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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\$ By knowing the formulae? \$\endgroup\$ – Bradman175 Jan 6 '17 at 7:31
  • \$\begingroup\$ What is the answer? @Bradman175 \$\endgroup\$ – salsabil.raisa Jan 6 '17 at 7:32
  • \$\begingroup\$ Read my answer. \$\endgroup\$ – Bradman175 Jan 6 '17 at 7:43
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    \$\begingroup\$ You should ask this question on Mathematical Forum. I am afraid it will boils down to some theory of approximation of convex manifolds, or something of this sort. \$\endgroup\$ – Ale..chenski Jan 6 '17 at 8:05
  • \$\begingroup\$ @DaveTweed-- the Euclidean Algorithm will give you an answer. The request was for a minimal answer \$\endgroup\$ – Scott Seidman Jan 6 '17 at 14:24
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I never faced this particular question before. But faced with it now, I'd start out (after mentally verifying that a bridge arrangement would be pointless -- and it is) by asking myself what, in parallel with \$8\:\Omega\$ would yield \$5\:\Omega\$?

It turns out that the answer is \$13\frac{1}{3}\:\Omega\$.

Then I'd wonder about what might make that. Well, if I had \$8\:\Omega\$ already, then I'd need another \$5\frac{1}{3}\:\Omega\$ to get that total. Well, luckily \$16\:\Omega\vert\vert 8\:\Omega\$ makes that.

So that's my answer. It's not a general purpose algorithm to get from \$X\$ to \$Y\$, exactly. But it's a thought process to find an answer here. And it suggests an algorithm (discussed below.)

The answer I'd give is:

$$\left(\left(\left(8\:\Omega+8\:\Omega\right)~\vert\vert~8\:\Omega\right)+8\:\Omega\right)\vert\vert ~8\:\Omega$$

or,

schematic

simulate this circuit – Schematic created using CircuitLab

The algorithm might be to realize that 8 and 5 are relatively prime and that it is likely that you'll need to reach their product, \$8\cdot 5=40\$, in order to find an answer. Intuitively, this makes some sense thinking that you are probably looking for a least common multiple of the two values. So it does strongly suggest that 5 resistors will be the minimum you can use here.

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  • \$\begingroup\$ I did not understand the algorithm part. \$\endgroup\$ – salsabil.raisa Jan 6 '17 at 8:47
  • \$\begingroup\$ @Mathematics It comes from the idea of conductance, which is \$\frac{1}{R}\$. You are combining such fractions. So you need to look for the LCM (I think.) However, I didn't actually develop an algorithm there. Just pointed in a direction where I think one might be found. \$\endgroup\$ – jonk Jan 6 '17 at 8:48
  • \$\begingroup\$ Good answer. A more formal description of the algorithm can be found here. \$\endgroup\$ – Dave Tweed Jan 6 '17 at 12:28
  • \$\begingroup\$ @DaveTweed Yes, I was thinking about the use of continued fractions, in fact. Nice to see it discussed by you. \$\endgroup\$ – jonk Jan 6 '17 at 19:27
  • \$\begingroup\$ +1 to your bank. What about more convoluted interconnect topologies? Like 3D cubes, with horizontal cross-links? \$\endgroup\$ – Ale..chenski Jan 6 '17 at 23:38
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You can find it mathematically. It only works for rational numbers though.

Note that this does not mean it's the solution with the least amount of resistors required.

You first need to find the highest number that when multiplied by any positive integer, it can equal both numbers.

So far for 8 and 5, it's only 1.

For 8 to reach 1. You need 8/1 which is 8.

Thus you put 8 resistors in parallel.

Now you have this.

enter image description here

Then you need to put these jumble of resistors in series to add to the amount of resistance you want. Since each jumble of resistors is 1, 5/1 is 5 so we need 5 jumbles of them.

Now you have this abomination.

enter image description here

Congrats you got your desired resistance... now count them up yourself.

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  • \$\begingroup\$ Obviously your solution is not optimal, since one simple answer is "10", not 40. The other answer is "13". \$\endgroup\$ – Ale..chenski Jan 6 '17 at 8:01
  • \$\begingroup\$ Using your algorithm it is obvious that the problem has infinite number of solutions. So it is quite natural to ask for a minimum. \$\endgroup\$ – Ale..chenski Jan 6 '17 at 8:08
  • \$\begingroup\$ What if the question asked the least possible number of resistors? \$\endgroup\$ – salsabil.raisa Jan 6 '17 at 8:27
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    \$\begingroup\$ @Mathematics Then you'd better ask that on a math site, as Ali suggested. Because the most reasonable solution an electronic engineer can come up with is: "go buy a 5 ohm resistor, for christ sake!". \$\endgroup\$ – dim Jan 6 '17 at 8:35
  • \$\begingroup\$ It is possible to use less resistors. If you made 1 Ω from 8 parallel resistors, you need to add 4 Ω to get 5 Ω. It is easy to get 4 Ω, just two 8 Ω resistors in parallel. That is one solution using 10 resistors, but a solution with only 5 resitros was shown here too. \$\endgroup\$ – Uwe Jan 6 '17 at 13:57
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It's possible but I cannot tell you with the fewest resistors possible.

16 ohm parallel to 8 ohm would get you close at 5,3 ohm. \$5,3 = R||2R\$. Depending on the tolerance it can already be a good answer.

Series: \$Rt = R1 + R2\$

Parallel: \$Rt = \frac{R1 \cdot R2}{R1 + R2}\$ also written as \$Rt = R1||R2\$

The easiest and exactly 5 is \$Rt = R||R + R||R||R||R||R||R||R||R\$ where R = 8 ohm. 2R parallel gives R/2. 8R parallel gives R/8. In this case a total of 5 ohm.

I would write a program to check the first 100k combinations of parallel and series resistors.

My best with trial and error is. \$5,05 = R||3R||4R\$

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