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 shows in Fig. Initially, the current in the inductor is zero. Now, switch `S` is closed.

The maximum current in the inductor is

Options

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

A

Explanation

When current is maximum `(di)/(dt) = 0`
`:.` e.m.f. across `L = 0`, so potential difference across the capacitor will be same.
From the law of conservation of charge on plates `2` and `3`,
`3CV + CV = 6 CV_(0) - CV_(0)`
`rArr V = (5V_(0))/(4)`
Loss in energy of capacitor = energy stored in inductor
`rArr (1)/(2) CV_(0)^(2) + (1)/(2) 3C(2V_(0))^(2) - (1)/(2) xx 4CV^(2) = (1)/(2) LI^(2)`
`rArr I = (3)/(2) V_(0) sqrt((3C)/(L))`