Re: infinite square lattice of resistors

[Ludwik Kowalski <KowalskiL@MAIL.MONTCLAIR.EDU>]

Let me try the last case (R along the "body diagonal" for a single
cubical cell of identical resistors). Each of the 12 resistors is r.
See below.

"Carl E. Mungan" wrote:

> Okay, somebody let me in on the trick for the resistance across a single
> diagonal of the infinite square lattice of say one-ohm resistors. I'd like
> both the elegant solution from symmetry which Leigh implied exists, and I'd
> like to know the formal Fourier method to go between any pair of junctions --
> or at least a sketch of the basic idea behind the latter if it's very
> complicated.
>
> I can see that the infinite lattice is symmetric across a diagonal and
> hence if I can find the resistance of one of these half planes, I would
> just need to need to halve that to get the desired answer. I also remember
> the trick for the infinite-in-one-dimension ladder of resistors. But
> neither of these ideas seems to help me make much progress on Leigh's
> problem.
>
> For that matter, what's the resistance along a single edge rather than a
> diagonal? Or what are the resistances for a simple cubic lattice along an
> edge, face diagonal, and body diagonal?

The answer is R=5*r/6. The challenge is to explain the reasoning without
the benefit of good drawings. Draw a cube in 3-dim perspective. Assume
there is one r along each of 12 edges. Select a diagonal passing through the
center. Label it end points as A and H. The current comes from A and exits
at H. Make the point A red and H green.

The current from A flows into three resistors. Label other edges of these
resistors as B, C and D. Make B, C, D yellow. Likewise, before coming
to H the current flows through three resistors. Label other edges of these
resistors as E, F and G. Make them blue. Due to symmetry potentials in
all yellow points are identical. You can thus connect B, C and D with one
triangular sheet of copper (zero resistance) without disturbing any current.
Likewise E, F and G can be connected to another triangular piece of
copper. The total R is not affected by the presence of virual (or real)
plates when all r are identical.

Now recognize that the resistors AB, AC and AD are connected in parallel
with each other; each starts at A and ends on the BCD plate. Replace them
by R1=r/3.  Likewise replace the resistors HE, HF and HG by one resistor
R2=r/3. What is left? Six resistors, each going from the plate BCD to the
plate EFG. Replace them by R3=r/6, because they are in parallel with each
other. The total resistance is thus equal to R1+R3+R2=r/3+r/6+r/3=5*r/6.
Ludwik Kowalski