Effective resistance between two points using symmetry

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Point D and C are equipotrntial so you can remove wire DC and see how the circuit look like

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Jahnavi said:
Thanks for replying .

This is exactly what I would like to understand [emoji2] .

Please assume this is a 2D circuit .

If I draw a line perpendicular to AB , the left and right parts are mirror image , this make points D and C symmetric . But symmetric points do not necessarily have to be equipotential . Right ?

How can then we argue that points C and D are equipotential .
Path symmetry indicates that D and C are equipotential

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Jahnavi said:
Thanks !
Hello again,

I kept quiet because I was obviously making it worse for you in your frustration -- but I couldn't think of a good way to help you out without giving it all away. The picture I had in mind was, of course

Circuit schematic with nodes A (top left) and B (bottom left). A 2 Ω resistor connects A to B directly. A middle branch has a 1 Ω resistor from A to node D and a 1 Ω resistor from D to B. A right branch has a 1 Ω resistor from A to node C and a 1 Ω resistor from C to B. A wire links D and C horizontally.


and once you have it in front of you like this, the path is clear (thanks to the fact that all four on the right have the same resistancce)
(note: in many cases you can't even get it as 'flat' as this and there are one or more crossovers left to make life edifficult)​

and now I have a question for you:

with hindsight, what would have been a good hint that could have helped you effectively at that point ? (after all, you will definitiely not be the last one to get stuck with a problem like this).and as a side note: you see that I don't agree with
ehild said:
The original circuit cannot be drawn in-plane
 
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  • Circuit schematic with nodes A (top left) and B (bottom left). A 2 Ω resistor connects A to B directly. A middle branch has a 1 Ω resistor from A to node D and a 1 Ω resistor from D to B. A right branch has a 1 Ω resistor from A to node C and a 1 Ω resistor from C to B. A wire links D and C horizontally.
    upload_2018-4-11_13-31-35.png
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Hi BvU ,

BvU said:
and now I have a question for you:

with hindsight, what would have been a good hint that could have helped you effectively at that point ?

Not hint , but if you could have been a little more expressive and had written - "Please try my suggestion and you would clearly see how the two points C and D are equipotential using the same technique (drawing a line perpendicular to AB) you are trying to employ in the OP " . That would have worked :smile:

AND you could have added a smiley in post#6 :wink: .

I was under the impression that you had overlooked my concern in the OP and were trying to suggest an alternative strategy to simplify the circuit .

By no means this is an excuse .I was clearly wrong in not working as per your suggestion .

But then , this is the limitation with any online platform . It doesn't precisely portray your thinking/feeling :smile: .
 
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Thanks, will :smile: more often !
 
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