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I'm trying to calculate the resistance between A and B where each resistor between them has resistance R. According to my key, the middle, vertical, resistor has no voltage across it, and thus is negligible in the calculation. Why is this the case? How can we "see" that it will have 0 voltage? enter image description here

It is stated that symmetry implies identical currents. But why would the following circuit be impossible? enter image description here

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You can see this by realizing the upper and lower path are completely symmetric. Thus the current through each of them must be the same. And if the same current is flowing through resistors with the same resistance (the two diagonal ones on the left), the voltage drop across them is exactly the same. Therefore, the topmost and bottom-most corner are at the same potential, and the vertical resistor is connecting points of equal potential, resulting in no current across it.

In general, it's good advice to take advantage of symmetries in a problem. If a problem is symmetric under some operation (in this case, flipping the diagram upside down), the solution very often must also be symmetric under that same transformation. Be careful though, there are also cases where symmetry spontaneously breaks (phase transitions for example), but looking for symmetries is good practice.

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  • $\begingroup$ Why does symmetric paths imply identical currents? $\endgroup$
    – MrPillow
    Commented May 19, 2021 at 23:26
  • $\begingroup$ Which one of the identical paths do you think should carry more current? $\endgroup$
    – noah
    Commented May 19, 2021 at 23:27
  • $\begingroup$ Say, the top one. (See the edited post) $\endgroup$
    – MrPillow
    Commented May 19, 2021 at 23:37
  • $\begingroup$ A silly question now would be, why it should be that and not exactly the reverse (the whole diagram flipped upside down). $\endgroup$
    – noah
    Commented May 19, 2021 at 23:43
  • $\begingroup$ Well, it could have been the bottom one, too. But that it could have been either one doesn't necessarily imply that this is impossible, no? To show that it is impossible, mustn't you prove that the second picture is impossible? $\endgroup$
    – MrPillow
    Commented May 19, 2021 at 23:44
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In your bottom diagram, let's say that the voltage at $B$ is zero, for convenience.

Let's label the top point as $x$, and the bottom point as $y$.

Ohm's law states that the voltage at $x$ must be

$$\frac{IR}{6}$$

And the voltage at $y$ must be

$$\frac{IR}{2}$$

But then this would mean that the current through the middle resistor must be

$$\frac{\frac{IR}{6}-\frac{IR}{2}}{R} = -\frac{I}{3}$$

(Negative meaning it would flow from $y$ to $x$).

Which it can't be, because our beginning assumption is that it was

$$\frac{I}{6}$$

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