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:

Step-by-step solution

The energy stored in the capacitor is given by, U c = 1 2 C V 2 . The energy stored in the inductor is given by. U L = 1 2 L I 2 . In L C circuit there is no loss of energy in the form of heat so, using energy conservation, U c i + U L i = U c f + U L f ⇒   1 2 × 0.2 × 10 - 6 × 10 2 + 0 = 1 2 × 0.2 × 10 - 6 × 5 2 + 1 2 × 0.5 × 10 - 3 I 2 I =   3 × 10 - 1 A = 0 .17   A