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Imagine if I placed resistors of 1 ohm each in shape of tertahedal. What would the total resistance be between A and D?

I can't find any website that gives me a good explaination. Tetrahedal Cirucit

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closed as off-topic by jinawee, Brandon Enright, centralcharge, John Rennie, Emilio Pisanty Jan 20 '14 at 12:37

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You mean: the resistance between any of the two vertices (like $A$ and $B$), correct? – Vibert Jan 19 '14 at 21:34
Yes, between A and D. – Vaishnavi Jan 19 '14 at 21:36

First redraw the schematic in 2D form so that it is easier to see the circuit:

Now the solution is obvious from inspection. It's just the parallel combination of R1, R2+R3, and R4+R5. You don't even need a calculator.

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It would be better if you explain, how you arrived at the above modified circuit, sir. – Immortal Player Jan 19 '14 at 22:55
You do understand that a wire has the same potential everywhere (until it meets a resistor) right? – nervxxx Jan 19 '14 at 23:03
@nervxxx: The electric potential at a point is the amount of electric potential energy that a unitary point charge would have when located at that point. At different point electric potential will be different, irrespective of point where we take. So, I don't think so it would be same until a resistor is reached. If there is a potential difference between two points then only charge flows, if we assume, potential to be same till resistor is reached, then how could we expect current in between? – Immortal Player Jan 19 '14 at 23:31
If you had formed the tetrahedron out of six, 1 meter lengths of fine wire with a resistance of 1 Ohm/meter, there would be no visible distinction between conductors and resistors, but the circuit diagram of Olin Lathrop above would still be correct (and very useful!) The length and shape of conductors in a wiring diagram need not bear any relationship to reality. Only the connectivity matters. – DJohnM Jan 20 '14 at 0:08
@Vinay: There is no need to explain more how the tetrahedron was drawn in 2D above. That REALLY should be obvious, especially when looking at the right diagram in the question. We have to assume at least some minimal intelligence of the reader when writing answers here. – Olin Lathrop Jan 20 '14 at 0:16

Let's say you are looking at the resistance between A and D. Temporarily remove (in your head) the resistor across CB. Look at the paths ABD and ACD. They are equivalent. So the potential at point B and C are equal, by symmetry.

Now when you put the resistor CB back, because the potential difference across BC is 0, no current flows across it. So the original setup looks as though the resistor CB is missing.

You can easily solve for the resistance from there.

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so, it's (1/1+1 + 1/1+1 + 1/1)^-1 – Vaishnavi Jan 19 '14 at 22:00
@nervxxx.According to you, by symmetry potential at A,C,D should be same as at B, thus current should also not flow through AB and BD also. I don't think that would give correct answer. – Immortal Player Jan 19 '14 at 22:06
@VINAY So, can you give me your answer please? – Vaishnavi Jan 19 '14 at 22:11
@VINAY what are you talking about? You put a potential difference across A and D. Obviously the potentials at A and D are different. Don't put words in my mouth. – nervxxx Jan 19 '14 at 22:23
@nervxxx. According to you by symmetry potential at A and B should be same isn't it? – Immortal Player Jan 19 '14 at 22:25

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