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Homework Help: Potential across a capacitor in an RC circuit

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    http://img14.imageshack.us/img14/4822/2116b.png [Broken]

    The circuit has been connected as shown in the figure for a “long” time.
    What is the magnitude of the electric potential across the capacitor?
    Answer in units of V.

    2. Relevant equations
    Kirchoff's rules
    [tex]V = IR[/tex]
    [tex]V(t) = V_{0} e^{-t/\tau}[/tex]
    [tex]C = \frac{Q}{C}[/tex]
    [tex]Q(t) = Q_{f}(1-e^{-t/\tau)[/tex]


    3. The attempt at a solution
    http://img215.imageshack.us/img215/7758/2116.png [Broken]

    That's how I redrew the circuit. But then I got confused. Does the capacitor short circuit the rest of the circuit (i.e., does no current flow in the 15 ohm and 48 ohm resistors)?

    If not, I don't get what happens to (I - i1) and i1 once they get to the junction with the capacitor.

    Help please?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Feb 25, 2010 #2

    collinsmark

    User Avatar
    Homework Helper
    Gold Member

    No. The capacitor might act as a short the very instant the switch is closed, if the capacitor's initial charge was 0. But then when the capacitor gains charge, the potential across its terminals changes until it reaches steady-state. Once the charge in the capacitor reaches steady-state, the potential across its terminals does not change either.

    Without giving you the answer, here is a big hint. If the potential (i.e. voltage) across a capacitor's terminals does not change, meaning the capacitor's charge is not changing either, what does that tell you about the current flowing through the capacitor? :wink:
     
  4. Feb 25, 2010 #3
    Ah!!! Thank you, I get it :D
     
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