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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