A condenser of capacity C is charged to a potential difference of `V_1`. The plates of the condenser are then connected to an ideal inductor of inductance L. The current through the inductor when the potential difference across the condenser reduces to `V_2`​ is :-

Options

A. `[(C(V_1-V_2)^2)/L]^(1/2)`
B. `(C(V_1^2-V_2^2))/L`
C. `(C(V_1^2+V_2^2))/L`
D. `[(C(V_1^2-V_2^2))/L]^(1/2)`

Answer

D

Explanation

In case of oscillatory discharge of a capacitor through an inductor , charge at instant t is given by
`q=q_0 cos omegat`
where `omega=1/sqrt(LC)`
`therefore cos omegat=q/q_0=(CV_2)/(CV_1)=V_2/V_1(because q=CV)` ….(i)
Current through the inductor
`I=(dq)/(dt)=d/(dt)(q_0cos omega t)=-q_0omega sin omegat`
`|I|=CV_1 1/sqrt(LC) [1-cos^2 omegat]^(1//2)`
`=V_1sqrt(C/L)[1-(V_2/V_1)^2]^(1//2)=[(C(V_1^2-V_2^2))/L]^(1//2)` {(using (i))}