# Capacitors in series?

1. Nov 15, 2013

### amiras

I found this problem in the book and the exact problem is not important, but I don't understand the picture. Book claims that after voltage source is disconnected and switch closed these capacitors in picture are parallel although I tend to think they are in series. (I agree they are in parallel as long as voltage source is connected)

The problem says that after capacitor C1 is charged the power supply (voltage) is disconnected. Also, switch in the middle is insulating handle, charge can only flow between the two upper terminals and between two lower terminals. After switch is closed book claims that capacitors are in parallel.

Isn't this circuit now is equivalent to this?

2. Nov 15, 2013

### ShreyasR

We don't like definitions of anything... But it is important... So how do you define "series" and "parallel"? You know pictorially what is series and parallel. I can see that... So just go back and think... what is the definition of series and parallel... Then think, how are they different? Once you are done with this, think: which category does this example fall into...? Your question will be answered!

3. Nov 15, 2013

### Tanya Sharma

Hi amiras...

When the capacitors are connected in series, the charge is the same on both. It is not the case now, as the left capacitor is charged, and after that connected to the uncharged capacitor .There is a net charge on the plates in contact. After connecting the plates, there is a charge transfer from one plate to the other till the potential becomes identical on both plates. This situation - equal potential difference and different charge on the connected capacitors - corresponds to parallel connection.

Had they been in series,the charge on both the capacitors would have been same ,which is not the case here.Both the top plates will have positive charges .If they were in series ,then positive plate of first capacitor would have been connected with negative plate of the other .In other words ,if in series ,the net charge on the top plates of both the capacitors would have been zero.

Hope this helps.

4. Nov 17, 2013

### CWatters

Most people would treat them as being in parallel but actually they are both in series and in parallel. The problem can be solved looking at it either way.

Yes probably.

However you should label the terminals so we can be sure you haven't turned one of the capacitors around.

5. Nov 17, 2013

### Drakkith

Staff Emeritus
I don't see how. The first picture clearly shows the voltage source between the upper and lower parts of the circuit.

Amiras, see the attached picture. I've added in the voltage source so it becomes apparent why the two caps are in parallel.

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6. Nov 17, 2013

### davenn

no, I disagree, its not showing a "voltage source" its only showing a potential difference between those 2 points. there is no battery or other PSU indicated as permanently connected
(A PSU may have been connected and disconnected giving that potential difference)

Dave

7. Nov 17, 2013

### Drakkith

Staff Emeritus
Hmm. I see your point. I thought the voltage source was located in the middle of the circuit where the arrows are prior to disconnection.

8. Nov 17, 2013

### Drakkith

Staff Emeritus
I still don't see how they could be in series. Post 3 seems to explain it pretty well. Is there something I'm missing?

9. Nov 17, 2013

### Okefenokee

If two things are in series then the same current must pass through each of them. If two things are in parallel then they must have the same Voltage across them. Here you have a special situation where both things are true. Imagine using a meter to measure amperage and voltage and you'll see what I mean.

10. Nov 17, 2013

### ShreyasR

Exactly... So for analysis, one must consider both cases: potentials across them are equal, and currents through them is same!

11. Nov 18, 2013

### CWatters

What others have said. When the switch is made KCL requires the current flowing to be the same in both capacitors. There is nowhere else for it to go :-)

Personally I agree it's easier to visualise them as being in parallel but in this case it's both.

12. Nov 18, 2013

### CWatters

No not the charge, the change in the charge.

The change in charge is the same because the same current flows through both caps. The absolute charge can be different because they start with different charges.

In this case that's true but makes no difference.

How does changing the initial voltage effect the circuit topology? It can't.

You can charge two caps up any way you want and connect them any way around you want. The same current/charge will flow in both caps. You can treat them as being in parallel or series it makes no difference.

I will agree that it might be easier to think of them as being in parallel if the positives are connected together and series if positive is connected to negative but in reality they are both.

13. Nov 18, 2013

### nasu

There is the fact that the relationship between the final voltage V and the total charge (Qo) is
Qo=V(C1+C2). So the equivalent capacitance is C1+C2.
This is consistent with the "parallel" label.

If you consider it series, it will be an exception to the general rule.

14. Nov 19, 2013

### CWatters

Agreed. I'm not suggesting the final capacitance between those two nodes is anything other than C1+C2.

I refer you to this thread where Ehild makes the same point as me..

Last edited: Nov 19, 2013
15. Nov 19, 2013

### CWatters

Ok here is how to analyse it as a series circuit. I admit that it is simpler to treat it as a parallel circuit but that's not the point. The point is that both approaches give the same answer when you only have two components connected like this.

Let

VC1 = Initial voltage on C1
VC2 = Initial voltage on C2

When the switch is made a quantity of charge ΔQ flows around the circuit. As they are in series the same charge flows through both capacitors so for C1..

ΔQ = ΔVC1 * C1

and for C2

ΔQ = ΔVC2 * C2

Equate..

ΔVC1 * C1 = ΔVC2 * C2 ..... (1)

After the switch is closed (made) applying KVL around the loop gives..

Final VC1 + Final VC2 = 0

(VC1 - ΔVC1) + (VC2 - ΔVC2) = 0

But VC2 = 0 because C2 is assumed to be initially discharged.

(VC1 - ΔVC1) - ΔVC2 = 0 ............ (2)

Solving these two gives..

ΔVC2 = VC1 * C1/(C1+C2)

16. Nov 19, 2013

### nasu

I agree with ehild's comment.
You don't need to assume any of them do analyse the circuit.

In your analysis you use conservation of charge which is a general principle and not a consequence of "series connection". In series connection it applies to the inner connection between capacitors but not at the ends of the series. So it does not follow automatically from the series assumption but from the nature of the junction, from the physical behavior.

I was not trying to say that is more parallel than series but that is depends on what feature you consider as defining the series/parallel connection.
This is a problem in which you have to understand the basic principles on which the calculations for series/parallel are made, rather than just apply blindly the results.

17. Nov 19, 2013

### CWatters

Err.. The circuit is symmetrical. Turn it upside down and the "ends" become the "inner".