SSC JE Electrical Engineering 2014 Set 2 Paper

Question : 112 of 200

Marks: +1, -0

A 150 V d.c. motor of armature resistance 0.4 Ω has back emf of 142 V. The armature current is :

Solution:

∵

V = E_b + I _aR_a

∴

I_ a ={V- E_b}/R_a

={150-142}/0.4=20A

∴ Option (c) is correct.

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