Questio about Vo of two capacitors in series

[windmillkity]

Two capacitor in series

Initially, s1 open, and s2 connected to ground.
Then s1 close, after enought time, open s1 and switch s2 to 10v
What is vo?

 

[jlee]

Re: Two capacitor in series

If initially (before "s1 open, and s2 connected to ground" ) both capacitors have no charges,
then the final result is Vo = 7.5V

 

[ptmsl]

Re: Two capacitor in series

Are you talking about ideal devices or real world capacitors? In the case of real world capacitors the output voltage will depend more on the leakage resistance of the two capacitors more than their values. For ideal devices the charge will devide the voltage if the capacitors have equal values.

 

[windmillkity]

Re: Two capacitor in series

Initially no charge.
How you get 7.5V? More detail please. Thx

If initially (before "s1 open, and s2 connected to ground" ) both capacitors have no charges,
then the final result is Vo = 7.5V

Added after 5 minutes:

you mean if the capacitors have equal value, the Vo is 5V or...? Thx

I was asked this question in an interview. I gave a answer, the intervier told another one which he thought it would be correct. I doubt it. so I ask here.

Are you talking about ideal devices or real world capacitors? In the case of real world capacitors the output voltage will depend more on the leakage resistance of the two capacitors more than their values. For ideal devices the charge will devide the voltage if the capacitors have equal values.

 

[jlee]

Re: Two capacitor in series

I assumed S1 and S2 are perfect switch... no leakage.

The final answer should be 7.5V because:

1. Initially, there was no charge on both capacitors. So voltage across both capacitors were zero.

2. After " s1 close, after enought time ", the right-side capacitor was charged to 5; accumulated charge on the top place is positive charge Q = 5*C. The left-side capacitor wass floating... no effect, no charge accumulated.

3. Now net charge on the top plates of two capacitors = 5*C. Consider conversation of charges.

4. After " open s1 and switch s2 to 10v", there was -2.5*C negative charge on the top plate of the left capacitor, and there was +7.5*C positive charge on the top plate of right capacitor. Total net charge was conserved. Therefore, the final Vo= +7.5V.


What is the answer your interviewer told you? If you think 7.5V is correct, please give me a help. Thanks.

 

[windmillkity]

Re: Two capacitor in series

Thx for your reply, I am going to post "it is helped" for you even though your answer is different from that interviewer's.

I got one question about your solution. Why "the left-side capacitor wass floating... no effect, no charge accumulated"?

Appreciate it.

I assumed S1 and S2 are perfect switch... no leakage.

The final answer should be 7.5V because:

1. Initially, there was no charge on both capacitors. So voltage across both capacitors were zero.

2. After " s1 close, after enought time ", the right-side capacitor was charged to 5; accumulated charge on the top place is positive charge Q = 5*C. The left-side capacitor wass floating... no effect, no charge accumulated.-------WHY?

3. Now net charge on the top plates of two capacitors = 5*C. Consider conversation of charges.

4. After " open s1 and switch s2 to 10v", there was -2.5*C negative charge on the top plate of the left capacitor, and there was +7.5*C positive charge on the top plate of right capacitor. Total net charge was conserved. Therefore, the final Vo= +7.5V.


What is the answer your interviewer told you? If you think 7.5V is correct, please give me a help. Thanks.

 

[jlee]

Re: Two capacitor in series

The left capacitor was floating because the switch was open, then no charge can be accumlated on this capacitor. Think in this way, when you try to put positive charge on the top plate, the same amount of negative charge has to be on the bottom plate... but there is no electrical path to provide the negative charge on the bottom plate. As a result, positive charge doesn't accumlate on the top-plate... thus this floating capacitor has no effect on the voltage.

Please tell us what your interviewer's answer is, we will see if it makes sense.

Added after 12 minutes:

oh sorry, I misread your question. The correct answer is Vo=10V.

I thought the S2 was open when you are charging up the right capacitor with S1 closed. Now I read your question again, both S1 and S2 are closed when you are charging up the capacitors.

Anyhow, the same theory applies on the situation:

Before S2 switches from 0V to 10V, positive charges accumulated on the top plates of two capacitors: Q = 5*2C. After switching S2 to 10V, no charge was across the left capacitor, all charge moved to the right capacitor, when total net charge was conserved. Q = 10*C = Vout*C, so Vout = 10V.

 

[020170]

Re: Two capacitor in series

jlee, you are right.

also, you're right, your explanation is great

congratulation

 

[montage3000]

Two capacitor in series

the answer is 10v

 

[windmillkity]

Re: Two capacitor in series

thx, I posted "helped me" for you, I am still confused about one detail.

After switching S2 to 10V, no charge was across the left capacitor, all charge moved to the right capacitor, when total net charge was conserved.

Why?

Appreciate it.



Before S2 switches from 0V to 10V, positive charges accumulated on the top plates of two capacitors: Q = 5*2C. After switching S2 to 10V, no charge was across the left capacitor, all charge moved to the right capacitor, when total net charge was conserved. Q = 10*C = Vout*C, so Vout = 10V.[/quote]

 

[eecs4ever]

Re: Two capacitor in series

" After switching S2 to 10V, no charge was across the left capacitor, all charge moved to the right capacitor, when total net charge was conserved.

Why?"

It happend to work out this way ... ( i'll show using algebra)

When S1 is closed, 2*5C = 10 C amount of charge is stored on the top node.

After S1 is open, this charge has no path to flow, so the amount of charge must be conserved when we close S2 .

We know Q = CV
Qinital = 10 C
Qafter = (vo-10v) * C + vo * C

set Qinitial = Qafter
(vo-10v) * C + vo * C = 10 C

=> vo = 10v

Charge on the left cap = C * V = C * (10v - 10v) = 0 .

now if the other end of S2 was connected to 20volts... then there will be charge on the left capacitor.

if S2 were connected to 20v then , we can again solve
Qinitial = Qafter
(vo-20v) * C + vo * C = 10 C
=> vo = 15v
now is a voltage drop across the left capacitor, meaning that there is charge on it.

 

[bob_lv]

Two capacitor in series

So, you mean, when S2 closed, there is no charge flow from 10V?

 

[windmillkity]

Re: Two capacitor in series

Thanks for your very helpful explanation. One more doubt

(vo-10v) * C + vo * C = 10 C

how you know it is Vo-10 not 10-Vo?

Thank you very much!

 

[safwatonline]

Re: Two capacitor in series

(vo-10v) * C + vo * C = 10 C

"
When S1 is closed, 2*5C = 10 C amount of charge is stored on the top node.

so i think that we are finding the charge consewrvation at the top node (Vo-10) and not (10-Vo).

 

[windmillkity]

Re: Two capacitor in series

I see.

Thank you very much!

 

[safwatonline]

Re: Two capacitor in series

"
if S2 were connected to 20v then , we can again solve
Qinitial = Qafter
(vo-20v) * C + vo * C = 10 C
=> vo = 15v
now is a voltage drop across the left capacitor, meaning that there is charge on it.

hello,
i cannot get this ??
if S2 is connected to 20v , why charging stop at Vout=15 , shouldnt the cap. continue to charge till Vout=20 . i thought that the cap will charge till V(s2)=Vout then no current pass (cause there is No potential diffrence , and charging stops)

also 020170 can u tell me what simulator u used ??
regards,
a.safwat

 

[020170]

Re: Two capacitor in series

I used circuitmaker 2000 professional

It is not good more than Hspice in the view of accuracy,

But if you just use in digital or some simple analog circuit like "two cap circuit",

It is better than Hspice or Cadence

thanks

 

[jonw0224]

Two capacitor in series

in response to bob_lv,

yes, charges flow when S2 is closed. You have to charge the left side capacitor from -5V to 0V. At the same time the charge flows to the right side capacitor (from the left) to charge it to 10V. They are correct, Vo is 10V.