1.

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 thepotential difference across the condenser reduces to V_2​ is :-

Answer»

`[(C(V_1-V_2)^2)/L]^(1/2)`
`(C(V_1^2-V_2^2))/L`
`(C(V_1^2+V_2^2))/L`
`[(C(V_1^2-V_2^2))/L]^(1/2)`

Solution :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))}


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