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Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield $15.00   Add to cart

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Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield

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Solution Manual For Power System Analysis and Design 7th Edition by J. Duncan Glover, Mulukutla S. Sarma, Thomas Overbye, Adam Birchfield

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  • October 20, 2024
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  • 2024/2025
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  • Analysis and Design 7th Edition
  • Analysis and Design 7th Edition
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An Instructor's Solutions Manual
To Accompany

Power System
Analysis And Design
Seventh Edition

J. Duncan Glover Mulukutla
S. SarmaThomas J. Overbye

, Contents
Chapter 2 1

Chapter 3 27

Chapter 4 71

Chapter 5 95

Chapter 6 137

Chapter 7 175

Chapter 8 195

Chapter 9 231

Chapter 10 303

Chapter 11 323

Chapter 12 339

Chapter 13 353

Chapter 14 379

,Chapter 2
Fundamentals

Answers To Multiple-Choice Type Questions
2.1 B 2.19 A
2.2 A 2.20 A. C
2.3 C B. A
2.4 A C. B
2.5 B 2.21 A
2.6 C 2.22 A
2.7 A 2.23 B
2.8 C 2.24 A
2.9 A 2.25 A
2.10 C 2.26 B
2.11 A 2.27 A
2.12 B 2.28 B
2.13 B 2.29 A
2.14 C 2.30 (I) C
2.15 A (ii) B
2.16 B (iii) A
2.17 A. (iv) D
A 2.31 A
B. B 2.32 A
C.
A
2.18 C




1
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

, 2.1 A1 = 530 = 5Cos30 + J Sin 30= 4.33 + J 2.5
(A
)

(B) = −3 + J4 Tan−1 = 5126.87 = 5ej126.87
4
A2 = 9 + 16
−3
(C) A3 = (4.33 + J2.5) + (−3 + J4 ) = 1.33+ J 6.5 = 6.63578.44
(D) A4 = (530)(5126.87) = 25156.87 = −22.99 + J9.821
(E) A5 = (530) / (5 − 126.87) = 1156.87 = 1ej156.87

2.2 I = 400 − 30 = 346.4 − J200
(A
)
(B) I(T) = 5sin(T + 15) = 5cos(T + 15 − 90) = 5cos(T − 75)

( 2 ) − 75 = 3.536 − 75 = 0.9151− J3.415
I = 5

(C) I = ( 4 2 )  − 30 + 5 − 75 = (2.449 − J1.414) + (1.294 − J4.83)
= 3.743 − J 6.244 = 7.28 − 59.06

2.3 Vmax = 359.3 V; Imax = 100 A
(A
)
(B) V = 2 = 254.1v; I = 100 = 70.71a
359.3 2
(C) V = 254.115V; I = 70.71 − 85A
− J6 6 − 90
2.4 I = 100 = 10 = 7.5 − 90A
8 + J6 − J6
1
(A 8
)
I2 = I − I1 = 100 − 7.3 − 90 = 10 + J7.5 = 12.536.87A
V = I2 (− J6) = (12.536.87) (6 − 90) = 75 − 53.13V
(B)




2.5 (A) (T) = 277 2 Cos(T + 30) = 391.7 Cos(T + 30)V
2
© 2012 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

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