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JEE Advanced 2020Paper 2
Exam
JEE Advanced
Year
2020
Paper
Paper 2
Subject
Physics
Chapter
Electromagnetic Induction

JEE Advanced 2020 · Paper 2 · Physics · Electromagnetic Induction

The inductors of two LR circuits are placed next to each other, as...

The inductors of two LRLR 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 EMFEMF in the inductors by the time the currents reach their steady-state values is_______mJmJ.

The inductors of two LR circuits are placed next to each other, as shown in the figure. The values of the

Correct answer

0

Step-by-step solution

The inductors of two LR circuits are placed next to each other, as shown in the figure. The values of the

ε1=L1di1dt+Mdi2dt{ε}_{1} = {L}_{1}\dfrac{d{i}_{1}}{dt} + M\dfrac{d{i}_{2}}{dt}

dW1=ε1i1dtd{W}_{1} = {ε}_{1}{i}_{1}dt

=L1(di1)i1+M(di2)i1= {L}_{1}\left(d{i}_{1}\right){i}_{1} + M\left(d{i}_{2}\right){i}_{1}

dW2=L2(di2)i2+M(di1)i2d{W}_{2} = {L}_{2}\left(d{i}_{2}\right){i}_{2} + M\left(d{i}_{1}\right){i}_{2}

dW1+dW2=L1i1(di1)+L2i2(di2)+Md(i1⋅i2)d{W}_{1} + d{W}_{2} = {L}_{1}{i}_{1}\left(d{i}_{1}\right) + {L}_{2}{i}_{2}\left(d{i}_{2}\right) + Md\left({i}_{1} \cdot {i}_{2}\right)

W=12L1i12+12L2i22+M1i2W = \dfrac{1}{2}{L}_{1}{i}_{1}^{2} + \dfrac{1}{2}{L}_{2}{i}_{2}^{2} + {M}_{1}{i}_{2}

=55mJ= 55mJ,

Note : If mutual induction is taken in opposite direction then W=35mJW = 35mJ

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