Charles Link said:
Yes. And @.Scott gave the additional part that you need: You can separate the circuit at the middle, with the upper part in one loop with C and D, and the lower part in another loop with A and B. ## \\ ## Edit: The reason is by symmetry: the current going toward the middle from C must be the same as the current coming out of the middle and going to D, etc., because ## V_C-V_{Middle}=V_{Middle}-V_D ##.Jahnavi said:
Charles Link said:
See my "Edit:" additions to the previous post.## \\ ## Once you see that the current into the middle from C is the current that comes out of the middle to D, (and the same with that coming from A and into the middle, and that coming out of the middle and going to B), the middle connection is not necessary, because no current crosses the middle between the upper and lower halves.Jahnavi said:
Charles Link said:
. I am thinking more on this specifically .I essentially did the same thing as drawing a vertical line through the middle, with symmetry on both sides of the line.## \\ ## And when I saw your OP of setting ## V_C=V_D ##, although it was incorrect, it did get me looking for a symmetry about those two points... Notice also that ## V_A-V_{Middle}=V_{Middle}-V_B ##. ## \\ ## And, in addition, notice the resistance of the lower part is then ## R_{lower}= 15 \Omega || (15 \Omega +15 \Omega) ##, and the upper part is also a simple resistor configuration. ## \\ ## This problem reminded me of a related one I saw a number of years ago: Take a cube with ## R=1 \Omega ## on each of the 12 edges. Find the equivalent resistance between diagonal points of the diagonal that passes through the center of the cube (it would be easier to draw a picture). This one also has symmetry in it, but it's a more complex symmetry. (And maybe it would be good to give a hint on this, because when I solved it, I had already learned about a symmetry of a cube from solid state physics courses: About the diagonal that passes through the center, there is a 3-fold axis of symmetry: You can rotate the cube 120 degrees about this axis and you get the same thing. That means the 3 edges coming from what would be point A in this cube are equivalent. Similarly, the 3 edges going into point B are equivalent). And it came up in a google of it:Jahnavi said:This is the key point
I am curious to know as to how you found symmetry in this circuit .
As far as I am concerned , I drew a line perpendicular to the line joining AB and found that the left and right parts are mirror images of each other . C is mirror image of D .
I hope this is right .
What did you do to find symmetry ?
I think the correct term for the relation between their potentials, if you let the middle be potential zero, is the potentials of ## C ## and ## D ## are basically "antisymmetric" about the middle. If ## V_C=+V ##, with ## V_{middle}=0 ##, then ## V_D=-V ##.Jahnavi said:
