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JEE Main 2015 (10 Apr, Shift 1) · Self-Inductance & Energy

Self-Inductance & Energy question from JEE Main 2015 (10 Apr, Shift 1)

Q78JEE Main 2015 · 10 APR · SHIFT 1Easy

Self-Inductance & Energy

When current in a coil changes from 5 A to 2 A in 0.1 s, an average voltage of 50 V is produced. The self - inductance of the coil is:

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Answer: (b)

$L = \dfrac{e\,\Delta t}{\Delta I}$ with $\Delta I = 5-2 = 3\text{ A}$ .
$L = \dfrac{50\times0.1}{3} = \dfrac{5}{3}\text{ H}$
So the answer is (b) 1.67 H.

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