## In This post you will get answer of Q.

An electric motor runs on DC source of emf 200 V and draws a current of 10 A. If the efficiency be 40%, then the resistance of armature is

Solution:

## Input power $ =VI=200times 10 $

$ =2000W $

Output power $ =frac{40}{100}times 2000=800W $

Power loss in heating the armature

$ =2000-800 $

$ =1200W $

$ therefore $ $ {{I}^{2}}R=1200 $

or $ R=frac{1200}{{{I}^{2}}} $

$ =frac{1200}{10times 10} $

or $ R=12,Omega $