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RC circuit discharging

  1. Jun 13, 2007 #1
    1. The problem statement, all variables and given/known data

    [​IMG]

    3. The attempt at a solution

    I really don't know how to tackle this question, but here is my attempt anyhow (wrong I know! :yuck:)

    RC=5000*2*10^-6
    RC=0.01

    V=Vo*e^(-t/RC)
    0.99=e^(-t/0.01)
    ln(0.99)=-t/0.01
    t=-0.01*ln(0.99)

    t=1*10^-4 seconds :cry:

    The answer is 4.605*10^-2 seconds
     
  2. jcsd
  3. Jun 13, 2007 #2

    andrevdh

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    By the final voltage they mean zero volts (completely discharged). So its voltage at that stage will be one percent of [tex]V_o[/tex] (the left over from the starting voltage).
     
  4. Jun 13, 2007 #3
    Oh, so If I change my equation to

    0.01=e^(-t/0.01)
    ln(0.01)=-t/0.01
    t=-0.01ln(0.01)

    t=0.461 seconds

    Thanks!
     
  5. Jun 14, 2007 #4

    andrevdh

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    You can mark the voltage axis off in two ways, either in terms of the initial discharge voltage [tex]V_o[/tex] in which case the 100% will be at the top and the 0% will be at the bottom next to the 0 volt mark, or you could mark it off in terms of the final dicharge voltage, 0 volts. In this case 100% will be at the bottom and 0% will be next to [tex]V_o[/tex]. This means that the 99% of the final discharge voltage will be next to the 1% of the intial starting voltage.

    Also 0.01 means divide by a hundred so the answer will be 0.0461 seconds.
     
    Last edited: Jun 14, 2007
  6. Jun 17, 2007 #5
    Think I got you!

    [​IMG]
    There was an amendment because they forgot to add "Take C as 2μF"

    Better do one more to double check!
    For this one,

    RC=8000*2*10^-6
    RC=0.016

    V=Vo*e^(-t/RC)
    0.37=e^(-t/0.016)
    ln(0.37)=-t/0.016
    t=-0.016*ln(0.37)

    t=0.016 seconds
     
  7. Jun 18, 2007 #6

    andrevdh

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    It is not stated explicitly in the problem, but I assume that the capacitor is charging up when the switch is closed. So it starts out at zero and ends at Vo. The formula you are using is for discharge!
     
  8. Jun 18, 2007 #7
    Ok, I just realised!

    I think I'm on the right track now..

    V=1.5*e^(-t/RC)
    0.63=1.5*e^(-t/0.016)
    0.42=e^(-t/0.016)
    ln(0.42)=-t/0.016
    t=-0.016*ln(0.42)

    t=0.014seconds?
     
  9. Jun 18, 2007 #8

    andrevdh

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    No, the charging up formula is

    [tex]V = V_o(1 - e^{-\frac{t}{\tau}})[/tex]

    for the voltage over the capacitor, that is these two graphs (charge and discharge) are mirror images (mirror along the x-axis) of each other
     
    Last edited: Jun 18, 2007
  10. Jun 18, 2007 #9
    Ok, gotcha, just double checked with my notes.
    My only query is this.
    For v, do I substitute the value of 0.63 or 0.37?
     
  11. Jun 18, 2007 #10

    andrevdh

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    The final voltage over the capacitor will be that of the power supply. When the voltage over the capacitor (due to charges accumating on its plates pushed/pulled there by the battery) is equal to that of the battery current stop flowing in the circuit and charging stops.
     
  12. Jun 18, 2007 #11
    So, 0.63 it is!
     
  13. Jun 18, 2007 #12

    andrevdh

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    The final voltage over the capacitor, [tex]V_o[/tex], will be 1.5 volt. So 63% of it will be ....
     
  14. Jun 18, 2007 #13
    65% of 1.5volts = 0.945V

    0.945=1.5(1-e^-t/RC)
    0.945=1.5-1.5e^-t/RC
    1.5e^-t/RC=0.555
    e^-t/RC=0.37
    -t/RC=ln(0.37)
    t=-0.016ln(0.37)

    t=0.016 seconds
     
  15. Jun 18, 2007 #14

    andrevdh

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    That is what I get too.
     
  16. Jun 20, 2007 #15
    I'm quite confused as to how you determine if the circuit is discharging or recharging. I noticed that in the 2 examples above, the location of the capacitors are in different places, but at the same time, only the discharge formula is given on the formula sheets where both questions came from.
     
  17. Jun 20, 2007 #16

    andrevdh

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    In your first diagram there is now power source. So the capacitor will discharge throught the resistor (if the capacitor starts out with charge on its plates.)

    In the second diagram there is a battery, [tex]V_o[/tex], in the circuit which will pull/push charges from/onto the pates of the capacitor once the switch is closed.
     
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