JEE Main 2018PhysicsAlternating CurrentHardMCQ

JEE Main 2018 — Alternating Current Question with Solution

JEE Main 2018 (15 Apr)

Question

An ideal capacitor of capacitance 0.2 μF is charged to a potential difference of 10 V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5 mH. The current at a time when the potential difference across the capacitor is 5 V is:

Choose an option

Show full solutionCorrect option: B
Correct answer
B0.17 A

Step-by-step explanation

The energy stored in the capacitor is given by, Uc=12CV2. The energy stored in the inductor is given by. UL=12LI2. In LC circuit there is no loss of energy in the form of heat so, using energy conservation, Uci+ULi=Ucf+ULf

⇒ 12×0.2×10-6×102+0=12×0.2×10-6×52+12×0.5×10-3I2
I= 3×10-1A=0.17 A

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About this question

This is a previous-year question from JEE Main 2018, 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.