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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