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Physics Help Me Please

  1. Apr 26, 2005 #1
    A network of capacitors can be connected to a given battery in two fundamentally different ways-in serious and in parallel inlcuding any number of different combination of each. As a specific example take a seris set of capacitive blocks C1, C2, and C3 containing respectively one, two or three of the same capacitors in series in each cell. The total equivalent capacitance of the three cells in series must be one farad (1F). The capacitors may only be chosen from a slection of two, three, four , six, eight, nine or twelve farads. Parallel combinations within each cell may be used for other numbers. (Hint: this is comparable to solving for ni = 2, 3, 4, 6, 8, 9, 12 the following equation:) 1/n1 + 2/n2 + 3/n3 = 1 :confused:
     
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  3. Apr 26, 2005 #2

    chroot

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    Did you copy this problem down incorrectly? The word in red doesn't seem to agree with the rest of the problem.

    - Warren
     
  4. Apr 26, 2005 #3
    Yes I doubled checked it is right, basically what i get form this it there will be 3 capacitors in a series and those will be the 3 cells. Now in each cell there may be parallel capacitors..


    -----|C1|--|C2|--|C3|------
    | |
    | |
    |_________EMF___________|
     
    Last edited: Apr 26, 2005
  5. Apr 26, 2005 #4

    chroot

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    The question does not make any sense, then. The problem goes on to state

    and

    which would indicate that the capacitors inside each cell are wired in parallel.

    - Warren
     
  6. Apr 26, 2005 #5
    I copied the question precisely from my worksheet and the total capacitance of the circuit must be 1F
     
  7. Apr 26, 2005 #6
    Correct me if im wrong but for capacitors in series you do 1/C(eq)=1/C1+1/C2+1/C3 and if they are in parallel it is jsut C(eq)=C1+C2+C3
     
  8. Apr 26, 2005 #7

    chroot

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    Okay, well, I have no idea how the cells can be composed of 1, 2, or 3 capacitors in series and simultaneously be composed of capacitors in parallel, but let's just assume the capacitors are all in series.

    The first cell is just a single capacitor, so it's total capacitance is just n1.

    The second cell is composed of two identical capacitors wired in series, each of capacitance n2. Their total capacitance is:

    [tex]\left( \frac{1}{n_2} + \frac{1}{n_2} \right)^{-1} = \frac{n_2}{2}[/tex]

    Similarly, the capacitance of the thid cell, composed of three identical capacitors wired in series, each of capacitance n3, is just n3/3.

    The total capacitance of all three cells combined in series is

    [tex]\left( 1/n_1 + 2/n_2 + 3/n_3 \right)^{-1}[/tex]

    Does that make sense?

    - Warren
     
  9. Apr 26, 2005 #8
    Yes I follow that......So the (1/n1+2/n2+3/n3)^-1 would have to equal the total capacitance of the circuit which would be 1F
     
  10. Apr 26, 2005 #9
    I think I may not have explain myself the best but they in series and simultaneously be composed of capacitors in parallel. What you have are three cells which are you capacitors in series.....now you can have those capacitors to be either 2, 3, 4, 6, 8, 9, or 12. Now within those three cells you can have one or all three of those having another or many other capacitors in parallel with just that one individually.
     
  11. Apr 26, 2005 #10

    chroot

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    Yes, bertholf07. Can you find three values for n1, n2, and n3 to satisfy that equation? Hint: I wrote everything in twelfths to make the pattern easier to spot. In other words, write 2 as "12/6" and 3 as "12/4" and so on.

    - Warren
     
  12. Apr 26, 2005 #11

    chroot

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    This does not appear to be what the problem stated. The problem stated "take a seris set of capacitive blocks C1, C2, and C3 containing respectively one, two or three of the same capacitors in series in each cell."

    I should also mention that you don't need any additional capacitors in parallel to make the whole series combination equal to 1F.

    - Warren
     
  13. Apr 26, 2005 #12
    No I dont follow can you give me an example
     
  14. Apr 26, 2005 #13

    chroot

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    You should be able to find a set of values that work with just a bit of trial and error.

    - Warren
     
  15. Apr 26, 2005 #14
    ok let me try
     
  16. Apr 26, 2005 #15
    so how about this if n1=2 n2=8 n3=12
     
  17. Apr 26, 2005 #16

    chroot

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    Looks good, bertholf07. There are a bunch of different valid solutions.

    - Warren
     
  18. Apr 26, 2005 #17
    Thanks for your help Warren greatly appreciated
     
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