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Why does a capacitor discharge slowly?

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  1. Apr 1, 2016 #1
    1. The problem statement, all variables and given/known data
    Explain why a capacitor will discharge, although very slowly when there is high internal resistance?

    2. Relevant equations
    V=IR Q=V/C

    3. The attempt at a solution
    I am confused slightly by internal resistnace. Does that refer to regular resistance. From that high internal resistance results in large resistance therefore large p.d. therefore from Q=V/C V=QC so large voltage means that there will be large capacitance?
     
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  3. Apr 1, 2016 #2

    NascentOxygen

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    Hi. smiley_sign_welcome.gif

    Have you tried google, to see how a capacitor's internal resistance can be modelled?
     
  4. Apr 1, 2016 #3

    berkeman

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    Are they asking about internal leakage current discharging the cap, or is it connected to a circuit and is discharging through that?

    Capacitor internal DCR for external current flowing is usually not all that big, certainly not big enough to limit the external discharge rate in most circuits.
     
  5. Apr 1, 2016 #4

    berkeman

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    I see now from your attachment in your first try at this thread that you have quoted the question incorrectly. It is saying that a charged cap will discharge slowly when connected to a DVM with an internal resistance of 12MegOhm. That's not a capacitor internal resistance.

    Can you maybe make a PDF of that one question with its figures, and UPLOAD it as a PDF? You can use PrimoPDF or some other free PDF writer software to do that. :smile:
     
  6. Apr 1, 2016 #5
    I quote from what I have found:

    "The smaller the Resistance or the Capacitance, the smaller the Time Constant, the faster the charging and the discharging rate of the Capacitor, and vice versa."

    So since there is large internaal resistance then this will surely result in slow discharge rate of the capacitor?
     
  7. Apr 1, 2016 #6
    I definitely do not think it is with regarding internal leakage current. What I am sort of muddled with is that usually I get given questions with large resistance. Does a large internal resistance ensue in a large resistance I am sure it does, although it has been a while since I covered this.

    I will try to sure.

    Can I ask what a DVM is? Direct voltage something perhaps?
     
  8. Apr 1, 2016 #7

    berkeman

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    DVM = Digital Voltage Meter, or
    DMM = Digital Multi-Meter (basically the same thing).

    https://en.wikipedia.org/wiki/Voltmeter#Digital_voltmeter

    It is a digital meter that has a high input resistance for voltage measurements (typically >= 10MegOhms)
     
  9. Apr 1, 2016 #8
    Here is the question I am reffering too. I am not cheating on hwk or anything rather seeking help on a question. It is only worth one mark and what I am referring to is 2di.of the uploaded image, with the initial question attached.
     

    Attached Files:

  10. Apr 1, 2016 #9

    berkeman

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    So given what I said about the internal resistance of the DVM, why will the capacitor(s) discharge slowly when the DVM is measuring their voltage? Are you familiar with how RC circuits work?
     
  11. Apr 1, 2016 #10
  12. Apr 1, 2016 #11
    Since it will discharge through the voltemeter. Since there is high resistance, then capacitance decreases from the Time constant=RC therefore the high resistance will result in a slow discharge?
     
  13. Apr 1, 2016 #12

    berkeman

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    Mostly correct except for the typo I fixed. :smile:
     
  14. Apr 1, 2016 #13
    Thank you very much for your help.

    Just to consolidate though, the internal resistance causes the overall resistance to be large in this circuit?
     
  15. Apr 1, 2016 #14

    berkeman

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    Internal resistance of what? :smile:
     
  16. Apr 1, 2016 #15
    The internal resistance of the voltmeter which was very large.:)
     
  17. Apr 1, 2016 #16

    berkeman

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    Yes, exactly. Any DCR in any capacitor will be negligible compared to the many MegaOhm input/internal resistance of a voltmeter.
     
  18. Apr 1, 2016 #17

    berkeman

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    Basically the voltmeter and the capacitor are in series, so the main resistance in the circuit is the internal resistance of the meter. That and the value of the capacitance give you the RC time constant of the decay of the capacitor voltage.
     
  19. Apr 2, 2016 #18
    Thank you very much. You have been very helpful.
     
  20. Apr 2, 2016 #19
    After giving this some thought;

    V = V0 e^(-t/RC) so as resistance increases then voltage decreases? Doesn't this contradict what you said - or am I totally wrong.

    And what does DCR mean :/
    Digital something resistance?
     
  21. Apr 2, 2016 #20

    berkeman

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    Not exactly. The voltage v(t) across the capacitor decays with the time constant RC because the internal resistance of the DVM is across the capacitor when it is measuring the capacitor voltage. The time constant is RC, so a bigger capacitance means that the capacitor voltage takes longer to decay towards zero. And a bigger resistance (in this case in the DVM) means that the capacitor voltage takes longer to decay towards zero.
    DC Resistance. Capacitors and inductors have a parasitic DCR in real circuits, but that doesn't affect this problem here.
     
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