Consider a parallel pate capacitor of capactance C with partially conducting medium between its plates having a resistacne `R_c` if this capacitor is connected to a battery of emf `epsilon` and a resistor R as shown in fig.
a. charge on the capacitor as a function of time.
The switch R as a function of time.

Explanation

The equivalent circuit can be drawn as shown. Applying
Kirchhoff's law for the two loops,
`(i-i_(1))R_(C ) + iR = epsilon` (i)
`(q)/(C ) = R_C (i=i_(1))` (ii)
Eliminating I between (i) and (ii) we get
`(q)/(C ) = (1 + (R )/(R_(C))) + i_(1)R = epsilon`
For capacitor `C, (dq)/(dt) = i_(1)`
`(dq)/(dt) = (CE -q(1 + alpha))/(RC) ("where" alpha = (R )/( C))`
Solving, we get
`q = (EC)/(1 + alpha) (1 - e^(-t(1 +alpha)//RC))`
Adding (i) and (ii), we get
`(Q)/(C ) + iR = E or i = (CE - q)/(RC)`
Substituting for q, we get
`I = (E )/(R )[1 - (1)/(1 +alpha) (1 - e ^(-t(1 + alpha)RC))] ("where" alpha = (R )/(R_C))`