🚀 go-pugleaf

Regarding the resistance across the terminals A & B of a one-ohm
resistor that is part of an infinite square mesh of resistors:

I encountered this problem in my undergraduate days in India (perhaps
the textbooks followed there are a generation behind :-) )

Connect a current source of 1 ampere to the terminals A & B.  One
would expect the voltage differences between nodes far away from A&B
to be very small, and so it is a good approximation to connect a circle of
nodes  far away from A&B with a resistanceless wire and discard 
everything outside. Next, in this large but finite problem, superposition
can be used to replace the single current source by two current sources,
one between terminal A and our far-away equipotential; and the other
between our far-away equipotential and terminal B. Approximate
symmetry enables us to solve our two circuits and superpose them.
Since, in the original infinite circuit, voltage differences between 
distant nodes falls off sufficiently rapidly with distance from A&B
(how fast -- that's an exercise I leave for the reader :-)), our approxi-
mation improves as our equipotential tends to infinity.

In the case of the Bethe lattice , which is an infinite tree-like structure
with no circuits, this argument doesn't hold.  All the branches of the
tree that are connected to terminal A  (without going through terminal
B) are all at the same potential, which is different from the common
potential of all the nodes that connect to terminal B. So there is no 
"infinity" at which to connect current sources in the way we did the
first problem.

-arun gupta
ag3L@andrew.cmu.edu