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Capacitance in a network

  1. Feb 9, 2013 #1
    1. The problem statement, all variables and given/known data
    http://loncapa2.fsu.edu/enc/66/45d3639022485d2ba995cf84c579b584bffe5ace9851f3da3fbac5f5a1c7d2ee9dfed8743cf75459ef336ad7a1bbe176f7a0cf1a35a519ce0cef3c31febf9858404cac5a2cd3ace3851dbef61ee03a25f7a5c41a87b32e0f.gif

    Calculate the equivalence capacitance of the circuit shown in the diagram above; where C1 = 8.65 μF, C2 = 4.00 μF, C3 = 8.55 μF, C4 = 3.20 μF.


    2. Relevant equations

    Capacitance in a series: 1/Ctot = Ʃ(1/C)

    Capacitance in parallel: Ctot = ƩC

    3. The attempt at a solution

    Well, I attempted multiple ways to order each capacitor into a way to solve a formula. In my sixth attempt I presumed that C1 and C2 are in a series and are parallel to C3. I also believe C4 is parallel to C3, so here is my equation:

    C = (1/C1 + 1/C2) + C3 + C4

    Another idea that came into ferment was the concept of thinking that C2 and C3 are in series and are parallel to C1. C4 would be parallel to C2 and C3 so:

    C = C1 + (1/C2 + 1/C3) + C4

    I could go on, but since my professor never went over this I am thoroughly confused.
     
  2. jcsd
  3. Feb 9, 2013 #2

    ehild

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    Show the circuit diagram, please.

    ehild
     
  4. Feb 9, 2013 #3
  5. Feb 9, 2013 #4

    ehild

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    I can not open the link. It is not safe.

    Anyway, the equation you wrote "C = (1/C1 + 1/C2) + C3 + C4" is wrong.
    The resultant of the series capacitors is 1/(1/C1+1/C2).

    ehild
     
  6. Feb 10, 2013 #5

    CWatters

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    I can't see image by either method.

    Suggest you upload as an attachment?
     
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