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| Resistance across a 1k resistor cube. |
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| IanB:
--- Quote from: hamster_nz on April 22, 2018, 07:47:56 am ---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.. --- End quote --- My solution arose from observing the symmetry of the system. If all resistors have the same value then all parallel interior nodes of the network have the same voltage, which means they can be treated as if they were connected, which allows the network to be viewed as groups of parallel resistors. Following this observation the overall resistance is (1/3 + 1/6 + 1/3) · 1000 = 833⅓ |
| steverino:
--- Quote from: rs20 on April 22, 2018, 08:05:09 am --- --- 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.) --- End quote --- Relax my friend. I simply reduced parallel and series combinations. If my solution is incorrect, that's ok and I'll double check it in the morning. If it looks ok to me, I'll post a copy of my notes. If I find an error, I'll post that also and some kind soul can point out the error. Now, it's 1:27am and I'm relaxing enjoying a glass of brandy. Cheers! BTW, I just looked at Digikey, and I don't seem to be able to find any infinite resistors! :-DD |
| steverino:
quick request. Can somebody please post the ltspice file for the cube (i know, i'm lazy). I'll use it to check my work. Thanks. |
| rs20:
For those curious, I got Maple (a symbolic algebra solver thing) to solve the general case (resistors in the cube arrangement, but each resistor may have different, arbitrary values). I've attached a printout of the worksheet, but the TL;DR is: The formula for the resistance is 3 pages long, and has this form (where capital letters are the various resistances, in an weird arrangement explained in the PDF): total resistance = (A*U*G*F*C*V + K*B*A*E*C*F + ...) / (U*A*B*E*V + K*C*G*V*F + ...) Which is at least dimensionally consistent! In the case where four parallel resistors are equal (resistance = Rz) and the remaining 8 resistors are equal (resistance = Ry), then we find: total resistance = (Ry^2 + 3 * Ry * Rz + Rz ^ 2) / (2 * Ry + 4 * Rz) Notably, taking Ry or Rz to infinity brings the whole expression to infinity, as Ry and Rz both have higher powers in the numerator. Finally, setting all the resistances to 1000 gives an answer of 833.33..., which is reassuring :) Btw, hamster_nz, the above is not how I initially solved the simpler problem: I initially solved in exactly the way that Brumby described. Unfortunately, that method can't be applied in the case where the resistors have different values, as the underlying symmetry breaks down :( I'd be fascinated to hear if anyone has a non-maching-requiring method of solving the arbitrary case; I'll be the first to admit that I have no idea how to do it by hand. |
| RoGeorge:
OK, then, the cube was easy, try this one: What is the resistance between 2 nodes A and B, in an infinite 2 dimensional square grid of resistors, each resistor having the same value, R? >:D |
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