Kenbo; just a lucky guess!

Actually, I worked it out while roofing a dormer today and the sun must have addled my brains, ie. 1/3 + 1/3 = 1/2 . DOH!
Still, only 1/6 of an ohm out!!! [Linked Image]

This is how I figured it:

Let each corner be designated a letter,
A,B,C,D,E,F,G,H
Each resistor can then be identified by a pair of letters ie. 'AD'

From point A we get, [from my layout]
AD+AH+AE, in parallel, = 1/3 ohm
Call this the first leg.

PLUS:
HC+CB =2ohm parallel 2 ways = 1 ohm
HG+GB =2ohm

EF+FB =2ohm parallel 2 ways = 1 ohm
EG+GB =2ohm

DC+CB =2ohm parallel 2 ways = 1 ohm
DF+FB =2ohm

But these 3 routes are also in parallel to complete the 2nd leg;

Total for second leg = 1/3 ohm

1/3 + 1/3 = 2/3 ohm total resistance

Alan, Class Dunce of 1948!

ps. I just spotted I got 15 resistors, [some used twice], so I give in!






[This message has been edited by Alan Belson (edited 09-20-2006).]

[This message has been edited by Alan Belson (edited 09-20-2006).]

Wood work but can't!