Answer Posted / sudipta

Starter is connected in series with the armature. Now, this
armature current is given by Ia=(v-Eb)/Ra. Here, v is
terminal voltage, Eb is back emf, Ra is armature resistance.
In the case of an induction motor,at the time of
starting,back emf is zero. So, Ia=v/Ra. As Ra is very small,
Ia is too high. So,a resistance(R) is connected in series
with the armature. So, the starting current decreases and
motor gets protection from overcurrent.

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