[rec.puzzles] Resistor Puzzle

awpaeth@watcgl.waterloo.edu (Alan Wm Paeth) (11/26/88)

Scenario: A resistor decade allows the selection of one through ten ohms.
Typical internals are ten one-ohm resistors and a 10-pole switch. A more part-
efficient design (ignoring switch complexity) might use four resistors of
binary weights (1,2,4,8) thereby extending the range to fifteen ohms.

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PUZZLE: using four resistors construct a "decade" good for (at least) the
range 1 to 16 ohms, inclusive. No restrictions on wiring or resistor values.
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Reminder: resistors in series sum values, eg: (4+4) = 8. Resistors in parallel
sum their inverse values (1/a+1/b=1/sum), eg: (4|4) = 2.

     /Alan Paeth
     Computer Graphics Laboratory
     University of Waterloo

awpaeth@watcgl.waterloo.edu (Alan Wm Paeth) (12/02/88)

In <6951@watcgl.waterloo.edu> awpaeth@watcgl.waterloo.edu (Alan Paeth) writes:
>-----------------------------------------------------------------------------
>PUZZLE: using four resistors construct a "decade" good for (at least) the
>range 1 to 16 ohms, inclusive. No restrictions on wiring or resistor values.
[make that a "double" decade, good from 0-19 ohms]
>-----------------------------------------------------------------------------

I know only one solution. One can nearly make the 1-10 ohm decade with just
three resistors, wired in the following fashion (+ is series, | is parallel):

resistors: 2, 3, 6.
wiring: 1. 2|3|6 2. 2 3. 3 4. 2+(3|6) 5. 2+3 6. 6 7. ? 8. 2+6 9. 3+6 11. 2+3+6

this leaves holes at 7 and 10 ohms. Adding a 7 ohm resistor to extend the range
causes the problem to crop up again at 14 ohms; adding a 10 ohm resistor widens
the range by 10+[1..6,8..9,11] or to 19 ohms continuous, sans seven. The gap at
seven ohms can be finessed as ((10+2)|6)+3 -- this is the most intricate net.

In general, the identities 2a|2a = a, 3|6 = 2, 6|12 = 4, 4|12 = 3 are helpful.

    /Alan Paeth
    Computer Graphics Laboratory
    University of Waterloo