By observation of what you have indicated as the connections there is no
direct connection between corner 0 and 3 so the total resistance from
0 to 3 is the resistance 1-2 +the resistance 1 -3 for a sum of 5/3
ohms. There is a way to deal with this using the Zbus matrix but, in
this case it
would be use of a pile driver where a tack hammer works.
The problem that you mention
http://xkcd.com/356/
is not trivial and even the case with the resistance between adjacent
points in an infinite grid is not trivial by use of expansion of series
parallel
or star delta conversions. The latter and various fully interconnected
polygons become trivial through use of nodal analysis and superposition
along with consideration of symmetry. Such methods deal with the
topology and variations on them may make the solution of the cited
problem easier.
So far I have found establishing a routine to be very messy-not a
programming but a conceptual approach problem.
Don
On 07/01/2013 2:36 PM, Raul Miller wrote:
http://xkcd.com/356/ proposes a problem involving finding the
resistance in a case involving an infinite number of resistors.
Before I can reason about that graph, I'd want a systematic way of
reasoning about much more trivial cases.
So... here's a representation of a square with four corners:
Resistors connect corner 0 with corner 1, corner 1 with corner 2, and
both corners 1 and 2 with corner 3. Unnamed corner pairs are not
connected. All resistors are 1 ohm. The expected net resistance
between corner 0 and corner 3 should be 1 ohm.
1 ((, |.&.>)0 1;0 2;1 3;2 3)} <:%=i. 4
0 1 1 _
1 0 _ 1
1 _ 0 1
_ 1 1 0
The values in this table represent the resistor connecting the two
corners whose row/column indices represent the locations in the
network.
But.. how would I express this calculation without encoding the grid
into the calculation itself? (In other words, so that any such
resistance graph is allowed -- though it is ok to assume that the
matrix describing the connections is symmetric.)
This is probably fairly trivial for anyone with an EE background, but
I'm not quite seeing how to approach this.
Thanks,
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