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Capacitors In Series

  1. Mar 1, 2013 #1
    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. jcsd
  3. Mar 1, 2013 #2
    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
  4. Mar 1, 2013 #3

    Doc Al

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    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.
     
  5. Mar 1, 2013 #4
    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.
     
  6. Mar 2, 2013 #5
    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?
     
  7. Mar 2, 2013 #6

    Doc Al

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    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.
     
  8. Mar 2, 2013 #7
    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....
     
  9. Mar 2, 2013 #8
    So if I put two different capacitors in series, they will accumulate the same charge?
     
  10. Mar 2, 2013 #9

    Doc Al

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    Sure.
     
  11. Mar 2, 2013 #10
    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
     
  12. Mar 2, 2013 #11

    stevendaryl

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    They have to, because any charge on one plate of one capacitor must be taken from the other plate of the other capacitor.
     
  13. Mar 2, 2013 #12

    Doc Al

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    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?
     
  14. Mar 2, 2013 #13

    stevendaryl

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    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. [itex]Q_B + Q_C = [/itex] the charge on the first capacitor [itex] = 0[/itex].
    2. [itex]Q_D + Q_E = [/itex] the charge on the second capacitor [itex] = 0[/itex].

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

    1. [itex]Q_C + Q_D = [/itex] the charge on the isolated set of points C and D [itex]=0[/itex].

    Together, these imply that [itex]Q_B = Q_D[/itex].
     
  15. Mar 2, 2013 #14
    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.
     
  16. Mar 2, 2013 #15

    stevendaryl

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    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 [itex]I = \dfrac{dQ}{dt}[/itex]
    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 [itex]QC[/itex] where Q is the charge on the capacitor, and [itex]C[/itex] is the capacitance.
    9. The voltage change across a resistor is equal to [itex]IR[/itex] where R is the resistance and [itex]I[/itex] is the current in the wire.
    10. The voltage change across an inductor is equal to [itex]L \dfrac{dI}{dt}[/itex]
     
  17. Mar 2, 2013 #16

    stevendaryl

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    I'm not sure exactly what the stumbling block is: the voltage drop across a capacitor is given by [itex]V = QC[/itex]. For capacitors in series, the Qs are the same, but the Cs are different, so the Vs are different.
     
  18. Mar 2, 2013 #17
    So, if the capacitance of each capacitor was the same, then electric potential would be the same at each capacitor?
     
  19. Mar 2, 2013 #18

    Doc Al

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    Sure.
     
  20. Mar 2, 2013 #19
    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.
     
  21. Mar 2, 2013 #20

    Doc Al

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    What do you mean?

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