JEE Mains · Physics · STD 12 -6. Electromagnetic induction
For the given circuit the current \(i\) through the battery when the key in closed and the steady state has been reached is .....\(A\)
- A \(6\)
- B \(25\)
- C \(10\)
- D \(0\)
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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