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Capacitor RC value

  1. Jan 14, 2010 #1
    Dear Experts,

    I charged a capacitor that is in series with a resistor using a DC power source.
    In practice, does a capacitor charge/discharge as per the calculated RC based on formula?
    For example, does the capacitor really fully charge / discharge at 5RC?

    Last edited: Jan 14, 2010
  2. jcsd
  3. Jan 14, 2010 #2
    Yes and Yes. If you wait 5RC, the voltage across the capacitor will be very close to its final value.
  4. Jan 14, 2010 #3
    The really strict answer is no, although as Corneo said, the value will be very close after 5RC. This applies to charge and discharge equally, but for the moment, let's think about discharging.

    When a capacitor is discharging into a resistor, the voltage will decay as follows:

    After 1RC the voltage will be 36.79% of the original value.
    After 2RC the voltage will be 13.53% of the original value.
    After 3RC the voltage will be 4.98% of the original value.
    After 4RC the voltage will be 1.83% of the original value.
    After 5RC the voltage will be 0.67% of the original value.
    After 6RC the voltage will be 0.25% of the original value.

    This really goes on forever, but note that by 5RC the voltage is below 1% of its starting point.
    Similarly on charging, 5RC takes you to within less than 1% of the final (or more correctly asymptotic) value.
  5. Jan 14, 2010 #4


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    To add to Adjusters explanation, here's a graphic look at the charge/discharge curve.
    http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html" [Broken]

    Attached Files:

    Last edited by a moderator: May 4, 2017
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