In the circuit shown (figure), switch $S_2$ is closed first and is kept closed for a long time. Now,  $S_1$ is closed. Just after that instant, the current through  $S_1$ is 

Just before  $S_1$is closed, the potential difference across capacitor 2 is  $2\epsilon .$ 

Just after $S_1$ is closed, the potential differences across capacitors 1 and 2 are 0 and  $2\epsilon ,$respectively. Applying KVL to loop ABCD immediately after $S_1$ is closed.

$\epsilon =-i{R}_1+0+2\epsilon ,i=\frac{\epsilon }{R_1}towardsleft$