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Homework Help: Circuit formula problem

  1. Jan 20, 2007 #1
    Question:

    A 0.25 x 10^-6 F capacitor is charged to 50V. It is then connected in series with a 25 ohm resistor and a 100 ohm resistor and alowed to discharge completely. How much energy is dissipated by the 25 ohm resistor ?


    Attempt:

    I found a formula in the textbook related to this question:

    The resistors dissipate energy at the rate:

    PR = I^2R = (Delta V)^2 / R


    I don't know what to do next .... and if I'm using the right formula. Please somebody help.:confused: :confused: :confused:
     
  2. jcsd
  3. Jan 21, 2007 #2

    mjsd

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    Homework Helper

    ok... conservation of energy
    ask yourself, how much energy was stored in capacitor initially, then how do u think the energy are dissipated between the 25 and 100 ohm resistor? remember everything is in series... and you already have [tex]P=I^2 R[/tex]
     
  4. Jan 21, 2007 #3
    Hi, Do you calculate the energy by this formula;

    Q=CV
    Q= (0.25 * 10^-6) ( 50)
    Q = 0.0000125 C
     
  5. Jan 21, 2007 #4

    mjsd

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    Homework Helper

    Your Q means "charge" not energy

    formula for energy stored in a capacitor is given in most books or can easily googled....
    [tex]\displaymath{U = \frac{1}{2}CV^2}[/tex]
     
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