Effective resistance between two points using symmetry

[Jahnavi]

Homework Statement

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 .

 

[BvU]

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

 

[Jahnavi]

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 .

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

Yes .

How should I use this to simplify the circuit ?

 

[BvU]

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

 

[Jahnavi]

Sorry . I don't understand what you are trying to convey .

 

[BvU]

What can I do to make you draw two horizontal lines on a piece of paper ?

 

[BvU]

To help you get started:

 

[Jahnavi]

Could someone help me in this problem using symmetry .

@tnich , @gneill , @ehild

 

[ehild]

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?

 

[tnich]

Could someone help me in this problem using symmetry .

@tnich , @gneill , @ehild

Sorry. Symmetry is not going to help you on this problem.

 

[Jahnavi]

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 .

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 .

 

[Jahnavi]

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 ?

 

[ehild]

Thanks for replying .

This is exactly what I would like to understand .

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?

 

[tnich]

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.

 

[tnich]

They don't look symmetric to me. Try it with a mirror.

Oh, sorry, I see the symmetry now.

 

[Jahnavi]

Oh, sorry, I see the symmetry now.

Ok . Please reply to my above post .

 

[vela]

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.

 

[tnich]

Ok . Please reply to my above post .

You should try @BvU's suggestions in posts #2, 4, 6 and 7.

 

[Jahnavi]

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 .

 

[tnich]

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.

 

[vela]

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.

 

[tnich]

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.

 

[Jahnavi]

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 .

 

[tnich]

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.

 

[willem2]

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 V_battery - V. If the permutation does not move a point, it must be at potential V_battery/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 V_battery and C and D are not moved by the permutation and must be at V_battery/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 R_ca = R_da and R_cb = R_db, 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.

  A B C D
A - 2 1 1
B 2 - 1 1
C 1 1 - 1
D 1 1 1 -

 

[Jahnavi]

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?

What would happen if you interchanged the blue and red triangles, would be anything different?

Thank you .

 

[Jahnavi]

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 .

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

Neither am I

 

[Jahnavi]

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 !

 

[ehild]

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.

 

[Jahnavi]

Thank you so much ehild