An electric motor runs a `D.C.` source of e.m.f. `200 V` and draws a current of `10A`. If the efficiency is `40%`, then ressistance of the armature is:

Options

A. `5Omega`
B. `12Omega`
C. `120Omega`
D. `160Omega`

Answer

B

Explanation

Here `eta=(e)/(E)xx100`
`(40)/(100)=(e)/(E)=(4)/(10)=(2)/(5)`
`e=(2E)/(5)=(2xx200)8/(5)=80V`
But `I+(E-e)/(R )`
:. `10=(200-80)/(R )`
:. `R=12Omega`