Physics · Electromagnetic Induction

JEE Main 2026 — 5 April, Morning Shift — Question 23

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 $U_1$ and that stored in the $L_2$ inductor is $U_2$ . The ratio $U_1/U_2$ is (Ignore the mutual inductance if any)

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

Assuming current I through L₁, then current divides equally through L₂ and L₃ (each I/2). Energy in L₁: $\frac{1}{2}L I^2$ , in L₂: $\frac{1}{2}L (I/2)^2 = \frac{1}{8}L I^2$ , in L₃: same. Total $U_1 = \frac{1}{2}LI^2 + \frac{1}{8}LI^2 + \frac{1}{8}LI^2 = \frac{3}{4}LI^2$ . $U_2 = \frac{1}{8}LI^2$ . Ratio = $(\frac{3}{4})/(\frac{1}{8}) = 6$ .

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Self-Inductance and Mutual Inductance and Energy Density