

Resistancenetwork Problems  Fred Curtis
 Feb 2000 [Last annotated November 2004] 
I became interested in resistance network problems from the recurring questions posted to the rec.puzzles archive. In particular, the resistance between two adjacent nodes on an infinite 2D rectangular lattice of 1ohm resistors was given as ½ohm. The solution posted in the rec.puzzles FAQ (section "physics/resistors.p") was roughly
I was so skeptical of this reasoning that I wrote a C++ program, surl.cc, to examine the convergence in everlarger finite lattices. Not surprisingly, the FAQ gave the correct answer.
I used the same reasoning (see References for more on the superposition principle) on some nonrectangular infinite lattices and found the following results. Simulations in everlarger subsets of the lattices produced results which converged to the values predicted by the superposition principle:
I don't see why the symmetry in currents flowing through the edges joining the source vertex would stay when the current sink changes from being an infinitely large circle to an adjacent vertex  in fact it seems counterintuitive since the sink at the adjacent vertex seems to introduce a clear asymetry.
For example, consider an infinite halfplane (take an infinite lattice and throw away everything with a y coordinate less than 0). All the resistors along the bottom edge look equivalent, but we can't deduce anything about the resistance across one of these using the superposition principle. See Further Directions below for hints the result is 2/pi.
Even in the case of an infinite lattice where a current source is placed at a single vertex (and the sink is ignored or assumed to be a circle at infinity), there are no predictions about current flows in any edges except the 4 edges around the vertex.
Because the superposition principle only ever makes predictions for the resistance in one edge, and because the reasoning isn't strong, it seems plausible to me that the principle is just a coincidence. That is, the principle is a just story that agrees with observed results but doesn't explain the results. Contrast this with the works listed in the References below which mathematically prove the results.
There are some experiments I could perform (see Further Directions below) that might convince me to treat the superposition principle as more than a mathematical coincidence.
Note: [2004 November] I came across József Cserti's 1999 paper [see References], which deals with infinite lattice resistance problems. The techniques used in Cserti's paper [and, I'm guessing, many of the papers cited by Cserti] involve far more integral calculus than I can understand.
A general method in analysing finite lattices is to locate equipotential points and join them, yielding a simpler network.
When I read the rec.puzzles FAQ there was [November 2004: & still is] no general solution to the resistance between opposite corners of an arbitrary hypercube of 1ohm resistors; solutions were only given for 3D and 4D hypercubes. This problem seemed quite tractable  just combinatorics. The same reasoning about symmetry for the cube case can be applied to the arbitrary ddimensional case: joining the equipotential points results in d layers containing d(d1 choose i) resistors in parallel in the i'th layer, giving an overall resistance of (sum from i=0 to d1 of 1/(d1 choose i)) / d. The values for the first few d are:
d  2  3  4  5  6  7  8  9  10  11 
resistance  1  5/6  2/3  8/15  13/30  151/420  32/105  83/315  73/315  1433/6930 
As d gets larger, the resistance tends toward 2/d [more accurately, 2/d + 2/(d1)^2] ; the dense layers in the middle become relatively perfect conductors compared to the outermost layers.
The resistance between adjacent vertices of a ddimensional hypercube is (2^d  1)/d(2^(d1)) ohm, which has a nice geometric interpretation of (1lessthannumberofvertices)/(number of edges). This value, like the oppositevertices resistance, converges to 2/d.
A simplex is a ddimensional analogue of the tetrahedron. It has d+1 vertices,
each of which is joined to all the others by an edge. If a voltage v is applied
across any two vertices of a ddimensional simplex, the potential at each
of the remaining vertices is v/2 by symmetry. Joining equipotential vertices
gives a net resistance of (1/(d1) series 1/(d1)) parallel 1
= 2/(d1) parallel 1
= 2/(d+1) ohms.
The cross polytope is the ddimensional analogue of the octahedron. There are 2d vertices, two on each coordinate axis, each vertex joined to every other vertex except its opposite on the same axis.
If a voltage v is applied between opposite vertices, the remaining vertices are at a potential of v/2 by symmetry, yielding a resistance of 1/(2d2) series 1/(2d2) = 1/(d1) ohms.
If a voltage v is applied between adjacent vertices, the remaining vertices *not opposite* to the test points are at a potential of v/2 by symmetry yielding a simplified network:
AB \ /  \ /   \ /   \ /   v/2   / \   / \  / \ 0v
where the edge resistances are 1 ohm and the diagonal resistances are each 1/(2d4). Further symmetry indicates the potential at B is (vA). The resistance of the network, by tedious algebra, is (d0.5)/(d(d1)), which also yields the correct result when d=2. The potential at vertex A is v(d1)/(2d1).
A similar but more tractable model (test points at (0,N) and (N,0), all points on lines parallel to y=x connected to make them equipotential) has a diagonal resistance of (1 + 1/2 + 1/3 + ... 1/N). This underestimates the simulated diagonal resistance of an NxN lattice by about 10% for smaller N, rising to 13% for N=80.
N  Diagonal resistance on NxN lattice (by simulation)  1 + 1/2 + ... 1/N (From altered model with equipotential diagonals) 

1  1  1 
2  1.5  1.5 
3  1.857  1.833 
10  3.133  2.929 
30  4.450  3.995 
80  5.673  4.965 
Some things which would be interesting to investigate are:


