Question about electrical networks

[neverevernever]

TL;DR Summary

What is the total current flow in an electrical network?

Suppose we have a connected electrical network with only resistors. We also assume that each vertex is connected to at least $k$ vertices, where $k\geq2$. Suppose all resistors in the network has resistance between 1 and 2. Suppose we flow 1 unit of current from vertex $a$ to vertex $b$. Then, for each edge of the network, there will be some current flowing. I'm interested in the sum of the absolute value of the current in each edge. I conjecture that this sum will be bounded by some constant no matter how large is the network. Is this true? For simplicity, we can consider completely connected network first.

There is also some restriction on $a$ and $b$. We assume they are close to each other in the network.

 

[berkeman]

Summary:: What is the total current flow in an electrical network?

Suppose we have a connected electrical network with only resistors. We also assume that each vertex is connected to at least $k$ vertices, where $k\geq2$. Suppose all resistors in the network has resistance between 1 and 2. Suppose we flow 1 unit of current from vertex $a$ to vertex $b$. Then, for each edge of the network, there will be some current flowing. I'm interested in the sum of the absolute value of the current in each edge. I conjecture that this sum will be bounded by some constant no matter how large is the network. Is this true?

Welcome to the PF.

Could you post a sketch or schematic of the circuit you are envisoning? I'm having trouble visualizing it from your verbal description. Thanks.

 

[neverevernever]

Welcome to the PF.

Could you post a sketch or schematic of the circuit you are envisoning? I'm having trouble visualizing it from your verbal description. Thanks.

Sure! I'm new to here. What I'm considering is a graph with banded adjacency matrix, which means that for entries close to the diagonal they are non zero and for entries far away from the diagonal they are 0. The entries in the adjacency matrix is just the conductance of that edge.

 

[berkeman]

Sure! I'm new to here.

You can add a PDF or JPG image to your posts using the "Attach files" link at the lower left of the Edit windows.

Also, if you'd like to post math equations (including matrices), you can click on the LaTeX Guide link, also in the lower left of the Edit window.

 

[anorlunda]

At first, I thought power grid. But the network you describe has no connections between any node and ground. What is the voltage source? How many sources and where are they in the net?

@berkeman is right, a picture of the network topology would be very helpful.

You might ask how your network differs from the classical infinite ladder circuit that is frequently seen in homework problems.

 

[neverevernever]

At first, I thought power grid. But the network you describe has no connections between any node and ground. What is the voltage source? How many sources and where are they in the net?

@berkeman is right, a picture of the network topology would be very helpful.

You might ask how your network differs from the classical infinite ladder circuit that is frequently seen in homework problems.

View attachment 263369

The source is placed at node $a$ and $b$

 

[berkeman]

The source is placed at node $a$ and $b$

Wakarimasen...

 

[DaveE]

OK, Here's an idea. I haven't really thought this through all the way. Plus, I'm having a hard time with the sum of abs(I) concept. But at least it's something to discuss.

1) The only source of current in the network is between a & b, = 1 unit.
2) Imagine a surface S that separates the network into two groups of vertices so that a is in one group and b is in the other. S cuts edges.
3) Conservation of charge implies that the sum of the currents flowing through that surface is equal to 0. Note: not the sum of magnitudes. This is be a problem.
4)The surface S can be defined to include any arbitrary set of non-a vertices in the b group and vice-versa. Or, if you choose, any set of edges can be cut by S. (This needs to be proved I think, and I can't do topology!)
5)If there is any set of edges (which includes a-b) that have a sum greater than zero this violates conservation of charge when you choose a surface that cuts just those edges.

So, certainly not a proof, but maybe the start of something?

edit: S must either be closed or infinite. i.e.ever vertex must be defined as on one side or the other.

 

[DaveE]

More of a math problem than physics, I think.

 

[DaveE]

OK, another avenue to consider. Since you want to know about the absolute value sum.

Assuming the network is finite (which you didn't say, BTW), the power input to the network is I²⋅Rab. Rab is the driving point resistance ; the resistance measure from vertex a to vertex b with the ab edge cut. If the network is finite then Rab is finite and the power input is finite. The power dissipated in the resistive network is the sum of each edge current squared times the edge resistance and it must equal the input power, so it must be finite. Therefore, there can't be a set of edges that have infinite power, so the sum of the current magnitudes must be finite.

I think this approach may also work for an infinite network, but it hurts my brain to work on problems like that.

 

[neverevernever]

OK, another avenue to consider. Since you want to know about the absolute value sum.

Assuming the network is finite (which you didn't say, BTW), the power input to the network is I²⋅Rab. Rab is the driving point resistance ; the resistance measure from vertex a to vertex b with the ab edge cut. If the network is finite then Rab is finite and the power input is finite. The power dissipated in the resistive network is the sum of each edge current squared times the edge resistance and it must equal the input power, so it must be finite. Therefore, there can't be a set of edges that have infinite power, so the sum of the current magnitudes must be finite.

I think this approach may also work for an infinite network, but it hurts my brain to work on problems like that.

Your argument will give a bound involving the number of resistors in the network, which means the bound will grow with the number of resistors. However, I believe the bound should be independent of that.

 

[DaveE]

Your argument will give a bound involving the number of resistors in the network, which means the bound will grow with the number of resistors. However, I believe the bound should be independent of that.

Yes, I think you are correct.

Rab will decrease in value with the network size, but will be balanced by the increase in sum terms. These should roughly cancel, IMO. But I certainly haven't shown that. These aren't developed proofs, just ideas to start with.