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Resistance of infinite nested triangles

  1. Mar 27, 2009 #1
    1. The problem statement, all variables and given/known data
    Here is an interesting problem... there is a wire bent in the shape of an equilateral triangle, side length = a and resistivity = rho.

    In the center of this triangle is another equilateral triangle (inverted, side = a/2, resistivity = rho) and so on into infinity. What is the overall resistance between points A and B in terms of a and rho?

    circuit.jpg

    2. Relevant equations

    R = (rho * length)/area

    3. The attempt at a solution

    I started by using the equation for resistivity, R = (rho * length)/area, but I wasn't sure if area would apply here. We aren't given any information about the wire beyond the shape and length. So I'm really asking for help in determining a good starting point... I don't know of any other equations that would incorporate rho and length.
     
    Last edited: Mar 27, 2009
  2. jcsd
  3. Mar 28, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Mtnbiker! Welcome to PF! :wink:

    I don't think you can solve this on the information given. :frown:

    I suggest you say "let the resistance be R/a times length", and carry on from there. :smile:
     
  4. Mar 28, 2009 #3
    Re: Welcome to PF!

    Hi... thanks for the welcome, I'm glad to be here.

    I agree with you regarding keeping the area incorporated in the answer. However, I'm still struggling with what exactly would the length be (first triangle is 3a, second triangle is 3a/2, then 3a/4 and so on...). There is a point of diminishing returns, so I need to find that point.

    Thanks for the input!
     
  5. Mar 28, 2009 #4

    tiny-tim

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    Hi Mtnbiker! :smile:

    Yes, you're correct … obviously each triangle has sides half the length of the next one out.

    So assume there are n triangles, start from the inside, and work your way outward …

    at each stage, get rid of one triangle and calculate the equivalent resistances along the three sides of the next triangle. :wink:
     
  6. Mar 28, 2009 #5

    Dick

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    If the network is infinite you can use self similarity. Call the overall resistance between two vertices on the first inner triangle R. Now you have a simple network with three wires of resistance R and and six of resistance a*rho/2. Solve that for the resistance beween A and B in terms of R. Then realize that the outer triangular network is the same as the inner triangular network, but twice as big. So the resistance from A to B is also just 2R. Equate the two values and solve for R.
     
  7. Mar 28, 2009 #6
    Thanks guys!
     
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