In given circuit capacitor initially uncharged. Now at $$t = 0$$  switch $$S$$ is closed then current given by source at any time $$t$$ is

Solution :

$$\eqalign{ & Q = \frac{{EC}}{2}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right) \cr & {I_2} = \frac{{\varepsilon C}}{{2r}}{e^{\frac{{ - t}}{\tau }}} = \frac{\varepsilon }{R}{e^{\frac{{ - t}}{\tau }}} \cr & {I_1} = \frac{{Vc}}{R} = \frac{{\frac{\varepsilon }{2}}}{R}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right) \cr & {I_1} = \frac{\varepsilon }{R} - {e^{\frac{{ - t}}{\tau }}} + \frac{\varepsilon }{{2R}} - \frac{\varepsilon }{{2R}}{\varepsilon ^{\frac{{ - t}}{\tau }}} \cr & \frac{\varepsilon }{{2R}}\left( {1 + {e^{\frac{{ - t}}{\tau }}}} \right) = \frac{\varepsilon }{{2R}}\left( {1 + {e^{\frac{{ - 2t}}{{CR}}}}} \right) \cr} $$