InterviewSolution
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InterviewSolution
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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)` isA. `((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)` |
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Answer» 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)` |
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