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 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)` .....(i) 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)` (using (i)) |
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