The inductance of a coil is $L = 10\,\,{\rm{H}}$ and resistance $R = 5\,\,{\rm{\Omega }}$. If applied voltage of battery is ${\rm{10 V}}$ and it switches off in ${\rm{1 \,milli\,second}}$, find induced emf of inductor

$2 \times {10^{ - 4}}\,{\rm{V}}$

JEE PHYSICS MCQ: ELECTRO MAGNETIC INDUCTION

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Application +4 | -1

The inductance of a coil is

L = 10\,\,\mathrm{H}

and resistance

R = 5\,\,\mathrm{\Omega }

. If applied voltage of battery is

\mathrm{10 V}

and it switches off in

\mathrm{1 \,milli\,second}

, find induced emf of inductor

View answer and explanation

Correct option: A

2 \times {10^4}\,\mathrm{V}

Option A (

2 \times {10^4}\,\mathrm{V}

) is correct because the induced emf is calculated using the formula e = L(di/dt).

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