Q. The armature of a DC motor has a resistance of 20 $\Omega.$ It draws a current of 1.5 A when run by 220 V DC. The value of peak emf induced in it will be
A) 150 V
B) 170 V
C) 190 V
D) 180 V
Explanation
(c) Input power $P_{1} = 220 \times 1.5 = 330\ $W
Loss of power $i^{2}R = \left( \frac{3}{2} \right)^{2} \times 20 = 45\ $W
Output power, $P_{0} = 330 - 45 = 285\ $W
$\therefore\ \ $Peak emf induced, $V_{0} = \frac{P_{0}}{i} = \frac{285}{1.5} = 190$ V
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.