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. >> >> > ------------------------------**------------------------------**---------- > For information about J forums see > http://www.jsoftware.com/**forums.htm<http://www.jsoftware.com/forums.htm> > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
