Medium MCQ +4 / -1 PYQ · JEE Mains 2020

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 $\left( {{1 \over n}} \right)$ times of its maximum value, is :

  1. A ${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)$
  2. B ${L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)$ Correct answer
  3. C ${L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)$
  4. D ${L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)$

Solution

P.E. in inductor, $U = {1 \over 2}L{I^2}$<br><br>$U \propto {I^2}$<br><br>${U \over {{U_0}}} = {\left( {{I \over {{I_0}}}} \right)^2}$<br><br>${1 \over n} = {\left( {{I \over {{I_0}}}} \right)^2}$<br><br>$I = {{{I_0}} \over {\sqrt n }}$<br><br>We know, $I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)$<br><br>${{{I_0}} \over {\sqrt n }} = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)$ <br><br>$\Rightarrow$ ${e^{ - {{Rt} \over L}}}$ = 1 - ${1 \over {\sqrt n }}$ <br><br>taking ln &amp; solving we get,<br><br>$t = {L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)$

About this question

Subject: Physics · Chapter: Alternating Current · Topic: AC Circuits: R, L, C

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