Actually I have to apologize because what I indicated is not correct. I
think that Raul actually questioned the point when he inquired about equal
currents in the branches between the "squares" . If there is a current
of 1/12 A entering the corner of the second square and 3/20 A out
something is wrong. I cannot assume all current out at some distance
from the origin is the same as all points on the squares are not at the
same potential. There is a symmetry but -close but no cigar.
However this has nothing to do with the superposition approach.
You mention two nodes A and B each with 4 resistors and ao tied to bo,
a1 tied to b1 and in this case A and b appear to be diagonally opposite.
and where the resistors are tied there is a node- you are looking at two
of the resistor paths at that node.
Let's look at the simpler case of the resistance between adjacent nodes
| |B
-----------|----------------|----------|---------------|
A | ||\ /|
|__< 1A__________| \<1A__o___<1A
I want to solve the case of current in at A and out at B. A well known
linear circuit theorem is superposition (b(x+y) =bx+by)
I can replace the current source between A and B with two sources
connected at some arbitrary point- one injecting current into A and the
other into B . The net current into A is the same as before and the net
current out of B is the same as before If this arbitrary reference is
far enough away from A and B The net current into the reference point is
0. Choosing this reference as infinity ensures symmetry.The current
doesn't approach 0 because this reference is at infinity-
Suppose I do inject a unit current of 1A in at node A and out at infinity.
In other words I am treating the two sources independently.
This by itself results in a current of 1/4A between A and B with a
corresponding voltage drop.
Now remove this source and inject -1A at B which leads to 1A in at infinity.
Again we have a 4 way split and 1/4 of this current flows between A and
B- in the same direction as before.
There will be currents elsewhere in the grid but I am not concerned with
these.
Now consider both these sources at the same time -as the system is
linear- we can add the results of the two . This results in 1A in at A
and 1A out at B and also a net current of 0 at infinity- just as in the
original case. in particular the current AB is 0.25 +0.25 =0.5A for a
net voltage A to B of 0.5Vor RAB=0.5 ohms.
This superposition is a well established circuit technique.
Now as to the case that I think you are describing.
| |B
------- |-------bo---|-------
|ao |b1
----|------------------------
A | a1 |
Current in at A and current out at B
1)Take current in at A from reference at infinity gives 0.25A in a1 and
a0. at the bo-ao junctin there are 4 resistors so current splits 3 ways.
Assume for the sake of argument that the split is equal so 1/12 A flows
toward B
resultant voltage Vab= 1/4 +1/12 =1/3 V. Ditto for a1-b1
2)now take current out of B to reference.
1/4A in bo and b1 in the same direction as before. Also 1/12 A in ao and
a1 in the same direction as before
current in ao =1/4+1/12 =1/3 =current in a1 ,bo and b1. Going the ao-bo
route gives 1/3A in 2 ohms for Vab=2/3
total current in at A and out at B is 1A so V/I =2/3 ohm
However, for the case of A and B being a chess move apart. one can use
multiple current sources such that in the end, there will be
cancellation at the in between nodes. The problem is that of just how
current divides at the in between nodes- roughly 1/12 and 1/20 as an
initial guess (but using 1/24 and 1/40 gives a result close to 0.633 .
Here is a finite example.
Consider a pentagon with full cross connections between all vertices.
What is the resistance seen between any 2 vertices?
you could use a Y matrix with one node -say no.5 as reference so that
y=:4 4$4 _1 _1 _1 _1
y
4 _1 _1 _1
_1 4 _1 _1
_1 _1 4 _1
_1 _1 _1 4
%.y
0.4 0.2 0.2 0.2
0.2 0.4 0.2 0.2
0.2 0.2 0.4 0.2
0.2 0.2 0.2 0.4
The resistance seen between any pair of nodes is 0.4 ohms
Now try it this way for nodes 0 and 1
inject 1A into 0 with all other nodes grounded (i.e the current source
is from ground to 1)
This results in 1/4A in each branch from node 1 and 1/4A out to ground
from each of
nodes 1 to 5.
Now inject -1A into node 1 with nodes 0, 2 3 4 5 grounded.
this gives 1/4A in each branch 0-1 2-1 3-1 4-1 and into the grid at
these nodes
sum the two cases and in particular there will be 5/4A in at 0 and out
at 1 with a current of 1/2A between 0 and 1
resulting in R01=2/5 =0.4 ohms. In other words a hand solution which
takes less time than figuring out y and inverting- particularly in pre-J
time.
Don
On 29/01/2013 10:30 AM, Keith Park wrote:
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.
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