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

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In a symmetrical circuit, the lack of current between junctions AF and DG is attributed to the equal potential at both points, resulting from horizontal symmetry. The resistances in both branches of the circuit are equal, leading to no potential difference between AF and DG. Consequently, no current flows through the resistor connecting these junctions. While the circuit exhibits left-right symmetry, the key factor is the equal resistance that maintains the same potential at both junctions. Therefore, the assertion that there is no current between AF and DG holds true.
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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!
 
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Hi sparkle123! :smile:

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

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