Question

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

Answer 1

Answer: (A)

Amount of magnetic flux linked with inductor is $\phi=L i$
Now, the emf induced in the inductor is given by
$e=-\frac{d \phi}{d t}=-\frac{d}{d t}(L i)$
or $e=-L \frac{d i}{d t}$
or $|e|=L \frac{d i}{d t}$
Here, induced current
$=\frac{V}{R}$
$=\frac{10}{5}$
$=2 A$
Circuit switches off in $1$ millisecond or
$d t=1 \times 10^{-3} s$ and $L=10 H$
$\therefore $ Induced emf in inductor is
$|e|=10 \times \frac{2}{1 \times 10^{-3}}$
$=2 \times 10^{4} V$