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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
Power System Analysis And Design
Course
Power System Analysis and Design

Content preview

Solutions Manual
To Accompany


Power System Analysis And Design

Seventh Edition




J. Duncan Glover
Mulukutla S. Sarma
Thomas 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. A (Iv) D
B. B 2.31 A
C. A 2.32 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 (A A1 = 530 = 5Cos30 + J Sin 30= 4.33 + J 2.5
)
−1 4
(b) A2 = −3 + J4 9 + 16 Tan = 5126.87 = 5ej126.87
= −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

A5 = (530) / (5 − 126.87) = 1156.87 = 1ej156.87
(e)

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

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

(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 (A Vmax = 359.3 V; Imax = 100 A
)
(b) V = 359.3 2 = 254.1v; I = 100 2 = 70.71a
(c) V = 254.115V; I = 70.71 − 85A
− J6 6 − 90
2.4 (A I1 = 100 = 10 = 7.5 − 90A
) 8 + J6 − J6 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

(B) I = V / 20 = 13.8530A
I(T) = 19.58cos(T + 30)A




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
Power System Analysis and Design
Course
Power System Analysis and Design

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