Question 150 of 153

For the given circuit the current $i$ through the battery when the key in closed and the steady state has been reached is

A$10A$B$6A$C$25A$D$0A$

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Answer: (A) $10A$

In a steady state, the inductor behaves as a conducting wire. So, the equivalent circuit becomes,

$\dfrac{1}{{R}_{eq}}=\dfrac{1}{{R}_{1}}+\dfrac{1}{{R}_{2}}+\dfrac{1}{{R}_{3}}=\dfrac{1}{3}+\dfrac{1}{3}+\dfrac{1}{3}=1\Rightarrow {R}_{eq}=1\Omega$
so, the circuit becomes,

$\Rightarrow I=\dfrac{30}{3}=10A$

JEE Main Physics practice question on Electromagnetic Induction (LR Circuit), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.