The 2 Ohm Problem
Someone recently asked about combining resistors in parallel, noting
that 1/Rtotal = 1/R1 + 1/R2 + ... Also, commercially available
resistors typically come in 12 standard resistances over each power
of 10. This is called the "E12 series", consisting of resistors with
the following values (ohms):
10 12 15 18 22 27 33 39 47 56 68 82
100 120 150 180 220 270 330 390 470 560 680 820
1000 1200 1500 1800 2200 2700 3300 3900 4700 5600 6800 8200
etc....
Actually there is an "E24 series" that provides 24 resistances
per decade, but those are less widely available. Also, the above
sequence extends down into fractional values, i.e., 8.2, 6.8, 5.6,
and so on, but for whatever reason the person wanted to stay with
just the families shown above, with 10 ohms being the minimum
individual resistor. In general, we seek to determine the ways
of combining distinct E12 resistors in parallel so as to yield an
integer overall resistance. The restriction to just these particular
E12 values adds some interest to the problem.
Of course, if 1/R equals a sum of unit fractions then R must divide
the product of the denominators, which implies that R cannot be
synthesized by means of 2 or more distinct E12 resistors in parallel
if R is divisible by any prime other than those that appear in the
E12 sequence, namely, 2, 3, 5, 7, 11, 13, 17, 41, and 43. (It's
interesting that this covers the first seven primes, even though
I don't think the E12 values were chosen with prime factors in
mind. My sources claim that the E24 values were defined as the
closest integers to 10r^k, k=0,1,2,..,23, where r is the 24th
root of 10, but this doesn't match all the values, so it isn't
clear to me what criterion was used.)
On the other hand, even if R is entirely composed of "E12 primes" it
may not be possible to synthesize it by just TWO resistors in parallel.
It may require three or more. I believe the only "fundamental"
integer values of R that are expressible as a sum of just two unit
fractions with E12 denominators are those shown below:
1/R = 1/A + 1/B
R A B
----- ------ -------
6 10 15
20 22 220
30 33 330
72 120 180
88 120 330
99 180 220
102 120 680
108 180 270
132 220 330
975 1000 39000
1125 1200 18000
1476 1800 8200
1485 2700 3300
6875 10000 22000
9375 10000 150000
16875 18000 270000
17875 33000 39000
103125 150000 330000
193875 330000 470000
I call these "fundamental" because you can multiply any of these
by any power of 10. For example, since 1/99 = 1/180 + 1/220 it's
obvious that 1/9900 = 1/18000 + 1/22000, and so on. However,
although the above solutions are fundamental in this sense, they
are not all "primitive", because they may share common E12 factors.
For example, the solution [103125,150000,330000] is just 15 times
the solution [6875,10000, 22000].
Anyway, to create other fundamental solutions you have to allow
more terms, i.e., more resistors in parallel. Here's a summary
of the "most economical" solutions I've found for the first few
values of R:
R denominators
--- -------------------------------------------------------------
3 10 12 15 22 33 220 330
4 10 12 15
5 10 12 100 150
6 10 15
7 10 56 100 120 150
8 10 100 120 150
9 18 22 180 220
10 10
11 12 220 330
12 12
13 15 100 3900
14 33 100 120 150 180 220 330 1000 1500 1800 2200 5600 12000 18000
15 15
16 18 180 1200 1800
17 33 68 120 330 680 1200 6800
18 18
19 impossible
20 22 220
21 33 100 330 560 1000 1200 1500
22 22
23 impossible
24 33 120 330
25 33 150 330
26 39 100 390 3900
27 27
etc.
This raises some interesting questions. First, as suggested in the
title of this post, there seems to be a "2 ohm problem". It's easy
to see that you can't synthesize 1 ohm from any number of distinct
E12 resistors in parallel, because the infinite sum of the inverses
of all these resistors implies that the least resistance that can
be produced by such a network is
31873589748
----------- ohms = 1.764 ohms
18062619415
On the other hand, this doesn't rule out the possibility that 2 ohms
could be produced. However, I haven't been able to see how to do
it, nor to prove that it's impossible. So that is the "2 Ohm Problem",
i.e., find a combination of distinct E12 resistors that give exactly
2 ohms when connected in parallel.
Another question is whether my synthesis for 14 ohms is optimal,
or if it be done with fewer than 14 resistors.
In general, is it possible to express 1/R as a sum of inverses
of distinct "E12 integers" for any R that's composed entirely of
E12 primes? Is there a succinct characterization of the integer
resistance values that can be synthesized by a parallel arrangement
of distinct E12 resistors?
Peter Montgomery used Maple to perform exact rational arithmetic,
and found one solution for 2 ohms. If we define the resistance
decade scale factors
r0 = 1 r4 = 1/10000
r1 = 1/10 r5 = 1/100000
r2 = 1/100 r6 = 1/1000000
r3 = 1/1000
then we have the 2-ohm solution
1/2 = 1/10 + 1/12 + 1/15 + 1/18 + 1/100 + 1/1000 + 1/1500
+ (r3 + r4 + r5)/27
+ (r0 + r1 + r3 + r6)*(1/22 + 1/33)
+ (r0 + r1 + r2 + r3 + r4 + r5)/39
+ (r0 + r3 + r4)/47
+ (r0 + r1 + r3 + r4)/56
+ (r0 + r1 + r2 + r4)/68
+ (r0 + r1 + r2 + r3 + r4)/82;
This was found by picking the bottom five lines, to use some of the
available resistors while eliminating the primes 13, 47, 7, 17, 41
from the denominators. For example, 10011 = 47 * 213 is divisible
by 47, so 1/47 + 1/47000 + 1/470000 = 213/10000 has denominator
coprime to 47. Then remove powers of 2 and 5 from the denominator
of the sum. Finally, express 7/600 as a sum of reciprocals, which
gives 1/100 + 1/1000 + 1/1500.
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