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Resistor grid

  1. Nov 1, 2016 #1
    1. The problem statement, all variables and given/known data
    We have a resistor grid with 6x6 squres. ( figure below) If we know the voltage between ##O## and ##A##, what is the voltage between:
    ##a) O## and ##B##
    ##b) C## and ##D##
    ##c) E## and ##F##
    Every side has the same resistance.
    Resistor grid.png
    Attempt for solution:

    Because of the symmetry of the circuit , the equivalent circuit looks like this:
    Resistor grid 2.png
    Now every resistance is half its initial value because of the parallel connection.

    From here on I am stuck . I mean I don't really like applying kirchhoff's laws in situations like this. So I checked the solution and there the author connected the grid like this "from symmetry":
    Resistor grid 3.png
    And after this:
    Resistor grid 4.png
    Resistor grid 5.png
    I don't really think that these points that the author connected are equipotential especially in the first and third picture of his solution...
    Am I right?
     
    Last edited: Nov 1, 2016
  2. jcsd
  3. Nov 1, 2016 #2

    gneill

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    Staff: Mentor

    How are the potentials in the grid created? That will establish your symmetry. Is the voltage OA a measurement or the result of placing a voltage source between them? Is there a source that we don't see energizing the grid?
     
  4. Nov 1, 2016 #3
    The way I see this grid is like a superposition of two grids with two independent voltage sources ##V_- ## and ##V_+## which are in A and O. The voltage is created by passing a current trough a resistance. So is see now that the current in A has to exit in the opposite point. And than superposed with O creates same potential.
    Is it right?
     
  5. Nov 1, 2016 #4
    I think I got it. I had this feeling that as the current gets further away from A , is diverging more and more, but is diverging until it reaches the "O diagonal" of the square and than it converges again so it is symetric.
     
    Last edited: Nov 1, 2016
  6. Nov 1, 2016 #5

    gneill

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    I'm not following your description. No doubt you have a good picture of what you're describing in your mind. Perhaps you can make a sketch to illustrate?
     
  7. Nov 1, 2016 #6
    Resistor grid 6.png
    What I was trying to say is that if we consider only point A as a potential source, than , from symmetry , we can fold the square grid twice ( connecting N-M and P-A)
     
  8. Nov 1, 2016 #7

    gneill

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    A potential difference requires two points: You measure the potential at a given point with respect to some other point. Similarly for voltage sources, the potential difference is established between its two terminals. Both terminals must be connected to the circuit for the source to create potential differences within it. So if your point A is connected to the + terminal, where is the - terminal of the source connected? At point P?

    If your source connections are A and P then I agree that the grid can be folded to join N-M. But P-A would not work. Any folding you do must preserve the distinct connection points of the source.
     
  9. Nov 1, 2016 #8
    Ok , so the second point is P. You are right that P-A would not work. So, what is going on ? Is the solution from the book okay?
     
  10. Nov 1, 2016 #9

    gneill

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    I don't know what the author was thinking when he made his symmetry choice. Was there any other information given about the distribution of potentials other than the O-A voltage? For example, if it were given that point P has the same potential as point A with respect to O then that would change how we look at things.
     
  11. Nov 1, 2016 #10
    Actually there is a thing that the author says and I didn't notice. It says that the frame of this grid is made of superconducting material and only what is in the interior of the square has restistance.
     
  12. Nov 1, 2016 #11
    And I think that this means that A and P have the same potential , because if we were to apply kirchhoff's law for voltage taking the loop to be the square frame, than we would obtain ##V_A=V_P## . Am I right?
     
  13. Nov 1, 2016 #12

    gneill

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    Ah! Big difference! Yes, if the entire frame is constrained to be at the same potential then that makes a huge difference in your symmetry options.

    All the corners will be at the same potential and you can fold the circuit along either diagonal or even the centerlines.
     
  14. Nov 1, 2016 #13

    epenguin

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    Homework Helper
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    Could you please reproduce or quote verbatim the entire problem so we can advise on that and on any misunderstandings of this problem which sounds intriguing and even a bit fundamental?
     
  15. Nov 1, 2016 #14
    So the major misunderstanding here is that the square frame is actually made of superconducting material, so the entire frame is a equipotential region.
     
  16. Nov 1, 2016 #15

    gneill

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    Staff: Mentor

    Yes. As often is the case, the devil is in the details :smile:
     
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