🚀 go-pugleaf

In article <2798@ucl-cs.uucp> O.MostynOwen@cs.ucl.ac.uk writes:
>
>There's a problem I was told about years ago which I never managed to solve:
>
>You've got an infinite 2D grid if 1 ohm resistors. What's the resistance
>across 1 of them (no it's not 1ohm, 0 or infinity).
>
>              |                  |                   |
>         ------------1ohm-----------------1ohm------------
>              |                  |                   |
>              |                  |                   |
> .....      1ohm               1ohm                1ohm     .....
>              |                  |                   |
>              | A                | B                 |
>         ------------1ohm-----------------1ohm------------
>              |                  |                   |
>              |                  |                   |
>>
>I'd be grateful if someone laid this one to rest since it keeps on coming
>back to haunt me.

It's a superposition trick.  Imagine attaching a current source
between points A and B, and forcing 1 ampere through the circuit.
There's a theorem (I forget the name) which says that this is
equivalent to the sum of two separate experiments of forcing 1 amp
into A and of sucking 1 amp out of B.  The first experiment would push
1/4 amp into the resistor from A to B, and so would the second
experiment!  So we find that 1/2 amp flows through the resistor, hence
the voltage between A and B is 1/2 volt.  Then because R = E/I, we get
the resistance must be 1/2 volt over 1 amp at the terminals, so the
resistance is 1/2 ohm.  Wonderful example of a sort of circular reasoning.