An ideal capacitor of capacitance 0.2uF 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:

(a) 0.34 A

(b) 0.25 A

(c) 0.17 A

(d) 0.15 A

c

Using Energy Conservation, 

Initial energy stored in capacitor = Final energy stored in capacitor + Energy stored in inductor

$\Rightarrow I = \sqrt{\frac{C(V_i^2-V_f^2)}{L}} = 0.17A $