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 $
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 $