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 Homework Statement:
 A battery(electromotive force: V), two capacitors (capacitance: Ca,Cb), two switches (S1, S2), and a resistor are connect as shown in figure below. Initially, both switches are open and no charge in stored in either capacitor. Next, S1 is closed and, after sufficiente time elapses, is opened again. Afterwards, S2, is closed while S1 remains opened. As a result, a current begins to flow through the resistor. After sufficient time elapses the current stops flowing through the resistor. What is the Joule heat generated in the resistor during the time from when S2 is closed until the current stops flowing through the resistor?
 Relevant Equations:

Q = C.V
Epe = QV/2
V = R.i
P = V.i
q'a + qb = qa
To solve this question first I calculated the potential energy the capacitor A stored. It's equal a: Ca.V²/2. Ok, so when switch S1 is open and S2 is closed I calculated the equivalent capacitance as if they were in series > 1/Ceq = 1/Ca + 1/Cb > Ceq = (Ca.Cb)/(Ca+Cb). So I used the formula: CeqV²/2 > (Ca.Cb.V²)/2(Ca+Cb) and I found the awnser. The problem is, my friend said that the capacitors are in parallel, but I solved by series, who is right? Which is the best way to solve this question? Thanks