JEE Main 2026 Apr 5 Shift 1, Physics Q25: Inductance

JEE Main2026Apr 5, Shift 1Physics

In the given circuit below inductance values of $L_{1}$ , $L_{2}$ and $L_{3}$ are same. The magnetic energy stored in the entire circuit is ( $U_{t}$ ) and that stored in the $L_{2}$ inductor is ( $U_{l}$ ). ( $U_{t} / U_{l}$ ) is ______.

(Ignore the mutual inductance if any)

Your answer

Solution

Let each inductor have inductance $L$ and the total line current be $I$ . Then $L_{1}$ carries $I$ , and by symmetry $L_{2}$ and $L_{3}$ each carry $I /2$ .

Total stored energy:

$U_{t} = \frac{1}{2} L I^{2} + \frac{1}{2} L (\frac{I}{2})^{2} + \frac{1}{2} L (\frac{I}{2})^{2} = \frac{1}{2} L I^{2} + \frac{1}{4} L I^{2} = \frac{3}{4} L I^{2}$ .

Energy in $L_{2}$ alone: $U_{l} = \frac{1}{2} L (\frac{I}{2})^{2} = \frac{1}{8} L I^{2}$ .

Ratio: $\frac{U _{t}}{U _{l}} = \frac{3/4}{1/8} = 6$ .

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JEE Main 2026 Apr 5 Shift 1, Physics Q25: Inductance

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