I had a useless metallic wire of total resistance 6 0 Ω , 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 ?
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This problem is original.
This question is part of the set Platonic Electricity .
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Nice explanation (+1) solved it the same way
Or it can be generalized that the resistance between any two adjacent nodes of a n-sided 3D figure is 2 × Number of edges Resistance of wire .
In this question, the octahedral has
1
2
edges and total resistance of wire is
6
0
Ω
.
So, resistance =
2
×
1
2
6
0
=
2
5
=
2
.
5
.
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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
This is something new ;). But, Ashish, we are talking about opposite nodes. Please check it.
sir is there formula for equivalent resistance along two opposite sides
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First of all, let the octahedral circuit get stable, so that we can analyze it ;)
Now, as R ∝ l , each resistor has a resistance of 1 2 6 0 = 5 Ω .
We apply a voltage to any two opposite nodes,say E and F . By symmetry of the circuit, we can easily observes that points A , B , C , D are equipotential with respect to nodes E and F . Therefore, the circuit can be redrawn to two resistors in series, each with resistance 4 5 Ω .
∴ The Equivalent resistance is 2 5 Ω .