/min) = 1620 s 5. 15 parts 3600 −s
× = 4500 parts/h
b. 0.8 h × (3600 s/h) = 2880 s 12 −s h
7. 27 −−
mi 1.609 km 1−−−
gal
c. ⎛ 3600 s ⎞ ⎛ 60 s ⎞ × × = 11.5 km/liter
⎜2h− × h ⎟ + ⎜3 −−−
min ×
min ⎟⎠
+ 47 s = 7427 s −−−
gal −−
mi 3.785 liters
⎝ − ⎠ ⎝ −−−
9. 18−̊ 1 rev 60 −s
d. 35 hp × (746 W/hp) = 26 110 W × × = 150 rpm
0.02 −s 360−̊ 1 min
e. 1 hp 11. 60 mi 1− h
1827 W × = 2.45 hp × 500 −s × = 8.33 mi
746−−
W h− 3600 −s
f. 23 rev. × 360˚/rev = 8280˚ 13. 2000 −−yd 0.914 m 1 h−
× × = 0.508 m/s
3. a. 1m h− −−
yd 3600 s
1.2 m × 70 cm × = 0.84 m2 15.
100 −−
cm 3 km 8
× h+
5 km
× 1.25 h +
4 km 12
× h = 7.45 km
h 60 h h 60
b. 1⎛ 1m ⎞
cm ×
25 −− (0.5 m) = 0.0625 m2
2 ⎜⎝ cm ⎟⎠
100 −−
17. 2 km 15 5 km 18 h 2.5 km
× h+ + h+ ×t3 = 2.85km
h 60 h 60 h
c. ⎛ 1m ⎞⎛ 1m ⎞
cm ×
⎜10 −− cm ×
⎟ ⎜25 −− Thus, t3 = 0.34h = 20.4 minutes.
⎝ 100 −−
cm ⎠⎝ 100 cm ⎟⎠
−−
19. $0.43 60 min
⎛ 1m ⎞ 3
Machine 1: × = $25.80/h
cm × min h
⎜80 −− ⎟ = 0.02 m
⎝ 100 −−
cm ⎠ $200
d. 3
Machine 2: = $25.00/h (This one is cheaper to
8h
4π ⎛ 2.54 −−
cm 1m ⎞ operate. Buy it.)
⎜10−− × × = 0.0686 m3
cm⎟⎠
in
3 ⎝ −−
in 100 −−

1.4 Power of Ten Notation

21. a. 8.675 × 103 25. a. (4 × 103)(5 ×10–2)2 = (4 × 103)(25 × 10–4)
b. 8.72 × 10 –3 = 100 × 10–1 = 10
c. 1.24 × 103 b. (4 × 103)(–5 × 10–2)2 = (4 × 103)(25 × 10–4) = 10
d. 3.72 × 10–1 c. (6 × 10)2 36 × 102
= = 3.6 × 103
e. 3.48 × 102 (10 × 10−1) 1
f. 2.15 × 10–7 d. (50)−2(2.5 × 106)(6 × 103) (2.5)(6) × 106 × 103
g. 1.47 × 101 −1 2
=
3
(1 × 10 )(2 × 10 ) (5 × 10)2(1 ×103)(4 × 10−2)
23. a. 2
1.25 × 10
= 1.25 × 10−1
1 × 103 15 × 109 15 × 109
= =
b. 8 × 104 25 × 102 × 103 × 4 × 10−2 100 × 103
= 8 × 107
1 × 10−3 = 0.15 × 106 = 15 × 104
c. 3 × 104 (−0.027) ⁄3(−0.2)2 (−0.3)(−2 × 10−1)2
1
e.
= 2 × 104 × 10−6 = 2 × 10−2 =
(24)˚ × 10−3 1 × 10−3
6
1.5 × 10
d. (16)(21.8) × 10−7 × 106 (−0.3)(4 × 10−2)
= 2.05 × 104 = = −12
(14.2)(12) × 10−5 1 × 10−3

1
2 Chapter 1 Introduction

27. i. (8.42 × 102)(1.4 × 10–3) = 11.79 × 10–1 = 1.179 33. 3.47 × 105 km
t= = 1.16 s
ii. 299 792.458 km/s
3.52 × 10−2
= 0.445 × 101 = 4.45
7.91 × 10−3
35. 3.73 × 104 m3 3600 s 1 liter
× ×
Direct computation for these examples is less work.
1s 1h 1 × 10−3 m3

29. 6.24 × 1018 = 13.4 × 1010 liters/h

31.
6.24 × 1018 electrons
× 10.03 × 103 s
1s
= 62.6 × 1021 electrons

1.5 Prefixes
37. a. kilo, k 51. 5000 km
Radio signal: t = = 16.68 ms
b. mega, M 299 792.458 km/s
c. giga, G 5000 × 103 m
Telephone signal: t = = 33.33 ms
d. micro, l 150 m/ls
e. milli, m ∴ Radio signal arrives first by 16.65 ms.
f. pico, p 53. a. V 50 V
R= = = 2.083 kΩ. (Since V and I are
39. a. 1.5 ms I 24 mA
specified as exact, you can use as many digits as
b. 27 ls
you like.)
c. 350 ns
b. Vmax 50.1 V
41. a. 150 × 103 V; 0.15 × 106 V Rmax = = = 2.096 kΩ = 2.10 kΩ
Imin 23.9 mA
b. 0.33 × 10–3 W; 33 × 10–5 W
when rounded to 3 digits
43. a. 330 V + 150 V + 200 V = 680 V
Vmin 49.9 V
b. 60 W + 100 W + 2.7 W = 162.7 W Rmin = = = 2.071 kΩ = 2.07 kΩ
Imax 24.1 mA
45. 1500 W = 1.5 × 103 W = 1.5 kW
when rounded to 3 digits.
47. I3 = I1 + I2 + I4 = 12 A + 150 A + 25 A = 187 A
The actual value of R lies somewhere between
49. 39 mmfd = 39llF = 39 × 10−6 ×10−6 = 39 × 10−12F 2.07 kΩ and 2.10 kΩ.
= 39 pF 55. See CD in the back of the book.

1.6 Circuit Diagrams

57. Same as Figure 1–7(a) of the text.
Chapter

2
Voltage and Current

2.1 Atomic Theory

1. There are of the order of 1023 free electrons per 3. Q1Q2
cm3 at room temperature in copper. Original: F1 = k
r12
3 3 6 3
a. 1 m = (100 cm) = 10 cm . Thus, the number of k(2Q1)(3Q2) (2)(3) ⎡ Q1Q2 ⎤
electrons is New: F2 = 2
= 2 ⎢
k 2 ⎥ = 24 F1
⎛ r1 ⎞ ⎛ 1 ⎞ ⎣ r1 ⎦
1023 electrons ⎜ ⎟ ⎜ 2⎟
N= × 106 −−
cm3 = 1029 electrons ⎝2⎠ ⎝ ⎠
3
−−
cm ∴ Force increases by a factor of 24.
b. Volume = 5. a. It has a lot of free electrons. This results from
having few (e.g., 1) electrons in its valence shell
πd2 π
× l = (0.163 cm)2(500 cm) = 10.4 cm3 b. Inexpensive and easily formed into wires.
4 4
c. Has a full valence shell. Therefore, no free electrons.
1023 electrons
N= cm3
× 10.4 −− d. The electrical force is so great that electrons are
3
−−
cm torn from their parent atoms. This movement of
electrons constitutes a current. We see the effect as
= 10.4 × 1023 electrons a lightning discharge.

2.2 The Unit of Electrical Charge: The Coulomb

7. a. 9 × 109(1 × 10−6 C)(7 × 10−6 C) 11. 9 ×109Q1(5 Q1)
F= 0.02 N =
(10 ×10−3)2 (0.5)2
= 630 N (repulsive) 0.02 = 180 × 109 Q21
b. 9 × 109(8 × 10−6)(4 × 10−6) ∴ Q1 = 0.333 lC and Q2 = 1.67 lC, both (+) or
F=
(0.12)2 both (–).
= 20 N (attractive) 13. 19 × 1013 electron × 1.6 × 10–19 coulomb/electron
= 30.4 lC
c. 9 × 109(1.602 × 10−19)2 15. Q1 = – (14.6 × 1013 × 1.60 × 10–19) = – 23.4 lC
F=
(12 × 10−8)2 Q2 = 1.3 lC
= 1.60 × 10–14 N (replusive) Qfinal = Qinitial + Q1 + Q2
d. 9 × 109(1.602 × 10−19)2 5.6 lC = Qinitial – 23.4 lC + 1.3 lC
F=
(5.3 × 10−11)2 ∴ Qinitial = 27.7 lC (positive)
= 8.22 × 10–8 N (attractive)
e. Neutron is uncharged ∴ F = 0

9 × 109(4 × 10−6)Q2
9. 180 N = ∴ Q2 = 2lC
(2 × 10−2)2 (Attractive)

2.3 Voltage
17. W 360 J 21. W = QV = (0.5 × 10–6 C)(8.5 × 103 V) = 4.25 mJ
V= = = 24 V
Q 15 C
23. W 57 J
19. W
V= =
1200 J
= 2400 V Q= = = 4.75 C
Q 0.5 C V 12 V

3
4 Chapter 2 Voltage and Current

2.4 Current
25.
I=
Q 250 lC
= = 50 mA
33. 47 × 1019 electrons
t 5 ms Q= = 75.3 C
6.24 × 1018 electrons/C
27. Q = It = (16.7 mA)(20 ms) = 334 lC W 1353.6 J
V= = = 18.0 V
29. Q = (93.6 × 1012)(1.6 × 10–19) = 15 lC Q 75.3
Q 15 × 10−6 C Q 75.3 C
I= = = 0.966 A
I= = = 3 mA t 78 s
t 5 × 10−3 s
31. At t = 0, q0 = 20 C. At t = 1 s, q1 = 100 C.
∆q 100 C − 20 C 80 C
I= = = = 80 C/s = 80 A .
∆t 1s 1s

2.5 Practical DC Sources

35. a. ET = 1.47 + 1.61 + 1.58 = 4.66 V
b. ET = 1.47 + 1.61 – 1.58 = 1.50 V
37. capacity 1400 mAh
Life = = = 50 h
drain 28 mA
39. From Figure 2-15, capacity at 5˚C is 90% of its value
MAX at 25˚C. Therefore, capacity = 0.9 Max = 81
Ah. Thus, Max = 81/0.9 = 90 Ah. At –15˚C, capac-
ity = 0.65 Max = 0.65 (90) = 58.5 Ah. Thus, life ≈
58.5 Ah/5 A = 11.7 h.

2.6 Measuring Voltage and Current

41. BBoth 45. Meter 1: Reading is 7 A, Meter 2: Reading is – 7 A
4. The voltmeter and ammeter are interchanged.

2.7 Switches, Fuses, and Circuit Breakers

47. When a fuse “blows,” it becomes an open circuit
with source voltage across it. The voltage rating
tells you how much voltage you can use the fuse
with so that it does not arc over when it blows.
Chapter

3
Resistance

3.1 Resistance of Conductors

1. a. (2.825 × 10−18 Ω ⋅ m)(100 m) 5.
R=
qL
L=
RA
R= −3 2
= 3.60 Ω A ρ
π(0.5 × 10 m)
b. −8 (10.3 Ω)(π)(0.40 × 10−3 m)2
2.825 × 10 Ω ⋅ m)(100 m) =
R= = 0.900 Ω 1.723 × 10−8 Ω ⋅ m
π(1.0 × 10−3 m)2
L = 300 m
c. (2.825 × 10−8 Ω ⋅ m)(100 m)
R= = 36.0 kΩ L = 986 ft
π(0.005 × 10−3 m)2 7. qL RA
R= q=
d. (2.825 × 10−8 Ω ⋅ m)(100 m) A L
R= = 36.0 mΩ
π(0.05 × 10−2 m)2 (3.0 Ω)(π)(0.25 × 10−3 m)2
=
3. ⎛ 0.3048 m ⎞ 6.00 × 10−2 m
L = 250 ft = 250 ft ⎜ ⎟ = 76.2 m
⎝ 1 ft ⎠ = 982 × 10−8 Ω ⋅ m
qL (1.723 × 10−8 Ω ⋅ m)(76.2 m)
A= = Resistivity is less than q for carbon.
R 0.02 Ω
9. qL R⋅A
= 6.56 × 10−5 m2 R= q=
A L
⎛ 2.54 × 10−2 m ⎞
w = 0.25 inch = 0.25 inch ⎜ ⎟ (32 Ω)(π)(0.75 × 10−3 m)2
1 inch =
⎝ ⎠ 2500 m
= 0.00635 m
= 2.26 × 10−8 Ω ⋅ m
−5 2
A 6.56 × 10 m This alloy is not as good a conductor as copper.
h= = = 0.0103 m = 0.407 inch
w 0.00635 m

3.2 Electrical Wire Tables

11. ⎛ 16.2 Ω ⎞ 13. From Table 3-2, AWG 22 can handle about 2 A of
AWG 22 : R = (300 ft) ⎜ ⎟ = 4.86 Ω current. Since AWG 19 is 3 sizes larger, it should
⎝ 1000ft ⎠ handle about 4 A of current.
⎛ 8.05 Ω ⎞
AWG 19: R = (300 ft) ⎜ ⎟ = 2.42 Ω Since AWG 30 is ten sizes smaller than AWG 20
⎝ 1000 ft ⎠ (which can handle 3 A), AWG 30 can handle about
Diameter of AWG 19 is 1.42 times the diameter of 0.30 A.
AWG 22. 15. ⎛ 415 Ω ⎞
R=⎜ ⎟L = 550 Ω
1 ⎝ 1000 ft ⎠
The resistance of AWG 19 is the resistance of
2 ⎛ 1000 ft ⎞
AWG 22. L = (550 Ω) ⎜ ⎟ ≈ 1330 ft ≈ 405 m
⎝ 415 Ω ⎠

3.3 Resistance of Wires—Circular Mils

17. a. D = 0.016 in = 16 mil c. A = (0.25in)(6.0in)
2
A = 16 = 256 CM = (250mil)(6000mil)A

b. = 1.50 × 106sq mil

⎛ 1 in ⎞
D = 2.0 mm = 2.0 mm ⎜ ⎟ = 0.0787 in A = 1.91 × 106CM
⎝ 25.4 mm ⎠
= 1910 MCM
= 78.4 mil
A = 78.42 = 6200 CM

5
6 Chapter 3 Resistance

19. a. ⎛ CM ⋅ Ω ⎞ 27. a. Positive temperature coefficient since as

⎜ ⎟
qL ⎝1036 ft ⎠ (400 ft) temperature increases, resistance also increases.
R= = = 16.2 Ω
A 256 CM b. R2 = R1[1 + α1(T2 − T1)]
b. ⎛ CM ⋅ Ω ⎞ ⎛ R2 ⎞ ⎛ ⎞
⎜1036 ⎟ ⎜ − 1⎟ ⎜ 25 Ω − 1⎟
⎝ ft ⎠ (400 ft)
⎝ R1 ⎠ ⎝ 20 Ω ⎠
R= = 0.668 Ω
6200 CM α1 = = = 0.00385(˚C)−1
T2 − T1 85˚C − 20˚C
c. ⎛ CM ⋅ Ω ⎞
⎜1036 ft ⎟ c. Temperature intercept:
⎝ ⎠ (400 ft)
R= = 2.17 × 10−3 Ω 20˚ − T 85˚ − 20˚
1.91 × 106 CM =
20 Ω 25 Ω − 20 Ω
21. a. qL 20 Ω(85˚ − 20˚)
R= T=− + 20˚ = −240˚C
A 25 Ω − 20 Ω
⎛ CM ⋅ Ω ⎞ At T = 0˚C:
⎜10.36 ft ⎟
⎝ ⎠ (200 ft) 100˚ − (−240˚)
A= = 4144 CM R= (20 Ω) = 18.5 Ω
0.5 Ω 20˚ − (−240˚)
⎛ π sq mil ⎞
A = 4144 CM ⎜ ⎟ = 3255 sq mil At T = 100˚C:
⎝ 4 CM ⎠ 100˚ − (−240˚)
b. D=√ ⎯⎯A = 64.4 mil = 0.0644 in R= (20 Ω) = 26.2 Ω
20˚ − (−240˚)
23. D = 0.040 in = 40 mil 29. 260˚ − (−2270˚)
A = D2 = 1600 CM R= (15.2 Ω) = 16.8 Ω
a. 20˚ − (−2270˚)
⎛ π sq mil ⎞ 31. 70 Ω
A = 1600 CM ⎜ ⎟ = 1260 sq mil m= = 0.5 Ω/˚C
⎝ 4 CM ⎠ 140˚ C
b. qL R ⋅ A (12.5 Ω)(1600 CM) R = (0.5 Ω/˚C)T + 130 Ω
R= L= = CM ⋅ π ⎞
= 1930 ft
A ρ ⎛ −130 Ω
⎜10.37 ft ⎟ T= = −260˚C
⎝ ⎠ 0.5 Ω/˚C
25. At T = –30˚C:
−30˚ − (−236˚)
R= (50 Ω) = 40.2 Ω
20˚ − (236˚)
At T = 0˚C:
0˚ − (−236˚)
R= (50 Ω) = 46.1 Ω
20˚ − (−236˚)
At T = 200˚ C:
200˚ − (−236˚)
R= (50 Ω) = 85.2 Ω
20˚ − (−236˚)
3.5 Types of Resistors
33. a. Rab = 10 kΩ Rbc = 0
b. Rab = 8 kΩ Rbc = 2kΩ
c. Rab = 2kΩ Rbc = 8 kΩ
d. Rab = 0 Rbc = 10 kΩ

3.6 Color Coding of Resistors

35. a. 15 × 104 ± 10% = 150 kΩ ± 10%= 150 kΩ ± 15 kΩ c. 47 × 106 ± 5% = 47 MΩ ± 5% = 47 MΩ ± 2.4 MΩ
b. 28 × 0.1 ± 5% = 2.8 Ω ± 5% d. 39 × 100 ± 5% = 39 Ω ± 5%
= 2.8 Ω ± 0.14 Ω = 39 Ω ± 2Ω
(with a reliability of 0.001%) (with a reliability of 0.1%)

3.7 Measuring Resistance—The Ohmmeter

37. Connect the ohmmeter between the two terminals ohmmeter indicates an open circuit, the light bulb is
of the light bulb. If the resistance is measured to be burned out.
low (about 0 Ω), the bulb is functional. If the
Answers to Odd-Numbered Questions 7

39. AWG 24 has a resistance of 25.7 Ω/1000 ft

(Table 3-2)
Measure the resistance between the two ends of the
wire. Calculate the approximate length (in feet)
from:
R
L= (L−in feet)
0.0257

3.8 Thermistors
41. a. At T = 20˚C, R ≅ 380 Ω c. The thermistor has a negative temperature
b. At T = 40˚C, R ≅ 150 Ω coefficient since resistance decreases as
temperature is increased.

3.11 Conductance
43. a. 1 1 45. ⎛ 1 ft ⎞ ⎛ 104 Ω ⎞
G= = = 4.0 S R = (1000 m) ⎜ ⎟⎜ ⎟ = 341 Ω
R 0.25 Ω ⎝ 0.3048 m ⎠ ⎝ 1000 ft ⎠
b. 1 1
G= = 2.0 mS G= = 2.93 mS
500 Ω 341 Ω
c. 1
G= = 4.0 lS
250 kΩ
d. 1
G= = 0.08 lS
12.5 MΩ
Chapter

4
Ohm’s Law, Power, and Energy

4.1 Ohm’s Law

1. a. E 40 V b. At 40˚C, (using the method of Chapter 3),
I= = =2A ⎛ 274.5 ⎞
R 20 Ω R=⎜ ⎟ (20.8 Ω) = 22.43 Ω
b. 35 mV ⎝ 254.5 ⎠
= 7.0 A
5 mΩ 48 V
Thus, I = = 2.14 A
c. 200 V 22.43 Ω
= 5 mA
40 kΩ 17. ᐉ (2.825 × 10−8)(100 m)
R=q = = 14.39 Ω
d. 10 V A π
= 4lA (0.5 × 10−3 m)2
2.5 MΩ 4
e. 7.5 V
= 3 mA E = I R = (200 × 10−3 A)(14.39 Ω) = 2.88 V
2.5 × 103 Ω 19. This is a linear graph. Therefore, choose any V-I
f. 12 × 103 V point, for example, 80 V and 20 A. Thus,
= 6 mA V 80 V
2 × 106 Ω R= = =4Ω
I 20 A
3. a. V = I R = (1 × 10−3 A)(40 × 103 Ω) = 40 V 21. 100 V ⎛ 100 V ⎞
b. −6 3 Iold = Inew = 2 Iold = 2 ⎜ ⎟
(10 × 10 A)(30 × 10 Ω) = 0.3 V R ⎝ R ⎠
−3
c. (10 × 10 A)(4 × 104 Ω) = 400 V Also: Inew =
Enew
2R
d. (12 A)(3 × 10−2 Ω) = 0.36 V
2(100 V) Enew
5. E 120 V ∴ =
R= = = 96 Ω R 2R
I 1.25 A
Cancel R. ∴ Enew = 400 V
7. V = I R = (50 × 10−3 A)(560 Ω) = 28 V
23. πd2 π
9. I = G E = (0.2 S)(30 V) = 6 A A= = (0.1 × 10−3)2 = 0.7854 × 10−8 m2
4 4
11. V 33 V
R= = = 2.2 kΩ q = 99.72 × 10−8 Ω−m
I 15 mA
∴ Color code is Red, Red, Red qᐉ (99.72 × 10−8)(25)
R= = = 3174 Ω
13. From its color code, R = 22 Ω. Fuse blows at 1 A. A 0.7854 × 10−8
This corresponds to V = I R = (1 A)(22 Ω) = 22 V. 12 V
I= = 3.78 mA
15. ᐉ = (800 t)(3 inches/t) = 2400 in = 200 ft. 3174 Ω
From the wire table (Chapter 3), resistance is 104
Ω/1000 ft. Thus, R = (0.2) (104 Ω) = 20.8 Ω at
20˚C.
a. 48 V
At 20˚ C, I = = 2.31 A
20.8 Ω

4.2 Voltage Polarity and Current Direction

25. 45 V 90 V
+ +
(a) (c)

(b) (d)
4A 7A

8
Answers to Odd-Numbered Questions 9

4.3 Power
27. W 723 J 37. P 445 W
P= = = 3.19 J/s = 3.19 W I= = = 37.9 A
t 227 s V 12 V
29. P = EI = (12 V)(3 A) = 36 W 39. P = (3.56 hp)(746 W/hp) = 2656 W
31. P=I R2 41. V=√ ⎯⎯⎯
PR = √⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯
(2 W)(270 Ω = 23.2 V

⎯ √
√ ⎯⎯⎯
P 1200 W P 2W
I2 = = = 200 I = R = 270 Ω = 86.1 mA
R 6Ω
∴I=√ ⎯⎯⎯
⎯ = 14.1 A
200
43. Vmin = 95 V; Vmax = 105 V
33. V2
P= ∴ V2 = PR (95 V)2
R ∴ Pmin = = 361 W
25 Ω
V2 = (752 W)(3 Ω) = 2256
V = 47.5 V (105 V)2
Pmax = = 441 W
35. R = 10 000 Ω 25 Ω

⎯ √
√ ⎯⎯⎯⎯
P 0.25 W
I = R = 10 000 Ω = 5 mA

P 0.25 W
V= = = 50 V
I 0.005 A

4.4 Power Direction Convention

45. a. P = (12 V)(4 A) = 48 W → c. P = (8V)(16 A) = 128 W ←
b. P = (15 V)(2 A) = 30 W ← d. P = (30 V)(8 A) = 240 W →

4.5 Energy
47. a. W = Pt = (40 W) (9 h × 3600 s/h) 51. Per day: W = (200 W × 24 h) + (4000 W)(24 h) +
W = Pt = 1.296 × 106 W– s = 1.296 × 106 J (3600 W)(12 h) = 144 kWh
b. W = (40 W)(9 h) = 360 Wh Per year: WT = 365 × 144 kWh = 52 560 kWh
c. Cost = (0.360 kWh)($0.08/kWh) = 2.88 cents Cost at $0.10/kW = $5256
49. ⎛5⎞ 53. It takes 12 (1.75 min) = 21 minutes to toast the loaf.
W = (900) ⎜ ⎟ + (120)(8)(1.7) +
⎝ 60 ⎠ ⎛ 21 ⎞
W = (1100 W) ⎜ ⎟ = 385 Wh = 0.385 kWh
⎛ 36 ⎞ (120)2 ⎛ 24 ⎞ ⎝ 60 ⎠
(1100) ⎜ ⎟ + = 2.387 kWh
⎝ 60 ⎠ 288 ⎜⎝ 60 ⎟⎠ Cost = (0.385)($0.13) = $0.05 = 5 cents

Cost = (2.387)($0.11) = $0.26

4.6 Efficiency
55. Pout = gT Pin = (0.76)(8640 W) = 6566 W
Pout 50 kW
Pin = = = 51.5 kW 6566 W
g 0.97 ∴ Pout = = 8.8 hp
746 W/hp
57. Pout = Pin – Ploss = 1100 W – 190 W = 910 W Pm
63. 1600 W g1 0.75 1100 W
Pout 910 W 1100 W 1467
η= × 100% = × 100% = 82.7% Pm = = 1467 W = = 1.97 hp
Pin 1100 W 0.75 746
65. Wout = Pt = (35 kW)(24 h) = 840 kWh
59. Pin = (120 V)(15 A) = 1800 W
Wout 840 kWh
Pout = (0.89)(1800 W) = 1602 W Win = = = 1527 kWh
g 0.55
Pout = 1602 W/(746 W/hp) = 2.15 hp Cost = (1527 kWh)($0.09/kWh) = $137.45

61. gT = g1 × g2 = (0.95)(0.80) = 0.76

Pin = (48 V)(180 A) = 8640 W
10 Chapter 4 Ohm’s Law, Power, and Energy

4.7 Nonlinear and Dynamic Resistances

67. a. 25 V c. Because, for this resistor, R varies with current.
From the graph, I = 2.5 A ∴ Rdc = = 10 Ω
2.5 A
b. 60 V
From the graph, I = 4.5 A ∴ Rdc = = 13.3 Ω
4.5 A

4.8 Computer-Aided Circuit Analysis

69. Use the following circuit. Set the source voltage 73. Following the procedure outlined in the text, build
and resistance as shown. Click the power ON/OFF the circuit on the screen. Double-click the default
switch and read the ammeter. voltage value beside the V1 source symbol and
change it to 12.9 V, then click OK. Similarly, set
V2 to 11.6 V. Double-click the 1k default
resistance value and set it to 0.12 ohm. Click the
New Simulation Profile icon, enter a name, click
Create, select Bias Point Analysis and click OK.
Click the Run icon and when simulation is
complete, close the inactive window that appears.
Click the Bias Point Current Display icon to read
71. Click the power ON/OFF switch to the ON posi- the current—see below.
tion, then operate switches A and B.

75. See below.

Chapter

5
Series Circuits

5.2 Kirchhoff’s Voltage Law

1. a. V = +(3 A)(10 Ω) = +30 V 5. a. V1 = 33 V – 16 V – 10 V = 7 V
b. V = –(6 A)(15 Ω) = –90 V b.
V2 =
12 W
=4V
3A
3. a. V = +(3 A)(15 Ω) = +45 V
V1 = 6 V – 9 V + 3 V + 4 V = 4 V
b. V = +(–4 A)(15 Ω) = –60 V
7. 36 W
c. V = +(6 A)(15 Ω) = +90 V V3 = = 12 V
3A
d. V = +(–7 A)(15 Ω) = –105 V V4 = 24 V – 10 V – 12 V = 2 V

5.3 Resistors in Series

9. a. RT = 3 kΩ + 2 kΩ + 5 kΩ = 10 kΩ b. VR1 = (300 W)(0.04 A) = 12 V
b. RT = 360 kΩ + 580 kΩ + 2000 kΩ VR3 = (250 W)(0.04 A) = 10 V
RT = 2940 kΩ c. V = 12 V + 4 V + 10 V = 26 V
RT = 2.94 MΩ 17. a. VR2 + VR3 = 16 V − 7.5 V = 8.5 V
c. R = 3900 Ω 4.7 kΩ
VR2 = (8.5 V) = 4.81 V
RT = 6(3900 Ω) = 23 400 Ω = 23.4 kΩ 4.7 kΩ + 3.6 kΩ
RT = 200 Ω + 400 Ω + 1000 Ω + 50 Ω = 1650 Ω 3.6 kΩ
11. a. VR3 = (8.5 V) = 3.69 V
4.7 kΩ + 3.6 kΩ
10 V
I= = 6.06 mA b. 8.5 V
1.65 kΩ I= = 1.02 mA
(4.7 kΩ + 3.6 kΩ)
b. RT = 1.2 kΩ + 3.3 kΩ + 5.6 kΩ + 0.82 kΩ +
c. 7.5 V
2.2 kΩ + 0.33 kΩ + 4.7 kΩ = 18.15 kΩ R1 + + 7.32 kΩ
1.02 mA
300 V
I= = 16.5 mA 19. a. RT = 120 Ω + 39 Ω + 78 Ω + 220 Ω = 457 Ω
18.15 kΩ
b. 36 V
I= = 0.0788 A = 78.8 mA
13. a.
I=√ ⎯⎯⎯⎯⎯
⎯⎯P = √ 100 mW
= 10 mA 457 Ω
R 1 kΩ
c. V1 = (120 Ω)(0.0788 A) = 9.45 V
b. 130 V
RT = = 13 kΩ V2 = 3.07 V
10 mA
V3 = 6.14 V
c. R = 13 kΩ – (3 kΩ + 4 kΩ + 1 kΩ) = 5 kΩ V4 = 17.33 V
d. V3 kΩ = (3 kΩ)(10 mA) = 30 V d. VT = 9.45 V + 3.07 V + 6.14 V + 17.33 V
V4 kΩ = 40 V VT = 35.99 V ≅ E
V1 kΩ = 10 V e. P1 = (0.0788 A)2(120 Ω) = 0.745 W
VR = 50 V P2 = 0.242 W
e. PT = (130 V)(10 mA) = 1300 mW P3 = 0.484 W
P3 kΩ = (10 mA)2(3 kΩ) = 300 mW P4 = 1.365 W
P4 kΩ = 400 mW f. R1 : 1 W
PR = 500 mW 1
R2 : W
4
PT = 300 mW + 400 mW + 100 mW + 500 mW
PT = 1300 mW (as required.) 1
R3 : W
2
15. a. 4V
I= = 0.04 A = 40 mA R4 : 2 W
100 Ω
11
12 Chapter 5 Series Circuits

g. PT = (36 V)(0.0788 A)
= 2.836 W
= 0.745 W + 0.242 W + 0.484 W + 1.365 W

5.5 Interchanging Series Components

23.
E 2V F
21. R5
IF
24 V 23.8 V
5 12 V
10 mA Circuit 1: I = = 0.15 A
80 Ω
5 2.02 k Ω
9V
Circuit 2: I = = 0.115 mA
78 kΩ
5.6 Voltage Divider Rule
25. Circuit 1: c. P1 = (20 mA)2(0.104 kΩ) = 41.7 mW
RT = 24 Ω P2 = 146.1 mΩ
⎛6Ω⎞ P3 = 292.2 mΩ
V6 Ω = ⎜ ⎟ 24 Ω = 6 V
⎝ 24 Ω ⎠
a. circuit (b)
V3 Ω = 3 V
V5 Ω = 5 V R2 R2
V2 = (50 V) = (50 V) = 6.25 V
R1 + R2 + R3 8 R2
V8 Ω = 8 V
V22 (6.25 V)2
V2 Ω = 2 V R2 = = = 244 Ω
P2 0.160 W
VT = 6 V + 3 V + 5 V + 8 V + 2 V =24 V = E
R1 = 4(244 Ω) = 977 Ω
Circuit 2: R3 = 3(244 Ω) = 732 Ω
RT = 4.3 kΩ + 2.7 kΩ + 7.8 kΩ + 9.1 kΩ = 23.9 kΩ b. V2 = 6.25 V
⎛ 4.3 Ω ⎞ V1 = 25.0 V
V4.3 kΩ = ⎜ ⎟(120 V) = 21.6 V
⎝ 23.9 kΩ ⎠ V3 = 8.75 V
V2.7 kΩ = 13.6 V c. (25.0 V)2
P1 = = 640 mW
V7.8 kΩ = 39.2 V 977 Ω
V9.1 kΩ = 45.7 V P3 = 480 mW
VT = 21.6 V + 13.6 V + 39.2 V + 45.7 V = 120.1 V P2 = 160 mW (given)
29. a. RT = 24(25 Ω) = 600 Ω
VT ≅ (170 V – 50 V)
120 V
VT= ET I= = 0.2 A
600 Ω
27. a. circuit (a) b. ⎛ 25 Ω ⎞
V=⎜ ⎟120 V = 5.0 V
24 V ⎝ 600 Ω ⎠
RT = = 1.2 kΩ = R1 + R2 + R3
20 mA c. (5 V)2
P= =1W
R1 + 3.5 R1 + 2(3.5 R1) = 1.2 kΩ 25 Ω
d. RT = 22(25 Ω) = 550 Ω
R1 + 0.104 kΩ
120 V
R2 = 0.365 kΩ I= = 0.218 A
550 Ω
R3 = 0.730 kΩ ⎛ 25 Ω ⎞
V=⎜ ⎟120 V = 5.45 V
b. ⎛ 0.104 kΩ ⎞ ⎝ 550 Ω ⎠
V1 = ⎜ ⎟ 24 V = 2.09 V
⎝ 1.2 kΩ ⎠ (5.45 V)2
P= = 1.19 W
V 2 = 7.30 V 25 Ω
e. Since each of the remaining bulbs now dissipates
V 3 = 14.61 V 20% more power, the life expectancy will decrease.
Answers to Odd-Numbered Questions 13

5.8 Voltage Subscripts

31. circuit (a) circuit (b)
Vab = V1 + V2 = 2.09 V + 7.30 V = 9.39 V Vab = 25.0 V
Vbc = V3 = 14.61 V Vbc = 6.25 V

33. circuit (a) circuit (b)

54 V ET = 6 V + 3 V = 9 V
I= = 3 mA
3 kΩ + 9 kΩ + 6 kΩ ⎛ 330 Ω ⎞
V330 Ω = ⎜ ⎟ 9 V = 2.97 V
V 3 kΩ = (3 mA)(3 kΩ) = 9 V ⎝ 330 Ω + 670 Ω ⎠
V9 kΩ = 27 V V670 Ω = 6.03 V
V6 kΩ = 18 V Va = 6.03 – 3 V = 3.03 V
Va = +(3 mA)(9 kΩ + 6 kΩ) = 45 V
5.9 Internal Resistance of Voltage Sources
35. a. VR = 14.2 V – 6.8 V = 7.4 V
6.8 V
I= = 0.068 A
100 Ω
7.4 V
Rint = = 109 Ω
0.068 A
b. ⎛ 200 Ω ⎞
VL = ⎜ ⎟14.2 V = 9.20 V
⎝ (109 Ω + 200 Ω) ⎠

5.10 Ammeter Loading Effects

37. circuit (a)
15 V
Iactual = = 0.375 mA
(10 kΩ + 12 kΩ + 18 kΩ)
15 V
Imeasured = = 0.374532 mA
(10 kΩ + 12 kΩ + 18 kΩ + 0.050 kΩ)
⎛ 0.375 mA − 0.374532 mA ⎞
loading error = ⎜ ⎟ × 100% = 0.125%
⎝ 0.375 mA ⎠
circuit (b)
0.15 V
Iactual = = 0.375 mA
(100 Ω + 120 Ω + 180 Ω)
0.15 V
Imeasured = = 0.333 mA
(100 Ω + 120 Ω + 180 Ω + 50 Ω)
(0.375 mA − 0.333 mA)
loading error = × 100% = 11.1%
0.375 mA

5.11 Circuit Analysis Using Computers

39.
14 Chapter 5 Series Circuits

41.
Chapter

6
Parallel Circuits

6.1 Parallel Circuits

1. a. A and B are in series 3.
D and E are in series
C and F are in parallel
b. B, C, and D are in parallel
c. A and B are in parallel
D and F are in parallel
C and E are in series
d. A, B, C, and D are in parallel

6.2 Kirchhoff’s Current Law

5. a. I1 = 5 A – 2 A = 3 A (downward) 9. a. I1 = 500 mA – 200 mA – 100 mA = 200 mA
I2 = 4 A – 3 A = 1 A (left) I2 = 500 mA
b. I1 = 6 A + 1 A = 7 A (right) I3 = 200 mA – 50 mA = 150 mA
I2 = 7 A – 2 A – 3 A = 2 A (upward) I4 = 200 mA
I3 = 4 A + 3 A = 7 A (downward) b. V = (25 Ω)(100 mA) = 2500 mV
c. I1 = 15 mA – 5 mA – 6 mA = 4 mA (right) V = 2.5 V
I2 = 5 mA + 4 mA + 6 mA +7 mA –2 mA c. 2.5 V
R1 = = 12.5 Ω
I2 = 20 mA (right) 200 mA
2.5 V
7. a. 50 V R3 = = 15.7 Ω
I1 = = 1.25 A 150 mA
40 Ω
2.5 V
5V R4 = = 50 Ω
I2 = = 0.0833 A 50 mA
60 Ω
I3 = 1.25 A – 0.0833 A = 1.167 A
I4 = 0.0833 + 1.167 A = 1.25 A
b. 5V
R3 = = 4.29 Ω
1.167 A

6.3 Resistors in Parallel

11. a. (4 Ω)(6 Ω) b. 1 1 ⎛ 1 1 ⎞
RT = = 2.4 Ω = − + = 2.22 mS
4Ω+6Ω R 30Ω ⎜⎝ 50Ω 90Ω ⎟⎠
b. 1 R = 450 Ω
RT = = 32 kΩ
1 1 1 15. a. 1 1 1 1 6.25 1
+ + = + + = = Ω
480 Ω 240 kΩ 40 kΩ RT R1 4R1 R1 R1 200
c. 1 5
RT = = 4.04 kΩ R1 = 1250 Ω
1 1 1
+ +
8.2 kΩ 10 kΩ 39 kΩ R2 = 5000 Ω
13. a. 1 1 1 R3 = 250 Ω
= − = 0.5 lS
R 400 kΩ 500 kΩ
R = 2.0 MΩ

15
16 Chapter 6 Parallel Circuits

b. 2A 21. 32 V
IR1 = = 0.40 A IT = = 2.0 mA
5 16 kΩ
2A I1 + 3I1 + 1.5(3I1) = 2.0 mA
IR2 = = 0.10 A
(5)(4) I1 = 0.2353 mA
c. I = 0.40 A + 0.10 A + 2.0 A = 2.5 A I2 = 0.7059 mA
17. a. VR1 = VR2 = (2 mA)(450 Ω) = 900 mV I3 = 1.0589 mA
b. 900 mV 32 V
I2 = = 4.5 mA R1 = = 136 kΩ
200 Ω 0.2353 mA
19. a. 1 32 V
RT = = 240 Ω R2 = = 45.3 kΩ
1 1 1 0.7059 mA
+ +
600 Ω 800 Ω 800 Ω 32 V
R3 = = 30.2 kΩ
b. 1 1.0589 mA
RT = = 9.392 kΩ
1 1 1 1 23. a. RT = 12.5 kΩ
+ + +
27 kΩ 33 kΩ 56 kΩ 47 kΩ b. RT = 0 Ω (short circuit)
c. RT1 = 6kΩ||3 kΩ = 2 kΩ c. RT = 75 Ω
RT2 = 9 k Ω||9 kΩ||9 kΩ = 3 kΩ 25. RT ≅ 30 Ω||30 Ω = 15 Ω
(2 kΩ)(3 kΩ)
RT = = 1.2 kΩ
2 kΩ + 3 kΩ

6.4 Voltage Sources in Parallel

27. 20 V
I= = 0.2 A
100 Ω
0.2 A
I1 = = 0.1 A = I2
2

6.5 Current Divider Rule

29. a. ⎛ 2Ω ⎞ ⎛ 300 Ω ⎞
I1 = ⎜ ⎟ (10 A) = 2 A I5 = ⎜ ⎟ 150 mA = 110 mA
Ω + Ω
⎝2 Ω + 8 Ω ⎠ ⎝ 109.1 300 ⎠
⎛ 8Ω ⎞ ⎛ 109.1 Ω ⎞
I2 = ⎜ I1 = ⎜ ⎟ 110 mA = 60 mA
⎟ (10 A) = 8 A ⎝ 200 Ω ⎠
⎝2 Ω + 8 Ω ⎠
⎛ 109.1 Ω ⎞
b. ⎛ 1 kΩ ⎞ I2 = ⎜ ⎟ 110 mA = 30 mA
I1 = ⎜ ⎟ (16 mA) = 4 mA ⎝ 400 Ω ⎠
⎝ 1 kΩ + 3 kΩ ⎠
⎛ 109.1 Ω ⎞
⎛ 3 kΩ ⎞ I3 = ⎜ ⎟ 110 mA = 20 mA
I2 = ⎜
1 kΩ + 3 kΩ ⎟ (16 mA) = 12 mA ⎝ 600 Ω ⎠
⎝ ⎠ 33. ⎛ R1 ⎞
31. a. 1 ⎜ R + R ⎟ 90 mA = 60 mA
RT = ⎝ 1 ⎠
1 1 1 1
+ + + ⎛ 24 Ω ⎞
4.7 kΩ 3.3 kΩ 1.0 kΩ 2.2 kΩ ⎜ 24 Ω + R ⎟ 90 mA = 60 mA
⎝ ⎠
= 0.508 kΩ
(24 Ω)(90 mA)
⎛ 0.508 kΩ ⎞ 24 Ω + R =
I1 = ⎜ 60 mA
⎟ 60 mA = 6.48 mA
⎝ 4.7 kΩ ⎠ R = 36 Ω − 24 Ω = 12 Ω
⎛ 0.508 kΩ ⎞ RT = 24 Ω||48 Ω||16 Ω = 8 Ω
I2 = ⎜ ⎟ 60 mA = 9.23 mA 35. a.
⎝ 3.3 kΩ ⎠ b. 12 V
I3 = 30.45 mA I= = 1.50 A
8Ω
I4 = 13.84 mA c. ⎛8Ω⎞
I1 = ⎜ ⎟ 1.50 A = 0.50 A
b.
R′T =
1
= 109.1 Ω ⎝ 24 Ω ⎠
1 1 1 I1
+ + I2 = = 0.25 A
200 Ω 400 Ω 600 Ω 2
⎛ 109.1 Ω ⎞ ⎛8Ω⎞
I4 = ⎜ ⎟ 150 mA = 40 mA I3 = ⎜ ⎟ 1.50 A = 0.75 A
⎝ 109.1 Ω + 300 Ω ⎠ ⎝ 16 Ω ⎠
Answers to Odd-Numbered Questions 17

d. ∑Iin = 1.50 A
∑Iout = 0.50 A + 0.25 A + 0.75 A = 1.50 A
∑Iin = ∑Iout

6.6 Analysis of Parallel Circuits

37. a. RT = 60Ω||100 Ω||75 Ω = 25 Ω 41. a. 48 V
R3 = = 4 kΩ
b. 240 V 12 mA
I= = 9.60 A
25 Ω 48 V
R4 = = 6 kΩ
240 V (50 mA − 30 mA − 12 mA)
I1 = = 4.0 A
60 Ω V2 (48 V)2
R1 = = = 2 kΩ
240 V P 1.152 W
I2 = = 2.40 A
100 Ω 48 V
RT = = 0.96 kΩ
240 V 50 mA
I3 = = 3.20 A
75 Ω 1
R2 = = 8 kΩ
I4 = 2.40A + 3.20 A = 5.60 A 1 1 1 1
− − −
c.
∑Iin = 9.60 A b.
0.96 kΩ 4 kΩ 6 kΩ 2 kΩ
1.152 W
IR1 = = 24 mA
∑Iout = 4.0 A + 2.40 A + 3.20 A + 5.60 A = 9.60 A 48 V

∑Iin = ∑Iout IR2 = 30 mA – 24 mA = 6 mA

IR4 = 50 mA – 30 mA –12 mA = 8 mA
d. P1 = (4.0 A)2(60 Ω) = 960 W
c. I1 = 12 mA + 8 mA = 20 mA
P2 = (2.40 A)2(100 Ω) = 576 W
I2 = 20 mA + 24 mA + 6 mA = 50 mA
P3 = (3.20 A)2(75 Ω) = 768W
d. P2 = (6 mA)(48 V) = 288 mW
PT = (240 V)(9.60 A) = 2304 W
P3 = (12 mA)(48 V) = 576 mW
PT = P1 + P2 + P3
P4 = (8 mA)(48 V) = 384 mW
39. a. 20 V
I1 = = 1.00 A 43. 1000 W
20 Ω I1 = = 8.33 A
120 V
I2 = 2.0 A
120 V
I3 = 5.0 A I2 = = 5.00 A
24 Ω
I4 = 4.0 A 120V
b. I = 1.0 A + 2.0 A + 5.0 A + 4.0 A = 12.0 A I3 = = 2.50 A
48 Ω
c. P1 = 20 W I4 = 7.50 A
P2 = 40 W IT = 15.83 A
P3 = 100 W The rated current of the fuse will be exceeded; the
P4 = 80 W fuse will “blow” open.

6.7 Voltmeter Loading Effects

45. a. ⎛ MΩ ⎞
Vmeas = ⎜ ⎟ 30 V = 20 V
⎝ 1 MΩ + 0.5 MΩ ⎠
b. Vab = 30 V
30 V − 20 V
loading effect = × 100% = 33.3%
30 V
47. ⎛ 50 kΩ ⎞
1.2 V = ⎜ ⎟E
⎝ 50 kΩ + 1000 kΩ ⎠
E = 25.2 V
18 Chapter 6 Parallel Circuits

6.8 Computer Analysis

49. 53.

51.
Chapter

7
Series-Parallel Circuits

7.1 The Series-Parallel Network

1. a. RT = R1 + R5 + [(R2 + R3)||R4] 5. a.
b. RT = (R1||R2) + (R3||R4)
3. a. RT1 = R1 + [(R3 + R4)||R2] + R5
RT2 = R5
b. RT1 = R1 + (R2||R3||R5)
RT1 = R5||R3||R2
(R4 is shorted)

(one possible solution)

b.

(one possible solution)

7.2 Analysis of Series-Parallel Circuits

7. a. RT = 300 Ω + [4 kΩ||(200 Ω + 300 Ω + 500 Ω)] + 400 Ω = 1500 Ω
b. RT = 5.1 kΩ||[1.2 kΩ + {4.7 kΩ||(3.3 kΩ + 5.6 Ω)}] = 2.33 kΩ
9. Rab = 80 Ω + 40 Ω + [60 Ω||60 Ω||(10 Ω + 20 Ω + 30 Ω] = 140 Ω
Rcd = 10 Ω||[20 Ω + 30 Ω + (60 Ω||60 Ω)] = 8.89 Ω
11. a. RT = 100 Ω + [(560 Ω + 150 Ω)||(180 Ω + 220 Ω||(390 Ω + 910 Ω)]
= 100 Ω + 710 Ω||400 Ω||1300 Ω
= 313.78 Ω ≅ 314 Ω
⎛ 213.78 Ω ⎞
I1 = ⎜ ⎟ 63.7 mA = 19.2 mA
⎝ 560 Ω + 150 Ω ⎠
I2 = 63.7 mA – 19.2 mA = 44.5 mA
⎛ 213.78 Ω ⎞
I3 = ⎜ ⎟ 63.7 mA = 34.1 mA
⎝ 180 Ω + 220 Ω ⎠
⎛ 213.78 Ω ⎞
I4 = ⎜ ⎟ 63.7 mA = 10.4 mA
⎝ 390 Ω + 910 Ω ⎠
c. Vab = +(34.1 mA)(180 Ω +220 Ω) =13.6 V
b. 20 V Vbc = –(19.2 mA)(150 Ω) = –2.88 V
IT = = 63.7 mA
314 Ω

19
20 Chapter 7 Series-Parallel Circuits

13. a. RT = 1 kΩ + 2 kΩ + 5 kΩ||[1 kΩ + {6 kΩ||(3 kΩ + b. Vab = –(1.5 mA)(6 kΩ) = –9.0 V

6 kΩ)}] = 5.396 kΩ c. (18 V)2
___ PT = = 162 mW
28 V 2 kΩ
I1 = = 5.189 mA
5.396 kΩ P6kΩ = (1.5 mA)2(6 kΩ) = 13.5 mW
⎛ 5 kΩ ⎞ ___ ___
P3kΩ = (3.0 mA)2(3 kΩ) = 27.0 mW
I2 = ⎜ ⎟ 5.189 mA = 2.702 mA
⎝ 5 kΩ + 4.6 kΩ ⎠ P2kΩ = (4.5 mA)2(2 kΩ) = 40.5 mW
⎛ 6 kΩ ⎞ ___ ___
I3 = ⎜ ⎟ 2.702 mA = 1.081 mA P4kΩ = (4.5 mA)2(4 kΩ) = 81.0 mW
kΩ + 3 kΩ +
⎝ 6___ 6 kΩ ⎠
___ ___ 4
I4 = 5.189 mA − 2.702 mA = 2.486 mA PT=∑Pn = 162 mW
___ ___ ___
I5 = 2.702 mA − 1.081 mA = 1.621 mA n=1
___
I6 = 2.702 mA 15. circuit (b)
___ ___
b. Vab = +(5 kΩ)(2.486 mA) = 12.432 V a. RT = 10 Ω + 16 ||[5 Ω||(8 Ω + 4 Ω)] = 15.76 Ω
___ ___
Vcd = +(6 kΩ)(1.621 12V − 3 V
___ mA) = 9.729 V IT = I1 = = 0.571 A
c. PT = (28 V)(5.189 mA) = 145.3 mW 15.76 Ω
P1 = 26.9 mW P5 = 15.8 mW ⎛ 16 Ω ⎞
I2 = ⎜ ⎟ 0.571 A = 0.365 A
P2 = 7.3 mW P6 = 7.0 mW ⎝ 16 Ω + 9 Ω ⎠
⎛ 6Ω ⎞
P3 = 3.5 mW P7 = 53.9 mW I3 = ⎜ ⎟ 0.365 A = 0.122 A
⎝ 6 Ω + 12 Ω ⎠
P4 = 30.9 mW
7 I4 = 0.471 A − 0.122 A = 0.449 A
PT = ∑Pn = 145.3 mW b. Vab = –(0.365 A)(5 Ω) = –1.827 V
n=1 c. PT = (9 V)(0.571 A) = 5.14 W
15. circuit (a) P10Ω = (0.571 A)2 (10 Ω) = 3.26 W
a. P16Ω = (0.206 A)2(16 Ω) = 0.68 W
P5Ω = (0.365 A)2(5 Ω) = 0.67 W
P6Ω = (0.244 A)2(6 Ω) = 0.36 W
P8Ω = (0.122 A)2(8 Ω) = 0.12 W
P4Ω = (0.122 A)2(4 Ω) = 0.06 W
6
PT = ∑ Pn = 5.15 W
n=1

RT = 4 kΩ||(2 kΩ + 6 kΩ||3 kΩ = 2 kΩ
18 V
IT = = 9.0 mA
2 kΩ
18 V
I2 = = 4.5 mA
4 kΩ
18 V
I1 = = 4.5 mA
2 kΩ + (6 kΩ||3 kΩ)
⎛ 3 kΩ ⎞
I3 = ⎜ ⎟ 4.5 mA = 1.5 mA
⎝ 6 kΩ + 3 kΩ ⎠
7.3 Applications of Series—Parallel Circuits
17. Zener is on: IZ = 93.3 mA – 52.9 mA = 40.4 mA
V1 = 14.0 V PT = (24 V)(93.3 mA) = 2240 mW
24 V − 10 V PZ = 404.2 mW
I1 = = 93.3 mA
150 Ω P1 = 1306.7 mW
V2 = 2.06 V P2 = 109.2 mW
V3 = 7.94 V
P3 = 419.9mW
10 V
I2 = I3 = = 52.9 mA
39 Ω + 150 Ω
Answers to Odd-Numbered Questions 21

19. Rmin:
⎛ R ⎞
VL = ⎜ ⎟ 20 V = 5.6 V
⎝ R + 80 Ω ⎠
(20 V)R = (5.6 V)R + 448 V ⋅ Ω
448 V ⋅ Ω _
Rmin = 31.1 Ω
14.4 V
Rmax:
20 V − 5.6 V
I1 = = 180 mA
80 Ω
1W
IZ(max) = = 178.57 mA
5.6 V
IL(min) = 1.43 mA
5.6 V
Rmax = = 3.92 kΩ
1.43 mA
R = 31.1 Ω → 3900 Ω
21. Using KVL: (750 Ω)IE + 0.6 V + (7.5 kΩ)IB – 2 V = 0
But IE ≅ 100 IB
(750 Ω)(100IB) + (7.5 kΩ)IB = 1.4 V
1.4 V
IB = = 17.0 µA
82.5 kΩ
IE ≅ 1.70 mA ≅ IC
VB = (7.5 kΩ)(17.0 µA) – 2.0 V = –1.87 V
VCE = –16 V + (3.9 kΩ)(1.7 mA) + (0.75 kΩ)(1.7 mA) = –8.10 V
23. a. ID = 3.6 mA
VS = –(–2.0 V) = 2.0 V
b. 2.0 V
RS = = 0.556 kΩ = 556 Ω
3.6 mA
c. VDS = 15 V – (3.6 mA)(1.5 kΩ + 0.556 kΩ) = +7.6 V
25. ⎛ 4 kΩ ⎞
VB = ⎜ ⎟ (18 V) = 2.00 V
⎝ 4 Ω + 32 kΩ ⎠
VE = 2.00 – 0.7 V = 1.30 V
1.30 V
IE = = 3.25 mA ≅ IC
0.4 kΩ
VCE = 18 V – 3.25 mA (2.40 kΩ + 0.40 kΩ) = 8.90 V

7.4 Potentiometers
27. a. For R2 = 0 Ω : Vbc = VL = 0 V
For R2 = 10 kΩ:
⎛ 10 kΩ||10 kΩ ⎞
Vbc = VL = ⎜ ⎟ 36 V = 7.2 V
⎝ 20 kΩ + 10 k||10 kΩ ⎠
VL = 0 → 7.2 V
b. (2.5 kΩ||10 kΩ)
VL = × 36 V = 2.44 V
(2.5 kΩ||10 kΩ) + 20 kΩ + (10 kΩ − 2.5 kΩ)
If RL is removed:
7.5 kΩ
Vab = = 9.0 V
20 kΩ + 10 kΩ _ _
29. RT = (200 Ω||50 Ω) + (800 Ω||100 Ω) = 40 Ω + 88.8 Ω = 128.8 Ω
50 Ω||200
_ Ω × 24 V = 7.45 V
Vbc =
128.8 Ω
Vab = 24 V − 7.448 V = 16.6 V
22 Chapter 7 Series-Parallel Circuits

31. a. Vout(min) = 0 V
⎛ (10 kΩ||10 kΩ)
Vout(max) = ⎜ 120 V = 40 V
⎝ (10 kΩ||10 kΩ) + (R10 kΩ) + 10 kΩ
b. R2||10 kΩ
Vout = × 120 V
(R2||10 kΩ) + (R110 kΩ) + 10 kΩ
But R1 = 10 kΩ – R2
(R2)(10 kΩ)
× 120 V
R2 + 10 kΩ
20 V = + 10 kΩ
(R2)(10 kΩ) 10 kΩ − R2)(10 kΩ)
+
R2 + 10 kΩ 10 kΩ R2 + 10 kΩ
R2 = 3.82 kΩ RL = 500 Ω :
33. RL = 0 Ω : Vout = 0V (500 Ω||100 Ω)(20 V)
Vout = = 9.09 V
RL = 250 Ω : (500 Ω||100 Ω) + 100 Ω
(250 Ω||100 Ω)(20 V)
Vout = = 8.33 V
(250 Ω||100 Ω) + 100 Ω
7.5 Loading Effects of Instruments
35. a. 750 kΩ||200 kΩ ⎛ 3.9 Ω ⎞
5.0 V = × E = 0.4412E I2(loaded) = ⎜ ⎟ (24.1 mA
_750 kΩ||200 kΩ + 200 kΩ ⎝ 3.9 Ω + 6.8 Ω + 2 Ω ⎠
E = 11.3 V = 7.40 mA
b. Unloaded voltage: Measuring I3
750 kΩ _
VNL = ×11.3 V = 8.95 V 0.2 V
750 kΩ + 200 kΩ I1 = = 22.83 mA
5.6 Ω + 6.8 Ω||(3.9 Ω + 2 Ω)
c. 8.95 V − 5.00 V ⎛ 6.8 Ω ⎞
Loading effect = × 100% = 44.1% I3(loaded) = ⎜
8.95 V 3.9 Ω + 6.8 Ω + 2 Ω ⎟ 22.83 mA
⎝ ⎠
d. ⎛ 200 kΩ||200 kΩ ⎞ _ = 12.22 mA
V=⎜ ⎟ 11.3 V = 1.33 V
⎝ (200 kΩ||200 kΩ) + 750 kΩ ⎠ c. 0.2 V 0.2 V
37. a. Break the circuit between the 5.6 Ω resistor and the I1(NL) + =
5.6 Ω + 6.8 Ω + 3.9 Ω 8.0785 Ω
voltage source. Insert the ammeter at the break,
connecting the red (+) lead of the ammeter to the = 24.76 mA
positive terminal of the voltage source and the ⎛ 3.9 Ω ⎞
I2(NL) = ⎜ ⎟ 24.76 mA = 9.02 mA
black (–) lead to the 5.6 Ω resistor. ⎝ 2.9 Ω + 6.8 Ω ⎠
b. Measuring I1 ⎛ 6.8 Ω ⎞
I3(NL) + ⎜ ⎟ 24.76 mA = 15.73 mA
0.2 V ⎝ 3.9 Ω + 6.8 Ω ⎠
I1(loaded) =
2 Ω + 5.6 Ω + 6.8 Ω||3.9 Ω Loading(I1) =19.9%
= 19.84 mA Loading(I2) = 18.0%
Measuring I2 Loading(I3) = 22.3%
0.2 V
I1(loaded) =
5.6 Ω + (6.8 Ω + 2 Ω)||3.9 Ω
= 24.09 mA

7.6 Circuit Analysis Using Computers

39. 41.
Answers to Odd-Numbered Questions 23

43.

45.
Chapter

8
Methods of Analysis

8.1 Current Sources

1. VS = 18 V + 20 V = 38 V 7. Power provided by voltage source:
3. a. ⎛ 300 Ω ⎞ P = (15 V)(500 lA) - 7.5 mW
I3 = ⎜ ⎟ 20 mA = 12 mA
⎝ 200 Ω + 300 Ω ⎠
Power dissipated by resistors:
b. VS = (20 mA)(0.1 kΩ) + (12 mA)(0.2 kΩ) = 4.4 V P150kΩ = (300 lA)250 kΩ = 4.5 mW
V1 = (20 mA)(0.1 kΩ) = 2.0 V P150kΩ = (100 lA)2150 kΩ = 1.5 mW
5. 15 V Power absorbed by current source:
I50kΩ = = 300 lA
50 kΩ P = (0.1 lA)(15 V) = 1.5 mW

I150kΩ =
15 V
= 100 lA Pin = 7.5 mW = ∑Pout
150 kΩ
Note:
I1 = 300 lA + 100 lA = 400 lA
The current source is taking energy from the circuit
I2 = 400 lA + 100 lA = 500 lA rather than providing energy.

8.2 Source Conversions

9. a. 11. a. ⎛ 450 Ω ⎞
IL = ⎜ ⎟ 8 A = 7.2 A
⎝ 450 Ω + 50 Ω ⎠
b. E = (8 A)(450 Ω) = 3600 V
3600 V
IL = = 7.2 A
450 Ω + 50 Ω
13. a.
b.

b. 21.45 V − 16 V _
I= = 6.05 mA
330 Ω + 470 Ω + 100 Ω
_ _
c. Vab = (6.05) mA)(0.1 kΩ) = 0.605 V

8.3 Current Sources in Parallel and Series

15. IT = 100 mA – 50 mA = 50 mA (down) 17.
RT = 3 kΩ||6 kΩ||8 kΩ = 1.6 kΩ
V2 = –(50 mA)(1.6 kΩ) = –80 V
⎛ 1.6 kΩ ⎞ E = (3 mA)(2.4 kΩ) = 7.2 V
I1 = −⎜ ⎟ 50 mA = −26.7 mA
⎝ 3 kΩ ⎠ 8 V − 7.2 V
I3 = = 0.133 mA
2.4 kΩ + 1.6 kΩ + 2.0 kΩ
Vab = –7.2 V – (0.133 mA)(2.4 kΩ) = –7.52 V

24