Question

A capacitor of capacitance $C_{1}$ is charged to a potential $V$ and then connected in parallel to an uncharged capacitor of capacitance $C_{2}$ .The final potential difference across each capacitor will be

• A. $\frac{ C _{2} V }{ C _{1}+ C _{2}}$
• B. $\frac{C_{1} V}{C_{1}+C_{2}}$
• C. $\left(1+\frac{ C _{2}}{ C _{1}}\right)$
• D. $\left(1-\frac{ C _{2}}{ C _{1}}\right) V$

Answer 1

Answer: (B)

$\frac{C_{1} V}{C_{1}+C_{2}}$

Answer 2

When a charged capacitor of capacitance $C_{1}$ is connected in parallel to an uncharged capacitor of capacitance $C_{2}$, the charge is shared between them till both attain the common potential which is given by,
common Potential $=\frac{q_{1}+q_{2}}{C_{1}+C_{2}}$
$=\frac{C_{1} V+0}{C_{1}+C_{2}}$
$=\frac{C_{1} V}{C_{1}+C_{2}}$