I recently had to take test in basic electronics. I didn't get one question right, but I don't quite understand why.

How many 120Ω resistors are at minimum required to get a resistance of 80Ω?

The possible answers to this question are 2, 3, 4 and 6. The only answer I can come up with is 6, with the resistors arranged as seen bellow. But 6 isn't the correct answer.


How many resistors are required and to arrange them?


simulate this circuit – Schematic created using CircuitLab

I only know the very basics of electronics, so I hope my thoughts are correct.

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    \$\begingroup\$ @Autistic would 120 & 120 in paralell not be 60? \$\endgroup\$ – Marius Schär Nov 21 '15 at 22:27
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    \$\begingroup\$ maybe Autistic is being Artistic \$\endgroup\$ – Marla Nov 21 '15 at 22:31
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    \$\begingroup\$ The number is three. Deducing the combination is left as an exercise to the reader... but there are only so many possibilities. \$\endgroup\$ – Chris Stratton Nov 21 '15 at 22:31
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    \$\begingroup\$ This it the type of problem that can defeat us all. Sometimes the simplest solution sits in front of us. I encourage questions like this. I truly enjoy watching in an interview, this type of question. Martin, don't feel bad . . I myself have been lost on this type. We get locked in by our own limits \$\endgroup\$ – Marla Nov 21 '15 at 22:41
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    \$\begingroup\$ I was meaning 120 in paralell with 2 series 120 ohm resistors . \$\endgroup\$ – Autistic Nov 22 '15 at 4:04

120 || (120 + 120) If two 120 in parallel give 60, you want one of the branches to be a little higher, so... that's the next thing to try.

And the method is true in general for getting a 2/3-valued resistor using just a bin of the same kind. And in general for solving problems like this, it's worth remember that the equivalent resistance of two parallel resistors is less than that of either of the branches. You can also get 3/4 (so 90) for example by adding one more to a branch.

N.B.: Thanks to Massimo Ortolano's paper, now I know that what I've done above following just intuition is that I basically followed the search path indicated below in the Stern–Brocot tree:

enter image description here

  • \$\begingroup\$ Wow, thanks for that! It'd be really useful if they taught this simple method in the class.. \$\endgroup\$ – Marius Schär Nov 21 '15 at 22:58
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    \$\begingroup\$ The point of education is often to trigger discovery, not to simply tell you things. \$\endgroup\$ – Chris Stratton Nov 21 '15 at 23:03
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    \$\begingroup\$ Also codegolf.stackexchange.com/questions/20438/… \$\endgroup\$ – Fizz Nov 23 '15 at 0:36

A direct solution can be found through the application of continued fractions.

If what you have is 120Ω and what you want is 80Ω, write down the fraction:

$$\frac{80\Omega}{120\Omega} = 0.6667$$

Since the integer part is zero, you'll be starting by putting resistors in parallel. Invert the fractional part:

$$\frac{1}{0.6667} = 1.5$$

This tells you that you'll have 1 resistor in parallel with some number of resistors in series. Invert the fractional part again:

$$\frac{1}{0.5} = 2.0$$

This tells you that you need 2 resistors in series. Since there is no fractional part at this point, you're done.

The answer is a total of 3 resistors.

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    \$\begingroup\$ Resistor combinations by continued fractions.... neat. \$\endgroup\$ – Jasen Nov 22 '15 at 1:41
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    \$\begingroup\$ Do you think this algorithm gives the minimum solution [in numbers of resistors] in general? Looks like there's a recent paper about the topic, but it seem to be an education-oriented review. Can't see mention of minimality. \$\endgroup\$ – Fizz Nov 22 '15 at 6:03
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    \$\begingroup\$ Also math.stackexchange.com/questions/14645/… Note that the accepted answer is actually incorrect! \$\endgroup\$ – Fizz Nov 22 '15 at 7:25
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    \$\begingroup\$ @RespawnedFluff: no, generally, it doesn't give a minimum solution. Using the continued fraction expansion yields a solution composed of parallel and series combinations only, but, in general, solutions with fewer resistors can be found by taking into account also bridge-connected resistors. It can be shown that, for planar networks, the problem is equivalent to that of filling rectangles with integer-sided squares. If then one considers also non planar networks, probably solutions with even fewer elements can be found. \$\endgroup\$ – Massimo Ortolano Nov 22 '15 at 18:37
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    \$\begingroup\$ For the purpose of [better] keyword finding, the solution that Dave indicated is based on the Stern–Brocot tree approximation of a real number. I found this out reading Massimo Ortolano's paper, which is also freely available on arxiv, by the way. \$\endgroup\$ – Fizz Nov 22 '15 at 21:22

You can change your solution by swapping serial and parallel:


simulate this circuit – Schematic created using CircuitLab

You can then group R2, R3, R5 and R6 into a single 2x2 group:


simulate this circuit

And those 4 \$120\Omega\$ resistors make a single \$120\Omega\$ resistor:


simulate this circuit

  • \$\begingroup\$ This is the same as what user92407 said 3 hours eariler, albeit with a diagram. \$\endgroup\$ – Dave Tweed Nov 22 '15 at 16:04
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    \$\begingroup\$ Nevertheless I find the addition useful; it is actually using the equivalent geometric tiling problem indicated by Massimo Ortolano. The four resistances that can be replaced form a [bigger] square. \$\endgroup\$ – Fizz Nov 22 '15 at 21:12

Take your solution but without a central point in the middle: you can rearrange this as three parallel sections of 120+120 Ohm each (connecting the middle points does not make a difference since they all are at the same voltage). Now two of the three parallel 120+120 Ohm sections combine into 120 Ohm again, so you can replace those 4 resistors from the two parallel groupings with a single one, leaving just one 120 Ohm resistor parallel to 120+120 Ohm.

There is a plethora of solutions proving the correctness of this solution once you have it. But this rearrangement shows how to find it without reverting to mathematical trial and error.

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    \$\begingroup\$ Actually it does involve trial and error [in general]. There is no known solution for the problem of minimally tiling a rectangle with integer squares that doesn't involve an exhaustive search. There are some heuristics though that prune the solution tree, but they don't guarantee a minimal solution. \$\endgroup\$ – Fizz Nov 23 '15 at 1:15

Elaborating on @RespawnedFluff's answer, one way to find this is to think in the following way:

  1. What resistors do I have, ok 120.
  2. What do I need to make, 80
  3. What equations do we know? Well the two resistors in series or parallel are the simplest starting points. Clearly series doesn't help immediately - that would increase the resistance, not reduce it. So we will need to try parallel. We know the equations:

$$\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}=\frac{R_1 + R_2}{R_1 R_2}$$

  1. So maybe lets start with that:

$$\begin{align} \frac{R_1 R_2}{R_1 + R_2} &= 80\\\\ 80R_1 + 80R_2 &= R_1 R_2\\\\ R_2 &= \frac{80R_1}{R_1-80} \end{align}$$

  1. So can you find any combination that suits? Well, start with \$R_1=120\$ and then see what value \$R_2\$ needs to be. Can you make that value easily? In this case yes, so great.

  2. For other values, if you can't get a value immediately, you may need to try either the same approach as above iteratively to find the value for \$R_2\$. If that fails to work, you could also try changing \$R_1\$ - maybe two in series or parallel, and then try again for \$R_2\$.

This approach is quite iterative, but in this case it would have quickly found both the answer you got (using 6 resistors), and also the answer @RespawnedFluff got (using 3 resistors).

If you were trying to grow the resistance (i.e. the required resistance is larger than your available value), you basically do the same thing, but start with a larger available resistance, or split the larger resistance up in series chunks and solve for them (e.g. if you wanted \$180\Omega\$, you could pick a chunk of \$120\Omega\$ and \$60\Omega\$).

You may wonder how the method would have gotten to your answer - given that yours has 3 parallel branches, whereas this approach uses two. Well, in calculating \$R_2\$ above, iteratively, you would introduce \$R_2\$ being a parallel branch, which topologically is the same as if there were 3 branches to begin with.

  • \$\begingroup\$ Corrected my answer. I'd totally messed up my explanation. \$\endgroup\$ – Tom Carpenter Nov 21 '15 at 23:27
  • \$\begingroup\$ If one resistor branch is fixed, this is easy to solve (or determine that there's no [integer] solution). I'm still not sure how to solve even with two branches, never mind in general. It's a more complicated diophantine equation. \$\endgroup\$ – Fizz Nov 22 '15 at 1:15
  • \$\begingroup\$ The problem is probably NP-complete as far enumeration goes: arxiv.org/ftp/arxiv/papers/1004/1004.3346.pdf \$\endgroup\$ – Fizz Nov 22 '15 at 9:05

Basic resistance in series and resistance in parallel logic. Very simple..

We know the output is 80Ω, so solving the problmm is easy. Try the formula for parallel resistance: $$\frac{1}{Rp} = \frac{R1+R2}{R1\cdot R2}$$ Therefore.. $$Rp = \frac{R1\cdot R2}{R1+R2}$$ Here put \$Rp = 80\Omega\$.

Now as we know we have resistors of only 120Ω. Put \$R1 = 120\Omega\$

Solving ... you will get \$R2 = 240\Omega\$.

But we cannot use a 240Ω resistor here as it's said that we have only 120Ω resistors. So instead of 240Ω, we will use 120Ω + 120Ω (in series) in parallel with a single 120Ω resistor.

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    \$\begingroup\$ This is the same thing that Tom Carpenter said 11 hours earlier. Let's try to avoid duplicating answers. \$\endgroup\$ – Dave Tweed Nov 22 '15 at 16:05

protected by W5VO Nov 23 '15 at 21:46

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