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

SOLUTION MANUAL
FOR POWER
SYSTEM ANALYSIS
AND DESIGN 7TH
EDITION BY J.
DUNCAN GLOVER,
MULUKUTLA S.
SARMA, THOMAS
OVERBYE, ADAM
BIRCHFIELD

, 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

, 2.1 (A A1  530  5Cos30  J Sin 30 4.33  J 2.5
)
4
9  16 Tan
1
(b) A2  3  J4  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
(e) A  530 / 5  126.87  1156.87  1ej156.87

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 I  100  10  7.5  90A
)
8  J6  J6
1
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