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 :
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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 :

G1 Physics · Accepted Answer

By conservation of energy, ½CV² = ½LI(max)² Therefore, I(max) = V√(C/L) = 12 × √[(100 × 10⁻⁶)/(6.4 × 10⁻³)] = 12 × √(1/64) = 12/8 = 1.5 A