Yes- it does look at an infinite grid with an ideal current source- and
as such- is not a real world problem.
For an ideal current source- voltage approaching infinity when the load
approaches infinity is quite alright.
Valid circuit theory allows me to do this as well as apply superposition.
also note that injecting 1A at node a with a source between a and the
'edge' of the grid means that 1A is flowing out at the 'edge'.
injecting -1A at node b results in -1A flowing out at the 'edge' By
superposition I am summing 1 +(-1) =0 A at the edge- however far from a
and b this 'edge' is. the total current between a and b in this case is
1A (1A in at a and -1A out at b) and all currentsand voltages in the
grid will be just as if the source had been connected between a and b.
The only reason for considering an infinite grid is to take advantage of
symmetry- that is current into a given node flows from that node equally
in 4 directions. This makes life much simpler. A unit current will
cause 1/4 amp in each of these directions -hence a drop of 1/4 volts in
each branch from that node to adjacent nodes so that each of the
adjacent nodes is at the same potential-- and could be tied together by
a 0 impedance ring to form a 'super' node. External to the ring there
will be 12 outgoing leads and by symmetry the current outgoing will be
1/12A for a voltage drop of 1/12V to the next group of adjacent nodes.
Again these can be tied into a super node but this time with 12 +8 =20
outgoing paths for an added voltage drop in these of 1/20 volt.
So far we have 1/4 (1 + 1/3 +1/5 .. and this goes on with added 1/7 +1/9
etc.
This series diverges slowly.
Now repeating with the current out at the second node but a doubling
of voltage between the two points of concern.
When you used the y matrix- what was the location of the 2 points under
consideration and where was the reference node for which no row or
column exists? To simulate the infinite grid it is necessary to
consider all the edge nodes as being connected together as the
reference- just as is done for superposition. Using a single node as
reference leads to a different problem.
Don
On 27/01/2013 1:53 PM, Keith Park wrote:
The method of finding the resistance between the two nodes of an infinite
grid of resistances (Don&Kathy Kelly) is erroneous. The method fails
because a one ampere current flowing into the grid produces an infinite
voltage. The superposition of the two cases amounts to adding minus
infinity to plus infinity which does not produce a reliable result. By
inverting the Y matrix for grids of up to 100 by 100 I have found that the
resistance approaches 0.636 ohms
On Fri, Jan 11, 2013 at 6:20 PM, Don & Cathy Kelly <[email protected]> wrote:
It does assume that the reader knows what it is all about. For learning it
is veryy skimpy.
A more thorough reference is:
http://nptel.iitm.ac.in/**courses/Webcourse-contents/**
IIT-KANPUR/machine/ui/chap3.**pdf<http://nptel.iitm.ac.in/courses/Webcourse-contents/IIT-KANPUR/machine/ui/chap3.pdf>
The advantage of the Ybus is that it is easy to build -just choose a
reference node and go with the process at the other nodes. In power systems
the reference is generally the ground.
I have found a reference which discusses the Ybus and Zbus methods and the
way to build the latter without inversion of a large matrix.
Here is a simple example-- a square of 4 1 ohm resistors (conductance =1
Siemen (mho)
1 o---/\/\--o 2 At each node there are 2 resistors each
of 1 so the Y00 =Y11 =Y22
| | Between 2 and 1 there is a coupling of 1S
so Y01=_1 Y02=0 as there is no
/ / direct connection.
\ \
| |
0 o---/\/\--o reference
] y=:>2 _1 0;_1 2 _1;0 _1 2
2 _1 0
_1 2 _1
0 _1 2
this gives
]z=:%.y
0.75 0.5 0.25
0.5 1 0.5
0.25 0.5 0.75
The resistance from node j to the reference is the Zjj term
The resistance from node j to node k is zjj+zkk-2zjk
nodes (buses) that are not of interest can be eliminated by simply
removing the row and column corresponding.
Don.
On 11/01/2013 6:47 AM, Raul Miller wrote:
On Thu, Jan 10, 2013 at 11:07 PM, Don & Cathy Kelly <[email protected]> wrote:
A wee bit of a mix up between what I said and what Aai said (no 'quotes'
appeared to distinguish between the two).
the 9:23 message was from Aai but the
f=: 13 : '+/-: %>: +:i.y'
and the result corresponds to a specific immediate "calculator' solution
that I gave
]R100K=:+/0.5*%1+2*i.100000
I had not put it in the form that you present and I thank you for the
re-programming as a function-Gee-I can interpret it- I must be learning
something!!!.
Note that
R=:13 :'+/0.5*%1+2*i.y'
would be equally valid.
As for an example of the Z-Bus method I gave a small example in a post
at
6:12PM yesterday
I can give more examples but the practical ones I have on hand do involve
complex impedances.
Wiki gives little information so it appears that I will have to put
together
some notes that I have - it appears that this approach is something known
mainly in power system analysis.
This is the best I have found so far on line and it is inadequate:
http://en.wikipedia.org/wiki/**Impedance_parameters#The_Z-**
parameter_matrix<http://en.wikipedia.org/wiki/Impedance_parameters#The_Z-parameter_matrix>
Inadequate because it assumes the typical application is for a 2-port
network.
What do you think of
http://en.wikipedia.org/wiki/**Ybus_matrix<http://en.wikipedia.org/wiki/Ybus_matrix>?
Thanks,
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