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!!!

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).]