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A $12 \mathrm{~V}$ battery connected to a coil of resistance $6 \Omega$ through a switch, drives a constant current in the circuit. The switch is opened in $1 \mathrm{~ms}$. The emf induced across the coil is $20 \mathrm{~V}$. The inductance of the coil is :
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The correct answer is:
$10 \mathrm{mH}$
Induced emf, $\varepsilon=-\mathrm{L} \frac{\mathrm{dI}}{\mathrm{dt}} \Rightarrow 20=-\mathrm{L} \frac{(0-2)}{10^{-3}}$ Hence, inductance of the coil.
$\mathrm{L}=10 \mathrm{mH}$
$\mathrm{L}=10 \mathrm{mH}$
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