Question

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 the current i(t) in the circuit.

Answer 1

Let $q_0$ and q be the instantaneous charges on $C_0$ and C, respectively. Applying $KVL$ to the circuit, we have

$\frac{q_0}{C_0}+\frac{q}{C}+iR=\epsilon$

Differentiating this equation, we get

$\frac{1}{C_0}\frac{d{q}_0}{dt}+\frac{1}{C}\frac{dq}{dt}+R\frac{di}{dt}=0$

or $i[\frac{1}{C_0}+\frac{1}{C}]=-R\frac{di}{dt}ori\frac{C+C_0}{C{C}_0}=-R\frac{di}{dt}\left(\right(wherei=\frac{d{q}_0}{dt}=\frac{dq}{dt})$

or Integrating this expression, we have

$\stackrel{i\left(t\right)}{\underset{i_0}{\int }}\frac{di}{i}=\stackrel{t}{\underset{0}{\int }}\frac{dt}{R{C}_{eq}}or[\log e^i{]}_{i_0}^{i\left(t\right)}=-\frac{t}{R{C}_{eq}}$

or $i\left(t\right)=i_0{e}^{-\left(t/R\right)C_{eq}}$ (ii)

where $i_0$ is the initial current.

Further, $i_{0}R+\frac{q_0}{C_0}=\epsilon$or $i_0=[E-\frac{q_0}{C_0}]/R$ (iii)

Substituting $i_0$ kom Eq. (iii) into Eq. (ii), we get

$i\left(t\right)=\left[\frac{E-\left(q_0/C_0\right)}{R}\right]e^{-\left(\frac{1}{R{C}_{eq}}\right)},where{C}_{eq}=\frac{C{C}_0}{(C+C_0)}$