Question

The inductance of a coil is $L=10H$ and resistance $R=5\Omega$. If applied voltage of battery is $10V$ and it switches off in $1$ millisecond, find induced electromotive force 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

Answer 1

The correct option is A $2\times {10}^{4}V$

According to question.....................

$\begin{array}{c}\phi =Li\\ or,e-\frac{d\phi }{dt}=-\frac{d}{dt}\left(Li\right)\\ or,e=-L\frac{di}{dt}\\ Inducedcurrent=\frac{V}{R}=\frac{10}{5}=2A\\ Circuitswitchesoffin1millisecond\\ or,dt=1\times {10}^{-3}s\\ and,L=10H\\ \therefore Inducedempininductoris\left|e\right|=10\times \frac{2}{1\times {10}^{-3}}=2\times {10}^{4}V\\ sotheoptionisA.\end{array}$