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Exam (elaborations)

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

Institution
Analysis And Design 7th Edition
Course
Analysis and Design 7th Edition

Content preview

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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Institution
Analysis and Design 7th Edition
Course
Analysis and Design 7th Edition

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