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Homework Help: Capacitance of circuit & RC time constant

  1. Feb 15, 2009 #1
    1. The problem statement

    http://s685.photobucket.com/albums/vv212/dewdr0p714/?action=view&current=5circuittakehometest1.jpg

    1) Determine the equivalent capacitance of the circuit

    2) How many RC time constants are needed for the charge on the capacitors to reach 75% of their final charge?



    2. Relevant equations

    q= q0 (1 - e^-t/RC)
    q=CV


    3. The attempt at a solution

    I tried to find the equivalent capacitance....this was my method but I am not sure if it is correct.

    Equiv Capac = (15*10^-6)(10*10^-6) + (10*10^-6)(10*10^-6) + (15*10^-6)(10*10^-6) / (10*10^-6) + (10*10^-6)

    Equiv Capac = 20*10^-6 Farads

    For the RC time constants question, I think I need the capacitance to be able to solve it and with the capacitance that I got, I could find q thru the equation q=CV....after that I am unsure about how to proceed with question #2. Does the q get inputted for q or q0?
     
    Last edited: Feb 16, 2009
  2. jcsd
  3. Feb 15, 2009 #2

    Delphi51

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    Sorry to keep you waiting, but no one can see your picture until a member of the staff authorizes it. If you are in a hurry, you could describe the circuit, or post your picture to a service such as photobucket.com and then give us a link to it.
     
  4. Feb 16, 2009 #3
  5. Feb 16, 2009 #4

    Delphi51

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    I got the same answer for the total capacitance.
    For the 2nd part, Q=CV says 75% charge is 75% Voltage, so don't you just have to look at the factor (1 - e^-t/RC) ?
    You could try t = RC, 2RC, 3RC, etc until you the factor is .75 or more.
    Better, of course to make the factor equal to .75 and solve for t. Logs?
     
  6. Feb 16, 2009 #5
    Hmm I don't quite understand what you mean. I get the part about q being proportional to v so .75q~.75v. But being that the equation is q=q0(1-e^-t/RC), what would q0 be? And the whole RC time constant thing confuses me completely...like how can t=RC?
     
  7. Feb 16, 2009 #6

    Delphi51

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    Okay, you'd rather work with q= q0 (1 - e^-t/RC).
    If you put in t = 0, you get q = 0.
    put in t = RC and you get q = .63 q0, so 63% charged.
    At t = 2RC, you get q = .86 q0, so 86% charged.

    Set q = .75q0 to get .75 = (1 - e^-t/RC)
    Solve that for t to get somewhere between 1RC and 2RC where it is 75% charged.
     
  8. Feb 16, 2009 #7
    thank you for your help i figured it out! =]
     
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