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JEE Mains · Physics · STD 12 -Electromagnetic induction
- JEE Mains 2020
- Hard
- MCQ
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
Pick an option to check your answer.
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 \(\frac{ I _{ max }}{\sqrt{ n }}\)
\(\frac{I_{\max }}{\sqrt{n}}=I_{\max }\left(1-e^{-R t / L}\right)\)
\(e ^{- Rt / L }=1-\frac{1}{\sqrt{ n }}=\frac{\sqrt{ n }-1}{\sqrt{ n }}\)
\(-\frac{ Rt }{ L }=\ln \left(\frac{\sqrt{ n }-1}{\sqrt{ n }}\right)\)
\(t =\frac{ L }{ R } \ln \left(\frac{\sqrt{ n }}{\sqrt{ n }-1}\right)\)
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