JEE Mains · Physics · STD 12 -6. Electromagnetic induction
The current \((i)\) at time \(t =0\) and \(t =\infty\) respectively for the given circuit is
- A \(\frac{18 E }{55}, \frac{5 E }{18}\)
- B \(\frac{10 E }{33}, \frac{5 E }{18}\)
- C \(\frac{5 E }{18}, \frac{18 E }{55}\)
- D \(\frac{5 E }{18}, \frac{10 E }{33}\)
Answer & Solution
Correct Answer
(D) \(\frac{5 E }{18}, \frac{10 E }{33}\)
Step-by-step Solution
Detailed explanation
At \(t=0,\) current through inductor is zero, hence \(R _{ eq }=(5+1) \|(5+4)=\frac{18}{5}\) \(i _{1}=\frac{ E }{18 / 5}=\frac{5 E }{18}\) At \(t=\infty\), inductor becomes a simple wire and now the circuit will be as shown in figure hence…
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