A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6 . 4 mH inductor to…

A capacitor of capacitance 100 μF is charged to a potential of 12 V and connected to a 6.4 mH inductor to produce oscillations. The maximum current in the circuit would be :
  1. 3.2 A
  2. 1.5 A
  3. 2.0 A
  4. 1.2 A

Solution

By energy conservation, it follows that

12CV2=12LImax2   ...1

Equation (1) implies that

Imax=CLV=100×10-66.4×10-3×12=128=1.5 A

Asked in: JEE Main 2024 (29 Jan Shift 1)

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