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Resistance across a 1k resistor cube.
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Kirr:
For verifying solutions you can try my online solver: http://kirr.homeunix.org/electronics/resistor-network-solver/

It solves the network by eliminating nodes one by one using star-mesh transform, until only terminals are left. (See examples, which include resistor cube).

A similar quiz: Find resistance across a 4D cube, then also resistance across an edge (singel resistor) of a 4D cube.
hamster_nz:

--- Quote from: Kirr on April 22, 2018, 07:37:54 am ---For verifying solutions you can try my online solver: http://kirr.homeunix.org/electronics/resistor-network-solver/

It solves the network by eliminating nodes one by one using star-mesh transform, until only terminals are left. (See examples, which include resistor cube).

A similar quiz: Find resistance across a 4D cube, then also resistance across an edge (singel resistor) of a 4D cube.

--- End quote ---

I'll have to sketch them out.

I would be interested in hearing who solved this by which means, from LTspice, to changing to an equivalent structure that is obvious to solve..

My first solution was to split all the resistors in half and bisect the design, half way between the test points. Analysis is then a piece of cake.
steverino:
The solution I gave assumes real values for the resistors - not open circuit.  Perhaps arbitrary was the wrong term.  Correct in this case, no?
rs20:

--- Quote from: steverino on April 22, 2018, 07:56:36 am ---The solution I gave assumes real values for the resistors - not open circuit.  Perhaps arbitrary was the wrong term.  Correct in this case, no?

--- End quote ---

So we have two facts:
-- Your formula works when all the resistances are completely equal
-- Your formula totally fails as some of the resistances approach infinity

And you're still confidently asserting that it's correct for all arbitrary combinations of finite resistances? That's very very bold! The parallel and series resistance formulae always give sane results even when faced with infinite resistances, and yet it very specially breaks down just for you? Hmm...

Anyway, your formula is still definitely wrong. You can take my previous argument, replace "infinity" by "one gigaohm", and the same  disagreement between common sense and your formula will result. Your formula allows the current to jump freely between the three nodes one step away from the + terminal, and the three nodes one step away from the - terminal. In a real cube, this isn't allowed. This is why your formula gives absurdly low values when very high value resistors are placed in such a way that they provide a unavoidable barrier to current in a real cube.

I've got Maple v11.0 to provide a solution (to the all-twelve-resistors-may-be-different-to-each-other case), and the resulting formula is literally 6 pages long. This doesn't prove much on its own, though (I might have made a mistake in the system of equations I provided, and it's also possible that this 6 page formula could be greatly simplified.)
Brumby:
Equipotential shorting method - valid because of symmetry.

Yellow lines are connecting equipotential points.  Because they are equipotential, joining them will not affect circuit operation.


Section A is 3 resistors in parallel = 333.3 Ω
Section B is 6 resistors in parallel = 166.7 Ω
Section C is 3 resistors in parallel = 333.3 Ω
A + B + C in series = 333.3 + 166.7 + 333.3= 833.3 Ω
30 seconds of mental arithmetic (but I admit I wrote down the three intermediate values).
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