Effective resistance between two points using symmetry

In summary, @tnich is trying to solve the homework equation by symmetry, but ends up not being able to. @gneill and @ehild offer no help, but suggest following @BvU's suggestions.
  • #1
Jahnavi
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Homework Statement


circuit2.jpg


Homework Equations

The Attempt at a Solution



I am trying to solve this question by symmetry .If I draw a line perpendicular to the line joining AB then points C and D are symmetric . Now I am unsure how to proceed further .
 

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  • #2
Try to flatten the diagram: a top horizontal line is A, a bottom one is B. There are three paths from top to bottom: AB, ADB and ACB
 
  • #3
BvU said:
Try to flatten the diagram: a top horizontal line is A, a bottom one is B.

I am interpreting that the circuit is already flattened out . The dotted lines are showing that branches AC and BD are not meeting anywhere .

BvU said:
There are three paths from top to bottom: AB, ADB and ACB

Yes .

How should I use this to simplify the circuit ?
 
  • #4
Jahnavi said:
How should I use this to simplify the circuit ?
As I indicated: a top horizontal line is A, a bottom one is B.
Have the three paths well separated, but going straight down from the top line to the bottom one
 
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  • #5
Sorry . I don't understand what you are trying to convey .
 
  • #6
What can I do to make you draw two horizontal lines on a piece of paper ?
 
  • #7
To help you get started:

upload_2018-4-10_18-3-54.png
 

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  • #8
  • #9
upload_2018-4-10_19-2-9.png


The red path is out of the plane, but it is equivalent with the blue path otherwise.What does that tell you about the potentials at D and C?
 

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  • #10
Jahnavi said:
Could someone help me in this problem using symmetry .

@tnich , @gneill , @ehild
Sorry. Symmetry is not going to help you on this problem.
 
  • #11
ehild said:
View attachment 223752

The red path is out of the plane, but it is equivalent with the blue path otherwise.What does that tell you about the potentials at D and C?

Thanks for replying .

This is exactly what I would like to understand :smile: .

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 .
 
  • #12
tnich said:
Sorry. Symmetry is not going to help you on this problem.

Thanks for replying .

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 ?
 
  • #13
Jahnavi said:
Thanks for replying .

This is exactly what I would like to understand :smile: .

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 so not necessarily have to be equipotential . Right ?

How can then we argue that points C and D are equipotential .
What would happen if you interchanged the blue and red triangles, would be anything different?
 
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  • #14
Jahnavi said:
Thanks for replying .

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 so not necessarily have to be equipotential . Right ?
They don't look symmetric to me. Try it with a mirror.
 
  • #15
tnich said:
They don't look symmetric to me. Try it with a mirror.
Oh, sorry, I see the symmetry now.
 
  • #16
tnich said:
Oh, sorry, I see the symmetry now.

Ok :smile: . Please reply to my above post .
 
  • #17
If you simply followed @BvU's suggestion to redraw the circuit in a single plane, the symmetry is easier to see, and the answer becomes obvious.
 
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  • #18
Jahnavi said:
Ok :smile: . Please reply to my above post .
You should try @BvU's suggestions in posts #2, 4, 6 and 7.
 
  • #19
I am not saying that the circuit is not symmmetric . All I would like to understand is that why points D and C are equipotential .

If you remember in the previous thread you had given a very nice symmetry argument . But points L and Q , although being symmetric were not equipotential .

Please help me understand why D and C are equipotential in this problem .
 
  • #20
Jahnavi said:
I am not saying that the circuit is not symmmetric . All I would like to understand is that why points D and C are equipotential .

If you remember in the previous thread you had given a very nice symmetry argument . But points L and Q , although being symmetric were not equipotential .

Please help me understand why D and C are equipotential in this problem .
Redraw the circuit as @BvU has suggested. Then you will see it.
 
  • #21
We're not saying that you're saying the circuit doesn't possesses a symmetry.

In analyzing a circuit, one of the most useful tools you have is redrawing the circuit in a way that it's easier to work with. I'm not sure why you're so resistant (pun not intended, but I'll go with it) to the idea.
 
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  • #22
Jahnavi said:
If you remember in the previous thread you had given a very nice symmetry argument . But points L and Q , although being symmetric were not equipotential .
Remember in the previous thread, we redrew the circuit to make the symmetric obvious. We replaced one resistor with two resistors in series.
 
  • #23
tnich said:
Remember in the previous thread, we redrew the circuit to make the symmetric obvious. We replaced one resistor with two resistors in series.

But redrawing the circuit didn't change the circuit . Both were completely equivalent .

I am trying to apply the same idea here but can't distinguish why in the previous thread L and Q although being symmetric were not equipotential and why D and C here are equipotential .
 
  • #24
Jahnavi said:
But redrawing the circuit didn't change the circuit . Both were completely equivalent .
And in this case, also, redrawing the circuit does not change it.
 
  • #25
There are 2 different situations.
1. You permute some or all of the nodes in the circuit, and you end up with the same circuit.
In this case, if a permutation sends a node to any other node, the voltage of these 2 nodes must be the same. That is the case here with C and D.
2. You permute some or all of the nodes. and you end up with the same circuit, but with the battery reversed. In that case if one point is at V, the other point will be at Vbattery - V. If the permutation does not move a point, it must be at potential Vbattery/2. This works for points C and D here as well. If
you swap A and B and not C and D you get the same circuit except for a reversed battery voltage. A and B swap between 0 and Vbattery and C and D are not moved by the permutation and must be at Vbattery/2.
In general this kind of problem isn't about rotational or mirror symmetry at all, but you just need to check if you can exchange 2 (or more) points and keep the same circuit.
You can do this if the connections from these 2 (or more) points to the rest of the circuit are the same.
In this case Rca = Rda and Rcb = Rdb, and these resistances are all of the connections from CD to the rest of the circuit.
If you draw the adjacency matrix of the circuit, you can exchange columns C and D and Rows C and D and you end up with the same matrix.
Code:
  A B C D
A - 2 1 1
B 2 - 1 1
C 1 1 - 1
D 1 1 1 -
 
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  • #26
ehild said:
View attachment 223752

The red path is out of the plane, but it is equivalent with the blue path otherwise.What does that tell you about the potentials at D and C?
ehild said:
What would happen if you interchanged the blue and red triangles, would be anything different?

Thank you .
 
  • #27
vela said:
In analyzing a circuit, one of the most useful tools you have is redrawing the circuit in a way that it's easier to work with.

Thanks for your suggestion .

vela said:
I'm not sure why you're so resistant (pun not intended, but I'll go with it) to the idea.

Neither am I :smile:
 
  • #28
BvU said:
As I indicated: a top horizontal line is A, a bottom one is B.
Have the three paths well separated, but going straight down from the top line to the bottom one

Thanks !
 
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  • #29
The oeiginal circuit cannot be drawn in-plane, but it has a mirror plane of symmetry (the yellow plane)The rectangles are the resistors Quite bad picture, but I hope you understand.
upload_2018-4-11_6-34-29.png
 

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  • #30
Thank you so much ehild :smile:
 
  • #31
Point D and C are equipotrntial so you can remove wire DC and see how the circuit look like

Sent from my Mi A1 using Tapatalk
 
  • #32
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

Sent from my Mi A1 using Tapatalk
 
  • #33
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

upload_2018-4-11_13-31-35.png


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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  • #34
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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  • #35
Thanks, will :smile: more often !
 
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<h2> What is effective resistance?</h2><p>Effective resistance is a measure of the resistance between two points in an electrical circuit, taking into account all the resistors and their connections between those two points.</p><h2> How is effective resistance calculated?</h2><p>Effective resistance can be calculated using the symmetry method, which involves identifying symmetrical points in the circuit and using their resistances to determine the overall effective resistance.</p><h2> Why is the symmetry method used to calculate effective resistance?</h2><p>The symmetry method is used because it simplifies the calculation process and reduces the number of equations that need to be solved, making it a more efficient method for determining effective resistance.</p><h2> Can the symmetry method be used for any circuit?</h2><p>Yes, the symmetry method can be used for any circuit as long as there are symmetrical points present. However, it may not always be the most efficient method for calculating effective resistance in more complex circuits.</p><h2> Are there any limitations to using the symmetry method for calculating effective resistance?</h2><p>One limitation of the symmetry method is that it can only be used for circuits with symmetrical points. Additionally, it may not be accurate for circuits with non-linear components or for circuits with multiple sources of electricity.</p>

FAQ: Effective resistance between two points using symmetry

What is effective resistance?

Effective resistance is a measure of the resistance between two points in an electrical circuit, taking into account all the resistors and their connections between those two points.

How is effective resistance calculated?

Effective resistance can be calculated using the symmetry method, which involves identifying symmetrical points in the circuit and using their resistances to determine the overall effective resistance.

Why is the symmetry method used to calculate effective resistance?

The symmetry method is used because it simplifies the calculation process and reduces the number of equations that need to be solved, making it a more efficient method for determining effective resistance.

Can the symmetry method be used for any circuit?

Yes, the symmetry method can be used for any circuit as long as there are symmetrical points present. However, it may not always be the most efficient method for calculating effective resistance in more complex circuits.

Are there any limitations to using the symmetry method for calculating effective resistance?

One limitation of the symmetry method is that it can only be used for circuits with symmetrical points. Additionally, it may not be accurate for circuits with non-linear components or for circuits with multiple sources of electricity.

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