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Instructor’s Resource and Solutions Manual for Introductory Circuit Analysis, 11th Edition by Robert

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Review fundamental electrical engineering concepts with Introductory Circuit Analysis, 11th Edition by Robert. This resource supports the study of circuit theory, voltage, current, resistance, Ohm’s law, Kirchhoff’s laws, series and parallel circuits, and circuit analysis techniques. It is useful for students and instructors working through electrical circuit problems and reinforcing essential engineering principles. Use alongside the textbook and course materials to review solution methods, strengthen problem-solving skills, and support introductory circuit analysis coursework.

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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




s ResourceSolution
Instructor’s
Manual
ResourceSolution
to accompany Introductory
Manual to accompany
Page
Circuit
1https://online2pdf.com/delete-header-and-footer-in-pdf#
Analysis
Introductory
Eleventh
Circuit
Edition
Analysis
RobertEleventh
L. Boylestadhttps://online2pdf.com/delete-header-and-footer-in-pdf#
Edition Robert L. Boylestad.pdfhttps://online2pdf.com/delete-header-

, Page 2 of 354https://online2pdf.com/delete-header-and-footer-in-pdf#
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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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Instructor’s
Manual
ResourceSolution
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Manual to accompany
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Circuit
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Analysis
Introductory
Eleventh
Circuit
Edition
Analysis
RobertEleventh
L. Boylestadhttps://online2pdf.com/delete-header-and-footer-in-pdf#
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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

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Instructor’s
Manual
ResourceSolution
to accompany Introductory
Manual to accompany
Page
Circuit
3https://online2pdf.com/delete-header-and-footer-in-pdf#
Analysis
Introductory
Eleventh
Circuit
Edition
Analysis
RobertEleventh
L. Boylestadhttps://online2pdf.com/delete-header-and-footer-in-pdf#
Edition Robert L. Boylestad.pdfhttps://online2pdf.com/delete-header-

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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

s ResourceSolution
Instructor’s
Manual
ResourceSolution
to accompany Introductory
Manual to accompany
Page
Circuit
4https://online2pdf.com/delete-header-and-footer-in-pdf#
Analysis
Introductory
Eleventh
Circuit
Edition
Analysis
RobertEleventh
L. Boylestadhttps://online2pdf.com/delete-header-and-footer-in-pdf#
Edition Robert L. Boylestad.pdfhttps://online2pdf.com/delete-header-

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