Resistance of infinite nested triangles

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The discussion revolves around calculating the overall resistance between two points in a structure of infinitely nested equilateral triangles made of wire. The initial approach involves using the resistivity formula, but participants note the lack of sufficient information regarding area and length. A suggestion is made to express the resistance in terms of a variable and to analyze the structure layer by layer. It is emphasized that the triangles' dimensions decrease geometrically, allowing for self-similarity in the resistance calculations. Ultimately, the resistance can be derived by equating values from different layers of the triangle network.
Mtnbiker
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Homework Statement


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


Homework Equations



R = (rho * length)/area

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.
 
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Welcome to PF!

Mtnbiker said:
… 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?

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.

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:
 


tiny-tim said:
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:

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!
 
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:
 
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.
 
Thanks guys!
 
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