JEE Main 2024 Physics question
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:
Answer: (C)
Solution
= Qω = CV/√(LC) = V√(C/L) = 12 × √(100 × 10⁻⁶/6.4 × 10⁻³) = 12 × 0.125 = 1.5 A.