Question

A capacitor $C_1$ is charged to a potential difference $V$. The charging battery is then removed and the capacitor is connected to an uncharged capacitor $C_2$. The potential difference across the combination is

• A. $V\frac{C_1}{C_1+C_2}$
• B. $V\frac{C_2}{C_1+C_2}$
• C. $V\frac{C_1{C}_2}{C_1+C_2}$
• D. $\frac{V}{C_1+C_2}$

Answer 1

Answer: (A)

Initial the charge $Q=C_{1}V$.
After removing the battery the both capacitors are in parallel. So total capacitance $C=C_1+C_2$
Let the potential difference across the combination is $V^{\prime }$
now charge $Q^{\prime }=C{V}^{\prime }=(C_1+C_2)V^{\prime }$
As the total charge is conserved so $Q=Q^{\prime }\Rightarrow C_{1}V=(C_1+C_2)V^{\prime }$
$\therefore V^{\prime }=\frac{C_{1}V}{C_1+C_2}$