Another useless wire!

I had a useless metallic wire of total resistance 60 Ω 60 \Omega , and then I cut it into twelve equal parts and formed a regular octahedron. What is the resistance between any two opposite nodes in Ω \Omega ?

Image credit:Wikipedia


This problem is original.

This question is part of the set Platonic Electricity .


The answer is 2.5.

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1 solution

Swapnil Das
Jun 4, 2016

First of all, let the octahedral circuit get stable, so that we can analyze it ;)

Now, as R l R\propto l , each resistor has a resistance of 60 12 = 5 Ω . \frac{60}{12}=5 \Omega.

We apply a voltage to any two opposite nodes,say E E and F F . By symmetry of the circuit, we can easily observes that points A , B , C , D A,B,C,D are equipotential with respect to nodes E E and F F . Therefore, the circuit can be redrawn to two resistors in series, each with resistance 5 4 Ω \frac{5}{4}\Omega .

\therefore The Equivalent resistance is 5 2 Ω \frac{5}{2} \Omega .

Nice explanation (+1) solved it the same way

Ashish Menon - 5 years ago

Or it can be generalized that the resistance between any two adjacent nodes of a n-sided 3D figure is Resistance of wire 2 × Number of edges \dfrac{\text{Resistance of wire}}{2 × \text{Number of edges}} .

In this question, the octahedral has 12 12 edges and total resistance of wire is 60 Ω 60\Omega .
So, resistance = 60 2 × 12 = 5 2 = 2.5 \dfrac{60}{2×12} = \dfrac{5}{2} = \color{#3D99F6}{\boxed{2.5}} .

Ashish Menon - 5 years ago

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Hmm, never knew about the formula, thanks anyways.

Do you know its proof?

Swapnil Das - 5 years ago

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Nono, I dont know the proof, I just observed that a 3 sided 3D figure has x, and 4sided one y, a 5sided one z and 6 sided one w and generalized XD

Ashish Menon - 5 years ago

This is something new ;). But, Ashish, we are talking about opposite nodes. Please check it.

Abhay Tiwari - 5 years ago

sir is there formula for equivalent resistance along two opposite sides

Deepansh Jindal - 5 years ago

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