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Calculating voltage and current of RC Circuit

  1. Mar 11, 2014 #1
    1. The problem statement, all variables and given/known data

    Attached.

    2. Relevant equations



    3. The attempt at a solution

    Attached.

    I am looking at the circuit at t<0. I need to calculate v(0) but I'm not sure how to apply nodal analysis when you have a capacitor in the circuit. Do capacitors follow ohm's law?
     

    Attached Files:

  2. jcsd
  3. Mar 11, 2014 #2

    gneill

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    Staff: Mentor

    For the instant t = 0+, a capacitor "looks like" an ideal voltage source with the value of the then current voltage on the capacitor.
     
  4. Mar 11, 2014 #3
    Oh, I just realized the question is asking for t≥0. Opps.

    I've redrew the circuit for t≥0 but I'm not quite sure I understand what you mean. How does the capacitor look like a voltage source if it's not in the circuit diagram? How do I get the voltage of the capacitor if it's not in the circuit diagram?

    I've attached my new attempt. Disregard the t<0 part.
     

    Attached Files:

  5. Mar 11, 2014 #4

    gneill

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    If the voltage on the capacitor at time t=0+ happens to be zero then it's okay to replace the capacitor with a wire (short circuit) as you've done. After all, an ideal voltage source of 0 V is identical to a wire (short circuit). But if you run across a case where the capacitor happens to have some initial charge on it, then replace it with a voltage source for that initial instant. If you need to find the capacitor voltage for more more than that instant, it's okay to model the capacitor as an ideal voltage source in series with an uncharged capacitor (initial voltage = 0).

    attachment.php?attachmentid=67526&stc=1&d=1394554394.gif
     

    Attached Files:

  6. Mar 11, 2014 #5
    So in this case, is the voltage on the capacitor zero at t=0+ or is there some initial charge? It is zero because there is no source when t<0, right?

    From your diagram, how would I be able to calculate v(∞)? Capacitors act as open circuits to DC, right? So isn't that an open circuit?
     
  7. Mar 11, 2014 #6

    gneill

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    Right.

    Yup. The capacitor will charge until the voltage across it exactly balances the driving voltage (no potential difference means no more current flowing into the capacitor). So the thing to do there is to remove the capacitor and find out what the voltage is at the open terminals where it was removed.
     
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