Q. The inductance of a coil is L = 10 H and resistance $R = 5\Omega.$ If applied voltage of battery is 10 V and it switches off in 1 millisecond, find induced emf of inductor
A) $2 \times 10^{4}\ V$
B) $1.2 \times 10^{4}\ V$
C) $2 \times 10^{- 4}\ V$
D) None of these
Explanation
(a) The emf induced in the inductor is given by
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |e| = L\frac{di}{dt}\ $
Here, induced current =$\frac{V}{R} = \frac{10}{5} = 2A$
Circuit switches off in 1 millisecond
Or $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ dt = 1 \times 10^{- 3}s$
and $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ L = 10H$
$\therefore$ Induced emf in inductor is
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ |e| = 10 \times \frac{2}{1 \times 10^{- 3}} = 2 \times 10^{4}V\ $
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.