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JEE Mains · Physics · STD 12 -Electromagnetic induction
  • JEE Mains 2021
  • Hard
  • Numerical

For the given circuit the current \(i\) through the battery when the key in closed and the steady state has been reached is .....\(A\)

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Answer & Solution

Correct Answer

(C) \(10\)

Step-by-step Solution

Detailed explanation

In steady state, inductor behaves as a conducting wire.

So, equivalent circuit becomes

\(\frac{1}{{R}_{{eq}}}=\frac{1}{3}+\frac{1}{3}+\frac{1}{3}=1\)

\(\Rightarrow {R}_{{eq}}=1\, \Omega\)

\(\Rightarrow\) Circuit becomes

\(\Rightarrow {i}=\frac{30}{3}=10\, {A}\)

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