Question

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 capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combined system is

• A. $\frac{C}{4}\left(V_{1}^{2} - V_{2}^{2}\right)$
• B. $\frac{C}{4}\left(V_{1}^{2} + V_{2}^{2}\right)$
• C. $\frac{C}{4}\left(V_{1} - V_{2}\right)^{2}$
• D. $\frac{C}{4}\left(V_{1} + V_{2}\right)^{2}$

Answer 1

Answer: (C)

Total energy of two identical capacitors before connection.
$E_{1}=\frac{1}{2}CV_{1}^{2}+\frac{1}{2}CV_{2}^{2}$ .....(i)
When negative end of one capacitor is connected to negative end of other capacitor and positive to positive, then such type of connection is known as parallel connection.
Hence, equivalent capacitance after connection is given by,
$C_{e q}=C+C$
$C_{e q}=2C$
Common potential, $V=\frac{T o t a l \, c h a r g e}{T o t a l \, c a p a c i t a n c e}$
$=\frac{Q_{1} + Q_{2}}{C + C}$
$=\frac{C V_{1} + C V_{2}}{2 C}$
$=\frac{V_{1} + V_{2}}{2}$
$\therefore $ Total energy after connections
$E_{2}=\frac{1}{2}\cdot C_{e q}\cdot V^{2}$
$=\frac{1}{2}\cdot 2C\left(\frac{V_{1} + V_{2}}{2}\right)^{2}$
$=\frac{C}{4}\left(V_{1} + V_{2}\right)^{2}$
$\therefore $ Decrease in energy $=E_{1}-E_{2}$
$=\frac{1}{2}CV_{1}^{2}+\frac{1}{2}CV_{2}^{2}-\frac{C}{4}\left(V_{1} + V_{2}\right)^{2}$
$=\frac{C}{4}\left[2 V_{1}^{2} + 2 V_{2}^{2} - \left(V_{1}^{2} + V_{2}^{2} + 2 V_{1} V_{2}\right)\right]$
$=\frac{C}{4}\left[V_{1}^{2} + V_{2}^{2} - 2 V_{1} V_{2}\right]=\frac{C}{4}\left(V_{1} - V_{2}\right)^{2}$