In the adjacent circuit, the capacitor 'A' of capacitance '3C' and the capacitor 'B' of capacitance 'C' are initially charged by potential drops `4_(varepsilon)` and `8_(varepsilon)` respectively and connected through and inductor of inductance 'L' with zero initial current as shown. The switch 'S' is closed at `t = 0`. then the maximum current through the inductor is

Options

A. `varepsilonsqrt((3C)/(L))`
B. `varepsilonsqrt((6C)/(L))`
C. `varepsilonsqrt((12C)/(L))`
D. `varepsilonsqrt((15C)/(L))`

Answer

C

Explanation

When the cuurent through the inductor is maximum,
`V_(L) = L(di)/(dt) = 0`
Now, using energy conservation
`(1)/(2)Li_(max)^(2) = ((1)/(2) xx 3C xx 16epsilon^(2) + (1)/(2)C xx 64epsilon^(2)) - (1)/(2) (3C + C) 25epsilon^(2)`
`(1)/(2)Li_(max)^(2) = 56Cepsilon^(2) - 50Cepsilon^(2)`
`(1)/(2)Li_(max)^(2) = 6Cepsilon^(2)`
`:. i_(max) = epsilon sqrt(12C)/(L)`