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- Newsgroups: sci.physics
- Path: sparky!uunet!cs.utexas.edu!tamsun.tamu.edu!zeus.tamu.edu!dwr2560
- From: dwr2560@zeus.tamu.edu (RING, DAVID WAYNE)
- Subject: Re: ... an infinite mesh of 1ohm resistors ...
- Message-ID: <30JUL199217365829@zeus.tamu.edu>
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- Sender: news@tamsun.tamu.edu (Read News)
- Organization: Texas A&M University, Academic Computing Services
- References: <1992Jul27.210947.5820@fs7.ece.cmu.edu> <1992Jul28.000844.27051@mixcom.com> <1992Jul28.141407.17816@galois.mit.edu> <24327@castle.ed.ac.uk>
- Date: Thu, 30 Jul 1992 22:36:00 GMT
- Lines: 27
-
- graeme@castle.ed.ac.uk (G J Ackland) writes...
- >This source / sink argument is all very nice.
- >Doesn't it give the wrong answer for a Bethe lattice though?
- >
- >__/ \__
- > \__/
- >__/ \__/
- > \ / \
- >
- >where the absence of cicuits mean the solution is trivially 1 ohm, but
- >the three-way split of the current appears to give 2/3ohm using this
- >source/sink argument.
-
- Well, it depends how you measure current. Do you put your ammeter in series
- with the central 1ohm resistor, or in series with the power source? In the
- former case you would get (Vin-Vout)/1ohm. In the latter case you get
- (2Vin-Vout)/1ohm assuming the ammeter is in series with the Vin terminal.
- There is more current here because some of it goes to infinity.
-
- So for example, if Vin = +1/2 and Vout = -1/2 then 3/2 amps flow in,
- 3/2 amps flow out, and the central resistor carries 1amp. If Vin = 1,
- Vout = 0, then 2amps go in, 1amp comes out, and the central resistor
- carries 1amp. Voltages are given relative to the voltage at infinity of
- course.
-
- Dave Ring
- dwr2560@zeus.tamu.edu
-