Apologies, I see that my explanation of the puzzle could have been clearer. By a two-terminal network I mean that if a voltage is applied across the two terminals then a non-zero current flows through every resistor. So of course that disqualifies your attempted answer. [Note that it also disqualifies networks that contain balanced bridge configurations in which the central (redundant) resistor is connected between points of equipotential]. The puzzle is my own, of recent origin. It's a variation on the self-replicating resistor set theme. Funsters with a taste for electrical puzzles may also like to try this one: http://www.leesallows.com/index.php?page_menu=Varia&pagename=Electronic%20pu.... Lee At 04:48 AM 12/6/2014, you wrote:
Lee Sallows <Lee.Sal@inter.nl.net> wrote:
... I do have another in the form of a second resistor problem. Only this time I'm not going to reveal whether it is unsolved or not.
I have a set of 7 resistors, all distinct in value. You choose one. Following which I take the remaining 6 resistors and connect them together to form a two-terminal network whose total resistance is the same as that of the resistor you selected. I can do this whichever one you choose. What are these seven resistors?
1 = (p 2 3 6) 2 = (p 3 6) 3 = (s 1 (p 4 6 12)) 4 = (p 6 12) 5 = (s 1 2 (p 3 6)) 6 = (s 1 2 3) 12 = (s 1 2 3 6)
Resistors not mentioned are lying around loose in the two-terminal network. Or, if you prefer, only one side is soldered to anything, and it doesn't matter what.
Is that the same as your solution? Who first came up with this puzzle, and when? Thanks.
I solved it by hand, but not before wasting time trying to figure out how to write a program to loop over all possible networks of six or fewer resistors.
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