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## Homework Statement

Both diodes are ideal.

**Answers:**

(a) Vi

(b) -Vi

(c) -2Vi

(d) none of the above

(a) Vi

(b) -Vi

(c) -2Vi

(d) none of the above

## Homework Equations

Q=CV

## 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?

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