When the current changes from +2 A to -2 A in 0.05 second, an e.m.f. of 8 V is inducted in a coil. The coefficient of self- induction of the coil is (a) 0.2 H (b) 0.4 H (c) 0.8 H (d) 0.1 H

Answer: (d). L = (e Δ t)/(Δ I) with Δ I = 2-(-2) = 4 A. L = (8×0.05)/(4) = 0.1 H So the answer is (d) 0.1 H.

When the current changes from +2 A to -2 A in 0.05 second, an e.m.f… | JEE Main 2003 (AIEEE)

Q82JEE Main 2003 · AIEEEEasy

Self-Inductance & Energy

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

Show answer and solution

Answer: (d)

$L = \dfrac{e\,\Delta t}{\Delta I}$ with $\Delta I = 2-(-2) = 4\text{ A}$ .
$L = \dfrac{8\times0.05}{4} = 0.1\text{ H}$
So the answer is (d) $0.1\text{ H}$ .

Report a mistake

Self-Inductance & Energy question from JEE Main 2003 (AIEEE)

More questions on Self-Inductance & Energy