In the circuit shown in figure, switch S is closed at time t = 0. find the current through the cell at any time t.

 

Current in the branch of capacitor at any instant is $I_C=\frac{dq}{dt}=\frac{CV}{3}\left[-e^{-t/3RC}\right]\left[-\frac{1}{3RC}\right]=\frac{V}{9R}{e}^{-t/3RC}$

Let I be the current through the cell at any instant ‘t’.

 Applying KVL to the loop ABEFA,

 $I\left(6R\right)+I_c\left(R\right)+\frac{q}{C}-V=0$

$I\left(6R\right)+\frac{V}{9R}{e}^{-t/3RC}\left(R\right)+\frac{1}{C}\times \frac{CV}{3}\left(1-e^{-t/3RC}\right)-V=0$

$I\left(6R\right)=V\left[1-\frac{1}{9}{e}^{-t/3RC}-\frac{1}{3}\left(1-e^{-t/3RC}\right)\right]$

$I=\frac{V}{6R}\left[\left(1-\frac{1}{3}\right)+e^{-t/3RC}\left(\frac{1}{3}-\frac{1}{9}\right)\right]$

$I=\frac{V}{6R}\left[\frac{2}{3}+\frac{2}{9}{e}^{-t/3RC}\right]=\frac{V}{9R}+\frac{V}{27R}{e}^{-\frac{t}{3RC}}$