Page 288 - Solutions Manual to accompany Electric Machinery Fundamentals

(c) Input power
(d) P AG
(e) P conv
(f) P out
(g) 
ind
(h)  load
(i) Efficiency 
SOLUTION The equivalent circuit of the motor is shown below

I 1
R 1 jX 1 j0.5X 2
{ R
+ 1.30 ? j2.01 ? j2.01 ?

0.5Z F j0.5X M 5 . 0 s 2 Forward
j105 ?

{
V = 220?0° V jX 2
j105 ? { j2.01 ?

j0.5X M 5 . 0 R 2 Reverse
0.5Z B 2  s {

-

The impedances Z and Z are:
B
F
R s  jX  jX 
/
Z  2 2 M
F
R 2 / s  jX  2 jX M
The slip s  0.05 , so R 2  / s  1.73  / 0.05 20.6  
 j 20.6   105  2.01 j
Z  F  19.12  j 5.654

20.6  j 2.01 j 105
   R   /2 s  2  jX   jX
Z  2 M
B
2  R   /2 s  jX  2 jX M
The slip s  0.05 , so R 2   / 2 s     1.73   / 2 0.05  0.887 
 2.01 j 105  j

0.887 
Z  B  0.854  j 1.979 
0.887  j 2.01 j 105
(a) The input stator current is
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