Thus, $I_{1}=\frac{6}{2}=3 A, I_{2}=0$ and $I_{3}=0$

Long time after the switch $S$ is closed, charging of the capacitors will stop and current will pass through resistors $8 \Omega$ and $4 \Omega$

Applying Kirchhoffs law for loop $1$ and $2,$

$2 I_{1}+8 I_{2}=6 \Rightarrow I_{1}+4 I_{2}=3 \ldots . .(1)$

and $-2 I_{2}+4 I_{3}=0 \ldots(2)$

also $I_{1}=I_{2}+I_{3}=I_{2}+(1 / 2) I_{2}=(3 / 2) I_{2} \ldots(3)$

From $(1)$ and $(3),$ $(2 / 3) I_{2}+4 I_{2}=3 \Rightarrow I_{2}=0.6 A$

Using $(3),$ $I_{1}=(3 / 2) 0.6=0.9 A$ and $I_{3}=0.9-0.6=0.3 A$