Q. During a current change from 2 A to 4 A in 0.05 s, 8 V of emf is developed in a coil. The coefficient of self-induction is
A) 0.1 H
B) 0.2 H
C) 0.4 H
D) 0.8 H
Explanation
(b) $e = L\frac{dI}{dt}\ L = \frac{edt}{dI} = \frac{8(0.05)}{(4 - 2)} = \ 0.2\ $H
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.