I'll get into the math straight away $C_1, C_2, C_3$ each are charged to $q$. The energy is $$\frac{q^2}{2\left(\frac{1}{\frac{1}{C_1}\frac{1}{C_2}\frac{1}{C_3}}\right)}$$ $$\frac{1}{2}\left(\frac{q^2}{C_1} + \frac{q^2}{C_2} + \frac{q^2}{C_3}\right)$$  If I then connect them in parallel, the energy now is $$\frac{1}{2}\left(\frac{9q^2}{C_1 + C_2 + C_3}\right)$$ They are obviuosly not equal. Can someone tell me where the difference in energy goes (and also how it gets dissipated.)
The question at hand is:  "Two capacitors of capacitances $C_1$ and $C_1$ have charge $Q_1$ and $Q_2$. How much energy, $\Delta w$, is dissipated when they are connected in parallel. Show explicitly that $\Delta w$ is non-negative."  I'm confused about what the physical situation is. I took the assumption that these capacitors were somehow pre-charged, and then connected to each other in parallel without a voltage source. However, I don't understand how this would work. If a circuit is composed of only 2 elements, I don't see how they could be in any arrangement but series. Nevertheless, I tried solving it that way:  Two capacitors in parallel have the same voltage drop. Charge will be redistributed to make it the same voltage for both. Let $Q_1'$ and $Q_2'$ be the charges on the capacitors after they are connected. Now, picture the equivalent capacitor  $C_{eq} = C_1 + C_2 =$ $\frac{Q_1' + Q_2'}{V_f}$   conservation of charge:  $Q_1' + Q_2'= Q_1 + Q_2$,        $C_{eq} =$ $\frac{Q_1 + Q_2}{V_f}$  $V_f = $$\frac{Q_1 + Q_2}{C_1 + C_2}$  The initial energy of the capactiors is:  $U_0 = $$\frac{Q_1^2}{2C_1}$$+\frac{Q_2^2}{2C_2}$  $U_f = $$\frac{1}{2}$$(C_1 + C_2)V_f^2 = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$  $\Delta U$$ = $$\frac{(Q_1 + Q_2)^2}{2(C_1 + C_2)}$-$\frac{Q_1^2}{2C_1}$$-\frac{Q_2^2}{2C_2}$  However, I don't see how this is necessarily non-negative.
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