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Homework Help: Why do weaker capacitors have the the largest voltage acros

  1. Jul 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A voltage is applied across the capacitor network in the image shown Below (if it loaded I can only see a small panel with IMG on it), which of these individual capacitors has the highest voltage across it

    2. Relevant equationshttps://www.facebook.com/photo.php?fbid=1042796789135446&set=a.1042796642468794.1073741831.100002154049134&type=3&size=640%2C960: is in the attached file below ( I hope its
    visible )

    3. The attempt at a solution: I thought 2 micro-farad would have the largest voltage across it, because its the largest capacitor. But the answer is 530 pf which is the smallest capacitorn

    Attached Files:

  2. jcsd
  3. Jul 20, 2016 #2


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    Notice, the capacitors are connected in series.
    What does this tell you about the current through each capacitor and the amount of charge on each capacitor?
    Then remember Q=CV or V= Q/C. So what does that tell you about the voltage on each capacitor?
  4. Jul 20, 2016 #3
    I would assume that voltage would decrease as it pass down the capacitors in series , the same as it would with resistors, I don't think the capacitors would increase the voltage in series, because that is what i though would originally happen, as capacitors are designed to store energy and this is why I chose the larger capacitor to have the higher voltage across it in series. also it how do I do that equation V=Q/C if i'm only given the capacitance (C)
  5. Jul 20, 2016 #4
    unless the larger capacitors are storing more of the voltage in its electrostatic field than, the weaker capacitors so its voltage across is weaker. this would explain why the weakest capacitor has the highest voltage across it. Is this a correct statement
  6. Jul 20, 2016 #5


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    1- I would suggest, in general, you don't talk about voltage as an absolute value in circuits - only as a difference between two points, such as the ends of a component. That voltage can go up or down as you go along. V=IR , so if the resistances are different, then the voltages can be different, in proportion to their resistances.
    2 - I asked about the current, that is the flow of charge, through the series of capacitors.
    If you want to look at series resistors, then what can you say about the current through series resistors? What does that tell you about the charge flowing through each resistor?
    Then, what can you say about the current through series capacitors and what does that tell you about the charge flowing through each capacitor.

    Lets forget the voltages until we get that sorted.

    1 - In this context capacitors are not weak or strong: they have a larger or smaller capacitance.
    2 - What do capacitors store? Clue: not voltage. 2nd clue: Q=CV
    3 - If you charge a capacitor, then disconnect it from the circuit, you can change its capacitance by moving the plates closer or further apart or by changing the dielectric. In each case the capacitance and the voltage will change as you do this, but something will remain constant.

    On these diagrams, what can you tell me about I1, I2, I3, I4, & I5? What can you tell me about V1, V2, V3, V4, & V5?
  7. Jul 20, 2016 #6


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    By the way, it seems you are required to make an assumption about the initial charge on the capacitors. Assume that initially, before the voltage is applied, that each capacitor's charge is zero.

    The creator of the question really should have specified the initial charge as part of the problem statement, and shame on them for not doing so. This problem isn't really solvable without that information. So in lieu of that, just assume that the initial charge is 0 on all capacitors, just before the voltage is applied.

    Now go back to @Merlin3189's questions. Given the series configuration, what is the relative current through one capacitor compared to that of any other capacitor? What does that tell you about the relative charge on any individual capacitor compared to any other capacitor?
  8. Jul 21, 2016 #7
    [ voltage is measured at both the resistors and the capacitors]
  9. Jul 21, 2016 #8
    since the capacitors are in series the current should be equal through out the circuit at each capacitor. also if you have told me that each capacitors charge is zero then to find the voltage I use the formula V=Q/C
    but if Q = 0 wont that just mean all my voltage calculations will = 0 no matter what capacitance is divide by the charge e.g. 0/2x10^-6F=0 or 0/5.3x10^-10F=0
  10. Jul 21, 2016 #9


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    The INITIAL charge on the capacitors is zero. Then a voltage is applied and current (charge) flows. Then there is non-zero charge on the capacitors.

    Since the current is the same in all the series capacitors, the amount of charge accumulated on the capacitors must be the same. We don't know what this charge is, just that it is the same on every capacitor. (I think this is the crucial point you need to understand here.)

    Now think about Q=CV.
    Q is the charge and is the same for each capacitor.
    C is the capacitance and is different on each capacitor: some have higher capacitance, some lower.
    If Q is constant, do the big capacitances need big voltages and little capacitors little voltages to make CxV constant? Or do Big capacitances need small voltages and little capacitances need big voltages to make CxV constant?
  11. Jul 22, 2016 #10


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    Later when you study AC circuits you will discover that smaller capacitors have a higher impedance than larger capacitors.
  12. Jul 23, 2016 #11
    your right I did the calculation (I apologies for the late reply thinks have been busy lately) and little capaciors need big voltage across it to make CxV constant.
    2uF: 3/2x10^-6F=1.5x10^6 V
    530pF: 3/5.3x10^-12F= 5.66x10^11V
    so a much larger voltage is required for little capacitance capacitors to have constant charge with large capacitance capacitors.
  13. Jul 24, 2016 #12


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    At the risk of repetition consider that the whole arrangement ilustrated in your picture is overall electrically neutral. So positive charge on say the left plate of C1 is exactly equal to the negative charge on the far right plate of C4 (we suppose these connected to the terminals of a battery). Think too of a single charged capacitor – that too as well as overall neutral, is also locally as neutral as it can get. The positive charge is held opposite an equal negative charge separated by only a thin dielectric, a short distance. (The longer the distance the harder it is to hold the charge, the less is the capacitance as you may remember).

    We already commented the equal charges on the exterior plates, now look at the interior ones of the -||- parts. These are initially neutral. They are conductively isolated and so always stay overall neutral. But the positive charge of the left plate of C1 holds facing it an equal negative charge on the right plate, that is it pulls electrons there. Which have been pulled away from the right-hand plate of the overall neutral -||- structure, so that one, which is the left hand plate of C2, gets the same charge though opposite sign as the right plate of C1 - and so the same charge and sign as the left plate of C1. And so on - all the capacitors in series have the same charge. The 2μF capacitor has the same charge as the 10μF capacitor. Same charge but a fifth of the capacitance, it must have five times more voltage across it.
    Last edited: Jul 25, 2016
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