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

Explanation

At `t = 0`, the branch containing `L` will other infinite resistance while the branch containing the capacitor will be effectively a short circuit.
Hence,
`(i)_(t=0) = (E)/(R )`
Similarly, at `t = oo, L` will offer zero, resistance, where as `'c'` will be an open circuit.
Hence, effective resistance
`= R + (6R xx 3R)/(6R+3R) , (i)_(t=oo) = (e)/(3R)`
The required ratio , `= (e)/(R ) xx (3R)/(e) , = 3:1`