In the circuit shown (figure), switch S₂, is closed first and is kept closed for a long time, Now S₁, is closed. Just after that instant, the current through S₁, is

Just before S, is closed, the potential difference across capacitor 2 is 2$\mathrm{\epsilon }$. Just after S₁ is closed, the potential differences across A capacitors 1 and 2 are 0 and2$\mathrm{\epsilon }$,
respectively. Applying  KVL to loop ABCD immediately after S₁, is closed, 

$\mathrm{\epsilon }=-{\mathrm{iR}}_1+0+2\mathrm{\epsilon },\mathrm{i}=\frac{\mathrm{\epsilon }}{{\mathrm{R}}_1}$towards left