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JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A series \(L-R\) circuit is connected to a battery of emf \(V\). If the circuit is switched on at \(t =0\), then the time at which the energy stored in the inductor reaches \(\left(\frac{1}{n}\right)\) times of its maximum value, is
- A \(\frac{ L }{ R } \ln \left(\frac{\sqrt{ n }-1}{\sqrt{ n }}\right)\)
- B \(\frac{ L }{ R } \ln \left(\frac{\sqrt{ n }}{\sqrt{ n }+1}\right)\)
- C \(\frac{ L }{ R } \ln \left(\frac{\sqrt{ n }}{\sqrt{ n -1}}\right)\)
- D \(\frac{ L }{ R } \ln \left(\frac{\sqrt{ n }+1}{\sqrt{ n }-1}\right)\)
Answer & Solution
Correct Answer
(C) \(\frac{ L }{ R } \ln \left(\frac{\sqrt{ n }}{\sqrt{ n -1}}\right)\)
Step-by-step Solution
Detailed explanation
\(U _{\max }=\frac{1}{2} L I _{ max }^{2}\) \(i = I _{\max }\left(1- e ^{- Rt / L }\right)\) For \(U\) to be \(\frac{ U _{\max }}{ n } ;\) i has to be \(\frac{ I _{ max }}{\sqrt{ n }}\) \(\frac{I_{\max }}{\sqrt{n}}=I_{\max }\left(1-e^{-R t / L}\right)\)…
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