Question

A capacitor of capacity $C_1$ is charged by connecting it across a battery of e.m.f. $V_0$. The battery is then removed and the capacitor is connected in parallel with an uncharged capacitor of capacity $C_2$.The potential difference across this combination is

• A. $\frac{C_2}{C_1+C_2}.V_0$
• B. $\frac{C_1}{C_1+C_2}.V_0$
• C. $\frac{C_1+C_2}{C_2}.V_0$
• D. $\frac{C_1+C_2}{C_1}.V_0$

Answer 1

Answer: (B)

$\frac{C_1}{C_1+C_2}.V_0$

Answer 2

Common potential , V = $\frac{Total \, charge}{Total \, capacitance} = \frac{C_1 \,V_0}{C_1 + C_2}$