Two Identical Capacitors Have The Same Capacitance C, One Of Them Is Charged To Potential

SOL:

The initial energy of the combined system

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

By joining the two condensers in parallel, the standard potential we get,

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

Therefore, the final energy of the combined system.

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

Now decrease in energy =

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

= \( \frac{C}{2}[2(V_{1}^{2} + V_{2}^{2}) -(V_{1} + V_{2})^{2}] \)

=\( \frac{C}{4}[2V_{1}^{2} + 2V_{2}^{2}) – V_{1}^{2} – V_{2}^{2} – 2 V_{1} + V_{2}] \)

=\( \frac{C}{4}(V_{1} – V_{2})^{2}\)

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