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- Path: sparky!uunet!dtix!darwin.sura.net!Sirius.dfn.de!chx400!ugun2b!ugsc2a!cornell
- From: cornell@sc2a.unige.ch
- Newsgroups: sci.physics
- Subject: Re: ... an infinite mesh of 1ohm resistors ...
- Message-ID: <1992Jul29.144031.940@sc2a.unige.ch>
- Date: 29 Jul 92 12:40:31 GMT
- References: <1992Jul24.165217.16688@news.media.mit.edu>
- <1992Jul25.210947.12316@cs.yale.edu> <1992Jul27.210947.5820@fs7.ece.cmu.edu> <1992Jul28.000844.27051@mixcom.com>
- Organization: University of Geneva, Switzerland
- Lines: 35
-
- In article <1992Jul28.000844.27051@mixcom.com>,
- ttyytt@mixcom.com (Adam Costello) writes:
- >
- > The "solution"
- > involved imagining that you could inject an amp of current into a network
- > of resistors. No capacitors! Where is the charge supposed to go?
-
- Imagine a grid (2n+1)*(2n+1), with a perfect 1A current source in the middle
- AND perfect earths attached all around the edges. By symmetry,
- 1/4 amps flow through each of the resistors neighbouring the source.
- Now let n go to infinity. This is what is meant by "injecting 1A into
- an infinite network of resistors".
-
- All that is required for the current to be 1/4 A is for the (square)
- grid to have rotational symmetry order 4 around the source. As n-> infinity,
- it becomes less and less important for the sink to be attached precisely in
- the middle; you should satisfy yourself of this if you want to be sure the
- solution works. Then both the current source and the current sink mentioned
- in the solution may both be considered to be in the centre of the grid, so
- the total current in the resistor connecting the 2 terminals is 1/2 A.
-
- > Regardless of whether this can be done in the laboratory, there's still
- > the question of whether it can be done in the mind. Certainly, if the
- > network were finite, it would make no sense. Does the infinitude of the
- > network allow a source with no sink?
-
- Another way of doing the problem is to note that the capacitance of the
- network increases with size. Therefore, the initial 1/4 amps is able to flow
- for a time of order 1/n^2 before the potential of the grid increases
- significantly. Just take a series of measurements on successively larger
- and larger grids on a timescale that is smaller than this and you're laughing.
-
- > AMC
-
- Stephen Cornell.
-