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 wehnn 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 omega t `
where ` omega = (1)/(sqrt(LC))`
` therefore cos omega t = (q)/(q_0) = (CV_2)/(CV_1) = V_2/V_1 (because q = CV)`
Current through the inductor
`I = (dq)/(dt) = (d)/(dt) (q_0 cos omega t) = - q_0 omega sin omega t`
`|I| = CV_1 (1)/(sqrt(LC)) [1 - cos^2 omega t]^(1//2)`
` = V_1 sqrt(C/L) [1-(V_2/V_1)^2]^(1//2) = [ (C(V_1^2 - V_2^2))/(L)]^(1//2)`