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I have a problem where I have a cube with resistance $R$ along each edge. I am asked to find the potential between various nodes (corners) using symmetry and the fact that two nodes at the same potential can be collapsed to one point without changing the equivalent resistance between the two given nodes. Using this fact, I think I have solved the problem, and indeed I don't want help with the problem. However, I want to know why this fact holds. If I am drawing a circuit of the cube above, and the potential between, let's say, the upper far corner and all three adjacent corners is the same, why can I depict this as three resistors in parallel between two nodes? (I assume that there are terminals on two of the nodes between which I want to measure the potential).

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  • $\begingroup$ If you short-circuit 2 nodes of the same potential, how much current will run through the short-circuit? $\endgroup$ – Qmechanic Oct 18 at 1:07
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If I am drawing a circuit of the cube above, and the potential between, let's say, the upper far corner and all three adjacent corners is the same, why can I depict this as three resistors in parallel between two nodes?

If I understand you correctly, the reason is in order for the resistors to be in parallel, it is not sufficient that the value of the potential difference across each of the resistors is equal. It must be the identical potential difference. That requires each of the three resistors connected to one corner of the node terminate at their other end at the same node. Clearly that is not the case for the cube.

To put it another way. Suppose I have some arbitrary circuit with many resistors. I find that the voltage across two randomly selected resistors in the circuit happens to be equal. Does that automatically mean they are in parallel?

Hope this helps.

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