Q. A capacitor of capacity $C_{1}$ is charged upto $V$ volt and then connected to an uncharged capacitor of capacity $C_{2}.$ Then final potential difference across each will be
A) $\frac{C_{2}V}{C_{1} + C_{2}}$
B) $\left( 1 + \frac{C_{2}}{C_{1}} \right)V$
C) $\frac{C_{1}V}{C_{1} + C_{2}}$
D) $\left( 1 - \frac{C_{2}}{C_{1}} \right)V$
Explanation
(c) Common potential $V^{'} = \frac{C_{1}V + C_{2} \times 0}{C_{1} + C_{2}} = \frac{C_{1}}{C_{1} + C_{2}}.V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.