Question

In the circuit shown, which of the following statement is true, if the potential difference across the capacitors ${\text{C}}_1$ and ${\text{C}}_2$ is $50\text{V}$and $25\text{V}$ respectively.

• A. When $S_1$ is closed, $V_1=30\text{V}$, $V_2=20\text{V}$
• B. When $S_2$ is closed, $V_1=V_2=30\text{V}$
• C. When $S_1$ and $S_2$ both are closed $V_1=V_2=0$
• D. When $S_1$ and $S_3$ is closed, $V_1=50\text{V}$ and $V_2=25\text{V}$

Answer 1

The correct option is D When $S_1$ and $S_3$ is closed, $V_1=50\text{V}$ and $V_2=25\text{V}$

Let the charges on capacitor ${\text{C}}_1$ and ${\text{C}}_2$ be $q_1$ and $q_2$ respectively.

$q_1=C_1{V}_1=50\left(4\right)=200\text{PC}$

$q_2=C_2{V}_2=8\left(25\right)=200\text{PC}$

When $S_1$ is closed, the two capacitors remain isolated, and because the charge on both the capacitors are same, potential difference remains unchanged.

Similarly, when $S_2$ is closed, potential difference remains unchanged.

When $S_1$ and $S_2$ both are closed, still the two capacitors behave as isolated entities, and because the charge on them is same, the potential difference remains unchanged.

When switch $S_1$ and $S_3$ is closed, $c_1$ and $c_2$ are in series and same charge is accumulated on both $c_1$ and $c_2$. As we know that in series combination the charge remains the same.

So, potential across the two capacitors again remains unchanged.

Hence, option $\left(d\right)$ is the correct answer.