Why Do Capacitors Behave Differently in Series and Parallel Configurations?

[sdfsfasdfasf]

Homework Statement

.

Relevant Equations

Q = CV

Consider the above diagram. Once the first capacitor is charged, clearly it will have a voltage $E$. Then when the switch is flipped, the cell no longer matters (there is no complete circuit which goes through the cell), so we have the first capacitor connected to the second one, and it looks to me that they are in series. However my worked solution states that the two capacitors are in parallel, now I would agree if the circuit was connected to both capacitors and the cell (imagine we added another wire to the switch), but clearly when we connect both capacitors together, there is only one path for current to flow, hence they are in series?
Also when we are discharging a capacitor through a resistor, why must the voltage of the resistor match that of the capacitor? I get that when the capacitor is fully charged its voltage must match the resistor (they are in parallel), and that when we flick the switch the current will decrease exponentially, as will the charge on the capacitor.

 

[Steve4Physics]

Then when the switch is flipped, the cell no longer matters (there is no complete circuit which goes through the cell), so we have the first capacitor connected to the second one, and it looks to me that they are in series.

Two components connected to make an isolated loop can be regarded as being in series or in parallel. You choose for convenience.

If you are interested in current you treat the 2 components as being in series - necessarily carrying the same current.

If you are interested in voltage, you treat the 2 components as being in parallel - necessarily having the same voltage.

Also when we are discharging a capacitor through a resistor, why must the voltage of the resistor match that of the capacitor?

Connect the end of a lead from the capacitor to the end of a lead from the resistor. Call this connection point A.

Do the same with the other ends to give connnection point B, making a loop.

Assume the leads have negligible resistance.

Connect a voltmeter between A and B. Is the reading the voltage across the capacitor or the voltage across the resistor?

Edit - minor changes.

 

[flyusx]

Also when we are discharging a capacitor through a resistor, why must the voltage of the resistor match that of the capacitor? I get that when the capacitor is fully charged its voltage must match the resistor (they are in parallel)

There's no resistor in the diagram, but are you referring to a fully-charged capacitor connected to a resistor? If so, have you studied Kirchoff's voltage law (KVL) which is a special application of Faraday's law? The answer can be found there.
Faraday's law:
$$\oint\textbf{E}\cdot d\textbf{l}=-\frac{d}{dt}\iint\textbf{B}\cdot d^{2}\textbf{r}=0\text{ (in this case)}$$

 

[SammyS]

There's no resistor in the diagram, but are you referring to a fully-charged capacitor connected to a resistor? If so, have you studied Kirchoff's voltage law (KVL) which is a special application of Faraday's law? The answer can be found there.
Faraday's law:
$$\oint\textbf{E}\cdot d\textbf{l}=-\frac{d}{dt}\iint\textbf{B}\cdot d^{2}\textbf{r}=0\text{ (in this case)}$$

Faraday's Law ?

$\vec B$ field ??

These are NOT relevant to this thread.

 

[flyusx]

Faraday's Law ?

$\vec B$ field ??

These are NOT relevant to this thread.

The time-changing B-field is zero, so Faraday's law reduces to KVL (sum of voltage drops around a closed loop is zero). For a discharging resistor, the capacitor (with $\Delta V_{cap}$) is connected in a closed circuit with the resistor. Therefore, $$\Delta V_{cap}-\Delta V_{R}=0\rightarrow\Delta V_{cap}=\Delta V_{R}$$
Which answers his question.

 

[sdfsfasdfasf]

Two components connected to make an isolated loop can be regarded as being in series or in parallel. You choose for convenience.

If you are interested in current you treat the 2 components as being in series - necessarily carrying the same current.

If you are interested in voltage, you treat the 2 components as being in parallel - necessarily having the same voltage.


Connect the end of a lead from the capacitor to the end of a lead from the resistor. Call this connection point A.

Do the same with the other ends to give connnection point B, making a loop.

Assume the leads have negligible resistance.

Connect a voltmeter between A and B. Is the reading the voltage across the capacitor or the voltage across the resistor?

Edit - minor changes.

How can we choose which configuration they are in? Clearly the current in both capacitors is the same (indicative of series) and apparently the voltages both drop together (hence they are the same, indicative of parallel). Could you expand on this please?

 

[sdfsfasdfasf]

The time-changing B-field is zero, so Faraday's law reduces to KVL (sum of voltage drops around a closed loop is zero). For a discharging resistor, the capacitor (with $\Delta V_{cap}$) is connected in a closed circuit with the resistor. Therefore, $$\Delta V_{cap}-\Delta V_{R}=0\rightarrow\Delta V_{cap}=\Delta V_{R}$$
Which answers his question.

Is the same equation true in the case that the resistor is capacitor (as in my original post)

 

[Steve4Physics]

How can we choose which configuration they are in? Clearly the current in both capacitors is the same (indicative of series) and apparently the voltages both drop together (hence they are the same, indicative of parallel). Could you expand on this please?

In this question you are asked to show that the final voltage is given by $V = \frac {C_0}{C+C_0} E$.

Note that even though the voltmeter looks like it is reading the voltage across $C$, this voltage is the same as the voltage across $C_0$ because the capacitors are in parallel.

This is the final equilibrium state of the system. No current flows because the 2 voltages are equal and opposing. I=0. (It's like having 2 batteries of equal emfs but opposing polarities connected in a 'loop'.)

Thinking about the capacitors being in series is not a useful option. But thinking about them in parallel is (see below).

The strategy to solve the problem is this:

Find the charge on $C_0$ when the switch is in position A. Call it $Q_0$.

When the switch changes to position B, $Q_0$ gets redistributed - some remains on $C_0$ and the remainder goes to $C$. Say these final charges are $q_0$ and $q$.

The final voltage across $C_0$ is $\frac {q_0}{C_0}$. The final voltage across $C$ is $\frac qC$. These voltages are equal because of the parallelism. Using this equality of voltages and the fact that $Q_0 = q_0 + q$, you should be able to solve the problem.

 

[sdfsfasdfasf]

I can do the question fine. I am just confused on why the capacitors are in parallel. Surely when we are charging the first one, the second capacitor has nothing to do with anything (it may aswell not be there) because it is disconnected from the supply. Then when we flip the switch the cell no longer has anything to do (as its not connected) and hence we have a closed loop consisting of both capacitors. Now we both agree the voltages are equal (even though there is only one loop of wire I do not get how they are in parallel), so does the second capacitors voltage shoot up to the EMF of the cell the instant the switch is closed?

 

[nasu]

You are putting to much focus on the names (parallel or series). They have each of the two terminals connected together so they must end up with the same voltage in the equilibrium state. In the tranzient state they have different potential differences. This is their behaviour. They end up with the same voltage because they are connected between the same two points. It does not matter if you call them "in parallel" or not.
And they are not "in series" at any time because one of the conditions for series connection is for the capacitors to have the same charge which they don't.

 

[Steve4Physics]

Hope this isn't too long...

I can do the question fine. I am just confused on why the capacitors are in parallel.

Can you give your definition of 'in parallel'? Maybe you have an incorrect understanding. Also, see below.

Surely when we are charging the first one, the second capacitor has nothing to do with anything (it may aswell not be there) because it is disconnected from the supply.

Agreed.

Then when we flip the switch the cell no longer has anything to do (as its not connected) and hence we have a closed loop consisting of both capacitors.

Agreed.

Now we both agree the voltages are equal (even though there is only one loop of wire

We both agree that the final voltages across the capacitors are equal.

I do not get how they are in parallel),

Look at the (badly drawn) diagram below. Can the 2 capacitors be treated as being in parallel?

I hope you said yes - even though there are some unconnected wires. If you remove the unconnectinged top and bottom wires, nothing has changed electrically - the capacitors (conected in a loop) can still be treated as being in parallel, if that's useful. (And can be treated as being in series, if that's useful.)

so does the second capacitors voltage shoot up to the EMF of the cell the instant the switch is closed?

Maybe this is the sticking point. You’re thinking about an ideal (unrealistic) situation – zero resistance and therefore a zero time-constant. This would mean an infinite current flowing for an infinitesimal time! It woluld mean that the voltage across each capacitor is (at least) 3 different values, E, 0 and $V_{final}$, all at the same moment!

So, instead, you might like to consider what happens if the connecting wires are non-ideal, i.e. if they have some resistance.

In that case, the final configuration is 2 capacitors and 2 resistors in a series loop. The capacitors are not in parallel in that case. The answer to the problem would be the same but the change in voltages wouldn't be instantaneous.

By thinking about the capacitors finally being 'in parallel' we are taking the limiting case of the connecting wires' resistances tending to zero (if you have met the concept of limits).

It doesn't really matter whether you call the arrangement 'parallel' as long as you understand their final voltages must be equal..

 

[Steve4Physics]

You are putting to much focus on the names (parallel or series). They have each of the two terminals connected together so they must end up with the same voltage in the equilibrium state. In the tranzient state they have different potential differences. This is their behaviour. They end up with the same voltage because they are connected between the same two points. It does not matter if you call them "in parallel" or not.

Definitely agree. You beat me to it!

And they are not "in series" at any time because one of the conditions for series connection is for the capacitors to have the same charge which they don't.

For series, the charges don't have to be equal - it's the (instantaneous) currents which need to be equal.

For example, consider the case where you connect a charged capacitor, an uncharged capacitor and a resistor in series.

 

[nasu]

This indeed would be a more general way to define "series" connection. And it will result in the equal charge for the usual capacitors only (initially discharged) series connection. Nice.

 

[sdfsfasdfasf]

Right. I found the part that makes me confused. From what I understand a parallel circuit is one in which the current has more than one way to go, that is, some electrons go one way, some go the other, and they meet back up eventually. From where they split to where they come back traces out two branches, of which all components on one branch must have a total pd of equal amount to the other branch. The current in the two branches must sum to the current before and after the branching. Looking at your diagram, I can see the two branches, therefore the capacitors are in parallel. I get very very confused when you mention the ability to choose whether the two components are in parallel or series, that makes no sense to me. Surely its just one or the other? Why not?

 

[sdfsfasdfasf]

Also does the voltage across the second capacitor drop down from $E$ to $V$ or from $0$ to $V$?

 

[Steve4Physics]

Also does the voltage across the second capacitor drop down from $E$ to $V$ or from $0$ to $V$?

Take the real-life case where there is some resistance. Say the wires have a total resistance R.

I'll call the left capacitor ‘cap-1’ and the right capacitor ‘cap-2’.

The switch is flipped to B:

Cap 1’s voltage starts at E and then falls as cap-1 loses charge (since ‘V = Q/C’).

The voltage across R jumps from 0 to E because a current has suddenly started flowing through it. This voltage gradually drops as the current drops (since V = IR).

Cap-2’s voltage starts at 0 and rises as cap-2 gets charged (since ‘V = Q/C’).

Eventually, Voltage across cap-1 = Voltage across cap-2
The current has then reduced to zero so voltage across R =0

 

[Steve4Physics]

From what I understand a parallel circuit is one in which the current has more than one way to go, that is, some electrons go one way, some go the other, and they meet back up eventually. From where they split to where they come back traces out two branches, of which all components on one branch must have a total pd of equal amount to the other branch. The current in the two branches must sum to the current before and after the branching.

That's a very good description. Stick with it!

Looking at your diagram, I can see the two branches, therefore the capacitors are in parallel.

My diagram contained only 'dummy' branches - they wren't connected to anywhere so they were irrelevant! It was deliberate, to get you to think about what we mean by 'parallel'.

I get very very confused when you mention the ability to choose whether the two components are in parallel or series, that makes no sense to me. Surely its just one or the other? Why not?

That's not exactly what I said. If you have 2 components in an isolated loop (no other connections) you have a choice. You can say:

a) the compoents can be said to be in parallel (because they have the same voltage);

or

b) the components can be said to be in series (because they have the same current);

or, to add a 3rd new option

c) nothing at all!!! it really doesn't matter what you call it!

 

[sdfsfasdfasf]

Waaaaaaaaaaaaaaaaait. You are telling me that a closed loop means no supply?

 

[Steve4Physics]

Waaaaaaaaaaaaaaaaait. You are telling me that a closed loop means no supply?

I don't understand what you mean or to what you are referring.

You could either explain clearly or maybe it's time to end the thread?

 

[sdfsfasdfasf]

Where is the electrical supply in your image?

 

[Steve4Physics]

Where is the electrical supply in your image?

No specific supply is needed. It’s no different to the original problem when the switch is flipped to B. You are left with 2 capacitors connected together, with no supply, in a loop.

However, the purpose of the image (plus the accompanying description) was to show that 2 components which are in parallel could also be considered to be in series – if they are the only 2 components in an isolated loop.

I’ve explained as best as I can. So I won’t reply further to this thread. You can start a new thread, with a clear problem-statement, if you need to.