Question

In the circuit shown in figure initially key $K_1$ is closed and keys $K_2$ is open. Then $K_1$ is opened and $K_2$ is closed (order is important). [Take $Q_1^{\prime }$ and $Q_2^{\prime }$ as charges on $C_1$ and $C_2$ and $V_1$ and $V_2$ as voltage respectively. Then ]

• A. charge on $C_1$ gets redistributed such that $V_1=V_2$
• B. charge on $C_1$ gets redistributed such that $Q_1^{\prime }=Q_2^{\prime }$
• C. charge on $C_1$ gets redistributed such that $C_1{V}_1+C_2{V}_2=C_{1}E$
• D. charge on $C_1$ gets redistributed such that $Q_1^{\prime }+Q_2^{\prime }=Q^{\prime }$

Answer 1

Answer: (A), (D)

The correct options are
A charge on $C_1$ gets redistributed such that $V_1=V_2$
D charge on $C_1$ gets redistributed such that $Q_1^{\prime }+Q_2^{\prime }=Q^{\prime }$

When Ray $R_1$ is closed and Ray $R_2$ is open, the capacitor $C_1$ is charged by cell and when $R_1$ is opened and $R_2$ is closed, the charge stored by capacitor $C_1$ gets redistributed between $C_1$ and $C_2$

$\therefore$Charge on $C_1$ and $C_2$ in parallel combination,

$\therefore$their potential will be equal.

$\therefore$ $V_1=V_2$

Hence, Option $A$ and $D$ are correct.