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Combination of Capacitors and Equivalent Capacitance

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data
    Calculate the charge on C1 in Coulombs.


    2. Relevant equations
    umm q=CV



    3. The attempt at a solution

    http://answers.yahoo.com/question/index?qid=20120424171750AA1WxRv

    I have this same problem except my numbers are:
    C1= 6.00 μF, C2=3.10 μF, C3= 2.20 μF, C4= 1.30 μF, C5=4.40 μF, C6=6.50 μF
    and my Ceq = 4.994×10-6 F

    I dont understand when the person says: "Then you have to split up the 2.6 MC
    to assign parts of it to C1 and part of it to C6, proportionally."
    how do i do that?
     
  2. jcsd
  3. May 30, 2012 #2
    I really don't understand how to get the charge across C1 ??
     
  4. May 30, 2012 #3

    gneill

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    Staff: Mentor

    The given answer errs in that the charge on the capacitor network cannot possibly be in the MEGA coulomb range. MICRO coulombs would be indicated by the fact that the capacitances are on the order of microfarads and the supply is only 20V.

    The statement about splitting the charge is correct; how does charge divide between parallel capacitors (they have the same potential difference)?
     
  5. May 30, 2012 #4

    gneill

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    Staff: Mentor

    Voltages are ACROSS, currents are THROUGH, and charges are ON :wink:
     
  6. May 30, 2012 #5
    well i was thinking something like i can find the total charge by Ceq(V) where V=20 and Ceq =4.994x10^(-6)

    so i have a total charge now. looking at the whole thing, Ceq is for C16 in series with C2345 meaning that those two have the same charge but split voltage. so wouldnt the charges each just be (4.994x10^(-6))/2 and the voltage 10V for each?

    and when i do that i look further into C16. since C16 are parallel, they have the same voltage across but different charges. Q1 = C1(V) where V is 10 and Q6 = C6(V)? but thats not right
     
  7. May 30, 2012 #6

    gneill

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    Staff: Mentor

    Nope. Capacitors in series have the same charge. This must be so since it's current that "delivers" the charge, and series components MUST have the same current. This also means that the voltages need not be the same, since the same charge on different capacitances will yield different voltages.
    True; the voltage across them is not 20V/2. Use the charge that you found and apportion it according to the capacitance values.
     
  8. May 30, 2012 #7
    i dont know how!! im so lost
     
  9. May 30, 2012 #8
    i got it ;)
    thanks for your help
     
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