# Calculating Potential Difference in Series Capacitor Circuit

• flyingpig
In summary: Both terminals of each capacitor will be connected -- the capacitors end up in parallel (See attached figure). Which charges have access to each other via conduction paths once the connections are made? What will be the result?Why is it connected in parallel and not series?Because when the capacitors are connected in parallel, the charges on each capacitor can access each other through the conducting wires. This is different than when the capacitors are connected in series, where the charges on each capacitor are physically separated by the intervening wire.
flyingpig

## Homework Statement

A 3.0 µF capacitor charged to 40 V and a 5.0 µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other.

b. Find the potential difference across the capacitors.

## The Attempt at a Solution

Do you just add up the potentail differences? 18 + 40 = 58V?

flyingpig said:
A 3.0 µF capacitor charged to 40 V and a 5.0 µF capacitor charged to 18 V are connected to each other, with the positive plate of each connected to the negative plate of the other.

Find the potential difference across the capacitors.

## The Attempt at a Solution

Do you just add up the potentail differences? 18 + 40 = 58V?

Nope. There will be charge movement (and cancellation) that happens when the capacitors are connected. Sketch a diagram showing the plates with their charges before connection, then work out what charges will move where when they are connected.

Sum up the charges?

3.0 µF * 40V = 1.2 *10^-4C

5.0 µF * 18V = 9 * 10^-5 C

flyingpig said:
Sum up the charges?

3.0 µF * 40V = 1.2 *10^-4C

5.0 µF * 18V = 9 * 10^-5 C

Sum the charges taking into account their polarities on the capacitors. Did you draw the sketch that I suggested?

gneill said:
Sum the charges taking into account their polarities on the capacitors. Did you draw the sketch that I suggested?

I don't know how to draw it...

They are connected in a way that they can be added up right? It's from - to + to - to +

Can you draw one charged parallel plate capacitor?

[PLAIN]http://img641.imageshack.us/img641/2150/unledgkx.jpg

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Do the charges really sit off to the side of the plates?

No the plus and minus just indicate which plate is positive and negative.

Okay. So you've calculated the individual charges that exist on each capacitor. Draw two capacitors side by side (unconnected as yet) and label each plate with the charge that exists on it.

When the capacitor plates are connected as specified, what will happen to the charges?

Do they merge together to become one big charge? If so why is it leaking?

flyingpig said:
Do they merge together to become one big charge? If so why is it leaking?

Sorry, I don't understand your reference to "leaking".

Charges combine -- They add or cancel depending upon their signs. You do remember that a + unit charge will cancel with a - unit charge, right?

flyingpig said:
[PLAIN]http://img641.imageshack.us/img641/2150/unledgkx.jpg[/QUOTE]

While it is okay to show the polarity of the voltage across the capacitor as you have here, with a + and - it may lead to confusion. When you know the voltage difference, as we initially do here, it may be better to label one plate as +40v and the other as 0v.

Otherwise, I think you may be tempted to then proceed to wrongly label one plate as +40v and the other as -40v, and that would be wrong because it would describe a potential difference between the plates of 80 volts, not 40v.

A capacitor is only concerned with potential difference across its plates. That's all it sees, that's all it feels, that's all it knows. And that potential difference is wholly due to the charges on the plates. Q=C.V

Don't worry about doing any calculations until you have drawn and clearly labelled a couple of diagrams to show you understand the situation before and after. If you can't get the diagrams right, then you are wasting your time doing any calculations.

When the two capacitors are connected together, there will be a momentary spark. This indicates a brief flow of current through the connecting wires, and current denotes the movement of electrons through those wires.

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OKay so once the capacitors are hooked up the charges combine to become one current in one branch

so Q = 2.1 * 10^-4 C

flyingpig said:
OKay so once the capacitors are hooked up the charges combine to become one current in one branch

so Q = 2.1 * 10^-4 C

Can you draw a picture of that? How did you calculate Q?

Any current that flows will be very brief as the charges sort out their new distribution. Note that in the end both capacitors will still have a charge, it just won't be the same charge that they started out with.

[PLAIN]http://img37.imageshack.us/img37/2847/unledzz.jpg

I just added the charges from each plate. Should i divide it by 2 to average it out then?

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Both terminals of each capacitor will be connected -- the capacitors end up in parallel (See attached figure). Which charges have access to each other via conduction paths once the connections are made? What will be the result?

#### Attachments

• Fig1.gif
2.4 KB · Views: 1,352
Why is it connected in parallel and not series?

flyingpig said:
Why is it connected in parallel and not series?

With only two components, connected as shown, you could consider them connected both in series and in parallel if you wish. The net capacitance of the pair will be calculated using the parallel connection formula because the plates are in fact connected that way.

[PLAIN]http://img811.imageshack.us/img811/5116/unledws.jpg

They don't look parallel now.

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Looks aren't everything Drawing orientation on the page is irrelevant; Wires do nothing except provide connections, and their length, orientation, or twisty-turneyness make no difference -- everything connected by a single wire path in a circuit diagram is the same node of the circuit; Only the actual connections made are important.

Check how the component terminals in your circuit are connected. If both components' terminals are connected to the same nodes (so that they have shared potentials at both terminals), then the components are in parallel.

If the current through one is always equal to the current through the other, then they are in series. :)

NascentOxygen said:
If the current through one is always equal to the current through the other, then they are in series. :)

So... these two capacitors must be in series then?

#### Attachments

• Fig1.gif
1.3 KB · Views: 763
gneill said:
So... these two capacitors must be in series then?

No. Because if I trace the electron flow through one capacitor, it does not pass through the other. So the branches are not carrying the same current, so they cannot be in series.

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NascentOxygen said:
No. Because if I trace the electron flow through one capacitor, it does not pass through the other. So the branches are not carrying the same current, so they cannot be in series.

But that wasn't your claim. You said: "If the current through one is always equal to the current through the other, then they are in series." Clearly, both of these capacitors will always carry equal currents.

wait why is there a power supply now...?

gneill said:
But that wasn't your claim. You said: "If the current through one is always equal to the current through the other, then they are in series." Clearly, both of these capacitors will always carry equal currents.

It's the same statement, I just worded it differently for the benefit of others who might be hoping to learn something from this exchange.

You are paying insufficient heed to the implications of my "always" qualification. If, in your new arrangement, I were, for example, to change one of the capacitors to a different value, then the currents through both elements will no longer be equal. Thus demonstrating that in parallel branches the currents are not always equal. Were the elements in series, then my setting a new current through one element will always see an identical current flowing through the other. Always means always.

NascentOxygen said:
It's the same statement, I just worded it differently for the benefit of others who might be hoping to learn something from this exchange.

You are paying insufficient heed to the implications of my "always" qualification. If, in your new arrangement, I were, for example, to change one of the capacitors to a different value, then the currents through both elements will no longer be equal. Thus demonstrating that in parallel branches the currents are not always equal. Were the elements in series, then my setting a new current through one element will always see an identical current flowing through the other. Always means always.

We're arguing semantics and fiddly bits. If you rely on unstated assumptions in a definition, then confusion reigns. The point I was making was that in the particular instance of the circuit shown, for the component values shown, that your definition, taken at face value and applied literally, would force one to conclude that the capacitors were in series.

Returning to the OP's problem (flyingpig), he will find it convenient to use the "leads connected to the same pair of nodes" definition for parallel connections, and proceed along those lines. The capacitors, taken to be in parallel, will allow him to calculate the charge remaining on the pair, the net capacitance, and thus the potential difference.

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flyingpig said:
wait why is there a power supply now...?

Sorry, flyingpig, this is just part of a tangential discussion that sprang up regarding how best to define "series" and "parallel" connections for components.

As I stated previously, the two capacitors in this situation are the only components in the circuit, and the argument can be made that they are connected BOTH in parallel AND in series. Analysis can proceed using either assumption. Your choice.

## 1. How do I calculate potential difference in a series capacitor circuit?

To calculate potential difference in a series capacitor circuit, you need to use the formula V = Q/C, where V is the potential difference, Q is the charge stored in the capacitor, and C is the capacitance of the capacitor. You can also use the formula V = IR, where I is the current flowing through the circuit and R is the total resistance.

## 2. What is the relationship between capacitance and potential difference in a series capacitor circuit?

The potential difference across a series capacitor circuit is directly proportional to the capacitance of the capacitor. This means that as the capacitance increases, the potential difference also increases, and vice versa.

## 3. How does the placement of capacitors affect potential difference in a series circuit?

In a series capacitor circuit, the placement of capacitors does not affect the potential difference. This is because the potential difference is the same across all components in a series circuit.

## 4. Can the potential difference in a series capacitor circuit ever be negative?

No, the potential difference in a series capacitor circuit can never be negative. This is because potential difference is a measure of the difference in electrical potential between two points, and a negative potential difference would mean that the second point has a higher potential than the first, which is not possible in a closed circuit.

## 5. How does the potential difference change if a capacitor is added or removed from a series circuit?

If a capacitor is added to a series circuit, the potential difference across each capacitor will decrease, as the total capacitance in the circuit has increased. Conversely, if a capacitor is removed from the circuit, the potential difference across each capacitor will increase. However, the total potential difference in the circuit will remain the same, as it is determined by the total charge and capacitance in the circuit.

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