In a coil, the current is changed from 12 A to 24 A in 1 s. The self-induced emf of the coil is 24 V. The change in energy in the coil is

432 J

The correct answer is Option (1) → 432 J

Given:
Initial current, $I_1 = 12\,A$
Final current, $I_2 = 24\,A$
Time, $t = 1\,s$
Induced emf, $e = 24\,V$

Self-induced emf is given by:
$e = L \frac{dI}{dt}$

$L = \frac{e \, t}{\Delta I} = \frac{24 \times 1}{24 - 12} = 2\,H$

Change in magnetic energy stored:

$\Delta U = \frac{1}{2}L(I_2^2 - I_1^2)$

$\Delta U = \frac{1}{2} \times 2 (24^2 - 12^2)$

$\Delta U = (576 - 144) = 432\,J$

∴ Change in energy in the coil = 432 J