In the given circuit below inductance values of $\mathrm{L}_{1}$, $\mathrm{L}_{2}$ and $\mathrm{L}_{3}$ are same. The magnetic energy stored in the entire circuit is $(\mathrm{U}_{\mathrm{t}})$ and that stored in the $\mathrm{L}_{2}$ inductor is $(\mathrm{U}_{l})$. $(\mathrm{U}_{t} / \mathrm{U}_{l})$ is ______. (Ignore the mutual inductance if any)

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Electromagnetic Induction

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)

Answer: 6

The three inductors are equal,

L_{1} = L_{2} = L_{3} = L

. The whole current

I

passes through

L_{1}

, then splits equally between the parallel inductors

L_{2}

and

L_{3}

, each carrying

\frac{I}{2}

. The total stored energy is

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}{8} L I^{2} + \frac{1}{8} L I^{2} = \frac{3}{4} L I^{2}

, while the energy in

L_{2}

alone is

U_{ℓ} = \frac{1}{2} L (\frac{I}{2})^{2} = \frac{1}{8} L I^{2}

. Therefore

\frac{U _{t}}{U _{ℓ}} = \frac{\frac{3}{4} L I ^{2}}{\frac{1}{8} L I ^{2}} = 6

JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Integer answer