In the given circuit (Fig), key `K` is witched on the at `t = 0`. The ratio of current `i` through the cell at `t = 0` to that at `t = oo` will be

Options

A. `3 : 1`
B. `1 : 3`
C. `1 : 2`
D. `2 : 1`

Answer

A

Explanation

At `t = 0`, the branch containing `L` will offer infinite resistance while the branch containing the capacitor will be effectively
a short circuit. Hence, `(i)_(t = 0) = (epsilon)/(R )` Similary at `t = oo, L` will
offer zero resistance whereas `C` will be an open circuit. Hence
effective resistance `= R + (6 xx 3R)/(6 + 3) = 3R (i)_(t = oo) = (epsilon)/(R )`
The required ratio `= (epsilon)/(R ) xx (3R)/(epsilon) = 3:1`