# Finding Equivalent Resistance

Discussion in 'Electronic Design' started by [email protected], Jun 1, 2005.

1. ### Guest

This is basic Electrical Engineering puzzle - hounting me since 1993. I
hope this is a correct place to ask this and such questions. Simple
mathematicle induction fails in this 2D arrangement.

If '+' denotes a join, find equivalent resistances between any two
adjacent and diagonally opposite "+" in the figure below. The mesh
extends infinitely in all the directions.

| | |
--+--R--+--R--+----
| | |
R R R
| | |
--+--R--+--R--+----
| | |

Extending the question, what could we say about an infinite triangular
or hexagonal mesh like this?

-Bhushit

2. ### Martin DeMelloGuest

Hint

Inject a 1A current into one node and out of the other. Use symmetry and
superposition to calculate the voltage drop between the nodes.

martin

3. ### John BaileyGuest

The paper: Infinite Resistive Lattices by Atkinson and van Steenwijk
provides somewhat more details on the Fourier Transform method of
solving periodic networks. It develops solutions for multi-
dimensional, trangular and hexagonal lattices.
http://www-th.phys.rug.nl/~atkinson/resist.pdf

Three things:
1) The solution seems to be missing something. It is a methodology
lacking an explaination of (a key detail in) why it works.

2) When I tried to open the page just now, it came up with an error
message.

3) I have a hard copy and probably the pdf file on my system. If you
cannot find the web version, message me for a copy.

John Bailey

4. ### Fred BloggsGuest

damned little chance of that ever becoming a topic of discussion amongst
these pathetic old trolls...

5. ### Terry PinnellGuest

FWIW, there was some discussion 13 years ago in sci.physics, here:
http://tinyurl.com/cvh8k

Quoting one extract from one post:
This is problem 2.27 of
" -Introduction to Modern Network Synthesis-, by M. E. Van Valkenburg
(Wiley, New York, 1960). Although it is intended that much of the
book be covered in a one-semester undergraduate course, "[we] assume
that the reader is familiar with the elementary methods of network
analysis."

The solution approach looks like the one Martin hinted at.

6. ### r.e.s.Guest

http://atkinson.fmns.rug.nl/public_html/resist.pdf
seems to work.

7. ### Joel KolstadGuest

This is almost never true of today's undergraduates in EE programs and nowhere
not ubiquitous among graduate students. I've looked at those old network
theory books and it's amazing how 'deep' the area goes!

(Granted, when you look at the types of problems engineers typically work on
today, it's hard to argue they should be reading Van Valkenburg rather than
whoever writes a good Java or Linux book or something...)

8. ### r.e.s.Guest

WLOG, take R = 1 and suppose a square lattice with unit spacing.
The paper cited by John Bailey doesn't say so (afaics), but
inspection of the table of values there suggests that between
opposite corners of a square of side n in this network, the
resistance is exactly R_nn = (2/PI)*sum(1/(2*k-1),k=1..n)
for n=0,1,2,... (Between adjacent lattice points it is 1/2.)

--r.e.s.

9. ### Jonathan KirwanGuest

That seems right for R_nn. But what about R_nm? As you say, at R_01
it is 1/2. Also, at R_12 for example, the exact answer is 4/PI-1/2.

An iterative algorithm I devised (not the first person to invent the
idea, by any stretch, but it is what I came up with when thinking
about the problem before) is to arbitrarily place a voltage of 1V at
the node under consideration (n,m) and 0V at the node of (0,0). I
then arbitrarily set all other nodes of some arbitrarily sized finite
grid (extending from -S to S in both dimensions) to an assumed guess
of 0.5V to start. I then go to alternate nodes on the finite grid,
first those spanning from 1-S to S-1, stepped by 2, and compute the
node voltage as the mean of the adjacent node voltages (up, down,
left, and right, in the case of a square grid.) Once that is done,
the other alternate nodes on the finite grid are then visited, those
spanning from S to S, stepped by 2, and compute the node voltages
there in the same fashion. (Just be sure not to do any such averaging
for the node at (0,0) or at (n,m).) Repeat the process over and over
until the voltages settle. At that point, the resistance can be
computed by assuming that current flows accurately based on the
potential differences between (0,0) and its adjacent squares. (Which,
of course, determines the current that the node at (n,m) must
inevitably supply.) Since your starting potential is 1V between them
and since you know the required current to be supplied by examining
the nodes next to (0,0), the resistance is 1/sum.

Jon

10. ### Guest

In alt.engineering.electrical wrote:

| This is basic Electrical Engineering puzzle - hounting me since 1993. I
| hope this is a correct place to ask this and such questions. Simple
| mathematicle induction fails in this 2D arrangement.
|
| If '+' denotes a join, find equivalent resistances between any two
| adjacent and diagonally opposite "+" in the figure below. The mesh
| extends infinitely in all the directions.
|
| | | |
| --+--R--+--R--+----
| | | |
| R R R
| | | |
| --+--R--+--R--+----
| | | |
|
|
| Extending the question, what could we say about an infinite triangular
| or hexagonal mesh like this?

For AC, this could be even more interesting based on the lengths between
nodes and the AC frequency involved.

11. ### r.e.s.Guest

[r.e.s. wrote ...
[>> R_nn = (2/PI)*sum(1/(2*k-1),k=1..n) for n=0,1,2,...
The OP asked only about points that were "adjacent and diagonally
opposite" (which would reasonably be interpreted as referring only
to R_11, or possibly to both R_11 and R_01) -- I mentioned the
apparent formula for R_nn (n=0,1,2,...) because of its simplicity.

Are you aware of any formulas of comparable simplicity in the more
general cases?

--r.e.s.

12. ### Jonathan KirwanGuest

Yes, but I don't recollect them, now. I only recall what I wrote.

Jon

13. ### Mark PGuest

Resistance (ideally or "ohmically") is independent of frequency.

14. ### operator jayGuest

Sure and why don't you include self and mutual inductances, and the
capacitances, for spice.

15. ### Don KellyGuest

Martin Demello has the right approach. However, you have to do it in two
steps. Inject current into a node with the return connection a "circle" at
infinity. The current splits equally in 4 directions. Now disconnect and
draw current from the other node with injection at infinity. Superimpose.

16. ### The PhantomGuest

This simple technique doesn't work for the case of a pair of adjacent *diagonal* nodes.
Work out the details and you'll see why.

It does work for a pair of adjacent "orthogonal" nodes, however.