Series and Parallel Circuits using symmetry

[psandadi]

Homework Statement

I want to find the effective conductance between points a and z using symmetry. Each edge has a unit conductance of 1. The middle are not vertex.

Relevant Equations

No equations needed

 

[phinds]

I want to find the effective conductance between points a and z using symmetry

Then you should do so, not ask us to do it for you.

Please read the forum rules, particularly the one about homework questions requiring that the OP show some work before we offer assistance.

 

[psandadi]

Then you should do so, not ask us to do it for you.

Please read the forum rules, particularly the one about homework questions requiring that the OP show some work before we offer assistance.

Yes, I forgot to mention. I understand that all the vertex except a and z are symmetrical. I do not know where to go after that

 

[DaveE]

I do not know where to go after that

How about google? My first search attempt was "symmetric circuit questions" (although, honestly, google added the "questions" part as I was typing). The first real link was this, which looks helpful. There are MANY others.

https://scienceres-edcp-educ.sites.olt.ubc.ca/files/2015/01/sec_phys_circuits_resistorshapes.pdf

Maybe some independent study will be of benefit. Come back and ask if you get stuck, but "I do not know where to go after that" sounds lazy to me. Just go somewhere and you'll probably learn what your next question should be. I think you might be stuck because you have given up; prove me wrong.

PS: If you are looking for us to answer your question for you, you are likely to be disappointed. But we are here to help.

 

[haruspex]

Are you familiar with the method of images in electrostatics? It creates a symmetry from a fact about a potential. You can apply that idea in reverse here.

 

[psandadi]

Are you familiar with the method of images in electrostatics? It creates a symmetry from a fact about a potential. You can apply that idea in reverse here.

I am actually a math guy. This question i came across was talking about the probability of reaching z before a starting from a and the formula is C_{eff}(a,z)/C(a) where C(a) is the sum of all the conductance that are neighboring to a.

 

[psandadi]

I am actually a math guy. This question i came across was talking about the probability of reaching z before a starting from a and the formula is C_{eff}(a,z)/C(a) where C(a) is the sum of all the conductance that are neighboring to a.

Can I use the fact that in a complete graph K_n, the effective conductance between any two vertices is (n-1)c/2 where n is the number of vertices and c is the conductance of each edge

 

[Steve4Physics]

Can I use the fact that in a complete graph K_n, the effective conductance between any two vertices is (n-1)c/2 where n is the number of vertices and c is the conductance of each edge

Are you sure $\frac {(n-1)c}2$ is correct?

For example, consider a complete graph with 3 vertices (A, B and C) with each edge having a conductance $c$.

Your formula gives $c_{\text {total, AB}} = \frac {(n-1)c}2 = \frac {(3-1)c}2 = c$.

But the conductance for edge AB alone is $c$. The total conductance between A and B must be greater than this because there exists an additional path (ACB) between A and B.

 

[haruspex]

Are you sure $\frac {(n-1)c}2$ is correct?

For example, consider a complete graph with 3 vertices (A, B and C) with each edge having a conductance $c$.

Your formula gives $c_{\text {total, AB}} = \frac {(n-1)c}2 = \frac {(3-1)c}2 = c$.

But the conductance for edge AB alone is $c$. The total conductance between A and B must be greater than this because there exists an additional path (ACB) between A and B.

Yes, it should be $nc/2$. You can go directly, 1, or via any of n-2 others for 1/2 each: $1+(n-2)/2=n/2$. The symmetry argument says there is no flow that goes via two or more others.

 

[psandadi]

I feel like this would be the answer. I, initially drew an imaginiary vertical axis through c and seeing that the network is symmetrical, b and d would have the same voltages. So, I joined b and d to a point x and then added the conductance because they are parallel. By looking at the second diagram, it is again symmetrical from the a,z line due to ehcih c and x have the same voltage. The c-x line disappears because it forms a self loop at the point and since it has a potential difference 0, it doesn't influence the effective conductance. Then, I found the effective conductance between a and z using series and parallel law. Let me know if this answer is wrong please.

 

[psandadi]

Yes, it should be $nc/2$. You can go directly, 1, or via any of n-2 others for 1/2 each: $1+(n-2)/2=n/2$. The symmetry argument says there is no flow that goes via two or more others.

ok. thank you