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 ohmCall 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 ohm1/3 + 1/3 =
2/3 ohm total resistanceAlan, 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).]