Solution Manual for Introductory Circuit Analysis – 14th Global Edition by Boylestad – Complete Step-by-Step Answers

SOLUTION MANUAL

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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




iii

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Chapter 1
1. 

2. 

3. 
d ft ⎢1⎤ ⎡ ⎢60 s ⎤ ⎥ ⎡ 60 min ⎤
4. = = 20,000
mi
10 s 5,280
ft ⎥ ⎢1min1 h ⎥ = 1363.64 mph
t ⎣ ⎦ ⎣ ⎦ ⎣ ⎦

⎡ 1 ⎤
5. 4 min = 0.067 h
h 

⎣ 60 min ⎦
= 31 = 29.05 mph
d
mi
1.067 h
t

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

7.

8.

9.

5 5
10. MKS, CGS, C ( F 32) (68 32) = 20
= 5
(36)
9 9 9
SI: K = 273.15 + C = 273.15 + 20 = 293.15

⎡ 0.7378 ft - lb ⎤
11. 1000 J 
⎥ = 737.8 ft-lbs
⎣ 1J ⎦

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

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13. a. 104 b. 106 c. 103 d. 10 3
e. 100 f. 10 1


3
14. a. 15 103 b. 30 10 c. 2.4 106 d. 150
103
4 10
e. 4.02 10 f. 2 10



Chapter 1 1