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. The maximum current in the indcutor is :A. `(3V_(0))/(2) sqrt((3C)/(L))`B. `V_(0) sqrt((3C)/(L))`C. `2V_(0) sqrt((3C)/L)`D. `V_(0) sqrt((C )/(L))`

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. The maximum current in the indcutor is :A. `(3V_(0))/(2) sqrt((3C)/(L))`B. `V_(0) sqrt((3C)/(L))`C. `2V_(0) sqrt((3C)/L)`D. `V_(0) sqrt((C )/(L))`
Answer» Correct Answer - A Loss in energy of capacitor = Energy store in inductor `rArr (1)/(2)VC_(0)^(2) + (1)/(2)3C(2V_(0))^(2)-(1)/(2) xx 4 CV^(2) = (1)/(2)LI` `rArr I = (3)/(2)V_(0)sqrt((3V_(0))/(L))`

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