That's right Bill !! I have never heard this mentioned anywhere and it just struck me one day. I was wondering if anyone else had noticed the coincidence. Good thinking there and keen perception as well ! How long have you known this ??
Donny; I couldn't begin to draw this so I'll have to trust my writing skills....
Three resistors in a straight line, R-1, R-2 and R-3 in series between points A and B. All are 300 ohms. Over the resistors, there is a jumper between point A and the junction of R-2 and R-3 which appears to short out R-1 & R-2. Below the resistors, there is a jumper between the junction of R-1 and R-2 to point B which appears to short out R-2 & R-3.
What is the equivilant resistance of RAB equal to ??
I worked at Tandy Corp. in the mid-90's where I coded 4 and 8 bit microprocessors for answering machines, telephones, etc. In those days we used 7-segment LEDs or LCDs for most displays, and rather than a 7-seg display driver we typically used I/O from the chip and coded our own character set. We always had less pins than we wanted to do the display, key scan, etc. (since less GPI/O = cheaper processors), so I ended up spending a lot of time figuring how to 'light the right segments', make 9k of code fit in 8k ROM, not miss any interrupts, all that stuff.
Somewhere at sometime in that painful process this tidbit of trivia surfaced...
Paul...your answer to my riddle...not quite that long! Try again.
In answer to your riddle, 100 ohms. The circuit you describe is actually three 300 ohm resistors in parallel...diagramed to look more complex. <FONT SIZE=1 COLOR="#8e236b"><p align=CENTER>[This message was edited by Donny Hinson on 10 August 2002 at 03:37 AM.]</p></FONT>
The answer is 300 ohms. Taking the path of least resistance current would flow from point A through the jumper located between
R2 and R3 (0 ohms) then through R3 (300 ohms)
to point B. It's still only 9 am what do I do for the rest of the day.
Bob; I'm afraid Donny saw through my smoke screen. Point A looks straight into R-1 and then via the lower jumper, gets to point B.
Via the upper jumper, point A at the junction of R-2 & R-3, looks to the left into R-2 and again throught the lower jumper to point B. To the right it sees R-3 and then directly to point B. So each time it sees a 300 ohm resistor and 300/3=100.
Donny; I must have been sleepy last night but maybe I rushed your riddle a bit. Give me a couple days to see if I can figure some easier solution out......
Closer, Paul, but no cigar! Actually, Marty is right on track, here. Yes, this (2002) is one of those upside-down years, and probably none of us will be around when it next happens...in 2112.
What is it they used to say, Donny ?? Can't see the forrest for the trees. <FONT SIZE=1 COLOR="#8e236b"><p align=CENTER>[This message was edited by Paul Graupp on 11 August 2002 at 10:45 AM.]</p></FONT>
And here I was lookin' up Norton & Thevenin again...
Multiply 12345679 x 8 on yer calculator & see whatcha get.<FONT SIZE=1 COLOR="#8e236b"><p align=CENTER>[This message was edited by Ray Minich on 05 August 2004 at 11:27 AM.]</p></FONT>