Removing connections between equipotential points in solving circuits?

[gregorspv]

Homework Statement

An octahedron is made of wire, where an edge has resistance R. What is the resistance between two adjacent points?

Relevant Equations

Kirchhoff's rules, in parallel and in series resistance.

A sketch of the setup and the equivalent circuit are attached.

I believe the correct way to solve this is to redraw the circuit as shown in Fig. 3 and then remove the connections between evidently equipotential points, which reduces the problem to a familiar setup of in parallel and in series resistors.

However, the circuit may also be redrawn as shown in Fig. 4 where there are also two lines connecting equipotential points, yet if one removes them, an incorrect solution is obtained.

How do I then know, when such connections are removable? I have encountered a statement that such a connection must lie on the circuit's axis of symmetry and this seems to do the trick in this example, but I don't understand why it works that way.

Please excuse the formatting, I'm sadly on mobile at the moment.

 

[haruspex]

the circuit may also be redrawn as shown in Fig. 4

Are you sure? Looks to me that you removed two current-carrying wires.

 

[gregorspv]

Oh, I did not word that correctly. In Fig. 4 the connections between equipotential points from Fig. 2 were removed. Perhaps then equipotentiality is not a valid criterion for removal, but rather the absence of current? Would you care to explain why there is no current in the wires removed in Fig. 3?

 

[haruspex]

Perhaps then equipotentiality is not a valid criterion for removal, but rather the absence of current

They're the same thing. The wires you removed in fig 3 do connect equipotential points, so carry no current. But in fig 4 it looks to me that you removed wires that do carry current, so are not connecting equipotential points. As a result the resistance increased.

 

[gregorspv]

I think I see what you're saying, but it still seems odd to me that there can be a voltage drop (potential difference) between two points on a perfect conductor. Or else how are the wires removed in Fig. 4 connecting points of different potential?

 

[haruspex]

I think I see what you're saying, but it still seems odd to me that there can be a voltage drop (potential difference) between two points on a perfect conductor. Or else how are the wires removed in Fig. 4 connecting points of different potential?

My statement that two points being equipotential implies no current flows between them is not correct when the connection has no resistance. The test should be whether they would still be equipotential were the wire removed.

So the flaw in your method is using the equipotentiality test on zero resistance wires while they are in place; it would justify removing all of them!

 

[gregorspv]

D'oh – it had occurred to me that my reasoning would lead to removing virtually any wire, but I could not have found a way out were it not for your edifying response. Thank you!

I find it remarkable that some people just happen to get this intuitively (the solution I was provided with does not comment on this step at all), whereas I just spent the last 20 minutes or so proving that the current is indeed zero. I believe the symmetry argument still rests on the assumption that one is intuitively able to figure out the current directions in a circuit which I don't think is necessarily a trivial task (especially when dealing with different resistors and other elements).

 

[haruspex]

I believe the symmetry argument still rests on the assumption that one is intuitively able to figure out the current directions in a circuit

No, it's easier than that.
The circuit in the upper figure in image 3 is symmetric about a horizontal line through the middle. (Where that cuts a resistor, split that into two in series.) Consequently, all points it cuts are at the average potential, so all at the same potential.
This is still true if the two horizontal lines that lie on that central line are removed, so no current flows in those and they can be removed.

 

[rude man]

Aren't 7 resistors in series and then connected in parallel with one resistor?
I must be mssing something

 

[haruspex]

Aren't 7 resistors in series and then connected in parallel with one resistor?
I must be mssing something

No. Which figure are you looking at?

 

[rude man]

I

No. Which figure are you looking at?

I confused "octagon" and "octahedron".