What is the Resistance Across a Resistor in an Infinite Square Mesh?

[AntonVrba]

Consider a infinite square mesh made up of 100 Ohm resistors. What will be the resistance measured across one 100 Ohm resistor somewhere in the middle of the mesh (i.e between a-b)


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The answer is surprisingly simple and takes just 2 lines to calculate/prove.

 

[Averagesupernova]

An infinitely large gridwork of resistors and you want to choose a resistor somewhere close to the 'middle' of it?

 

[Automagman]

yeah dude, the meer act of finding the middle is impossible. Change infinite to a defined value.

 

[Hurkyl]

Anywhere is in the middle. The problem is symmetric -- you should get the same answer for any individual link.

 

[Averagesupernova]

Off the top of my head I'd say zero ohms, or not, who knows...

 

[AntonVrba]

Come on folk, you can do better than that do not be so strict with the middle. You all can integrate from minus infinity to plus infinity.

By the way a integral is not part of the solution.

Hurkyl's comment is correct - symmetry is on the right track, and
super average nova's solution of zero is wrong

 

[whozum]

Wouldn't it just be 100 ohms? The individual resistor has the same value, the equivalent resistance of the entire matrix would be infinite I believe.

 

[Moo Of Doom]

But isn't the matrix like a huge parallel circuit? The infiniteness seems to imply zero, but it might converge to some other value. (I don't feel like doing the math at this point... too late.)

 

[whozum]

In a circuit with two resistors of resistance A = 1ohm and Resistance B = 2ohm, the equivalent resistance of the resistors is 1/(1+1/3) = 0.75ohms, however, resistor A still has resistance 1ohm, the value of 1ohm is used to find the current through that resistor when a DC voltage is applied (after the current is found).

Thats just my take, I don't know, there's probably more to it than this.

 

[Averagesupernova]

My current way of thinking is telling me it is close to 57 ohms.

 

[AntonVrba]

Supernova how you get to 57 is a mystery - but not correct and a lot closer than your previous guess zero.

HINT: instead of mathematics, calculators and arithmic series think Kirchhoff.

 

[dextercioby]

Consider a infinite square mesh made up of 100 Ohm resistors. What will be the resistance measured across one 100 Ohm resistor somewhere in the middle of the mesh (i.e between a-b)


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+-+-+-+-
| | | |
+-a-b-+-
| | | |
+-+-+-+-
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The answer is surprisingly simple and takes just 2 lines to calculate/prove.

According to the elementary laws of logics,the answer is 100$\Omega$...

Rephrase it.

Daniel.

 

[AntonVrba]

Dextercioby you exploited my weakness in language skills

Now I shall re-phrase and change my question:

Consider a infinite square mesh made up of 100 Ohm resistors. What will be the resistance measured across two adjacent nodes somewhere in the middle of the mesh, after removing the one 100 Ohm resistor joining these two nodes (i.e between a-b)


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+-a b-+-
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[minger]

I'm going to say 50 ohms. It's just a guess, but there is an answer. I remember back in my circuits days, we had homework problems 'similar' to this. The ones we had just had resistors arranged like a ladder though. However, I would think that this can be equated to an infinite number of "ladders" plated in parallel.

 

[kleinwolf]

According to the elementary laws of logics,the answer is 100$\Omega$...

Rephrase it.

Daniel.

This is not, because there are a lot of parallel resistors with the one between AB...but I cannot figure out how to compute them (for example there are 2 300 Ohm equivalent resitors, and so on...)

 

[ontadian]

From Kirchoff - "The algebraic sum of the voltages around any closed path is zero" and
"The algebraic sum of the currents toward any junction point is zero". Now, I notice that the original measurement was to be across one resistor, whilst later was described as a measurement after removing one resistor. In the latter case I think the answer would be 3r/2

 

[AntonVrba]

Now, I notice that the original measurement was to be across one resistor, whilst later was described as a measurement after removing one resistor. In the latter case I think the answer would be 3r/2

Don't get mislead by the latter question (phrased for dextercioby benefit) of one resistor being removed. If I know the resistance (impedance) between two points in a circuit and then remove a known resistor between those two points I immedialy can easely calculate the new equivalent resistance or impedance.

ontadian - 3r/2 you thought incorrectly

 

[Davorak]

Anwser:
minger was right:
Rab is 50 ohms
Reason:
Consider a electron current source at A with current I and a hole source at B with current I.
The current has 4 equeal resistant paths to flow along for both electrons and holes. Therefore 1/4 the current flows along each path.

In the link between A and B has I/4 current of electrons and I/4 current of holes. A total net current of I/2(through Rab) and a total current of I from node A to node B from all paths.

V = (I/2)*R



V = Itotal Rtotal
(I/2)*R = I Rtotal
(1/2)*R = Rtotal

Rtotal = R/2 = 100ohms/2 = 50ohms


A different of finding the anwser that works with capacitors, inductors, as well as reisitors is using
$$R = \frac{\rho l}{A}$$
Where l and A are discrete. And rho is the resistivity. Though I have left out the details.

I believe that this second method will even work for an arbitary node to an arbitary node though I have never proved such.

 

[AntonVrba]

Davorak - congratulations

 

[Davorak]

Davorak - congratulations

According to the elementary laws of logics,the answer is 100 ...

Rephrase it.

Daniel.

This is not, because there are a lot of parallel resistors with the one between AB...but I cannot figure out how to compute them (for example there are 2 300 Ohm equivalent resitors, and so on...)

Looks like a mix up in terminology here. Resistor can be a circuit element or it can be the resistance between two nodes. I have always observed the first definition as the most common one in which case nothing was wrong with AntonVrba original statement.

 

[ontadian]

Can someone explain as to why the formula R = pl/A is relevant in this case, as my understanding is that this refers to a calculation of the resistance of a conductor when the resistivity, the length and also the area of that conductor is known?.

 

[Davorak]

Ontadian's answer in white:
Consider one node in the infinite mesh of R resistors and connect it to a current source I. The four surrounding node receive I/4. Each of the four nodes make an equalpotential in the resistor network. Rather what we can consider an equalpotential since the two nodes are nto directly connected it is not technically an equalpotential. The resistance between the center node and the four surrounding nodes is
Rtot0-1=rho l/A = (R *(1))/4 = R/4

For this problem(where Ip(at node B) is positive current and In is negative current(ar node B))

VnAB = I pl/A = I R/4
VpAB = I pl/A = I R/4

VtotAB = I R/2
VtotAB/ItotAB = RtotAB = R/2

Now for two nodes apart straight line. A1-*-B1

The next equalpotential are the nodes 2 resistors away from the center.
So from the first equal potential and the second the resistance is:
Rtot1-2 = rho I/A = R*(1)/12 =R/12

Therefore
Rtot0-2 = (1/4 + 1/12)R

Vn1AB1 = I pl/A = I (1/4 + 1/12)R
VpA1B1 = I pl/A = I (1/4 + 1/12)R

Vtot = I(1/2 + 1/6)R

Vtot/Itot = Rtot = I(1/2 +1/6)R/Itot

Itot = I //still

Vtot/Itot= Rtot = I(1/2 +1/6)R

Solving the problem this way is a special case that takes advantage of the symmetry of the mesh. If the resistors are not all equal it seems likely that it would then be necessary to use Bessel functions or legendre polynomials to figure out a new symmetry.

 

[willib]

This is not, because there are a lot of parallel resistors with the one between AB...but I cannot figure out how to compute them (for example there are 2 300 Ohm equivalent resitors, and so on...)

to calculate parallel resistance use the recriprocal of the sum of the recriprocals

 

[jdlech]

What you are measuring is an infinite series of 2 parallel circuits measuring 300, 500, 700, 900, etc. ohms.
Therefore, you are effectively measuring 1/(1/150)+(1/250)+(1/350)...)

the answer should be 50 ohms, assuming your meter has a finite precision.