Physics · Electromagnetic Induction

JEE Advanced 2020 — Paper 2 — Question 15

The inductors of two LR circuits are placed next to each other, as shown in the figure. The values of the self-inductance of the inductors, resistances, mutual-inductance and applied voltages are specified in the given circuit. After both the switches are closed simultaneously, the total work done by the batteries against the induced EMF in the inductors by the time the currents reach their steady state values is $\_\_\_\_$ mJ .

Answer: 55

Numerical answer — enter this value.

Step-by-step solution

$\mathrm{dU}=\phi_{1} \mathrm{di}_{1}+\phi_{2} \mathrm{di}_{2}$

$\mathrm{U}=\frac{1}{2} \mathrm{~L}_{1} \mathrm{l}_{1}^{2}+\frac{1}{2} \mathrm{~L}_{2} \mathrm{I}_{2}^{2} \pm \mathrm{Ml}_{1} \mathrm{l}_{2}=55 \mathrm{~mJ}$

Answer key and solution verified before publishing.

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Exam: JEE Advanced 2020
Paper: Paper 2
Subject: Physics
Chapter: Electromagnetic Induction
Topic: Self-Inductance and Mutual Inductance and Energy Density