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.
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