$A$ capacitor of capacitance $C_0$ is charged to a potential $V_0$ and is connected with another capacitor of capacitance $C$ as shown. After closing the switch $S$,the common potential across the two capacitors becomes $V$. The capacitance $C$ is given by
- A$\frac{C_0(V_0 - V)}{V_0}$
- B$\frac{C_0(V - V_0)}{V_0}$
- C$\frac{C_0(V + V_0)}{V}$
- D$\frac{C_0(V_0 - V)}{V}$
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