Two capacitors $C_{1} \text { and } C_{2}$ are connected to resistor $R$ via two switchs $S_{1}$ and $S_{2}$ as shown in Fig. Capacitor $C_{1}$ is initially charged to a voltage $V_{1}=$ $15 \mathrm{~V}$. Both the switchs are then closed simultaneously. When the steady state is…

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Two capacitors $C_{1} \text { and } C_{2}$ are connected to resistor $R$ via two switchs $S_{1}$ and $S_{2}$ as shown in Fig. Capacitor $C_{1}$ is initially charged to a voltage $V_{1}=$ $15 \mathrm{~V}$. Both the switchs are then closed simultaneously. When the steady state is established, the charges on $C_{1}$ and $C_{2}$ respectively are

Aicharya · Accepted Answer

Initial charge as $C_{1}$ is $Q=C_{1} V_{1}=3 \mu \mathrm{F} \times 15 \mathrm{~V}=45 \mu \mathrm{C}$. When $S_{1}$ and $S_{2}$ are closed, this charge will flow from $C_{1}$ to $C_{2}$ until both the capacitors have the same potential difference $V$. Let $Q_{1}$ and $Q_{2}$ are the charges on $C_{1}$ and $C_{2}$ when steady state is established. Since $C_{1}$ and $C_{2}$ are in parallel, $V$ is the same in both, i.e., $\begin{aligned} V &= \frac{Q_{1}}{C_{1}} = \frac{Q_{2}}{C_{2}} \ \text{or } \frac{Q_{1}}{C_{1}} &= \frac{Q-Q_{1}}{C_{2}} \end{aligned}$ $\Rightarrow \quad \frac{Q_{1}}{3}=\frac{45-Q_{1}}{12}$ $\Rightarrow \quad Q_{1}=9 \mu \mathrm{C}$ and $Q_{2}=Q-Q_{1}=45-9=36 \mu \mathrm{C}$. So the correct choice is $(b)$.

Two capacitors \(C_{1} \text { and } C_{2}\) are connected to resistor \(R\) via two switchs \(S_{1}\) and…

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