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JEE Main 2019 (10 Jan, Shift 2) · Self-Inductance & Energy

Self-Inductance & Energy question from JEE Main 2019 (10 Jan, Shift 2)

Q76JEE Main 2019 · 10 JAN · SHIFT 2Medium

Self-Inductance & Energy

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -

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Answer: (d)

$L = \dfrac{e}{\Delta I/\Delta t} = \dfrac{25}{15} = \dfrac{5}{3}$ H
$\Delta U = \dfrac{1}{2}\times\dfrac{5}{3}\times(25^2 - 10^2) = \dfrac{5}{6}\times 525 = 437.5$ J
So the answer is (d) 437.5 J.

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