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Circuit with Capacitors

  1. Feb 12, 2012 #1
    1. The problem statement, all variables and given/known data

    What is Q5, the charge on C5? V of battery = 12 V.

    2. Relevant equations

    C=Q/V

    3. The attempt at a solution

    V1 + V5 + V(2,3) = Vb
    V1 + V5 + V4 = Vb

    I have three unknowns and 2 equations. :(
     

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  3. Feb 12, 2012 #2

    gneill

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    Staff: Mentor

    Look for the total equivalent capacitance instead. Then apply your Relevant Equation to find the charge that is pushed onto that capacitance by the voltage V.
     
  4. Feb 12, 2012 #3
    Great! I got that part right.

    I'm confused though on how to find Q2. I know that charge adds across a paralell circuit so Q(total) = Q2 + Q4, but I don't know Q4 either.
     
  5. Feb 12, 2012 #4

    gneill

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    Looks like more algebra will be involved :wink:

    You now know the total charge Q that will be put onto the equivalent capacitance comprising C2,C3, and C4. You can work out the equivalent capacitance of C2 & C3 alone (perhaps call it C23), then determine how to divide the total charge Q across the parallel C23 and C4.
     
  6. Feb 12, 2012 #5
    Better algebra than calculus. :)

    I know that C23 = 1.22F and that Q2 = Q3 = Q23

    I don't see how C23 is related to Q2, except for using the relevant equation. Also, I don't think that Q total divides evenly between C23 and C4 because I've already tried that.
     
  7. Feb 12, 2012 #6

    SammyS

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    Do you have any capacitance values?
     
  8. Feb 12, 2012 #7

    gneill

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    The charge won't divide evenly between the two. You need to determine how the charge distributes between two parallel capacitors (same voltage across both, different capacities).
     
  9. Feb 12, 2012 #8
    Yep! C1 = C5 = 6 μF, C2 = 1.6 μF, C3 = 5.1 μF, and C4 = 4.2 μF. The battery voltage is V = 12 V.

    Okay, I think I get it now.
     
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