No Current Between AF & DG in Symmetrical Circuit: True?

[sparkle123]

Due to the symmetry of the circuit there is no current through the resistor between the junctions (AF) and (DG) so the junctions (AF) and (DG) have the same potential.

I don't understand how this is true. The circuit is symmetrical left-right but not top-bottom. Isn't top-bottom symmetry what is required to eliminate current between AF and DG?
Thanks!

 

[I like Serena]

Hi sparkle123!

Due to the horizontal symmetry, all points in the circuit that are on the vertical center line have the same potential.
(There's as much resistance left as right of these points.)

When there is no potential difference, no current will flow.

 

[Delphi51]

The top branch with R1 and R1 leaves the AF junction half way between the potentials at E and B.

The bottom branch also has equal resistance on either side, splitting the potential to half way between E and B. The potentials at AF and DG are the same so there is no current from one to the other.

edit: didn't see your post before posting, I like Serena.
Great minds think alike?

 

[sparkle123]

Thanks I like Serena and Delphi51! :)
(yes, great minds do think alike!)

 

[asteriatic]

I would like to clarify that the statement is indeed true. The symmetry of the circuit refers to its left-right symmetry, which means that the circuit is identical when flipped horizontally. This symmetry ensures that the current flowing through the resistor between junctions (AF) and (DG) is equal in both directions, resulting in a cancellation of current and thus no net current between the two junctions. The top-bottom symmetry is not necessary for this phenomenon to occur. I hope this clarifies any confusion.