Verified Solution • 2025–26 Session moderate

QJEEM20Phbe61: JEE Main 2020 Physics - Class 12 JEEM20Phbe61 Solution: A series L-R circuit is connected to a battery of emf V. If the …

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

Multiple Choice Options

  • A LRln⁡(nn+1){L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n + 1}}} \right)
  • B LRln⁡(nn−1){L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)
  • C LRln⁡(n+1n−1){L \over R}\ln \left( {{{\sqrt n + 1} \over {\sqrt n - 1}}} \right)
  • D LRln⁡(n−1n){L \over R}\ln \left( {{{\sqrt n - 1} \over {\sqrt n }}} \right)

Step-by-Step Verified Solution

P.E. in inductor, U=12LI2U = {1 \over 2}L{I^2}

U∝I2U \propto {I^2}

UU0=(II0)2{U \over {{U_0}}} = {\left( {{I \over {{I_0}}}} \right)^2}

1n=(II0)2{1 \over n} = {\left( {{I \over {{I_0}}}} \right)^2}

I=I0nI = {{{I_0}} \over {\sqrt n }}

We know, I=I0(1−e−RLt)I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)

I0n=I0(1−e−RLt){{{I_0}} \over {\sqrt n }} = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)

⇒\Rightarrow e−RtL{e^{ - {{Rt} \over L}}} = 1 - 1n{1 \over {\sqrt n }}

taking ln & solving we get,

t=LRln⁡(nn−1)t = {L \over R}\ln \left( {{{\sqrt n } \over {\sqrt n - 1}}} \right)

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