Instructor’s Resource Solution Manual to Accompany Introductory Circuit Analysis 11th Edition Robert Complete Step-by-Step Solutions Guide 2025/ 2026 Electrical Engineering Problem Solving Resource for Mastery and Exam Success

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Robert L. Boylestad.pdf




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
Instructor’s
ResourceSolution
Manual toResourceSolution
accompany
Manual to
Introductory
accompany
ManualCircuit
to
Introductory
accompany
Analysis
Circuit
Introductory
Eleventh
Analysis
Edition
Circuit
Eleventh
Robert
Analysis
Edition
L. Boylestad.pdf
Eleventh
RobertEdition
L. Boylestad.pdf
Robert L. Boylestad.pdf

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Instructor’s
ResourceSolution
Instructor’s
ResourceSolution
Manual toResourceSolution
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Manual to
Introductory
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to
Introductory
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Analysis
Circuit
Introductory
Eleventh
Analysis
Edition
Circuit
Eleventh
Robert
Analysis
Edition
L. Boylestad.pdf
Eleventh
RobertEdition
L. Boylestad.pdf
Robert L. Boylestad.pdf




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
Instructor’s
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Manual to
Introductory
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to
Introductory
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Analysis
Circuit
Introductory
Eleventh
Analysis
Edition
Circuit
Eleventh
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Analysis
Edition
L. Boylestad.pdf
Eleventh
RobertEdition
L. Boylestad.pdf
Robert L. Boylestad.pdf

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Instructor’s
ResourceSolution
Instructor’s
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Introductory
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Edition
Circuit
Eleventh
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Analysis
Edition
L. Boylestad.pdf
Eleventh
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L. Boylestad.pdf
Robert L. Boylestad.pdf




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

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L. Boylestad.pdf
Robert L. Boylestad.pdf

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Instructor’s
ResourceSolution
Instructor’s
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Manual toResourceSolution
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Introductory
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Circuit
Eleventh
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Analysis
Edition
L. Boylestad.pdf
Eleventh
RobertEdition
L. Boylestad.pdf
Robert L. Boylestad.pdf




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
Instructor’s
ResourceSolution
Manual toResourceSolution
accompany
Manual to
Introductory
accompany
ManualCircuit
to
Introductory
accompany
Analysis
Circuit
Introductory
Eleventh
Analysis
Edition
Circuit
Eleventh
Robert
Analysis
Edition
L. Boylestad.pdf
Eleventh
RobertEdition
L. Boylestad.pdf
Robert L. Boylestad.pdf