1. The problem statement, all variables and given/known data Both diodes are ideal. Answers: (a) Vi (b) -Vi (c) -2Vi (d) none of the above 2. Relevant equations Q=CV 3. The attempt at a solution 1. For t=0 to Pi: Cap1 and Cap2 are in series and hence Q1=Q2, therefore Vc1 = Vc2. From this we get Vi = Vc1 + Vc2, hence Vc1 =Vc2 = Vi/2. 2. For t=Pi to 2Pi, The circuit should look like 3 batteries in series as such: , Therefore Vo = -(Vi + Vi/2 + Vi/2) = -2Vi, And so the answer is (c). BUT... Couldn't the circuit look like this too? And so Vo = - (1.5Vi - 0.5Vi) = -Vi Thus the answer is (b) ? BUT... We are just taking the voltage drop across the resistor, so Vo= Vc2 = 1/2Vi So the answer is (d). Which is correct and what exactly am I misunderstanding?