In the given circuit (Fig), key `K` is switched 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`

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, `(R )_(L=0)=R` and `(i)_(t=0)=(epsilon)/(R )`. Similarly at `t=oo`, `L` will offer zero resistance whereas `c` will be an open circuit. Hence effective resistance `=R+(6xx3R)/(6+3)=3R` and `(i)_(t=oo)=(epsilon)/(3R)`
`:.` The required ratio `=(epsilon)/(R )xx(3R)/(epsilon)=3 : 1`