Re: [Phys-l] resistors & nerd sniping - elegant

[Brian Whatcott <betwys1@sbcglobal.net>]

At 09:40 AM 3/17/2008, Dan M., you wrote this:

> resistors & nerd sniping
>
> check out
> http://xkcd.com/356/
>
> and hints for elegant solutions are welcome (no, I don't have one).
>
> Dan M

Elegant? You want elegant? Take a look at this spare analysis from
Uberpuzzler for the adjacent intersection case as proposed on Wu's Puzzle list

*************************
Infinite Resistor Network  « on: Nov 21st, 2002, 2:55am »
Imagine an infinitely extending grid of squares.
 Between every two adjacent intersection points is a resistor of 1 ohm.
 Compute the resistance between two adjacent intersection points.

[ wu ] : <http://www.ocf.berkeley.edu/~wwu>http://www.ocf.berkeley.edu/~wwu


 ****************************
Uberpuzzler responded like this...

Superimpose these two ways of applying current to this circuit:
Case 1:  Force 1 Amp into the circuit at point A while holding 
voltage zero at infinity.
 By symmetry, 0.25 Amp flows through the resistor between A and B
(going from A toward B).

Case 2: Take 1 Amp out of circuit at point B while holding voltage 
zero at infinity.
  By symmetry, 0.25 Amp flows through the resistor between A and B
(going from A toward B).

Superimposing these two situations gives 0.5 Amp flowing through the resistor
between A and B for the case of 1 Amp into the circuit at A and 1 Amp out of
the circuit at B.
 So voltage difference between A and B is 0.5 Volts =0.5 Amp* 1 Ohm.
Since 1 Amp flows through the whole circuit, the equivalent resistance
is 0.5 Volts/1 Amp= 0.5 Ohm.

 ******************************************

 Taken from this archive.
<http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi?board=riddles_hard;action=display;num=1037876137>

However, if you apply the same technique carelessly to the adjacent 
diagonal resistance,
you come up with 1/3 ohm rather than 1/pi ohm.


Brian Whatcott    Altus OK    Eureka!