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Question:

An ideal capacitor of capacitance 0.2µF is charged to a potential difference of 10V. The charging battery is then disconnected. The capacitor is then connected to an ideal inductor of self inductance 0.5mH. The current at a time when the potential difference across the capacitor is 5V, is:

0.25A

0.15A

0.34A

0.17A

Solution:

Using Energy Conservation,
Initial energy stored in capacitor = Final energy stored in capacitor + Energy stored in inductor
(1/2)CV²i = (1/2)CV²f + (1/2)Li²
C(V²i - V²f) = Li²
i = √[C(V²i - V²f)/L]
Here, Vi = 10V
Vf = 5V
C = 0.2µF
L = 0.5mH
On putting these values, we will get
i = 0.17A