What will be the resistance

1. Apr 18, 2005

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)

| | | |
+-+-+-+-
| | | |
+-a-b-+-
| | | |
+-+-+-+-
| | | |

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

2. Apr 18, 2005

Averagesupernova

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

3. Apr 18, 2005

Automagman

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

4. Apr 18, 2005

Hurkyl

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

5. Apr 18, 2005

Averagesupernova

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

Last edited: Apr 18, 2005
6. Apr 18, 2005

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

7. Apr 18, 2005

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.

8. Apr 19, 2005

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.)

9. Apr 19, 2005

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 dont know, theres probably more to it than this.

10. Apr 19, 2005

Averagesupernova

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

11. Apr 19, 2005

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.

12. Apr 19, 2005

dextercioby

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

Rephrase it.

Daniel.

13. Apr 20, 2005

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)

| | | |
+-+-+-+-
| | | |
+-a b-+-
| | | |
+-+-+-+-
| | | |

14. Apr 20, 2005

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 infinte number of "ladders" plated in parallel.

15. Apr 20, 2005

kleinwolf

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...)

16. Apr 20, 2005

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

17. Apr 21, 2005

AntonVrba

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

Last edited: Apr 21, 2005
18. Apr 21, 2005

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.

Last edited: Apr 21, 2005
19. Apr 21, 2005

AntonVrba

Davorak - congratulations

20. Apr 21, 2005

Davorak

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.

21. Apr 21, 2005

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?.

22. Apr 21, 2005

Davorak

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.

23. Apr 21, 2005

willib

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

24. May 4, 2005

jdlech

What you are measuring is an infinite series of 2 paralel 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.