Question

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
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 1

Correct Answer - D
Here, `q_(0) = CV_(1)` and `q = CV_(2)`
When a charged capacitor is connected to ideal inductor, the discharge of capacitor is oscillatory. The chagre on capacitor at an instant `t` is given by, `q = q_(0) sin omega` where `omega = (1)/(sqrt(LC))`.
Therefore, `sin omega = (q)/(q_(0)) = (CV_(2))/(CV_(1)) = (V_(1))/(V_(1))`
Current through inductor is
`I = (dq)/(dr) = (d)/(dt) (q_(0) sin omega t) = q_(0) omega cos omega`
`= q_(0) [1 - sin^(2) omega t]^(1//2)`
`= CV_(1) xx (1)/(sqrt(LC)) [1 - ((V_(2))/(V_(1)))^(2)]^(1//2)`
`= [(C(V_(1)^(2) - V_(2)^(2)))/(L)]^(1//2)`