How does mirror symmetry for resistance work?

[phantomvommand]

Homework Statement

Please see attached photo

Relevant Equations

Mirror symmetry, same potential nodes

The question is to find the resistance between AF (top 2 points). Let the far right unlabelled vertex of the pentagon be B.
Why can we say that points O and C are at the same potential? I get that both points O and C appear to be at 'half the path' if you consider AOF and ABCDF, and so potential at O = C. However, this is less clear if you consider the path AODF or ABCOF, and it is no longer obvious that O and C are at the same potential.
Why does the mirror symmetry method for resolving resistors work, despite the existence of other unsymmetrical paths?

 

[Delta2]

This must be some sort of triweird symmetry which unfortunately I don't really understand. @Baluncore can you put this on spice please and tell us what you get. Is it indeed that $V_O=V_C$?

I think all the resistors are meant to be of the same value.

 

[phantomvommand]

This must be some sort of triweird symmetry which unfortunately I don't really understand. @Baluncore can you put this on spice please and tell us what you get. Is it indeed that $V_O=V_C$?

I think all the resistors are meant to be of the same value.

I have a very indirect way of reasoning that V_O is equal to V_C. If you try drawing any path from A to F, say ABCOF, and suppose the current flows upwards from C to O, there exists another perfectly equal path, AOCDF, in which the current from C to O has the same magnitude as above but flows in the opposite direction. Thus, the currents cancel and i9 = 0, hence V_O = V_C. But this is extremely indirect, and my teacher stated immediately that V_O = V_C using the 'half the path' argument. However, I still cannot accept it, since 'half the path' is no longer intuitive when considering unsymmetrical paths.

 

[Steve4Physics]

Because of the symmetry, the potential at O must be half the emf. And the potential at C must be half the emf. So the potentials at O and C are equal. But if a (slightly) more rigorous explanation is needed, how about this…

Let ‘S’ be the axis of symmetry (through OC).

I think the cell’s emf is called $\mathscr E$. Take point F to be 0 volts so point A is $\mathscr E$ volts.

The question should state that each resistor (except $R_{OC}$) has the same value as its ‘partner’ resistor reflected in S. (A special case is when all resistors are equal, but having all resistors equal is not essential.)

Currents (apart from $i_9$) are in equal pairs ($i_1=i_4, i_2=i_3$, etc.). This can be seen from the symmetry about S and from noting that reversing the cell’s polarity will not affect the magnitudes of any currents.

Apply Kirchhoff’s 2nd (voltage) law to loop AOFA. By symmetry the potential at O, halfway round the loop, is $\frac {\mathscr E}{2}$.

Similarly for loop ABCDEFA, the potential at C, halfway round the loop, is also $\frac {\mathscr E}{2}$.

Edit - typo' corrected.

 

[vela]

However, I still cannot accept it, since 'half the path' is no longer intuitive when considering unsymmetrical paths.

You shouldn't expect it to be intuitive when you consider asymmetrical paths. You're essentially saying if you look at the problem one way, the answer is clear. But if you look at the problem in a complicated way, the answer is no longer obvious, so it can't be right. The whole point of taking advantage of the symmetry is to avoid having to do a complicated analysis.

 

[Baluncore]

This must be some sort of triweird symmetry which unfortunately I don't really understand.

If it helps, rotate the diagram counter-clockwise by 90°, then you should see that there is no voltage across Roc which is part of a bridge diamond. The geographical orientation should have been reduced to orthogonal before drawing a circuit diagram. Redraw the rotated circuit.
When you throw out Roc, or replace it with a short, the network simply collapses into series and parallel combinations.

Make Roc = open circuit.
Then Raf = 2R // 4R, yes, that is simply parallel = 1/(1/2 + 1/4) = 1/3/4 = 4/3.

 

[Delta2]

Make Roc = open circuit.
Then Raf = 2R // 4R, yes, that is simply parallel = 1/(1/2 + 1/4) = 1/3/4 = 4/3.

If I make OC=short circuit i get $R_{AF}=\frac{6}{5}R$.
That is I get that $R_{BO}=R_{OD}=R/2$ then this is connected in series with $R_{AB}$ (or $R_{FD}$ at the left part), so we have $R+\frac{R}{2}=3\frac{R}{2}$ there, which is in parallel with $R_{AO}$ (or $R_{OF}$ at the left part) so we have the resistance for this part is $\frac{3}{5}R$ and two parts of that in series , total $\frac{6}{5}R$.

 

[Baluncore]

Sorry, I did it in my head with my eyes closed. You are correct.

Roc is in a balanced bridge. Make Roc = infinite.
The horizontal link o-o can also = infinite.
Then for all three columns of two; 1+1=2.
The two columns on the right make 2//2=1, in series with 2 = 3.
3//2 = 6/5 = 1.2 ohm.