Two identical capacitors have the same capacitance ' C '. One of them is charged to potential V 1 and other to V 2 . The negative ends of capacitors are connected together. When positive

Question

Two identical capacitors have the same capacitance ' C '. One of them is charged to potential V 1 ​ and other to V 2 ​ . The negative ends of capacitors are connected together. When positive ends are also connected, the decrease in energy of the combined system is

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Initial energy of the system is given by:

$U_{i} = \frac{1}{2} C V_{1}^{2} + \frac{1}{2} C V_{2}^{2}$

When the capacitors are connected with their negative ends together and positive ends together, they are in parallel. The common potential $V$ is:

$V = \frac{Q _{1} + Q _{2}}{C _{1} + C _{2}} = \frac{C V _{1} + C V _{2}}{C + C} = \frac{V _{1} + V _{2}}{2}$

Final energy of the system is:

$U_{f} = \frac{1}{2} (2 C) V^{2} = C (\frac{V _{1} + V _{2}}{2})^{2} = \frac{1}{4} C (V_{1} + V_{2})^{2}$

The decrease in energy is:

$Δ U = U_{i} - U_{f}$

$Δ U = \frac{1}{2} C (V_{1}^{2} + V_{2}^{2}) - \frac{1}{4} C (V_{1}^{2} + V_{2}^{2} + 2 V_{1} V_{2})$

$Δ U = \frac{1}{4} C (2 V_{1}^{2} + 2 V_{2}^{2} - V_{1}^{2} - V_{2}^{2} - 2 V_{1} V_{2})$

$Δ U = \frac{1}{4} C (V_{1}^{2} + V_{2}^{2} - 2 V_{1} V_{2})$

$Δ U = \frac{1}{4} C (V_{1} - V_{2})^{2}$

Alternatively, using the direct formula for energy loss when two capacitors are connected in parallel:

$Δ U = \frac{1}{2} \frac{C _{1} C _{2}}{C _{1} + C _{2}} (V_{1} - V_{2})^{2}$

Substituting $C_{1} = C_{2} = C$ :

$Δ U = \frac{1}{2} \frac{C \cdot C}{C + C} (V_{1} - V_{2})^{2} = \frac{1}{4} C (V_{1} - V_{2})^{2}$

Answer: $\frac{1}{4} C (V_{1} - V_{2})^{2}$

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