The armature of a `DC` motor has `20Omega` resistance. It draws a current of `1.5 A` when run by `200 V DC` supply The value of back emf induced in it will be

Options

A. 150 V
B. 170 V
C. 190 V
D. 180 V

Answer

C

Explanation

Input power, `P_(1)=220xx1.5=330V`
Loss of power `i^(2)R=((3)/(2))^(2)xx20=45W`
Output power, `P_(0)=330-45=285W`
`therefore`Peak emf induced, `V_(0)=(P_(0))/(i)=(285)/(1.5)=190V`