1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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

    User Avatar
    Homework Helper

    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! =]
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Capacitance of circuit & RC time constant
Loading...