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!

RC Circuits, time to charge a capacitor

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Switch S in the figure below is closed at time t = 0, to begin charging an initially uncharged capacitor of capacitance C = 13.0 µF through a resistor of resistance R = 24.0 . At what time is the electric potential across the capacitor equal to that across the resistor?

    hrw7_27-52.gif


    2. Relevant equations

    i = dq/dt
    q = CE(1-e^-t/RC)

    3. The attempt at a solution

    I really don't even know where to start with this one
     
  2. jcsd
  3. Oct 17, 2011 #2

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    I assume that E is the EMF of the battery.

    Is q the charge on the capacitor at time, t ?

    How much charge would be on this capacitor of the potential difference across it was E ?
     
  4. Oct 17, 2011 #3
    the charge on the capacitor would be

    Q = CV = (13e-6)(E)

    if the potential difference were E
     
  5. Oct 17, 2011 #4

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    At the above time, how do these two potentials compare with E , the electric potential provided by the battery?
     
  6. Oct 17, 2011 #5
    I'm not sure i follow what you are saying? Would the combination of the 2 be equal to E?
     
  7. Oct 17, 2011 #6

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes, according to Kirchhoff.

    Since the two are equal, what is the electric potential across the capacitor ?
     
  8. Oct 17, 2011 #7
    Would it just be 1/2 E?
     
    Last edited: Oct 17, 2011
  9. Oct 17, 2011 #8
    where do i go from here?
     
  10. Oct 17, 2011 #9

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    If the electric potential on the capacitor is E/2, then how much charge is on the capacitor?
     
  11. Oct 17, 2011 #10
    the charge is Q = CV which would be 13uF * E / 2
     
  12. Oct 17, 2011 #11

    SammyS

    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    So, solve this equation for t when q = C(E/2)

    q = CE(1-e^-t/RC)
     
  13. Oct 17, 2011 #12
    C(E/2) = CE(1-e^-t/RC)
    1/2 = 1 - e^-t/RC
    e^-t/RC = 1/2
    -t/RC = ln(1/2)
    -t = ln(1/2) * RC
    t = -1 * ln(1/2) * RC

    t = -1 * -.6931 * 24 * 13e-6 = 0.000216 s = 0.216 ms

    Thanks again for the help, these forums (you especially) are fantastic
     
  14. Jan 7, 2012 #13
    Hello, may I ask what kind of equation q = CE(1-e^-t/RC) is? Does the problem give it as the equation which determines the time to charge of the capacitor through this RC circuit? How do they get to such an equation?
     
  15. Jan 7, 2012 #14
    If you check your capacitor equations you will find a more convenient expression for the voltage across a capacitor during charging.
    V = Vmax(1 - e^-t/RC) so you can calculate the voltage across the capacitor t sec after switch on.
    The charge equation is the same exponential form
    Q = Qmax(1-e^-t/RC)
    hope this helps
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook