Page 193 - Solutions Manual to accompany Electric Machinery Fundamentals

(a) If this motor is started from a 460-V infinite bus, how much current will flow in the motor at
starting?
(b) If transmission line with an impedance of 0.50 + j0.35  per phase is used to connect the
induction motor to the infinite bus, what will the starting current of the motor be? What will the
motor’s terminal voltage be on starting?

(c) If an ideal 1.4:1 step-down autotransformer is connected between the transmission line and the
motor, what will the current be in the transmission line during starting? What will the voltage be at
the motor end of the transmission line during starting?
SOLUTION

(a) The equivalent circuit of this induction motor is shown below:
I A
R 1 jX 1 jX 2 R 2

+ j3.209 ? 0.488 ?
0.54 ? j2.093 ?
   s
1
j51.12 ? jX M R  
V ? 2  s 
23.912 ?
-

The easiest way to find the line current (or armature current) is to get the equivalent impedance Z F of the
rotor circuit in parallel with jX M , and then calculate the current as the phase voltage divided by the sum
of the series impedances, as shown below.

I A
R 1 jX 1 jX F R F

+
0.54 ? j2.093 ?

V ?

-

The equivalent impedance of the rotor circuit in parallel with jX is:
M
1 1
Z  F 1 1  1 1  0.435  j 2.864 2.90 81.3    
jX M  Z 2 j 51.12   0.488  j 3.209

The phase voltage is 460/ 3 = 266 V, so line current I L is
V 266 0 V

 I  I  
L A
R  1 jX  1 R  F jX F 0.54   j 2.093   0.435   j 2.90 

 I  I 52.3  79 A
L A

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