Q. Two identical capacitors have the same capacitance$C$. One of them is charged to potential $V_{1}$ and the other to$V_{2}$. The negative ends of the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the system is
A) $\frac{1}{4}C(V_{1}^{2} - V_{2}^{2})$
B) $\frac{1}{4}C(V_{1}^{2} + V_{2}^{2})$
C) $\frac{1}{4}C\left( V_{1}^{\ } - V_{2}^{\ } \right)^{2}$
D) $\frac{1}{4}C\left( V_{1}^{\ } + V_{2}^{\ } \right)^{2}$
Explanation
(c) On sharing of charges loss in electrical energy,
$\Delta U = \frac{C_{1}C_{2}}{2\left( C_{1}{+ C}_{2} \right)}\left( V_{1} - V_{2} \right).$ In present case $C_{1} = C_{2} = C_{\ }$
$\therefore\ \Delta U = \frac{C^{2}}{2(2C)}\ \left( V_{1} - V_{2} \right)^{2} = \frac{1}{4}C\left( V_{1} - V_{2} \right)^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.