An ideal capacitor of capacitance `0.2muF` is charged to a potential difference of `10V`. The charging battery is than 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 :

Options

A. 0.15 A
B. 0.17 A
C. 0.34 A
D. 0.25 A

Answer

B

Explanation

According to energy conservation total energy remains constant hence we can write the following equation:
`(1)/(2)C_(i)V_(i)^(2)+ (1)/(2)(i)L_(i)I_(i)^(2)= (1)/(2)C_(f)V_(f)^(2)++ (1)/(2)(i)L_(f)I_(f)^(2)`
`(1)/(2)(0.2 xx 10^(-6) ) xx 10^(2) +0= (1)/(2)(0.2 xx 10^(-6)) xx 5^(2) + (1)/(2)(0.5 xx 10^(-3))I^(2)`
On solving we get I= 0.17A