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JEE Main 2020 (9 Jan, Shift 1) · Self-Inductance & Energy

Self-Inductance & Energy question from JEE Main 2020 (9 Jan, Shift 1)

Q72JEE Main 2020 · 9 JAN · SHIFT 1Medium

Self-Inductance & Energy

In a fluorescent lamp choke (a small transformer) 100 V of reverse voltage is produced when the choke current changes uniformly from 0.25 A to 0 in a duration of 0.025 ms. The self-inductance of the choke (in mH) is estimated to be ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 10

Self-induced emf $e = L\dfrac{\Delta I}{\Delta t}$ , so $L = \dfrac{e\,\Delta t}{\Delta I}$ .
$L = \dfrac{100 \times 0.025 \times 10^{-3}}{0.25} = 10^{-2}$ H
So $L = 10$ mH.

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