Q. Two condensers $C_{1}$ and $C_{2}$ in a circuit are joined as shown in figure. The potential of point $A$ is $V_{1}$ and that of $B$ is $V_{2}.$ The potentials of point $D$ will be
A) $\frac{1}{2}(V_{1} + V_{2})$
B) $\frac{C_{2}V_{1} + C_{1}V_{2}}{C_{1} + C_{2}}$
C) $\frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}}$
D) $\frac{C_{2}V_{1} - C_{1}V_{2}}{C_{1} + C_{2}}$
Explanation
(c) Charge on $C_{1} =$ charge on $C_{2}$
$\Rightarrow C_{1}\left( V_{A} - V_{D} \right) = C_{2}(V_{D} - V_{B})$
$\Rightarrow C_{1}\left( V_{1} - V_{D} \right) = C_{2}\left( V_{D} - V_{2} \right) \Rightarrow V_{D} = \frac{C_{1}V_{1} + C_{2}V_{2}}{C_{1} + C_{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.