Capacitors In Series

1. Mar 1, 2013

Bashyboy

Hello,

I've read about capacitors in series from two different textbooks, and both of them seem to implicitly state that the electric potential is different for each capacitor in series. Why is this so?

2. Mar 1, 2013

Crazymechanic

Well it may differ but not in a large amount , certainly you wont see that with your average low budget multimeter.
Have you heard about capacitor ESR? They usually try to match as close as possible capacitor in the power supplies of hi end pre amplifiers and other places which demand precision and quality.

The potential difference comes from the fact that the capacitors themselves tend to differ.Just like transistors , you can get two of the same model but measure them precisely and they won't be exactly the same that's why they match transistors in pairs for hi end amplifiers and other products.
It has alot to do with the quality and manufacturing process.ofcourse in theory two similar caps should act the same way but in reality they usually don't.
Electrolytic capacitors have the foil and dielectric soaked in electrolyte hence the name.When time goes one capacitor can be more hermetically sealed than the other and so evaporate faster and loose it's capacitance , there are other factors that contribute to the electric potential upon a capacitor.

Last edited: Mar 1, 2013
3. Mar 1, 2013

Staff: Mentor

Why would you think them to be the same? (Unless their capacitance were equal.) What will be the same though?

Similar idea for resistors in series. The voltage across each will not necessarily be equal.

4. Mar 1, 2013

Naty1

C = q/V right....

So with two 'C's in series, whatever current [charge [q]] that accumulates on
one accumulates on the other.....THAT'S ALL THAT IS, IN GENERAL EQUAL.

When they are in PARALLEL is when the voltages across capacitors is the same....if they
are different 'C', then they accumulate different q.

5. Mar 2, 2013

Bashyboy

Well, I am pretty certain that each capacitor in series has to have the same charge. So, the consensus is that, in theory, the voltage is the same across each capacitor in series, and that I gathered an incorrect impression from the two textbooks?

6. Mar 2, 2013

Staff: Mentor

Correct.
No. (Crazymechanic's response was off the mark.)
No. No reason for capacitors in series to have the same voltage unless they happen to have the same capacitance.

7. Mar 2, 2013

Naty1

of course that is correct....where would electrons go other than along the conductor....from one circuit element to the next in series.....they cannot appear and disappear willy-nilly....from that insight, and knowing q = CV you should be to derive and see for yourself what DocAl and I have been saying....

8. Mar 2, 2013

greswd

So if I put two different capacitors in series, they will accumulate the same charge?

9. Mar 2, 2013

Staff: Mentor

Sure.

10. Mar 2, 2013

Bashyboy

Well, Doc Al, my initial glance of your first reply left me quite confused by your seemingly rhetorical question. Obviously I am missing out on some conception, as to why the voltage isn't the same across capacitors that are in series. If someone could perhaps explain this, I'd be eternally grateful

11. Mar 2, 2013

stevendaryl

Staff Emeritus
They have to, because any charge on one plate of one capacitor must be taken from the other plate of the other capacitor.

12. Mar 2, 2013

Staff: Mentor

I thought you answered my 'rhetorical' question. What must be the same for each capacitor in series? Answer: The charge!

Once you understand that, you can then deduce that the voltage across each capacitor in series must be V = Q/C. Thus the voltage depends on the capacitance.
Again, what leads you to think that capacitors in series would have the same voltage?

13. Mar 2, 2013

stevendaryl

Staff Emeritus
A capacitor is basically two metal plates with a gap between them. So let's consider a simple circuit:

Let point A = the negative end of a battery. This point is connected by a wire to point B = the negative plate of the first capacitor. Nearby is point C = the positive plate of the first capacitor. This is connected by wire to point D = the negative plate of the second capacitor. Nearby is point E = the positive plate of the second capacitor. Finally, this is connected by wire to point F = the positive end of the battery.

Before the circuit was closed, the capacitors had no charge, and no current through them. When you close the circuit, current starts flowing. Specifically, electrons start flowing out of the battery at point A. These electrons build up at point B, causing negative charge to build up. This repels electrons in the plate at point C, making it positively charged. The electrons that leave C travel to point D, causing it to become negatively charged. This repels electrons in plate E, causing it to be positively charged. Then finally, the electrons that leave point E travel into the battery, completing the circuit.

The general rule of thumb is that each capacitor remains electrically neutral at all times. That means that:

1. $Q_B + Q_C =$ the charge on the first capacitor $= 0$.
2. $Q_D + Q_E =$ the charge on the second capacitor $= 0$.

In addition, since charge can't flow anywhere except between points connected by wire, then

1. $Q_C + Q_D =$ the charge on the isolated set of points C and D $=0$.

Together, these imply that $Q_B = Q_D$.

14. Mar 2, 2013

Bashyboy

Well, I suppose mathematically I can see that the electric potential can be difference for each capacitor in a series; I am just trying to figure out a more intuitive way of what's going on, and I just can't figure it out.

15. Mar 2, 2013

stevendaryl

Staff Emeritus
The rules for circuits are the following:

1. At each point in the circuit, there is a current.
2. At each juncture, the total current into that juncture must be zero.
3. At each capacitor, there is a charge.
4. The current flowing into the capacitor is given by $I = \dfrac{dQ}{dt}$
5. At each point in the circuit, there is a voltage.
6. Points connected by wires have the same voltage (neglecting the resistance in the wire).
7. Points connected by a battery have a voltage difference equal to the battery's voltage.
8. The voltage change across a capacitor is equal to $QC$ where Q is the charge on the capacitor, and $C$ is the capacitance.
9. The voltage change across a resistor is equal to $IR$ where R is the resistance and $I$ is the current in the wire.
10. The voltage change across an inductor is equal to $L \dfrac{dI}{dt}$

16. Mar 2, 2013

stevendaryl

Staff Emeritus
I'm not sure exactly what the stumbling block is: the voltage drop across a capacitor is given by $V = QC$. For capacitors in series, the Qs are the same, but the Cs are different, so the Vs are different.

17. Mar 2, 2013

Bashyboy

So, if the capacitance of each capacitor was the same, then electric potential would be the same at each capacitor?

18. Mar 2, 2013

Staff: Mentor

Sure.

19. Mar 2, 2013

Bashyboy

Oh, okay, I see. The impression that I was getting from my textbooks was that the electric potential HAD to drop because of the series configuration. But now I know that isn't true. Thank you, everyone.

20. Mar 2, 2013

Staff: Mentor

What do you mean?

The potential difference across each capacitor is what we are talking about. (Or so I thought.)

21. Mar 2, 2013

Staff: Mentor

Just to be clear, here's an example:

Imagine three identical capacitors hooked up in series. The entire string is attached to a 9 volt battery. Once they are fully charged, each capacitor will have a potential difference of 3 volts.

If the three capacitors were not the same, then the voltage across each would not be equal. But the total voltage across all three must still add up to 9 V.

Make sense?

22. Mar 2, 2013

sophiecentaur

This is the problematic word. We may all be talking at cross purposes here (not an uncommon situation and easily rectified by using diagrams and labels). For two identical capacitors, in series, the PD across each capacitor will be the same or the Potential (reference Earth) on the positive terminal of the 'upper' capacitor will be twice the Potential on the positive terminal of the 'lower' capacitor.

23. Mar 2, 2013

stevendaryl

Staff Emeritus
Yes, if the capacitors in series had the same capacitance, then the voltage drop across each capacitor would be identical. (I'm using "voltage drop" rather than "electric potential", because voltage is the terminology that's usually used in circuits. But it's the same thing as electric potential.)

24. Mar 2, 2013

stevendaryl

Staff Emeritus
This might be a confusion of terminology. Let's make this explicit:

In this picture, we have a 9 volt battery connected in series with two capacitors. The voltage drop--that is, the change in voltage--across the first capacitor is $Q C_1$. The voltage drop across the second capacitor is $Q C_2$. If the capacitors are identical, then these two voltage drops are equal. But that doesn't mean that the voltages are equal. The voltage at the positive plate of the first capacitor is $V_2$. The voltage on the negative plate of the first capacitor is $V_3$. The voltage on the positive plate of the second capacitor is $V_3$. The voltage on the negative plate of the second capacitor is $V_1$. If $C_1 = C_2$, then we would have:

$V_3 - V_2 = V_1 - V_3$

The voltage drops would be the same.

25. Mar 2, 2013

Crazymechanic

I apologize for my off the mark post yesterday I was kinda sleepy and I somehow thought that your talking about capacitors in parallel.

Well anyways ofcourse you don't get the same voltage on the first plate of the second capacitor.This circuit you draw is a voltage divider.A capacitor one.
They have the same ones with resistors too.
This kinda circut was used in the older smps power supplies in the primary mains circuit to switch from 220v to 110v

In theory the voltage drop across each capacitor assuming the capacitance is the same should be equal.Well the voltage you get is depending on where you measure with respect to the ground.
Or do you want to know why is the voltage divided in capacitors in series?