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Is there a mathematical way to know the answer? (or you can do it only by trial and error) Could you prove that it is possible or impossible mathematically?

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    \$\begingroup\$ It is possible to arrange them to get 6 ohm. Make sure you combine some in parallel and some in series. \$\endgroup\$ – Lior Bilia Apr 3 '14 at 11:42
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    \$\begingroup\$ Just use one resistor & keep the other 5 as spares. \$\endgroup\$ – oconnor0 Apr 3 '14 at 17:27
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    \$\begingroup\$ Normally you do this to increase the power rating. In that respect it would be best to use 4 and keep 2 as spares. \$\endgroup\$ – starblue Apr 3 '14 at 20:05
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    \$\begingroup\$ Correct me if i'm wrong : if you want all the 6 resistors to carry current, there is only two solutions (given on this page). The rest is either solutions using 4 resistors (6+6)//(6+6) with 2 resistors "not used" (like Andy aka answer) or solutions using 1 resistor with 5 others are not used. I don't think there is other possibilities. \$\endgroup\$ – tigrou Apr 9 '14 at 7:36
  • \$\begingroup\$ only connect one of the six resistors in your circuit and save your money (in other words, don't buy a large quantity of the same resistor just to make a crude way to get that one value of resistance). \$\endgroup\$ – user116345 Sep 10 '16 at 5:47
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schematic

simulate this circuit – Schematic created using CircuitLab

here R5//R1 series to R3 => 3 + 6 = 9 in one branch

R4 + R6 + R2 => 6 + 6 + 6 = 18 in 2nd branch

18 // 9 gives 6

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    \$\begingroup\$ To get this circuit think of a square of 9 resistors and collapse the square in the lower left corner into a single resistor. \$\endgroup\$ – starblue Apr 3 '14 at 20:01
  • \$\begingroup\$ @starblue can you make it more clear ? \$\endgroup\$ – tollin jose Apr 4 '14 at 11:59
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    \$\begingroup\$ If you arrange resistors in a square you get the same resistance value again, because n times parallel divides the resistance by n and n times in series multiplies by n. It doesn't matter whether you first connect in series or parallel, i.e. you may choose to connect nodes of the same potential or not, without changing the resistance value. In your example R3 could be expanded to a 2x2 square, then you would get a 3x3 square overall. You could then make it regular by adding connections. \$\endgroup\$ – starblue Apr 4 '14 at 14:27
  • \$\begingroup\$ ok you meant it is possible to make 6 ohm resistance using 9 six-ohm resistors. \$\endgroup\$ – tollin jose Apr 5 '14 at 3:01
  • \$\begingroup\$ He meant any square of identical resistors produce resistance identical to each resistor in the square. Thus by collapsing or expending squares you can avoid doing any calculations while you search for the resistor count you want. Doesn't really provide a rigorous algorithm for proving what resister counts would be impossible but it provides an elegant way to simplify the trial and error. It means needing to use 1 is the same as needing to use 4 or 9 or 16 ... \$\endgroup\$ – candied_orange May 20 '14 at 6:13
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Arrange 5 in your pocket, connect up one.

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    \$\begingroup\$ Here's a schematic. i.imgur.com/jypv2Ck.png \$\endgroup\$ – Vortico Apr 3 '14 at 20:02
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    \$\begingroup\$ @Vortico That's an antenna, might not be suitable for some designs. \$\endgroup\$ – Adam Davis Apr 5 '14 at 13:58
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What about these. Are they eligible or just cheats?: -

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ That is actually the same solution twice, you just positioned the resistors a bit different. Cheat or not, if all the resistors are identical your solution will take more current before burning than tollin's, despite two of the resistors not really doing anything here. \$\endgroup\$ – aaaaaaaaaaaa Apr 3 '14 at 18:20
  • \$\begingroup\$ @eBusiness muhuhahaha you foiled my cunning plan! \$\endgroup\$ – Andy aka Apr 3 '14 at 20:30
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    \$\begingroup\$ +1 This is the kind of circuit that would make you feel really bad when it's marked "wrong", because it probably satisfies the original problem statement perfectly. \$\endgroup\$ – Spehro Pefhany Apr 3 '14 at 20:43
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    \$\begingroup\$ Since R14 and R15 conduct no current, you can remove them from the circuit. And give them to me. \$\endgroup\$ – markrages Apr 4 '14 at 2:37
  • \$\begingroup\$ @markrages they are precision 100 watt wirewounds - too expensive to give away and what about postal charges LOL \$\endgroup\$ – Andy aka Apr 4 '14 at 7:44
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It is possible to arrange all possible topologies and calculate the resistance of each. Nice idea for programming homework.

Proving that something is possible requires only one example. In your case: one resistor between the two poles, all other resistors unconnected (or connected to one pole, etc).

Proving that something is impossible requires an ad-hoc proof or enumerating all possible topologies.

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  • \$\begingroup\$ Your proof that it's possible makes an assumption that they do not all need to be connected. A probably false assumption, since I doubt the OP is completely stupid. \$\endgroup\$ – OJFord Apr 5 '14 at 23:04
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    \$\begingroup\$ No such requirement was mentioned, hence the assumption that such a requirement exist seems more far fetched than assuming that the question is complete. And what exactly is connected? As I suggested, the remaining resistors could all be connected (with both leads) to one of the poles. \$\endgroup\$ – Wouter van Ooijen Apr 6 '14 at 7:12
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Another possibility would be:

(6//6//6) + 6//(6+6) = 2 + 6//12 = 2 + 4 = 6

schematic

simulate this circuit – Schematic created using CircuitLab

BTW, I did note that you're after a mathematical solution, but since I couldn't think of one, I offered this. It would certainly be possible to solve it algorithmically, with iterations, but a single mathematical solution may not be possible? Very interesting question.

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This problem is under constrained.. what does 'arranged' mean? Can you use one or four in series-parallel and short the left-over resistors?

It's not possible to have them share power equally, however it is possible to actively use all the resistors. Hint: calculate 1/( 1/9 + 1/18 )

If there is a straightforward mathematical way, I'm not aware of it.

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This appears to be related to:

https://mathoverflow.net/questions/66853/number-of-graphs-with-n-edges

which leads to just twelve graphs for six edges - quite a suprise to me. You will then need to measure n! node pairs.

Oh - I quickly came up with the 'leave 5 unconnected' (a definite cheat) and bridge (not a cheat) circuits. Kudos to the answers where all resistors carry current.

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  • \$\begingroup\$ Should that be $a(6) = 30$? (no mathjax here???) \$\endgroup\$ – copper.hat Apr 4 '14 at 22:33
  • \$\begingroup\$ @copper.hat use \$ for inline math, $$ sets it apart from text. \$inline\$ $$not inline$$ \$\endgroup\$ – OJFord Apr 5 '14 at 23:06

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