When the switch is closed, then initial current through 1 $\mathrm{\Omega }$ resistor is (see figure)

Initially cement through the capacitor will be  maximum
So 6 $\mathrm{\Omega }$ and 3 $\mathrm{\Omega }$ are in parallel total  resistance
$\begin{array}{l}\frac{6\times 3}{9}+1=3\mathrm{\Omega }\\ \mathrm{i}=\frac{12}{3}=4\mathrm{\Omega }\end{array}$