The switch S is closed at t = 0.The capacitor C is uncharged but $C_0$ has a charge $q_0$ at t = 0. Calculate i (t) in the circuit.

Let $q_0$ and q be the instantaneous charges on $C_0$ and C respectively. 

Applying KVL to the circuit, we have $\frac{q_0}{C_0}+\frac{q}{C}+IR=E .....\left(1\right)$

Differentiating this equation, we get $\frac{1}{C_0}\frac{d{q}_0}{dt}+\frac{1}{C}\frac{dq}{dt}+R\frac{dI}{dt}=0$

$\Rightarrow I\left(\frac{1}{C_0}+\frac{1}{C}\right)=-R\frac{dI}{dt}$

Where $I=\frac{{dq}_0}{dt}=\frac{dq}{dt}\Rightarrow \frac{dI}{I}=\frac{-dt}{R{C}_{eq}}$

Where $C_{\mathrm{eq}}=\frac{{\mathrm{C}}_0\mathrm{C}}{\mathrm{C}+{\mathrm{C}}_0}$

Integrating this expression, we have

${\int }_{{\mathrm{I}}_0}^{\mathrm{I}\left(\mathrm{t}\right)} \frac{\mathrm{dI}}{\mathrm{I}}=-{\int }_0^{\mathrm{t}} \frac{\mathrm{dt}}{{\mathrm{RC}}_{\mathrm{eq}}}\Rightarrow [l\mathrm{nI}{]}_{{\mathrm{I}}_0}^{\mathrm{I}\left(\mathrm{t}\right)}=\frac{-\mathrm{t}}{{\mathrm{RC}}_{\mathrm{eq}}}$

$\Rightarrow I\left(t\right)=I_0{e}^{\frac{t}{R{C}_{eq}}}.....\left(2\right)$

Where ${\mathrm{I}}_0$ is initial current

Further, ${\mathrm{I}}_0\mathrm{R}+\frac{{\mathrm{q}}_0}{{\mathrm{C}}_0}=\mathrm{E}\Rightarrow {\mathrm{I}}_0=\frac{\mathrm{E}-\frac{{\mathrm{q}}_0}{{\mathrm{C}}_0}}{\mathrm{R}}.....\left(3\right)$

Substituting ${\mathrm{I}}_0$ from (3) into eqn (2) we get

$I\left(t\right)=\left[\frac{E-\left({\mathrm{q}}_0/{\mathrm{C}}_0\right)}{\mathrm{R}}\right]{\mathrm{e}}^{\frac{-\mathrm{t}}{{\mathrm{RC}}_{\mathrm{eq}}}}$