Solution of Introductory Circuit Analysis [11th Edition]

Instructor’s Resource Manual
to accompany
Introductory
Circuit Analysis
Eleventh Edition
Robert L. Boylestad
Upper Saddle River, New Jersey
Columbus, Ohio

iii
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

Chapter 1 1
Chapter 1
1. −
2. −
3. −
4.
υ = t
d= ⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
h 1
min60
min1
s 60
ft 5,280
mi 1
s 10
ft 000,20 = 1363.64 mph
5. 4 min ⎥
⎦
⎤
⎢
⎣
⎡
min 60
h 1 = 0.067 h
υ = h 1.067
mi 31
=
t
d = 29.05 mph
6. a. ⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
s 60
min 1
min 60
h 1
mi
ft 280,5
h
mi 95 = 139.33 ft/s
b. t = ft/s 139.33
ft 60
=
υ
d = 0.431 s
c.
υ = ⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
=ft 5,280
mi 1
h 1
min60
min 1
s 60
s 1
ft 60
t
d = 40.91 mph
7. −
8. −
9. −
10. MKS, CGS, °C = )36(
9
5
)3268(
9
5
)32F(
9
5=−=−° = 20°
SI: K = 273.15 + °C = 273.15 + 20 = 293.15
11. ⎥
⎦
⎤
⎢
⎣
⎡
J 1
lb-ft 0.7378
J 1000 = 737.8 ft-lbs
12. 0.5 yd ⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
⎥
⎦
⎤
⎢
⎣
⎡
in. 1
cm 54.2
ft 1
in. 12
yd 1
ft 3 = 45.72 cm
13. a.
104 b. 106 c. 103 d. 10−3 e. 100 f. 10−1
14. a.
15 × 103 b. 30 × 10−3
c. 2.4 × 106 d. 150 × 103
e.
4.02 × 10−4 f. 2 × 10−10

Chapter 1
2
15. a. 4.2 × 103 + 48.0 × 103 = 52.2 × 103 = 5.22 × 104
b. 90 × 103 + 360 × 103 = 450 × 103 = 4.50 × 105
c. 50 × 10−5 − 6 × 10−5 = 44 × 10−5 = 4.4 × 10−4
d. 1.2 × 103 + 0.05 × 103 − 0.6 × 103 = 0.65 × 103 = 6.5 × 102
16. a. (102)(103) = 105 = 100 × 103
b. (10−2)(103) = 101 = 10
c. (103)(106) = 1 × 109
d. (102)(10−5) = 1 × 10−3
e. (10−6)(10 × 106) = 10
f. (104)(10−8)(1028) = 1 × 1024
17. a. (50 × 103)(3 × 10−4) = 150 × 10−1 = 1.5 × 101
b. (2.2 × 103)(2 × 10−3) = 4.4 × 100 = 4.4
c. (82 × 106)(2.8 × 10−6) = 229.6 = 2.296 × 102
d. (30 × 10−4)(4 × 10−3)(7 × 108) = 840 × 101 = 8.40 × 103
18. a. 102/104 = 10−2 = 10 × 10−3
b. 10−2/103 = 10−5 = 10 × 10−6
c. 104/10−3 = 107 = 10 × 106
d. 10−7/102 = 1.0 × 10−9
e. 1038/10−4 = 1.0 × 1042
f. 2 10/100 − = 101/10−2 = 1 × 103
19. a. (2 × 103)/(8 × 10−5) = 0.25 × 108 = 2.50 × 107
b. (4 × 10−3)/(60 × 104) = 4/60 × 10−7 = 0.667 × 10−7 = 6.67 × 10−8
c. (22 × 10−5)/(5 × 10−5) = 22/5 × 100 = 4.4
d. (78 × 1018)/(4 × 10−6) = 1.95 × 1025
20. a. (102)3 = 1.0 × 106 b. (10−4)1/2 = 10.0 × 10−3
c. (104)8 = 100.0 × 1030 d. (10−7)9 = 1.0 × 10−63
21. a. (4 × 102)2 = 16 × 104 = 1.6 × 105
b. (6 × 10−3)3 = 216 × 10−9 = 2.16 × 10−7
c. (4 × 10−3)(6 × 102)2 = (4 × 10−3)(36 × 104) = 144 × 101 = 1.44 × 103
d. ((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
22. a. (−10−3)2 = 1.0 × 10−6
b. 3
42
10
)10)(10( −
= 10−2/103 = 1.0 × 10−5
c. 4
4
4
26
4
223
10
10
10
)10)(10(
10
)10()10( −
−−
== = 1.0 × 10−8
d. 4
43
10
)10)(10(
−= 107/10−4 = 1.0 × 1011

Chapter 1 3
e. (1 × 10−4)3(102)/106 = (10−12)(102)/106 = 10−10/106 = 1.0 × 10−16
f.
[]
[][]
10
1
)10)(10(
1
10)10(
)10)(10(
34
322
3
22
== −
−
−
−
= 1.0 × 10−1
23. a. 4
222
103
)10()103(
×
× = (9 × 104)(102)/(3 × 104) = (9 × 106)/(3 × 104) = 3 × 102 = 300
b. 3
8
3
24
108
1016
)20(
)104(
×
×
=
× = 2 × 105 = 200.0 × 103
c. 4
8
22
24
104
1036
)102(
)106(
−− ×
×
=
×
× = 9.0 × 1012
d. 5
2
5
3/16
102
103
102
)1027(
×
×
=
×
×−
−
= 1.5 × 10−7 = 150.0 × 10−9
e. 4
8
4
26
4
223
102
1048
102
)103)(1016(
102
)103()104(
−−− ×
×
=
×
××
=
×
×× = 24.0 × 1012
f. (16 × 10−6)1/2(105)5(2 × 10−2) = (4 × 10−3)(1025)(2 × 10−2) = 8 × 1020
= 800.0 × 1018
g.
21/2
33 2 2 4
52
(310) 1.6010 (210)(810 )
(7 10 )
−−
−
⎡⎤⎡⎤⎡ ⎤
××××
⎣⎦⎣⎦⎣ ⎦
×
9421/2
10
51 6
10 10
(27 10 )(2.56 10 )(16 10 )
49 10
(69.12 10 )(4 10 ) 276.48 10
49 10 49 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 +3 +3
c. 50 × 105 = 5000 × 103 = 5 × 106 = 0.005 × 109
+2 −3 −3