Question

When the current changes from $+2A$ to $-2A$ in $0.05$ second, an e.m.f. of $8V$ is induced in a coil. The coefficient of self-induction of the coil is :

• A. $0.2H$
• B. $0.4H$
• C. $0.8H$
• D. $0.1H$

Answer 1

The correct option is D $0.1H$

lf $e$ is the induced $e$.m.f. in the coil, then $e=-L\frac{di}{dt}$

Therefore, $L=-\frac{e}{di/dt}$

Substituting values, we get $L=\frac{-8\times 0.05}{-4}=0.1H$