Question

Consider a parallel plate capacitor of capacitance $C$ with partially conducting medium between its plates having a resistance $R_c$. If this capacitor is connected to a battery of emf $\epsilon$ and a resistor $R$ as shown in figure, find
a. charge on the capacitor as a function of time.
b. current through $R$ as a function of time.

Answer 1

The equivalent circuit can be drawn as shown. Applying Kirchhoff's law for the two loops,

$(i-i_1)R_C+iR=\epsilon$ (i)

$\frac{q}{C}=R_C(i-i_1)$ (ii)

Eliminating i between (i) and (ii), we get

$\frac{q}{C}=[1+\frac{R}{R_C}]+i_{1}R=\epsilon$

For capacitor C, $\frac{dq}{dt}=i_1$

$\frac{dq}{dt}=\frac{CE-q\left(\right(l+\alpha )}{RC}(where\alpha =\frac{R}{R_C})$

Solving, we get

$q=\frac{EC}{1+\alpha }(1-e^{-t(1+\alpha )/RC})$

Adding (i) and (ii), we get

$\frac{Q}{C}+iR=E$ or $I=\frac{CE-q}{RC}$

Substituting for $q$, we get

$I=\frac{E}{R}[1-\frac{1}{1+\alpha }(1-e^{-t(1+\alpha )/RC}\left)\right](where\alpha =\frac{R}{R_C})$