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 the capacitors are connected together. When the positive ends are also connected, the decrease in energy of the combine 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)

$\frac{C}{4}\left( V_{1} - V_{2} \right)^{2}$

Answer 2

: Initial energy of the combined system

$U_{1} = \frac{1}{2}CV_{1}^{2} + \frac{1}{2}CV_{2}^{2} = \frac{C}{2}\left( V_{1}^{2} + V_{2}^{2} \right)$

On joining the two condensers in parallel, common potential.

$V = \frac{CV_{1} + CV_{2}}{C + C} = \frac{V_{1} + V_{2}}{2}$

$\therefore$Final energy of the combined system

$U_{2} = \frac{1}{2}(C + C)\left( \frac{V_{1} + V_{2}}{2} \right)^{2}$

Decrease in energy

$\Delta U = U_{1} - U_{2} = \frac{1}{2}C(V_{1}^{2} + V_{2}^{2}) - \frac{1}{2}(2C)\left( \frac{V_{1} + V_{2}}{2} \right)$

$= \frac{C}{2}\left\lbrack 2\left( V_{1}^{2} + V_{2}^{2} \right) - \left( V_{1} + V_{2} \right)^{2} \right\rbrack$

$= \frac{C}{4}\left\lbrack 2V_{1}^{2} + 2V_{2}^{2} - V_{1}^{2} - V_{2}^{2} - 2V_{1}V_{2} \right\rbrack = \frac{C}{4}(V_{1} - V_{2})^{2}$