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Calculating resistance of wired cubes

  1. Aug 6, 2010 #1
    1. The problem statement, all variables and given/known data
    This problem was posted in february in my country's math-physics magazine for high school students, as part of the competition. I couldn't do it then, and every now and again I try it, but still I can't do it, so any help on where to start would be very much appreciated.

    http://www.komal.hu/verseny/feladat.cgi?a=honap&h=201002&t=fiz&l=en"

    Scroll down and see problem P. 4231. Unfortunately for me, this site has excellent math, physics, and informatics problems, but they usually only provide the solution for the math and informatics ones.

    2. Relevant equations
    Not much. R = U/I may help and Kirchoff's equations.


    3. The attempt at a solution

    As I said, I didn't really get anywhere. I tried to draw down the whole thing as a circuit in 2D, but what I got wasn't pretty, with overlapping wires and all. Normally in problems like these, the idea I saw was that you write down the resistance after n number of cubes have been attached to either side, then the same for n+1, and you say as n nears infinity, they are equal. Normally that equation gives a value for the whole resistance. But I just couldn't write down those equations.

    Thanks for any help!
    -Tusike
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: Apr 25, 2017
  2. jcsd
  3. Aug 6, 2010 #2
    A very tricky one :uhh:
    You can easily see that each couple of symmetric points (e.g. C and D, E and F, G and H) have the same electric potential (see figure (1)). That means you can join the points in each couple together and get a 2-D circuit like in figure (2). Again, because of symmetry, potential at C-D = potential at E-F, there is no current running through the wire joining them, so you can omit it. Therefore, the problem becomes solving for the resistance of AP (the 2r resistor is excluded). As the circuit is periodic and infinite, we can consider RAP as in figure (3). From here, you can solve for RAP and then RAB.
     

    Attached Files:

  4. Aug 6, 2010 #3
    Okay, I see how you got figure (2), I even did that back then, but I didn't know how to continue. First of all: after getting figure (2), the red wires on your image would have the resistance of R (where R is the one in the original problem), and the black ones would have a resistance of R/2, right? Since we got them by combining two parallel R's.
    I'm not sure I understand how you got figure (3) from (2). because if you fold figure (2) on that AP line, which you can because of the symmetry in potentials, AP would be connected with that red wire, still a resistance of R, and the mentioned R(AP). So I would understand figure (3) if there was a red line connecting A and P:) In figure (3), this line would have a resistance of R, the red lines R/2, and the black ones R/4 right?
    And after getting R(AP), how does R(AB) come from that?

    Sorry for not understanding this right away, and thanks for the help so far!

    -Tusike
     
  5. Aug 6, 2010 #4
    Sorry for the confusion. I forgot to say r=R/2, it's just to make it look a bit better :biggrin: When I mentioned R(AP), I excluded the red wire (the 2r one) in order to make it cleaner, so we got something like in figure (3). You can include the red wire, but it doesn't matter. In figure (3), the red lines are R (or 2r), the black lines are R/2 (or r).
    And after solving for R(AP), you just have to redraw the circuit in some convenient way. It is no longer a complicated circuit, but just a simple one which contains series and parallel resistors, as we have removed the line joining C-D and E-F.
     
  6. Aug 6, 2010 #5
    wires.JPG

    I'm not sure you understood what I meant. Here is what I get from figure (2). B' and Q' are equipotential to B and Q, as shown by the purple curves. And the whole (3) is the infinite wire you get after connecting these equipotential points.

    EDIT: OK, I see what you meant:) It's things like these that make me wonder why I'm trying for the IPhO. By R(AP) I thought you meant the whole resistance between those two points. Of course, if you just look at the whole part to the left of points A and P, you get what you're figure (3) had, and the same on the other side's B and Q. Everything's perfectly clear now, thank you very much!

    -Tusike
     
  7. Aug 6, 2010 #6
    Oh sorry for the confusion again :uhh:
    I'll make it a bit clearer. You'll see that I have split (AP) into 2 parallel parts: one is the red wire, the other is (A'P') which corresponds to figure (3). The blue wires are just connecting wires, zero resistance.

    EDIT: I see you got it. Btw, I think if it's crucial to practice with these kinds of tricky problems to get into the IPhO team, then go for it. But if you don't have to, then you shouldn't dig into these things too much. Sometimes there is not much physics in these problems.
     

    Attached Files:

  8. Aug 6, 2010 #7
    Well every now and again tricky problems like these are fun:) plus it really did bug me a lot.
    Thanks again for the help!
     
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