Page 95 - Solutions Manual to accompany Electric Machinery Fundamentals

E
A




 V  jX I
 S A
I R I
A
A
A
By the Pythagorean Theorem,
E A 2  V   R I XI A sin    cos  2  S A cos   R I sin   XI 2
A A
A S
S

V   E  A 2  S A cos  RI sin   X I 2  RI cos   X I sin
A A
A S
S A

In this case,   25.84 , so cos  0.9 and sin  0.6512 .
A phasor diagram representing the situation at leading power factor is shown below:
E
A

jX I
S A 
I
A R I
  A A 
V

By the Pythagorean Theorem,

E A 2  V   R I XI A sin    cos  2  S A cos   R I sin   XI 2
A A
A S
S

V   E  A 2  S A cos  RI sin   X I 2  RI cos   X I sin
A S
S A
A A
In this case,   25.84 , so cos  0.9 and sin   0.6512 .
A phasor diagram representing the situation at unity power factor is shown below:
E
A
jX I
S A

I V R I
A  A A
By the Pythagorean Theorem,

E A 2   2 V  S A  X I 2

V   A 2  E S A  X I 2
0
0
In this case,   , so cos  1.0 and sin  .
The MATLAB program is shown below takes advantage of this fact.

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