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 .....\(A\)

Pick an option to check your answer.
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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