JEE Main 2015PhysicsAlternating CurrentHardMCQ

JEE Main 2015Alternating Current Question with Solution

JEE Main 2015 (04 Apr)

Question

An LCR circuit is equivalent to a damped pendulum. In an LCR circuit the capacitor is charged to Q0 and then connected to the L and R as shown below:

If a student plots graphs of the square of maximum charge QMax2 on the capacitor with time (t) for two different values L1 and L2(L1>L2) of L then which of the following represents this graph correctly? (plots are schematic and not drawn to scale)

Choose an option

Show full solutionCorrect option: C
Correct answer
C

Step-by-step explanation

Comparing to the damped pendulum. We write, mdvdt=-kx-bv, bv is resistive force,  Amplitude, A=A0e-bt2m. Comparing results, we can write

qc=+Ldidt+iR, as charge decreasing, 


qc=Ld2qdt2-dqdtR

A=qR=bm=L
q=q0e-Rt2Lq2=q02e-RtLq2=q02e-RtL.

Exponentially decreasing function, L1 graph has higher values than L2.

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About this question

This is a previous-year question from JEE Main 2015, covering the Alternating Current chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.