Two capacitors of capacitance `C` and `3C` are charged to potential difference `V_(0)` and `2V_(0)` respectively, and connected to an inductor of inductance `L` as shown in figure. Initially the current in the inductor is zero. Now, the switch `S` is closed. Potential difference across capacitor of capacitance `3C` when the current in the cirucit is maximum is :A. `(V_(0))/(4)`B. `(V_(0))/(4)`C. `(5V_(0))/(4)`D. `(7V_(0))/(4)`

Two capacitors of capacitance `C` and `3C` are charged to potential di

1.	Two capacitors of capacitance `C` and `3C` are charged to potential difference `V_(0)` and `2V_(0)` respectively, and connected to an inductor of inductance `L` as shown in figure. Initially the current in the inductor is zero. Now, the switch `S` is closed. Potential difference across capacitor of capacitance `3C` when the current in the cirucit is maximum is :A. `(V_(0))/(4)`B. `(V_(0))/(4)`C. `(5V_(0))/(4)`D. `(7V_(0))/(4)`
Answer» Correct Answer - C When current is maximum `(dI)/(dt) = 0` e.m.f of across `L = 0` so potential difference across the capacitor will be same. From the law of conservation of charge on plate `2` and `3`. `3CV + CV = 6CV_(0) - CV_(0) rArr V = (5)/(4)V_(0)`

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