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))]^(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 oscillatroy, 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))`
`:. cos omegat q/(q_(@))=(CV_(2))/(CV_(1))=(V_(2))/(V_(1))( :'q=CV)`………i
Current through the inductor
`I=(dq)/(dt)=d/(dt)(q_(0)cos omegat)=-q_(0) omega sin omega t
`|I|-CV_(1)1/(sqrt(LC))[1-cos^9(2)omega t ]^(1//2)`
`V_(1)sqsrt(C/L)[1-((V_(2))/(V_(1)))^(2)]^(1//2)=[(C(V_(1)^(2)-V_(2)^(2)))/L]^(1//2)`
[(using (i))]