Solucionario introducción al análisis de circuitos boylestad 11ed

Instructor’s Resource Manual
to accompany

Introductory
Circuit Analysis
Eleventh Edition

Robert L. Boylestad

Upper Saddle River, New Jersey
Columbus, Ohio

www.elsolucionario.net




LIBROS UNIVERISTARIOS
Y SOLUCIONARIOS DE
MUCHOS DE ESTOS LIBROS
LOS SOLUCIONARIOS
CONTIENEN TODOS LOS
EJERCICIOS DEL LIBRO
RESUELTOS Y EXPLICADOS
DE FORMA CLARA
VISITANOS PARA
DESARGALOS GRATIS.

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Contents
CHAPTER 1

1

CHAPTER 2

9

CHAPTER 3

13

CHAPTER 4

22

CHAPTER 5

29

CHAPTER 6

39

CHAPTER 7

52

CHAPTER 8

65

CHAPTER 9

86

CHAPTER 10

106

CHAPTER 11

124

CHAPTER 12

143

CHAPTER 13

150

CHAPTER 14

157

CHAPTER 15

170

CHAPTER 16

195

CHAPTER 17

202

CHAPTER 18

222

CHAPTER 19

253

CHAPTER 20

266

CHAPTER 21

280

CHAPTER 22

311

CHAPTER 23

318

CHAPTER 24

333

CHAPTER 25

342

TEST ITEM FILE

353

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iii


Chapter 1
1.

−

2.

−

3.

−

4.

υ=

5.

6.

d 20,000 ft ⎡ 1 mi ⎤ ⎡ 60 s ⎤ ⎡ 60 min ⎤
=
= 1363.64 mph
t
10 s ⎢⎣ 5,280 ft ⎥⎦ ⎢⎣1 min ⎥⎦ ⎢⎣ 1 h ⎥⎦

⎡ 1h ⎤
4 min ⎢
⎥ = 0.067 h
⎣ 60 min ⎦
d
31 mi
υ= =
= 29.05 mph
t 1.067 h
a.
b.
c.

95 mi ⎡ 5,280 ft ⎤ ⎡ 1 h ⎤ ⎡1 min ⎤
= 139.33 ft/s
h ⎢⎣ mi ⎥⎦ ⎢⎣ 60 min ⎥⎦ ⎢⎣ 60 s ⎥⎦
60 ft
d
t= =
= 0.431 s
υ 139.33 ft/s
d 60 ft ⎡ 60 s ⎤ ⎡ 60 min ⎤ ⎡ 1 mi ⎤
= 40.91 mph
υ= =
t
1 s ⎢⎣1 min ⎥⎦ ⎢⎣ 1 h ⎥⎦ ⎢⎣ 5,280 ft ⎥⎦

7.

−

8.

−

9.

−

10.

MKS, CGS, °C =

11.

⎡ 0.7378 ft - lb ⎤
1000 J ⎢
⎥ = 737.8 ft-lbs
1J
⎣
⎦

12.

⎡ 3 ft ⎤ ⎡12 in. ⎤ ⎡ 2.54 cm ⎤
0.5 yd ⎢
⎥⎢
⎥⎢
⎥ = 45.72 cm
⎣1 yd ⎦ ⎣ 1 ft ⎦ ⎣ 1 in. ⎦

13.

a.

104

14.

a.

15 × 103

b.

e.

4.02 × 10−4

f. 2 × 10−10

5
5
5
(°F − 32) = (68 − 32) = (36) = 20°
9
9
9
SI: K = 273.15 + °C = 273.15 + 20 = 293.15

b. 106

c. 103
30 × 10−3

d. 10−3
c. 2.4 × 106

Chapter 1

e. 100

f. 10−1

d. 150 × 103

1

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15.

a.
b.
c.
d.

4.2 × 103 + 48.0 × 103 = 52.2 × 103 = 5.22 × 104
90 × 103 + 360 × 103 = 450 × 103 = 4.50 × 105
50 × 10−5 − 6 × 10−5 = 44 × 10−5 = 4.4 × 10−4
1.2 × 103 + 0.05 × 103 − 0.6 × 103 = 0.65 × 103 = 6.5 × 102

16.

a.
b.
c.
d.
e.
f.

(102)(103) = 105 = 100 × 103
(10−2)(103) = 101 = 10

(103)(106) = 1 × 109
(102)(10−5) = 1 × 10−3
(10−6)(10 × 106) = 10
(104)(10−8)(1028) = 1 × 1024

17.

a.
b.
c.
d.

(50 × 103)(3 × 10−4) = 150 × 10−1 = 1.5 × 101
(2.2 × 103)(2 × 10−3) = 4.4 × 100 = 4.4
(82 × 106)(2.8 × 10−6) = 229.6 = 2.296 × 102
(30 × 10−4)(4 × 10−3)(7 × 108) = 840 × 101 = 8.40 × 103

18.

a.
b.
c.
d.
e.
f.

102/104 = 10−2 = 10 × 10−3
10−2/103 = 10−5 = 10 × 10−6
104/10−3 = 107 = 10 × 106
10−7/102 = 1.0 × 10−9

1038/10−4 = 1.0 × 1042
100 / 10 −2 = 101/10−2 = 1 × 103

19.

a.
b.
c.
d.

(2 × 103)/(8 × 10−5) = 0.25 × 108 = 2.50 × 107
(4 × 10−3)/(60 × 104) = 4/60 × 10−7 = 0.667 × 10−7 = 6.67 × 10−8
(22 × 10−5)/(5 × 10−5) = 22/5 × 100 = 4.4
(78 × 1018)/(4 × 10−6) = 1.95 × 1025

20.

a.
c.

(102)3 = 1.0 × 106
(104)8 = 100.0 × 1030

21.

a.
b.
c.
d.

22.

a.

(4 × 102)2 = 16 × 104 = 1.6 × 105
(6 × 10−3)3 = 216 × 10−9 = 2.16 × 10−7
(4 × 10−3)(6 × 102)2 = (4 × 10−3)(36 × 104) = 144 × 101 = 1.44 × 103
((2 × 10−3)(0.8 × 104)(0.003 × 105))3 = (4.8 × 103)3 = (4.8)3 × (103)3
= 110.6 × 109 = 1.11 × 1011
−3 2
−6
(−10 ) = 1.0 × 10
(10 2 )(10 −4 )
= 10−2/103 = 1.0 × 10−5
3
10
(10 −3 ) 2 (10 2 ) (10 −6 )(10 2 ) 10 −4
=
= 4 = 1.0 × 10−8
10 4
10 4
10
3
4
(10 )(10 )
= 107/10−4 = 1.0 × 1011
10 − 4

b.
c.

d.

b.
d.

(10−4)1/2 = 10.0 × 10−3
(10−7)9 = 1.0 × 10−63

Chapter 1

2

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e.

(1 × 10−4)3(102)/106 = (10−12)(102)/106 = 10−10/106 = 1.0 × 10−16

[(10 )(10 )]
[(10 ) ][10 ]
−2

2

f.

23.

a.

b.

2 2

−3

−3

=

1
1
=
= 1.0 × 10−1
−3
(10 )(10 ) 10
4

(3 × 10 2 ) 2 (10 2 )
= (9 × 104)(102)/(3 × 104) = (9 × 106)/(3 × 104) = 3 × 102 = 300
3 × 10 4
(4 × 10 4 ) 2 16 × 10 8
=
= 2 × 105 = 200.0 × 103
(20) 3
8 × 10 3

c.

(6 × 104 ) 2

36 × 108
=
= 9.0 × 1012
−2 2
−4
( 2 × 10 )
4 × 10

d.

(27 × 10 −6 )1 / 3 3 × 10 −2
= 1.5 × 10−7 = 150.0 × 10−9
=
2 × 10 5
2 × 10 5

e.

(4 × 10 3 ) 2 (3 × 10 2 ) (16 × 10 6 )(3 × 10 2 ) 48 × 10 8
= 24.0 × 1012
=
=
−4
−4
−4
2 × 10
2 × 10
2 × 10

f.

(16 × 10−6)1/2(105)5(2 × 10−2) = (4 × 10−3)(1025)(2 × 10−2) = 8 × 1020
= 800.0 × 1018
2

g.

1/ 2

⎡ (3 × 10−3 )3 ⎤ ⎡1.60 × 102 ⎤ ⎡(2 × 102 )(8 × 10−4 ) ⎤
⎣
⎦⎣
⎦ ⎣
⎦
(7 × 10−5 ) 2
(27 × 10−9 )(2.56 × 104 )(16 × 10−2 )1/ 2
49 × 10−10
(69.12 × 10−5 )(4 × 10−1 ) 276.48 × 10−6
=
=
49 × 10−10
49 × 10−10
= 5.64 × 104 = 56.4 × 103
=

+3
24.

a.

6 × 103 = 0.006 × 10+6
−3
−3

b.

4 × 10−3 = 4000 × 10−6
+3
−2

c.

+3

+3

50 × 105 = 5000 × 103 = 5 × 106 = 0.005 × 109
+2

−3

−3

Chapter 1

3

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−3

+5
d.

−3

30 × 10−8 = 0.0003 × 10−3 = 0.3 × 10−6 = 300 × 10−9
−5

+3

+3

−3
25.

a.

0.05 × 100 s = 50 × 10−3 s = 50 ms
+3
+3

b.

2000 × 10−6 s = 2 × 10−3 s = 2 ms
−3
−3

c.

0.04 × 10−3 s = 40 × 10−6 s = 40 μs
+3
+6

d.

8400 × 10−12 s ⇒ 0.0084 × 10−6 s = 0.0084 μs
−6
−3

e.

4 × 10−3 × 103 m = 4 × 100 m = 4000 × 10−3 m = 4000 mm
+3
100

f.

increase by 3

260 × 103 × 10−3 m = 0.26 × 103 m = 0.26 km
−3

26.

a.
b.
c.
d.

⎡ 60 s ⎤
1.5 min ⎢
⎥ = 90 s
⎣1 min ⎦
⎡ 60 min ⎤ ⎡ 60 s ⎤
0.04 h ⎢
⎥⎢
⎥ = 144 s
⎣ 1 h ⎦ ⎣1 min ⎦
⎡ 1 μs ⎤
0.05 s ⎢ −6 ⎥ = 0.05 × 106 μs = 50 × 103 μs
⎣10 s ⎦
⎡ 1 mm ⎤
0.16 m ⎢ −3 ⎥ = 0.16 × 103 mm = 160 mm
⎣10 m ⎦

Chapter 1

4

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27.

28.

e.

⎡ 1 ns ⎤
1.2 × 10−7 s ⎢ −9 ⎥ = 1.2 × 102 ns = 120 ns
⎣10 s ⎦

f.

⎡1 min ⎤ ⎡ 1 h ⎤ ⎡1 day ⎤
3.62 × 106 s ⎢
⎥ = 41.90 days
⎥⎢
⎥⎢
⎣ 60 s ⎦ ⎣ 60 min ⎦ ⎣ 24 h ⎦

a.

⎡10 −6 F ⎤ ⎡ 1 pF ⎤
−6
12
5
0.1 μF ⎢
⎥ ⎢ −12 ⎥ = 0.1 × 10 × 10 pF = 10 pF
1
F
μ
10
F
⎦
⎣
⎦⎣

b.

⎡100 cm ⎤
= 8000 × 10−3 cm = 8 cm
80 × 10−3 m ⎢
⎥
⎣ 1m ⎦

c.

⎡ 1 m ⎤ ⎡ 1 km ⎤
−5
60 cm ⎢
⎥⎢
⎥ = 60 × 10 km
⎣100 cm ⎦ ⎣1000 m ⎦

d.

⎡ 60 min ⎤
3.2 h ⎢
⎥
⎣ 1h ⎦

e.

⎡10−3 m ⎤ ⎡ 1 μ m ⎤
3
0.016 mm ⎢
⎥ ⎢ − 6 ⎥ = 0.016 × 10 μm = 16 μm

⎣ 1 mm ⎦ ⎣10 m ⎦

f.

⎡ 1m ⎤⎡ 1m ⎤
−4
2
60 cm2 ⎢
⎥ ⎢100 cm ⎥ = 60 × 10 m
100
cm
⎣
⎦⎣
⎦

a.

⎡ 1m ⎤
100 in. ⎢
⎥ = 2.54 m
⎣ 39.37 in. ⎦

b.

⎡12 in. ⎤ ⎡ 1 m ⎤
4 ft ⎢
⎥ = 1.22 m
⎥⎢
⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦

c.

⎡ 4.45 N ⎤
6 lb ⎢
⎥ = 26.7 N
⎣ 1 lb ⎦

d.

⎡ 1 N ⎤ ⎡ 1 lb ⎤
60 × 103 dynes ⎢ 5
⎥⎢
⎥ = 0.13 lb
⎣10 dynes ⎦ ⎣ 4.45 N ⎦

e.

⎡ 1 in. ⎤
150,000 cm ⎢
⎥
⎣ 2.54 cm ⎦

f.

⎡ 5280 ft ⎤ ⎡12 in. ⎤ ⎡ 1 m ⎤
0.002 mi ⎢
⎥⎢
⎥⎢
⎥ = 3.22 m
⎣ 1 mi ⎦ ⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦

⎡ 1 ms ⎤
6
⎢10− 3 s ⎥ = 11.52 × 10 ms
⎦
⎣

⎡ 60 s ⎤
⎢1 min ⎥
⎦
⎣

⎡ 1 ft ⎤
⎢12 in. ⎥ = 4921.26 ft
⎣
⎦

Chapter 1

5

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29.

⎡1 yd ⎤
5280 ft ⎢
⎥ = 1760 yds
⎣ 3 ft ⎦

⎡ 12 in. ⎤ ⎡ 1 m ⎤
5280 ft ⎢
⎥⎢
⎥ = 1609.35 m, 1.61 km
⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦

5280 ft,

m ⎡ 39.37 in. ⎤ ⎡ 1 ft ⎤ ⎡ 1 mi ⎤ ⎡ 60 s ⎤ ⎡ 60 min ⎤
s ⎢⎣ 1 m ⎥⎦ ⎢⎣12 in. ⎥⎦ ⎢⎣ 5280 ft ⎥⎦ ⎢⎣1 min ⎥⎦ ⎢⎣ 1 h ⎥⎦
= 670,615,288.1 mph ≅ 670.62 × 106 mph

30.

299,792,458

31.

⎡ 3 ft ⎤
100 yds ⎢
⎥
⎣1 yd ⎦

⎡ 1 mi ⎤
⎢ 5,280 ft ⎥ = 0.0568 mi
⎣
⎦

60 mi ⎡ 1 h ⎤ ⎡1 min ⎤
= 0.0167 mi/s

h ⎢⎣ 60 min ⎥⎦ ⎢⎣ 60 s ⎥⎦

t=

32.

33.

34.

d

υ

=

0.0568 mi
= 3.40 s
0.0167 mi/s

30 mi ⎡ 5280 ft ⎤ ⎡12 in. ⎤ ⎡ 1 m ⎤ ⎡ 1 h ⎤ ⎡ 1 min ⎤
= 13.41 m/s
h ⎢⎣ 1 mi ⎥⎦ ⎢⎣ 1 ft ⎥⎦ ⎢⎣ 39.37 in. ⎥⎦ ⎢⎣ 60 min ⎥⎦ ⎢⎣ 60 s ⎥⎦
50 yd ⎡ 60 min ⎤ ⎡ 3 ft ⎤ ⎡ 1 mi ⎤
= 1.705 mi/h
⎢
⎥
min ⎢⎣ 1 h ⎥⎦ ⎣1 yd ⎦ ⎢⎣ 5,280 ft ⎥⎦
3000 mi
d

⎡1 day ⎤
t= =
= 1760 h ⎢
⎥ = 73.33 days
υ 1.705 mi/h
⎣ 24 h ⎦
⎡1000 m ⎤ ⎡ 39.37 in. ⎤ ⎡ 1 ft ⎤ ⎡ 1 mi ⎤
10 km ⎢
⎥⎢
⎥⎢
⎥⎢
⎥
⎣ 1 km ⎦ ⎣ 1 m ⎦ ⎣12 in. ⎦ ⎣ 5280 ft ⎦
1 mi
d 6.214 mi
= 40.39 min
υ=
,t= =
1 mi
6.5 min
υ
6.5 min

= 6.214 mi

35.

⎡ 3 ft ⎤ ⎡12 in. ⎤
100 yds ⎢
⎥⎢

⎥ = 3600 in ⇒ 3600 quarters
⎣1 yd ⎦ ⎣ 1 ft ⎦

36.

60 mph:

100 mi
= 1.67 h = 1 h:40.2 min
υ 60 mi/h
d 100 mi
75 mph:
t= =
= 1.33 h = 1h:19.98 min
υ 75 mi/h
difference = 20.22 minutes

t=

d

=

Chapter 1

6

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37.

cm ⎤
⎡
[0.016 h ] ⎡⎢ 60 min ⎤⎥
d = υt = ⎢600
⎥
s ⎦
⎣ 1h ⎦
⎣

38.

⎡ 14 ft ⎤
d = 86 stories ⎢
⎥
⎣ story ⎦

υ=

39.

40.

⎡ 1m ⎤
⎢100 cm ⎥ = 345.6 m
⎦
⎣

⎡

⎤
⎢1 step ⎥
⎢ 9 ⎥ = 1605 steps
⎢ ft ⎥
⎣⎢ 12 ⎦⎥

d
d 1605 steps
⎡ 1 minute ⎤
⇒t = =
= 802.5 seconds ⎢
⎥ = 13.38 minutes
2 steps
t
υ
⎣ 60 seconds ⎦
second

⎡ 14 ft ⎤
⎡ 1 mile ⎤
d = (86 stories) ⎢
⎥ = 1204 ft ⎢
⎥ = 0.228 miles
story
⎣ 5,280 ft ⎦
⎣
⎦
min 10.7833 min
=
= 47.30 min/mile

mile 0.228 miles
5 min
1 mile ⎡ 5,280 ft ⎤ 1056 ft
,
=
⇒
mile
5 min ⎢⎣ 1mile ⎥⎦ minute

υ=

41.

⎡ 60 s ⎤
⎢1 min ⎥
⎦
⎣

1204 ft
d
d
= 1.14 minutes
⇒t = =
υ 1056 ft
t
min

a.

⎡ 1 Btu ⎤

−3
5 J⎢
⎥ = 4.74 × 10 Btu
⎣1054.35 J ⎦

b.

⎡ 1 gallon ⎤
24 ounces ⎢
⎥
⎣128 ounces ⎦

c.
d.

⎡ 14 ft ⎤
distance = 86 stories ⎢
⎥ = 1204 ft
⎣ story ⎦

⎡
⎤
1 m3
−4 3
⎢
⎥ = 7.1 × 10 m
⎣ 264.172 gallons ⎦

⎡ 86,400 s ⎤
5

1.4 days ⎢
⎥ = 1.21 × 10 s
1
day
⎣
⎦
⎡ 264.172 gallons ⎤ ⎡ 8 pints ⎤
1 m3 ⎢
⎥ ⎢1 gallon ⎥ = 2113.38 pints
1 m3
⎣
⎦⎣
⎦

42.

6(4 + 8) = 72

43.

(20 + 32)/4 = 13
(82 + 122 ) = 14.42

44.
45.

MODE = DEGREES: cos 50° = 0.64

46.

MODE = DEGREES: tan−1(3/4) = 36.87°

Chapter 1

7

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47.

( 400 /(6

2

)

+ 10) = 2.95

48.

205 × 10−6

49.

1.20 × 1012

50.

6.667 × 106 + 0.5 × 106 = 7.17 × 106

Chapter 1

8

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Chapter 2
1.

−

2.

a.

F= k

Q1Q2 (9 × 109 )(1 C)(2 C)
=
= 18 × 109 N
r2
(1 m)2

b.

F=k

(9 × 109 )(1 C)(2 C )

Q1Q2
= 2 × 109 N
=
r2
(3 m)2

c.

F= k

Q1Q2 (9 × 109 )(1 C)(2 C)
=
= 0.18 × 109 N
2
2
r
(10 m)

d.

Exponentially,

a.

r = 1 mi:

3.

r3 10 m
F

18 × 109 N
= 100
=
= 10 while 1 =
r1 1 m
F2 0.18 × 109 N

⎡ 5280 ft ⎤ ⎡12 in. ⎤ ⎡ 1 m ⎤
1 mi ⎢
⎥ = 1609.35 m
⎥⎢
⎥⎢
⎣ 1 mi ⎦ ⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦
kQ1Q2 (9 × 109 )(8 × 10−6 C)(40 × 10-6 C) 2880 × 10−3
=
=
F=
(1609.35 m) 2
2.59 × 106
r2
= 1.11 μN

b.

r =10 ft:
⎡12 in. ⎤ ⎡ 1 m ⎤
10 ft ⎢
⎥⎢
⎥ = 3.05 m
⎣ 1 ft ⎦ ⎣ 39.37 in. ⎦

kQ1Q2 2880 × 10−3 2880 × 10−3
=
=
F=
= 0.31 N
9.30
r2
(3.05 m) 2

c.

1 in. ⎡ 1 m ⎤
= 1.59 mm
16 ⎢⎣ 39.37 in. ⎥⎦

kQ1Q2
2880 × 10−3
2880 × 10−3
=
=
= 1138.34 × 103 N
(1.59 × 10−3 m) 2 2.53 × 10−6
r2
= 1138.34 kN

F=

4.

−

5.

F=

kQ1Q2
⇒r=
r2

kQ1Q2
=
F

(9 × 109 )(20 × 10 −6 ) 2
= 10 mm
3.6 × 104

Chapter 2

9

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6.

F=

a.
b.

kQ1Q2
kQ Q
⇒ 1.8 = 1 22 ⇒ kQ1Q2 = 4(1.8) = 7.2
2
r
(2 m)
7 .2
kQ1Q2
=
= 72 mN
2
r
(10) 2
Q1/Q2 = 1/2 ⇒ Q2 = 2Q1
7.2 = kQ1Q2 = (9 × 109)(Q1)(2Q1) = 9 × 109 2Q12

F=

( )

7 .2
7 .2
= Q12 ⇒ Q1 =
= 20 μC
9
18 × 10
18 × 109
Q2 = 2Q1 = 2(2 × 10−5 C) = 40 μC
W

1.2 J
=
= 3 kV
Q 0.4 mC

7.

V=

8.

W = VQ = (60 V)(8 mC) = 0.48 J

9.

Q=

W 96 J
=
=6C
V 16 V

10.

Q=

W 72 J
=
=8C
V

9V

11.

I=

Q 12 mC
=
= 4.29 mA
t
2.8 s

12.

I =

Q
312 C
= 2.60 A
=
t (2)(60 s)

13.

Q = It = (40 mA)(0.8)(60 s) = 1.92 C

14.

Q = It = (250 mA)(1.2)(60 s) = 18.0 C

15.

t=

16.

17.

Q 6 mC
=
=3s
I
2 mA

1C
⎡
⎤
21.847 × 1018 electrons ⎢
18
⎥ = 3.5 C
⎣ 6.242 × 10 electrons ⎦
Q 3.5 C
=
I=
= 0.29 A
t
12 s

Q = It = (4 mA)(90 s) = 360 mC
⎡ 6.242 × 1018 electrons ⎤

18
360 mC ⎢
⎥ = 2.25 × 10 electrons
1
C
⎦
⎣

Chapter 2

10

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18.

19.

20.

I=

Q
86 C
=
= 1.194 A > 1 A (yes)
t
(1.2)(60 s)

1C
⎡
⎤
0.84 × 1016 electrons ⎢
18
⎥ = 1.346 mC
⎣ 6.242 × 10 electrons ⎦
Q 1.346 mC
=
I=
= 22.43 mA
t
60 ms

a.

Q = It = (2 mA)(0.01 μs) = 2 × 10−11 C
⎡ 6.242 × 1018 electrons 1 Â
2 ì 1011 C


1C

electron
= 1.25 ì 108 Â = $1.25 ì 106 = 1.25 million

Q = It = (100 μA)(1.5 ns) = 1.5 × 10−13 C
⎡ 6.242 × 1018 electrons ⎤ ⎡ $1 ⎤
1.5 × 10−13 C ⎢
⎥⎢

⎥ = 0.94 million
1C
⎦ ⎣ electron ⎦
⎣
(a) > (b)
b.

21.

22.

Q = It = (200 × 10−3 A)(30 s) = 6 C
W 40 J
=
V=
= 6.67 V
Q 6C
⎡ 420 C ⎤
Q = It = ⎢
⎥ (0.5 min) = 210 C
⎣ min ⎦
W
742 J
V=
= 3.53 V
=
Q 210 C

W 0.4 J
=

= 0.0167 C
V
24 V
Q 0.0167 C
I=
=
= 3.34 A
t 5 × 10−3 s

23.

Q=

24.

I=

25.

Ah = (0.8 A)(75 h) = 60.0 Ah

26.

t(hours) =

Ah rating 200 Ah
=
=5A
t (hours)
40 h

Ah rating 32 Ah
=
= 25 h
I
1.28 A

Chapter 2

11

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27.

⎡ 60 min ⎤ ⎡ 60 s ⎤
6
40 Ah(for 1 h): W1 = VQ = V⋅I⋅t = (12 V)(40 A)(1 h) ⎢
⎥ ⎢1 min ⎥ = 1.728 × 10 J
1
h
⎣
⎦⎣
⎦
60
min
60
s
⎡

⎤⎡
⎤
6
60 Ah(for 1 h): W2 = (12 V)(60 A)(1 h) ⎢
⎥ ⎢1 min ⎥ = 2.592 × 10 J
1
h
⎦
⎣
⎦⎣
Ratio W2/W1 = 1.5 or 50% more energy available with 60 Ah rating.
⎡1 min ⎤ ⎡ 1 h ⎤
= I(16.67 × 10−3 h)
For 60 s discharge: 40 Ah = It = I [60 s] ⎢
⎥
⎢
⎥
⎣ 60 s ⎦ ⎣ 60 min ⎦
40 Ah
and I =
= 2400 A
16.67 × 10-3 h
⎡1 min ⎤ ⎡ 1 h ⎤
= I(16.67 × 10−3 h)
60 Ah = It = I [60 s] ⎢
⎥
⎢
⎥
⎣ 60 s ⎦ ⎣ 60 min ⎦
60 Ah

= 3600 A
and I =
16.67 × 10-3 h

I2/I1 = 1.5 or 50 % more starting current available at 60 Ah
28.

I=

3 Ah
= 500 mA
6.0 h

⎡ 60 min ⎤ ⎡ 60 s ⎤
Q = It = (500 mA)(6 h) ⎢
⎥⎢
⎥ = 10.80 kC
⎣ 1 h ⎦ ⎣1 min ⎦
W = QV = (10.8 kC)(12 V) ≅ 129.6 kJ

29.

−

30.

−

31.

−

32.

−

33.

−

34.

−

35.

36.

⎡ 60 s ⎤
4 min ⎢
⎥ = 240 s
⎣1 min ⎦
Q = It = (2.5 A)(240 s) = 600 C

Q = It = (10 × 10−3 A)(20 s) = 200 mC
W = VQ = (12.5 V)(200 × 10−3 C) = 2.5 J

Chapter 2

12

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Chapter 3
1.

2.

3.

a.

0.5 in. = 500 mils

b.

⎡1000 mils ⎤
0.02 in. ⎢
⎥ = 20 mils
⎣ 1 in. ⎦

c.

1
⎡1000 mils ⎤
in. = 0.25 in. ⎢
⎥ = 250 mils
4
⎣ 1 in. ⎦

d.

1 in. = 1000 mils

e.

3
⎡12 in. ⎤ ⎡10 mils ⎤
0.02 ft ⎢
⎢
⎥ = 240 mils
⎥
⎣ 1 ft ⎦ ⎣ 1 in. ⎦

f.

⎡ 1 in. ⎤
0.1 cm ⎢
⎥
⎣ 2.54 cm ⎦

a.

ACM = (30 mils)2 = 900 CM

b.

0.016 in. = 16 mils, ACM = (16 mils)2 = 256 CM

c.

1"
= 0.125" = 125 mils, ACM = (125 mils)2 = 15.63 × 103 CM
8

d.

⎡ 1 in. ⎤ ⎡1000 mils ⎤
= 393.7 mils, ACM = (393.7 mils)2 = 155 × 103 CM
1 cm ⎢
⎥
⎢
⎥
⎣ 2.54 cm ⎦ ⎣ 1 in. ⎦

e.

⎡12 in. ⎤ ⎡1000 mils ⎤
= 240 mils, ACM = (240 mils)2 = 57.60 × 103 CM
0.02 ft ⎢
⎥
⎢
⎥
⎣ 1 ft ⎦ ⎣ 1 in. ⎦

f.

⎡ 39.37 in. ⎤
2

3
0.0042 m ⎢
⎥ = 0.1654 in. = 165.4 mils, ACM = (165.4 mils) = 27.36 × 10 CM
⎣ 1m ⎦

ACM = (dmils )2 → dmils =

⎡1000 mils ⎤
⎢ 1 in. ⎥ = 39.37 mils
⎦
⎣

ACM

a.

d = 1600 CM = 40 mils = 0.04 in.

b.

d=

820 CM = 28.64 mils = 0.029 in.

c.

d=

40,000 CM = 200 mils = 0.2 in.

Chapter 3

13

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d.

d=

625 CM = 25 mils = 0.025 in.

e.

d=

6.25 CM = 2.5 mils = 0.0025 in.

f.

d = 100 CM = 10 mils = 0.01 in.

4.

0.01 in. = 10 mils, ACM = (10 mils)2 = 100 CM
l
(200′)
R = ρ = (10.37)
= 20.74 Ω

A
100 CM

5.

ACM = (4 mils)2 = 16 CM, R = ρ

6.

a.

A= ρ

⎛ 80′ ⎞
l
⎟⎟ = 544 CM
= 17⎜⎜
R
⎝ 2 .5 Ω ⎠

b.

d=

ACM = 544 CM = 23.32 mils = 23.3 × 10−3 in.

7.

8.

1 "
= 0.03125" = 31.25 mils, ACM = (31.25 mils)2 = 976.56 CM
32
l
RA (2.2 Ω)(976.56 CM)
=
R=ρ ⇒l=
= 3.58 ft
R
ρ
600

a.

ACM = ρ
d=

9.

larger

c.

smaller

a.

ρ =

l

(10.37)(300′)
=
= 942.73 CM
3.3 Ω
A

942.73 CM = 30.70 mils = 30.7 × 10−3 in.

b.

b.

10.

(150 ft)
l
= 92.81 Ω
= (9.9)
A
16 CM

Rsilver > Rcopper > Raluminum
l
(9.9)(10 ft)
=
= 99 Ω { ACM = (1 mil)2 = 1 CM
1 CM
A
l
(10.37)(50 ft)

Copper: R = ρ =
= 5.19 Ω { ACM = (10 mils)2 = 100 CM
100 CM
A
l
(17)(200 ft)
Aluminum: R = ρ =
= 1.36 Ω { ACM = (50 mils)2 = 2500 CM
2500 CM
A

Silver: R = ρ

RA (500 Ω)(94 CM)
= 47 ⇒ nickel
=
l
1000′

Chapter 3

14

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11.

a.

3" = 3000 mils, 1/2" = 0.5 in. = 500 mils
Area = (3 × 103 mils)(5 × 102 mils) = 15 × 105 sq. mils
⎡ 4 / π CM ⎤
5
15 × 105 sq mils ⎢
⎥ = 19.108 × 10 CM
1
sq
mil
⎣
⎦
R= ρ

b.

l
(10.37)(4′)
=
= 21.71 μΩ
A 19.108 × 105 CM

l
(17)(4′)
=
= 35.59 μΩ
A 19.108 × 105 CM
Aluminum bus-bar has almost 64% higher resistance.

R= ρ

c.

increases

d.

decreases

12.

l2 = 2l1, A2 = A1/4, ρ2 = ρ1
ρ 2 l2
ρl A
2l A
R2
A
= 2 = 22 1 = 1 1 =8
ρ
l
ρ1l1 A2 l1 A1 / 4
R1
11
A1
and R2 = 8R1 = 8(0.2 Ω) = 1.6 Ω
ΔR = 1.6 Ω − 0.2 Ω = 1.4 Ω

13.

A=

πd 2
4

⇒d=

4A

π

=

4(0.04 in.2 )

π

= 0.2257 in.

dmils = 225.7 mils
ACM = (225.7 mils)2 = 50,940.49 CM
l
ρ1 1
R1
ρlA
lA
A1
(ρ1 = ρ2)
=
= 11 2 = 1 2
R2 ρ l2
ρ 2l2 A1 l2 A1

2
A2
R l A (800 mΩ)(300 ft)(40,000 CM)
and R2 = 1 2 1 =
= 942.28 mΩ
l1 A2
(200 ft)(50,940.49 CM)
14.

a.

⎡1.260 Ω ⎤
#11: 450 ft ⎢
⎥ = 0.567 Ω
⎣ 1000 ft ⎦
⎡ 2.525 Ω ⎤
#14: 450 ft ⎢
⎥ = 1.136 Ω
⎣ 1000 ft ⎦

b.

Resistance: #14:#11 = 1.136 Ω:0.567 Ω ≅ 2:1

c.

Area:

#14:#11 = 4106.8 CM:8234.0 CM ≅ 1:2

Chapter 3

15

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15.

a.

⎡ 0.6282 Ω ⎤
#8: R = 1800 ft ⎢
⎥ = 1.13 Ω
⎣ 1000 ft ⎦

⎡ 6.385 Ω ⎤
#18: R = 1800 ft ⎢
⎥ = 11.49 Ω
⎣ 1000 ft ⎦

16.

17.

18.

b.

#18:#8 = 11.49 Ω:1.13 Ω = 10.17:1 ≅ 10:1

c.

#18:#8 = 1624.3 CM:16,509 CM = 1:10.16 ≅ 1:10

a.

A= ρ

b.

A= ρ

a.

A/CM = 230 A/211,600 CM = 1.09 mA/CM

b.

⎡
⎤
1.09 mA ⎢ 1 CM ⎥ ⎡1000 mils ⎤ ⎡1000 mils ⎤
= 1.39 kA/in.2
⎢π
⎥⎢
⎥
⎢
⎥
CM ⎢ sq mils ⎥ ⎣ 1 in. ⎦ ⎣ 1 in. ⎦
⎣4

⎦

c.

⎡ 1 in.2 ⎤
2
5 kA ⎢
⎥ = 3.6 in.
⎣1.39 kA ⎦

l (10.37)(30′) 311.1 CM
=
= 51,850 CM ⇒ #3
=
R
6 mΩ
6 × 10−3
but 110 A ⇒ #2
l (10.37)(30′) 311.1 CM
=
=
= 103,700 CM ⇒ #0
R
3 mΩ
3 × 10−3

1
⎛ 2.54 cm ⎞
in. = 0.083 in. ⎜
⎟ = 0.21 cm

12
⎝ 1 in. ⎠

πd2

(3.14)(0.21 cm) 2
= 0.035 cm2
4
4
RA (2 Ω)(0.035 cm 2 )
l=
=
= 40,603 cm = 406.03 m
ρ
1.724 × 10−6
A=

19.

a.

b.

=

1 " ⎡ 2.54 cm ⎤
⎡ 2.54 cm ⎤
= 1.27 cm, 3 in. ⎢
⎢
⎥

⎥ = 7.62 cm
2 ⎣ 1" ⎦
⎣ 1 in. ⎦
⎡12 in. ⎤ ⎡ 2.54 cm ⎤
4 ft ⎢
⎥⎢
⎥ = 121.92 cm
⎣ 1 ft ⎦ ⎣ 1 in. ⎦
l (1.724 x 10 -6)(121.92 cm)
R= ρ
= 21.71 μΩ
A
(1.27 cm)(7.62 cm)
R= ρ

A (2.825 × 10−6 )(121.92 cm)
=
= 35.59 μΩ
A
(1.27 cm)(7.62 cm)

Chapter 3

16

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c.

increases

d.

decreases

ρ

= 100 ⇒ d =

ρ

250 × 10−6
= 2.5 μcm
100

20.

Rs =

21.

R l (150 Ω)(1/ 2 in.)
= 0.15 in.
R = Rs l ⇒ w = s =
R
500 Ω
w

22.

a.

d = 1 in. = 1000 mils
ACM = (103 mils)2 = 106 CM
RA (1 mΩ)(106 CM)
ρ1 =
=
= 1 CM-Ω/ft
l
103 ft

b.

1 in. = 2.54 cm
π d 2 π (2.54 cm)2
A=
=
= 5.067 cm2
4
4
⎡12 in. ⎤ ⎡ 2.54 cm ⎤
l = 1000 ft ⎢
⎥⎢
⎥ = 30,480 cm
⎣ 1 ft ⎦ ⎣ 1 in. ⎦

d

ρ2 =

c.

23.

24.

25.

k=

100

=

RA (1 mΩ)(5.067 cm 2 )
=
= 1.66 × 10−7 Ω-cm
l
30,480 cm

ρ 2 1.66 × 10−7 Ω-cm
=
= 1.66 × 10−7
1 CM-Ω / ft
ρ1

234.5 + 10 234.5 + 80
,
=

2Ω
R2

R2 =

(314.5)(2 Ω)
= 2.57 Ω
244.5

236 + 0 236 + 100
=
0.02 Ω
R2
(0.02 Ω)(336)
= 0.028 Ω
R2 =
236
5
5
(°F − 32) = (32 − 32) = 0° (=32°F)
9
9
5
C = (70 − 32) = 21.11° (=70°F)
9
234.5° + 21.11° 234.5° + 0°
=
4Ω
R2
(234.5)(4 Ω)

= 3.67 Ω
R2 =
255.61

C=

Chapter 3

17

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26.

27.

28.

234.5 + 30 234.5 − 40
=
R2
0.76 Ω
(194.5)(0.76 Ω)
= 0.56 Ω
R2 =
264.5

243 + (−30) 243 + 0
=

0.04 Ω
R2
(243)(40 mΩ)
= 46 mΩ
R2 =
213
a.

b.

29.

a.
b.

30.

68°F = 20°C, 32°F = 0°C
234.5 + 20 234.5 + 0
=
0.002
R2
(234.5)(2 m Ω)
= 1.84 mΩ
R2 =
254.5
212°F = 100 °C
234.5 + 20 234.5 + 100
=
2 mΩ

R2
(334.5)(2 m Ω)
= 2.63 mΩ
R2 =
254.5
ΔR 2.63 m Ω − 2 m Ω 0.63 m Ω
=
=
= 7.88 μΩ/°C or 7.88 × 10-5 Ω/10°C
100° C − 20° C
80° C
ΔT
234.5 + 4 234.5 + t2
=
,
1Ω
1.1 Ω
234.5 + 4 234.5 + t2
=
,
1Ω
0.1 Ω

t2 = 27.85°C
t2 = −210.65°C

a.

K = 273.15 + °C
50 = 273.15 + °C

°C = −223.15°
234.5 + 20 234.5 − 223.15
=
R2
10 Ω
11.35
(10 Ω) = 0.446 Ω
R2 =
254.5

c.

F=

b.

K = 273.15 + °C
38.65 = 273.15 + °C
°C = −234.5°
234.5 + 20 234.5 − 234.5
=
R2
10 Ω
(0)10 Ω
=0Ω
R2 =
254.5
Recall: −234.5° =
Inferred absolute zero
R=0Ω

9
9
°C + 32 = (−273.15°) + 32 = −459.67°
5
5

Chapter 3

18

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31.

1
1
1
=
=
= 0.003929 ≅ 0.00393
Ti + 20°C 234.5 + 20 254.5

a.

α20 =

b.

R = R20[1 + α20(t − 20°C)]
1 Ω = 0.8 Ω[1 + 0.00393(t − 20°)]
1.25 = 1 + 0.00393t − 0.0786
1.25 − 0.9214 = 0.00393t
0.3286 = 0.00393t
0.3286
= 83.61°C
t=
0.00393

32.

R = R20[1 + α20(t − 20°C)]
= 0.4 Ω[1 + 0.00393(16 − 20)] = 0.4 Ω[1 − 0.01572] = 0.39 Ω

33.

Table: 1000′ of #12 copper wire = 1.588 Ω @ 20°C
5
5
C° = (F° − 32) = (115 − 32) = 46.11°C
9
9
R = R20[1 + α20(t − 20°C)]
= 1.588 Ω[1 + 0.00393(46.11 − 20)]
= 1.75 Ω

34.

ΔR =

35.

ΔR =

36.

−

37.

−

38.

#12: Area = 6529 CM

Rnominal
22 Ω
(200)(65° − 20°) = 0.198 Ω
(PPM)( ΔT) =
106
106
R = Rnominal + ΔR = 22.198 Ω
Rnominal
100 Ω
(100)(50° − 20°) = 0.30 Ω
(PPM)( ΔT) =
6
10

106
R = Rnominal + ΔR = 100 Ω + 0.30 Ω = 100.30 Ω

⎡ 2.54 cm ⎤
6529 CM = 80.8 mils = 0.0808 in. ⎢
⎥ = 0.205 cm
⎣ 1 in. ⎦
πd 2 π (0.205 cm)2
A=
=
= 0.033 cm2
4
4
1 MA
I=
[0.033 cm 2 ] = 33 kA >>> 20 A
cm2
d=

39.

−

40.

a.

41.

10 kΩ − 3.5 kΩ = 6.5 kΩ

2 times larger

b.

4 times larger

Chapter 3

19

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42.

6.25 kΩ and 18.75 kΩ

43.

−

44.

a.
b.
c.

560 kΩ ± 5%, 560 kΩ ± 28 kΩ, 532 kΩ ↔ 588 kΩ
220 Ω ± 10%, 220 Ω ± 22 Ω, 198 Ω ↔ 242 Ω

100 Ω ± 20%, 100 Ω ± 20 Ω, 80 Ω ↔ 120 Ω

45.

a.
b.
c.
d.

120 Ω = Brown, Red, Brown, Silver
8.2 Ω = Gray, Red, Gold, Silver
6.8 kΩ = Blue, Gray, Red, Silver
3.3 MΩ = Orange, Orange, Green, Silver

46.

10 Ω ± 20% ⇒ 8 Ω − 12 Ω ⎫
⎬ no overlap, continuance
15 Ω ± 20% ⇒ 12 Ω − 18 Ω ⎭

47.

10 Ω ± 10% ⇒ 10 Ω ± 1 Ω = 9 Ω − 11 Ω

⎫
⎬ No overlap
15 Ω ± 10% ⇒ 15 Ω ± 1.5 Ω = 13.5 Ω − 16.5 Ω ⎭

48.

a.
b.
c.
d.

621 = 62 × 101 Ω = 620 Ω = 0.62 kΩ
333 = 33 × 103 Ω = 33 kΩ
Q2 = 3.9 × 102 Ω = 390 Ω
C6 = 1.2 × 106 Ω = 1.2 MΩ

49.

a.

G=

1
1
=
= 8.33 mS
R 120 Ω

b.

G=

1
= 0.25 mS
4 kΩ

c.

G=

a.

Table 3.2, Ω/1000′ = 1.588 Ω
1
1
= 629.72 mS
G= =
R 1.588 Ω
A 6529.9 CM (Table 3.2)
or G =
=
= 629.69 mS (Cu)
ρl
(10.37)(1000′)

b.

G=

6529.9 CM
= 384.11 mS (Al)
(17)(1000′)

c.

G=

6529.9 CM
= 88.24 mS (Fe)
(74)(1000′)

50.

1
= 0.46 μS
2 .2 M Ω
Ga > Gb > Gc vs. Rc > Rb > Ra

Chapter 3

20

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51.

l
2
5
⎛ 2⎞
A1 = A1, l2 = ⎜1 − ⎟ l1 = 1 , ρ2 = ρ1
3
3
3
⎝ 3⎠

A
⎛ l1 ⎞
ρ1 1
A
G1
l1
ρ 2l2 A1 ⎜⎝ 3 ⎟⎠ 1 1
=
=
=
=
G2 ρ A2 ρ1l1 A2
⎛5 ⎞ 5
l1 ⎜ A1 ⎟
2
l2
⎝3 ⎠

A2 = 1

G2 = 5G1 = 5(100 S) = 500 S
52.

−

53.

−

54.

−

55.

a.

−50°C specific resistance ≅ 105 Ω-cm
50°C specific resistance ≅ 500 Ω-cm
200°C specific resistance ≅ 7 Ω-cm

b.

negative

c.

No

d.

ρ=

a.

Log scale:

10 fc ⇒ 3 kΩ
100 fc ⇒ 0.4 kΩ

b.

negative

c.

no—log scales imply linearity

d.

1 kΩ ⇒ ≅ 30 fc
10 kΩ ⇒ ≅ 2 fc
10 k Ω − 1 k Ω
ΔR
=
= 321.43 Ω/fc
30 fc − 2 fc
Δfc

56.

ΔΩ − cm 300 − 30 270 Ω − cm
=
=
≅ 3.6 Ω-cm/°C
ΔT
125 − 50
75° C

and

57.

a.

@ 0.5 mA, V ≅ 195 V
@ 1 mA, V ≅ 200 V
@ 5 mA, V ≅ 215 V

b.

ΔVtotal = 215 V − 195 V = 20 V

c.

5 mA:0.5 mA = 10:1
compared to
215 V: 200 V = 1.08:1

Chapter 3

ΔR
= −321.43 Ω/fc
Δ fc

21

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Chapter 4

1.

V = IR = (2.5 A)(47 Ω) = 117.5 V

2.

I=

V 12 V
=
= 1.76 A
R 6.8 Ω

3.

R=

V
6V
=
= 4 kΩ
I 1.5 mA

4.

I=

V
12 V
=

= 300 A
R 40 × 10−3 Ω

5.

V = IR = (3.6 μA)(0.02 MΩ) = 0.072 V = 72 mV

6.

I=

V
62 V
=
= 4.13 mA
R 15 kΩ

7.

R=

V 120 V
=
= 54.55 Ω
I 2.2 A

8.

I=

V
120 V
=
= 16 mA
R 7.5 kΩ

9.

R=

V 120 V
=
= 28.57 Ω
I 4.2 A

10.

R=

V
4.5 V
=
= 36 Ω
I 125 m A

11.

R=

V 24 mV

=
= 1.2 kΩ
I 20 μ A

12.

V = IR = (15 A)(0.5 Ω) = 7.5 V

13.

a.

R=

V 120 V
=
= 12.63 Ω
I 9.5 A

b.

⎡ 60 min ⎤ ⎡ 60 s ⎤
t = 1 h⎢
⎥⎢
⎥ = 3600 s
⎣ 1 h ⎦ ⎣1 min ⎦

W = Pt = VIt
= (120 V)(9.5 A)(3600 s)
= 4.1 × 106 J

14.

V = IR = (2.4 μA)(3.3 MΩ) = 7.92 V

15.

−

Chapter 4

22

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