Question

The capacitor $C_1$ in the figure shown initially carries a charge $q_o$. When the switches $S_1$ and $S_2$ are closed, capacitor $C_1$ is connected in series to a resistor R and a second capacitor $C_2$ which is initially uncharged. Find the charge flown through wires as a function of time :

• A. $q_o{e}^{-t/RC}+\frac{C}{C_2}{q}_o$
• B. $\frac{q_{o}C}{C_1}[1-e^{-t/RC}]$
• C. $q_o\frac{C}{C_1}{e}^{-t/RC}$
• D. $q_o{e}^{-t/RC}$, where $C=\frac{C_1{C}_2}{C_1+C_2}$

Answer 1

Answer: (B)

$\frac{q_{o}C}{C_1}[1-e^{-t/RC}]$

As no battery is connected to the circuit so when switches are closed, both capacitors are in parallel and total charge $q_0$ will distribute between them in direct ratio of capacity.
In steady state, the charge on $C_2$ is $q_{20}=\frac{C_2}{C_1+C_2}{q}_0$
Thus, charge on $C_2$ at time t is $q_2\left(t\right)=C_{20}(1-e^{-t/RC})=\frac{C_2}{C_1+C_2}{q}_0(1-e^{-t/RC})=\frac{q_{0}C}{C_1}(1-e^{-t/RC})$

where, $C=\frac{C_1{C}_2}{C_1+C_2}$
Initially $C_2$ is uncharged so, whatever is the charge on $C_2$ that is the charge flown through switches.