Just to be sure of my reasoning I am describing a simplified version of my problem
Two capacitors of capacitance $C_1$ and $C_2$ are connected in series to a battery of emf $V$ volts having some internal resistance.Capacitor $C_1$ is charged prior to being placed in the circuit with charge $q$ on either plates which is less than the final charge it would have held if placed in the circuit in uncharged state.How do I find time dependent rates for the charging of both capacitors separately ?
I know how to do that for a single capacitor and resistor in series with battery but here I have two capacitors and one is having some charge on it initially.
Initially I thought I will consider the charge present on $C_1$ and find potential difference across it and subtract it from battery emf to get a starting potential difference which charges both capacitors but I'm not sure whether this is valid or not , moreover $C_1$ will finally reach the amount of charge it can hold if it had been connected uncharged, what after that given that th other capacitor hasn't reached it's full charged state , how does it get charged then , because it's in series it has to pull charges from one plate of other capacitor if it has to charge but doing so will lead $C_1$ to overshoot it's maximum charge threshold (for the series circuit) . Someone explaining the step by step mechanism of this is highly needed.
1 Answer
Initially I thought I will consider the charge present on C1 and find potential difference across it and subtract it from battery emf to get a starting potential difference which charges both capacitors but I'm not sure whether this is valid or not
Yes, it's valid since you're simply changing the initial conditions. Proceeding as usual for the series RC circuit, simply add a term for the voltage across the second capacitor as so:
(1) KVL ($r$ is the internal resistance of the battery)
$$V = v_r(t) + v_{C1}(t) + v_{C2}(t)$$
(2) Differentiate with respect to time
$$0 = \dot v_r(t) + \dot v_{C1}(t) + \dot v_{C2}(t) = \frac{di}{dt}r + \frac{i}{C_1} + \frac{i}{C_2}$$
(3) Solve the differential equation
$$i(t) = i(0)\,e^{-t/rC_{eq}}$$
where
$$\frac{1}{C_{eq}} = \frac{1}{C_1}+\frac{1}{C_2}$$
and
$$i(0) = \frac{v_r(0)}{r}=\frac{V - v_{C_1}(0) - v_{C_2}(0)}{r}$$
Now that you have found the series current $i(t)$, you can find the voltage across each capacitor as so
$$v_{C1}(t) = v_{C1}(0) + \frac{1}{C_1}\int_0^ti(\tau)\,\mathrm{d}\tau$$
and similarly for $v_{C2}(t)$.
Calculate these voltages yourself and verify that their sum approaches $V$ as $t\rightarrow\infty$.
