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Combination of Capacitors

  1. Feb 18, 2012 #1
    1. The problem statement, all variables and given/known data
    A circuit is constructed with five capacitors and a battery as shown. The values for the capacitors are: C1 = C5 = 4.6 μF, C2 = 1.7 μF, C3 = 5.7 μF, and C4 = 2.7 μF. The battery voltage is V = 12 V. 3)
    What is Q5, the charge on capacitor C5?


    2. Relevant equations
    Parallel:
    Cequ=C1+C2
    Vequ=V1=V2
    Qequi=Q1+Q2

    Series:
    Cequ=1/c1+1/c2
    vequ=v1+v2
    Qequi=Q1=Q2

    C=Q/V

    3. The attempt at a solution

    I found the total capacitance for C234=4.009. Then I figured that C234 was in a series with C15.....Where C15 the parallel capacitors C1 and C5. Is this logic correct? I added them appropriately and got 2.79 for C12345. Knowing these two groups had equal charges I multiplied C234(Qtotal)=V234 and C15(Qtotal)=V15. Q total 12v*C12345. From there I figured solved for Q5.. Q5=(C5)*(V15) because C1 and C5 have the same voltage. I got Q5 to be 16.75microC which is wrong.
     
    Last edited: Feb 18, 2012
  2. jcsd
  3. Feb 18, 2012 #2
    Could you post a picture of the circuit? It's difficult to interpret the question without the diagram.
     
  4. Feb 18, 2012 #3
    I forgot to do that. Oops. Circuit.jpg
     
  5. Feb 18, 2012 #4

    SammyS

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    C1 is not in parallel with C5 .
     
  6. Feb 18, 2012 #5
    How is it not parallel with C5? I understand parallel to be this and thats what I see.... Capacitors.jpg

    So the total capacitance would be (1/C1234+1/C5)^-1 ?
     
  7. Feb 18, 2012 #6

    SammyS

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    If those capacitors were resistors, would you say they're in series or say they're in parallel ? (in either figure)
     
  8. Feb 18, 2012 #7
    Think of parallel and series in this way.

    To be in series, there is only one way for the current to get from start to finish, and that's through the circuit elements you're interested in. (A)

    However, when two things are in parallel the current can either go through one, or the other, and still end up completing the circuit (B)

    See my drawing

    Apply this to your circuit, what conclusion do you have to make when it comes to the current passing through C1 and C5
     

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  9. Feb 18, 2012 #8
    Oh I see... so a parallel series is when the current is split and can go through one of multiple paths. That helps a lot. I wish I had tried that possibility before resorting to the forum for help. Thanks.
     
  10. Feb 18, 2012 #9
    nice visualization. thanks!
     
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