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Behaviour of a RC circuit charging

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data

    A circuit is powered by a 10 V power supply and has a resistor of 500 ohms in series with a capacitor. After 4 seconds, the value of ln(1-(Vc/Vo)) is -2, where Vc is the voltage in the capacitor at a given time and Vo is 10 V.

    1) Find the capacitance.
    2) Find how long it takes for the potential difference across the capacitor to rise from 0 V to 4 V.

    2. Relevant equations

    Vc = Vo(1 - e^(-t/RC))
    Tau = RC

    3. The attempt at a solution
    ln(1-(Vc/Vo))/t = -2/4 = 0.5

    Let's try to figure out what this represents
    Vc = Vo(1 - e^(-t/RC))
    ln Vc = ln Vo(1 - e^(-t/RC))
    ln Vc = ln Vo + ln(1 - e^(-t/RC))
    ln Vc - ln Vo = ln(1 - e^(-t/RC))
    Vc/Vo = 1 - e^(-t/RC)
    e^(-t/RC) = 1 - Vc/Vo
    -t/RC = ln(1 - Vc/Vo)
    -1/RC = ln(1 - Vc/Vo)/t = -0.5

    So I have 1/RC = 0.5, therefore the time constant Tau = RC = 2 seconds.

    1) (500)(C) = 2 -> Capacitance = 0.004 F

    2) If -t/RC = ln(1 - Vc/Vo) then
    t = -RC*ln(1 - Vc/v0) = -2*ln(1-4/10) = 1.021 seconds.

    Is this correct? I've been told that I have a mistake somewhere, but I don't know where.
     
  2. jcsd
  3. Oct 24, 2011 #2

    ehild

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    This whole thing was not needed...

    Except for the missing parentheses, everything is correct.


    ehild
     
  4. Oct 25, 2011 #3
    Oops... sorry, I typed it up because I had written it, and I forgot the parenthesis :smile:

    I only had the formula for voltage when discharging, so I thought I had to figure out what ln(1-Vc/Vo) meant before I could use it. Thanks for going through it, I will ask my instructor again.
     
  5. Oct 25, 2011 #4

    ehild

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    You have a rounding error. T=1.02165 s which has to be rounded to t=1.022 s, if you give the result with 3 decimals.

    ehild
     
  6. Oct 25, 2011 #5
    Thanks, but it isn't an online homework problem, the instructor told me the working is incorrect. I'll ask him again when I next see him.
     
  7. Oct 25, 2011 #6
    You seem to be junior member like me.
    Have you read https://www.physicsforums.com/showthread.php?t=386951 [Broken].
    When i read your answer it was to cumbersome that's why i give you this link. If you use tex tags then it became easy to understand what you have written.
     
    Last edited by a moderator: May 5, 2017
  8. Oct 25, 2011 #7
    Thanks, I ought to learn LaTeX, but the equations I used didn't have many terms.
     
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