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Circuit Question

  1. Feb 10, 2008 #1
    This one has me stumped. Here is the circuit
    http://img90.imageshack.us/img90/6204/circuitdn1.jpg [Broken]

    I will go through all the 5 questions.

    A) Calculate the total capacitance of the circuit.
    18 uF=.5CV^2, 36uF=CV^2, 36uF/36v=1uColoumb. Correct?

    B) Calculate the current in the 10 ohm resistor.
    This is where we're stuck, there's going to be current coming off the capacitors, but which was will it flow? They have more potential difference than the battery, and I'm not sure how to figure that part out.

    We have - I(20 ohm + 10 ohm)=6v, which would be 6/30=I. But this is only one loop, and depending on where the current flows the 20 ohm resistor will have a different current.

    C) Calculate the voltage between A & B. (Forgot to add it, it is at the junction of the capacitors on either side.
    (20 ohm) (6/30A)=4v + 12v (capacitors in series) = 18v

    D) Calculate the charge stored on one plate of the 6 uF capacitor.
    Ue=.5QV = 12uF=QV, and 12uF/6v=2uC

    E)Wire is cut at p (Point b/w two capacitors), will the voltage increase, decrease, or remain the same?
    It would decrease, and it would be 4v after? (6/30A)(20Ohms)=4v.

    Thanks in advance, I have no idea on B.
    Last edited by a moderator: May 3, 2017
  2. jcsd
  3. Feb 12, 2008 #2
    Well here is my quick shot at the answer. Its been a while since I`ve done problems like this so bare with me.

    1) Capacitors are opposite to resistors. So add capacitors in series as you would add resistors in parallel.
    2) Well if the voltage supply is dc then it makes our problem easy. After the capacitors have fully charged I(10ohm)=Vtotal divided by R total.
    3)Again after the capacitors have fully charged you can treat the two resistors as a simple voltage divider. As well the voltage across the resistor in parallel with the capacitors will be the voltage that the capacitors charge to.
    4)Calculate the charge of the 6uf cap and just divide it by two.
    5) The voltage will obviously disappear for the resistor which now has one open lead. The capacitors will still remain charged since they are still connected to the voltage source. Voltage remains same basically.
    Well hopefully that answered some of your questions. Let me know if you need more help.
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