JEE Main202429 Jan 2024Morning ShiftPhysicsCapacitanceActual
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 :
Options
- A3 . 2 A
- B1 . 5 A
- C2 . 0 A
- D1 . 2 A
Correct answer
B. 1 . 5 A
Step-by-step solution
By energy conservation, it follows that 1 2 C V 2 = 1 2 L I max 2 . . . 1 Equation (1) implies that I max = C L V = 100 × 10 - 6 6 . 4 × 10 - 3 × 12 = 12 8 = 1 . 5 A