$A$ 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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