Both diodes are ideal.
(d) none of the above
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).
Couldn't the circuit look like this too?
And so Vo = - (1.5Vi - 0.5Vi) = -Vi
Thus the answer is (b) ?
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?