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Discharging capacitors

  1. Jul 12, 2011 #1
    Hi there,
    okay my Question is on discharging capacitors.
    the equation for instantanious voltage of a capacitor whilst dischargeing is : v=Vi*e^-t/R*C

    However im not sure how to find the time for a capacitor to completely discharge to zero volts.

    When I transpose for t
    t=-(R*C)*ln(v/Vi)

    and input v as zero (cap has completly discharge) we get an undefined answear.
    I havnt studied calculus and am not familiar with the concept of limits.

    how would I go about finding the precise time for a capacitor to discharge without using a normalised universal time constant curve to estimate the answear.

    thanks in advance
     
  2. jcsd
  3. Jul 12, 2011 #2

    sophiecentaur

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    Of course you get a dippy answer. With exponential processes, you Never get to zero. Each interval of RC seconds, the volts decrease by 1/e. You can't get to zero without an infinite value for the time.
    Of course, 'as near zero as dammit' would take a very finite time! (Engineer speaking)
     
  4. Jul 12, 2011 #3
    I understand the voltage will never actually reach absolute zero, however Im after the "practicle" time for the capacitor to reach "practicle" zero voltage.
     
  5. Jul 12, 2011 #4

    sophiecentaur

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    First decide on what is an acceptably low voltage for your purpose and then put it in your formula.
     
  6. Jul 12, 2011 #5

    uart

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    Hi Ruptures. As sophiecentaur has pointed out, it is the nature of the (negative) exponential function that it never reaches precisely zero in any finite time. A "practical" time depends upon just how close to zero you consider "practically zero", but two common choices are

    - 5%, which takes almost exactly 3 time constants, and

    - 1%, which takes approximately 5 times constants.
     
    Last edited: Jul 12, 2011
  7. Jul 12, 2011 #6
    Ahh 5 time constants that rings a bell. thanks uart
     
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