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Question: Answered & Verified by Expert
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:
PhysicsAlternating CurrentJEE Main
Options:
  • A 0.15 A
  • B 0.17 A
  • C 0.34 A
  • D 0.25 A
Solution:
1834 Upvotes Verified Answer
The correct answer is: 0.17 A

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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