Page 127 - Solutions Manual to accompany Electric Machinery Fundamentals

3VE
P   A sin  constant , so E sin  E sin
X S A 1 1 A 2 2

With a 5% decrease, E  14,115 V , and
A 2
 E  1  14,858 V 
 sin  1 A 1 sin   sin  sin 26.54  28.0
2  2   
E A 2    14,115 V 
Therefore, the new armature current is

E  V 14,115 28.0  8313 0 

I  A 2    3777  32.1 A
A
jX S j 2.07
(e) Repeating part (d):
With a 10% decrease, E A 2  13,372 V , and

 E  1 14,858 V 

 sin  1  A 1 sin  2   sin   sin 26.54   29.8
2
E   13,372 V 
A 2 
Therefore, the new armature current is
E  V 13,372 29.8  8313 0 


I A  A 2    3582  26.3 A
jX S j 2.07
With a 15% decrease, E A 2  12,629 V , and
 E  1  14,858 V 
 sin  1  A 1 sin  2   sin   sin 26.54   31.7
2
E A 2    12,629 V 
Therefore, the new armature current is


E  V 12,629 31.7  8313 0 

I  A 2    3414  20.1 A
A
jX S j 2.07
With a 20% decrease, E A 2  11,886 V , and
 E  1 14,858 V 

 sin  1  A 1 sin  2   sin   sin 26.54   34.0
2
E A 2    11,886 V 
Therefore, the new armature current is


E  V 11,886 34.0  8313 0 

I  A 2    3296  13.1 A
A
jX S j 2.07
With a 25% decrease, E A 2  11,144 V , and
 E  1  14,858 V 
 sin  1  A 1 sin  2   sin   sin 26.54   36.6
2
E A 2    11,144 V 
Therefore, the new armature current is


E  V 11,144 36.6  8313 0 

I  A 2    3224  5.4 A
A
jX S j 2.07
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