Resistance in an octahedron between adjacent vertices

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The discussion focuses on calculating the resistance between adjacent vertices in an octahedron formed by twelve identical wires, each with a resistance of 1.0 ohms. The user successfully determined the resistance between opposite vertices as R/2 using symmetry but struggles with the adjacent vertices. Suggestions include using Kirchhoff's laws and breaking the circuit into parallel and series combinations to simplify the calculations. The importance of showing attempts and reasoning in problem-solving is emphasized. The conversation highlights the complexity of the problem and the need for a structured approach to find the solution.
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Homework Statement


Twelve identical wires, each with resistance R = 1.0 ohms, are linked together to form a
octahedron. Find the resistance in an octahedron between opposite vertices and adjacent vertices( A,B for example)

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2. The attempt at a solution

I already found the resistance between opposite vertices (R/2) using symmetry and equipotential points but can't find any way to calculate the resistance between adjacent vertices. Do u have any tips for me?
 
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Hi Tazi. Welcome to Physics Forums.

One tip: Do not use text-speak on these forums ("u" rather than "you") . It's against the rules.

What have you tried so far? Show us your best effort.
 
My attempt (resistance between points A and B): potential in point E equals to potential in point F but I'm not sure what to do about points D and C. How to find potential in those points? I tried to use Kirchoff laws but didnt find anything.
 
Tazi said:
My attempt (resistance between points A and B): potential in point E equals to potential in point F but I'm not sure what to do about points D and C. How to find potential in those points? I tried to use Kirchoff laws but didnt find anything.
You can certainly get there using Kirchoff. Please show some attempt that way.
You can also go the more usual route of breaking it down into a nested combination of parallel and serial links. The first step, breaking it into two parallel paths, is obvious. After that it gets a bit tougher, but I think I see a way to use symmetry to simplify it.
 
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