Instructor’s Resource Solution Manual for Introductory Circuit Analysis 11th Edition by Robert Boylestad – Circuit Analysis Study Resource

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Instructor’s Resource Manual
to accompany


Introductory
Circuit Analysis
Eleventh Edition


Robert L. Boylestad




Upper Saddle River, New Jersey
Columbus, Ohio




Instructor’s
Instructor’s
ResourceSolution
ResourceSolution
Manual to
Manual
accompany
to accompany
Introductory
Introductory
Circuit Analysis
Circuit Analysis
EleventhEleventh
Edition Robert
EditionL.Robert
Boylestadhttps://online2pdf.com/remove-pdf-header-footer
1https://online2pdf.com/remove-pdf-header-footer
L. Boylestad.pdfhttps://online2pdf.com/remove-pdf-he

,€™s ResourceSolution
Page 2 of 354https://online2pdf.com/remove-pdf-header-footer
Manual to accompany Introductory Page
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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
Instructor’s
Instructor’s
ResourceSolution
ResourceSolution
Manual to
Manual
accompany
to accompany
Introductory
Introductory
Circuit Analysis
Circuit Analysis
EleventhEleventh
Edition Robert
EditionL.Robert
Boylestadhttps://online2pdf.com/remove-pdf-header-footer
2https://online2pdf.com/remove-pdf-header-footer
L. Boylestad.pdfhttps://online2pdf.com/remove-pdf-he

,€™s ResourceSolution
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Manual to accompany Introductory Page
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Chapter 1
1. −

2. −

3. −

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

⎡ 1h ⎤
5. 4 min ⎢ ⎥ = 0.067 h
⎣ 60 min ⎦
d 31 mi
υ= = = 29.05 mph
t 1.067 h

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

7. −

8. −

9. −

5 5 5
10. MKS, CGS, °C = (°F − 32) = (68 − 32) = (36) = 20°
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 ft ⎦ ⎣ 1 in. ⎦

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 1

Instructor’s
Instructor’s
ResourceSolution
ResourceSolution
Manual to
Manual
accompany
to accompany
Introductory
Introductory
Circuit Analysis
Circuit Analysis
EleventhEleventh
Edition Robert
EditionL.Robert
Boylestadhttps://online2pdf.com/remove-pdf-header-footer
3https://online2pdf.com/remove-pdf-header-footer
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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 = 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
−3 2 −6
22. a. (−10 ) = 1.0 × 10
(10 2 )(10 −4 )
b. = 10−2/103 = 1.0 × 10−5
10 3
(10 −3 ) 2 (10 2 ) (10 −6 )(10 2 ) 10 −4
c. = = 4 = 1.0 × 10−8
10 4 10 4 10
3 4
(10 )(10 )
d. = 107/10−4 = 1.0 × 1011
10 − 4




2 Chapter 1

Instructor’s
Instructor’s
ResourceSolution
ResourceSolution
Manual to
Manual
accompany
to accompany
Introductory
Introductory
Circuit Analysis
Circuit Analysis
EleventhEleventh
Edition Robert
EditionL.Robert
Boylestadhttps://online2pdf.com/remove-pdf-header-footer
4https://online2pdf.com/remove-pdf-header-footer
L. Boylestad.pdfhttps://online2pdf.com/remove-pdf-he