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- From: bweiner@ruhets.rutgers.edu (Benjamin Weiner)
- Newsgroups: sci.physics
- Subject: Re: ... an infinite mesh of 1ohm resistors ...
- Message-ID: <Jul.28.16.11.35.1992.4623@ruhets.rutgers.edu>
- Date: 28 Jul 92 20:11:35 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: Rutgers Univ., New Brunswick, N.J.
- Lines: 23
-
- Adam Costello writes:
- >But I think you are missing the point of the second post. 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? ...
-
- Argh, you nudniks. Imagine a very large square grid of resistors,
- say 1000 on a side. Now connect the edges of the network to ground.
- Consider point A, which is at the center of the network, and point B
- which is one resistor to the right.
- 1. Attach a 1 amp source to point A. The current flows out of A
- to the edges of the network (ground). By symmetry, 1/4 amp flows
- out through each resistor attached to A, so 1/4 amp goes to B.
- 2. Attach a 1 amp sink to point B (negative voltage). B is almost
- exactly in the center so again symmetry arguments apply, and 1/4 amp
- comes in from point A.
-
- Superposing the two gives the solution as before: 1 amp into A and out
- of B, 1/2 amp through resistor AB, so the resistance is 1/2 ohm.
- The answer to "where does the current go?" is: in the limit, it goes
- to infinity. That may seem odd, but remember that anything involving
- an infinite grid is already faintly ill-defined.
-
- Ben
-