The solution I proposed was an approach to exactly that problem.
Having an infinite grid actually makes the problem easier than in the
case of a finite grid because symmetry can be used.
I would not suggest this approach in an asymmetric situation.
Years ago, I ran into the easier adjacent node situation but never got
a handle on the the case with nodes that were not adjacent until now
The solution proposed came from drawing things out on a piece of graph
paper and noting that one could form equipotential squares at 45 degrees
to the grid squares as well as noting that for each of these- there are
n paths to the next one out and the total current is always 1A - the
symmetry due to the edge of the grid at infinity allows one to say that
the current in each of these paths is 1/n A.
Similar injection/superposition methods work for other configurations as
well. One such is a regular polygon with resistors between all vertices.
I first saw this in a problem in a grad circuits course- I spent a lot
of time trying to handle it--although it is simple enough using Z-bus
methods which I learned much later -but in the particular case involved
a 5 by 5 matrix in punch card and pre-calculator days..
The glaringly "oh shit!' solution to this used superposition- (paper &
pencil not needed but helpful)-there was a brighter student in the class..
Z-bus agrees-but involves a matrix inversion.
This and other cases were discussed some time ago on alt eng.electrical
or sci.physics.elecctromag
Don
On 28/01/2013 7:45 AM, Keith Park wrote:
Sorry, we don't seem to be discussing the same problem. The link
below is to the original problem which demands the solution to the
infinite case. The use of infinity is not just a convenience.
http://xkcd.com/356/ proposes a problem involving finding the
On Mon, Jan 28, 2013 at 12:11 AM, Don & Cathy Kelly <[email protected]> wrote:
Actually not
inject 1A at A with B open. and draws out 1A at the 'edge" (infinite
point or ring of all points equidistant from both A and B) which you can
call ground as it doesn't matter.
and finds the voltage drop between A and B.
Then inject -1A at B with A open and draws out 1A at the 'edge"
and find the voltage between A and B
Summing the two cases the result is the same as if the current source was
connected between A and B with the 'edge" open
there will be no current drawn out from the edge. It's easy to use
infinity but this is simply a way of stating equidistance from A and B in
order to call on symmetry. Errors will be in the order of 0.25% for
distance AB being about 1% of distances from both to the edge.
Don Kelly
On 27/01/2013 6:34 PM, Raul Miller wrote:
Ok, that clarifies your point. Except, in Kelly's method, it's not
tied to ground at infinity.
The problem is to determine the resistance between two nodes, A and B,
which are a finite distance apart. The method injects 1 amp of
current at A and grounds B.
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