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 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

Answer 1

Answer: (A)

Amount of magnetic flux linked with inductor is $\varphi =Li$
Now, the emf induced in the inductor is given by
$e=-\frac{d\varphi }{dt}=-\frac{d}{dt}(Li)$
or $e=-L\frac{di}{dt}$
or $|e|=L\frac{di}{dt}$
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
$|e|=10\times \frac{2}{1\times 10^{-3}}$
$=2\times 10^{4}V$