this is a random question that has no examples or explanations in my text book

how do I find any value if no values are given to me? plus it says "determine Vo" but every answer is a value for Vs which throws me off even more.

where do I begin with this one?

 

The output will be a function on the input.

Draw the two states as separate, complete circuits.

Pay attention to the charge polarity on the capacitors at the point where switch 1 is closed and switch 2 is opened, what do the series voltages add up to?

 

the capacitor is represented by two vertical bars. does the thicker bar represent the positive terminal?

 

Not necessarily. It is irrelevant to the exercise. Treat it as a non-polarized capacitor.

Draw the 2 different circuits and examine them.

Keep in mind that two capacitors connected in series share the voltage applied to them.

 

in the first part of the switching cycle the two capacitors are in series. so they would split the applied voltage. If Vs=10V then each cap would charge to 5V

the second part of the switching cycle puts both cap's in parallel with the load(Vo). because the voltage polarity is now applied to the other side of the cap, Vs would get added to the cap voltage. sticking with Vs=10V, we would add the 10V to the already charged 5V to get 15V. am I correct in my assumptions?

 

and even if I am correct, I still dont understand how I can pull values out of thin air to get one of the answers in the multiple choice

 

ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh.... the answers are actually in the form of (x)(Vs) not x = Vs.

so if Vs=10 and after the entire switching cycle the cap was equal to 15V
then
(1.5)(Vs) is my answer

 

Yes, that is the correct answer.

 

Hello,

Sorry, but 1.5 Vs is not the correct answer.
Take a good look at the direction of the capacitors when charging (switches 1 closed)
and when discharging (switches 2 closed).

Bertus

 

Hello,

Sorry, but 1.5 Vs is not the correct answer.
Take a good look at the direction of the capacitors when charging (switches 1 closed)
and when discharging (switches 2 closed).

Bertus

are you in agreement that both caps will be charged to 5V each during phase 1 of the switching cycle? this is assuming that Vs=10V

 

are you in agreement that both caps will be charged to 5V each during phase 1 of the switching cycle? this is assuming that Vs=10V

In position 1: The capacitors will be charged in series, each at \(0.5 \cdot V_s\) with the + potential being the left terminal in the drawing above.

In position 2: The two capacitors will be switched to parallel, with their "negative" terminals attached to the positive terminal of Vs. This will put the \(0.5 \cdot V_s\) of the capacitors in series with \(1 \cdot V_s\), for a total of \(1.5 \cdot V_s\)

 

In position 1: The capacitors will be charged in series, each at \(0.5 \cdot V_s\) with the + potential being the left terminal in the drawing above.

In position 2: The two capacitors will be switched to parallel, with their "negative" terminals attached to the positive terminal of Vs. This will put the \(0.5 \cdot V_s\) of the capacitors in series with \(1 \cdot V_s\), for a total of \(1.5 \cdot V_s\)

I concur... I was targeting that last post towards Bertus because he is in disagreement with what I think is the answer

 

After the capacitors are charged and disconnected from Ground, they ought to keep their load, shouldn't they. Bertus, why do you suggest they discharge at phase 2?

Starting from the + terminal of the battery, we cross the left capacitor from its - terminal to the + terminal. That means that the voltage level will increase by the amount of the voltage applied on the capacitor. Thus \(V_s+0.5\cdot V_s=1.5\cdot V_s\)

 

Hello,
Ok I see.

Bertus