In the circuit shown in Fig. ${\mathrm{C}}_1=2{\mathrm{C}}_2$ . Initially, capacitor C₁  is charged to a potential of V. The current in the circuit just after the switch S is closed is
 

Uncharged capacitor behaves as a short circuit just after closing the switch. But charged
capacitor behaves as the battery of e.m.f. V just after closing the switch.
$\therefore \mathrm{I}=\frac{{\mathrm{q}}_0}{{\mathrm{C}}_1\left(2\mathrm{R}\right)}=\frac{{\mathrm{q}}_0}{2{\mathrm{RC}}_1}=\frac{\mathrm{V}}{2\mathrm{R}}$