There seems to be some misunderstanding. What I said is that the currents do not remain equal when the two currents are applied. The 0.5 ohm result is therefore false.
On Tue, Jan 29, 2013 at 2:54 AM, Don & Cathy Kelly <[email protected]> wrote: > I will try and digest this a bit more before fully responding but here are > a few initial comments. > However, there seems to be a problem between your interpretation of > superposition than mine and a disagreement with your conclusions- which > requires more sleep and less wine. However, nothing that I have said is in > violation of any circuit theory or graph theory which is involved. > . > Initially we are faced with a current in at A and out at B. This can be > modeled by two current sources , one tied between A and an arbitrary > reference point. The second between B and the reference point. The net > current into the reference point is 0, > The system is linear so it is valid to say that the total sum of what > actually occurs in the network is the sum of the effects of the two source > applied independently. That is current in at A causes an effect at B and > elsewhere in the grid. Current out at B causes an effect similarly but it > wont be the > same in any particular part of the network. Superposition considers that > in a linear system, the actual effect at any point in the system with the > current in at A and out at B can be found by considering the two situations > independently and summing the results. > In this particular case, infinity is involved only because it is a > condition for use of symmetry. The approach results in the sum of currents > into the reference point is 0, whatever the reference point is but symmetry > involves, strictly speaking, the reference being at infinity (which becomes > an equipotential. > I note that you say that the current with both sources acting remains at > 0.25A in the 8 resistors- that is_not_ true. > The symmetry is there for each source by itself- but not for the > combination. What is true is that the voltage at B due to injection at A is > the same as the voltage at A due to excretion from B. and the current into > the reference is 0. > Nowhere is the "hidden infinity" a problem . I recognize that with one > source only- an infinite voltage is needed. Connect an ideal current > source to an infinite resistance and the same thing occurs. > > In the meantime I shall put together an example of a finite system using > the same technique and giving results agreeing with the Z-bus matricx > results. > > Don > > > ..On 28/01/2013 11:50 AM, Keith Park wrote: > >> The problem as I see it is that the method of obtaining the solution >> involves three assertions. The first assertion is that with one ampere fed >> into into a node, let`s call it node A, of an infinite grid the current >> divides equally between the four resistors connected to that node, let`s >> call the resistors a0,a1,a2 and a3. Because of the symmetry of the >> situation this assertion is certainly true. Similarly for the second >> assertion a current of one ampere pulled out of the opposite, node B, >> causes a current of 0.25 amperes to flow out of the four resistors >> connected to that node. This assertion is also certainly true. The four >> resistors are named b0,b1,b2and b3. Now a0 is connected to b0 and a1 to >> b1. >> The third assertion is that superposition may be invoked when the two >> currents are applied at the same time. The current in all eight resistors >> remains 0.25 amperes. Node A will therefore be at 0.5 volts above Node A. >> If we take the junction of a0 and b0 as the reference node (it is where >> we >> attach the negative terminal of our imaginary infinite impedance >> voltmeter) >> then node A is at 0.25 volts and node B is at -0.25 volts. Since a2 still >> has 0.25 amperes flowing through it there must be 0.25 volts across it so >> the voltage where it joins the rest of the grid must also be zero. So >> 0.25 >> amperes flowing into the grid produces zero voltage. The rest of the grid >> therefore displays zero resistance at that point. But there are only >> three >> other one-ohm resistors connected to a2 so the resistance can not be less >> than one third of an ohm and the voltage can not be less than 1%12 volts. >> So the third assertion leads to the conclusion that the voltage at this >> junction is both zero and non-zero, which is incorrect. The conclusion is >> that superposition fails because of the hidden infinity. The current a0 >> and a1 is greater than 0.25 amperes and the current through a2 and a3 is >> less than 0.25 amperes. >> >> >> >> >> >> >> >> On Mon, Jan 28, 2013 at 1:57 PM, Raul Miller <[email protected]> >> wrote: >> >> Actually, I think I see an inconsistency. Or, at least, an issue >>> which I do not understand. >>> >>> If I understand your model properly (and, I might not), you have the >>> same potential at all points on a square surrounding A (nodes which >>> have an equal manhattan distance from A). But this suggests that the >>> resistance is the same from all points on that square back to A. >>> >>> Can this be correct? >>> >>> Thanks, >>> >>> -- >>> Raul >>> >>> 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<http://www.jsoftware.com/forums.htm> >> >> > ------------------------------**------------------------------**---------- > 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
