1. Limited time only! Sign up for a free 30min personal 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!

Finding R in a RC circuit given battery, capacitance, and time.

  1. Nov 15, 2011 #1
    1. The problem statement, all variables and given/known data

    A 10.0 uF capacitor is charged by a 10-V battery through a resistance R. The capacitor reaches a potential difference of 4.00-V in a time interval of 3.00 s after charging begins. Find R.

    2. Relevant equations

    C = Q/ΔV
    q = εC(1-e(-t/RC)

    3. The attempt at a solution

    Im not really sure where to begin. Q is not given. But I know I can relate capacitance and potential difference using C = Q/ΔV. I cannot find q using q = εC(1-e(-t/RC) because R is not given. I am missing a variable and I do not know how to solve since q nor Q were given. I follow how my teacher got the equation of q = εC(1-e(-t/RC) and in all cases 2 variables are unknown. I am assuming I should set up a system of 2 equations to account for the variables, but I do not know how or where to begin. Could someone please help?
     
  2. jcsd
  3. Nov 15, 2011 #2
    The equation for charge at time t is the same as the equation for V at time t q = εC(1-e(-t/RC)
    Vt = V0 (1-e(-t/RC))
    You know V0 = 10, Vt = 4 and t =3. So you should be able to calculate (RC)
    You know C .....you should be OK
     
  4. Nov 15, 2011 #3
    Here is what I did:

    4 = 10(1-e(-3/RC))
    6/10 = e(-3/RC))
    Here is where I think I lose it

    6/10 = -ln(3) -ln(RC)

    My algebra is rusty :(
     
  5. Nov 15, 2011 #4
    You are OK, just tricky maths.
    4 = 10(1-e(-3/RC))
    4/10 = 1-e(-3/RC)
    0.4 = 1-e(-3/RC)
    0.4 -1 = -e(-3/RC)
    -0.6 = -e(-3/RC)
    0.6 = e(-3/RC)
    ln0.6 = -3/RC
    -0.51 = -3/RC
    RC = 3/0.51 = 5.9
    So5.9 = RC.... C = 10 x 10^-6F
    Gives R =5.9/10 x 10^-6
    R = 590,000 ohms
    Check that you know how to use your calculator ln and e^x
     
  6. Nov 15, 2011 #5
    Thank you so much! I knew I was using ln and e wrong (I was hoping it wouldnt come back to haunt me, but it found a way). Just so much to remember in so little time.
     
  7. Nov 15, 2011 #6
    in my posts e(-3/RC) means e^(-3/RC).
    I have not worked out how to post these numbers yet!!!!
     
  8. Nov 15, 2011 #7
    Right above where you enter these messages is a few options involving bold, italics, etc... if you look over more to the right you will see X2.

    Or you can type [.SUP.] "what you want supscripted" [./SUP] without the "." in the brackets.
     
  9. Nov 15, 2011 #8
    Brilliant.... thank you
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Finding R in a RC circuit given battery, capacitance, and time.
Loading...