Verified Solution • 2025–26 Session moderate

QJEEM15Ph267b: JEE Main 2015 Physics - Class 12 JEEM15Ph267b Solution: An circuit is equivalent to a damped pendulum. In an circuit…

An LCRLCR circuit is equivalent to a damped pendulum. In an LCRLCR circuit the capacitor is charged to Q0{Q_0} and then connected to the LL and RR as shown below :

JEE Main 2015 (Offline) Physics - Alternating Current Question 152 English

If a student plots graphs of the square of maximum charge (QMax2)\left( {Q_{Max}^2} \right) on the capacitor with time (t)(t) for two different values L1{L_1} and L2{L_2} \left( {{L_1} > {L_2}} \right) of LL then which of the following represents this graph correctly ?
(plots  are  schematic  and  niot  drawn  to  scale)\left( {plots\,\,are\,\,schematic\,\,and\,\,niot\,\,drawn\,\,to\,\,scale} \right)

Multiple Choice Options

  • A https://ik.imagekit.io/novatra/photos/ab7360e1d4d99d7157b91fdb9e042c31503f36eafe1601ef6a7e0fce5464ebff.webp
  • B https://ik.imagekit.io/novatra/photos/20c13075dcb36e2cac84abcfa90280f9822b50a588afa54313a73433a8afe5a9.webp
  • C https://ik.imagekit.io/novatra/photos/b0a83060785095f072c6b8a90547d45daa6fee730e923e051d890856db648e4f.webp
  • D https://ik.imagekit.io/novatra/photos/704ccb92538d4ec95938dfe0dbea493bed7783e6ff5def7df8fa9129cd995271.webp

Step-by-Step Verified Solution

From KVLKVL at any time tt

JEE Main 2015 (Offline) Physics - Alternating Current Question 152 English Explanation
qc−iR−Ldidt=0{q \over c} - iR - L{{di} \over {dt}} = 0

i=−dqdt⇒qc+dqdtR+Ld2qdt2=0i = - {{dq} \over {dt}} \Rightarrow {q \over c} + {{dq} \over {dt}}R + {{L{d^2}q} \over {d{t^2}}} = 0

d2qdt2+RLdqdt+qLc=0{{{d^2}q} \over {d{t^2}}} + {R \over L}{{dq} \over {dt}} + {q \over {Lc}} = 0

From damped harmonic oscillator, the amplitude is

given by A=A0e−dt2mA = {A_0}e - {{dt} \over {2m}}

Double differential equation

d2xdt2+bmdxdt+kmx=0{{{d^2}x} \over {d{t^2}}} + {b \over m}{{dx} \over {dt}} + {k \over m}x = 0

Qmax⁡=Q0e−Rt2L⇒Qmax⁡2=Q02e−RtL{Q_{\max }} = {Q_0}e - {{Rt} \over {2L}} \Rightarrow Q_{\max }^2 = Q_0^2e - {{Rt} \over L}

Hence damping will be faster for lesser self inductance.

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