The 2 Ohm Problem
The most common electrical resistors commercially available are a series of
twelve basic resistance values, multiplied by arbitrary powers of ten. 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....
The E12 series extends down into fractional values, i.e., 8.2, 6.8, 5.6, and
so on, but we will not consider these. There also exists an "E24 series" that
provides 24 resistances per decade, but those are less widely available.
Noting that the total resistance of a set of resistors R1, R2, ..., Rn
combined in parallel is
1 1 1 1
---- = ---- + ---- + ... + ----
Rtot R1 R2 Rn
we might ask which integer-valued resistances can be produced by combining
distinct integer-valued resistances in parallel from the standard E12 series.
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 it's doubtful that the E12 values were chosen with
prime factors in mind. According to the commercial literature, 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, although this doesn't match all the values.
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. 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 we can multiply any of them by any
power of 10. For example, since 1/99 = 1/180 + 1/220 it's obvious that
1/990 = 1/1800 + 1/2200, 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].
To create other fundamental solutions we must allow more than two resistors
in parallel. Here's a summary of the "most economical" solutions I've found
for the first several 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. For example, is the above expression
for 14 ohms optimal, or can 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 (greater than 1) that is composed entirely of E12 primes?
It's easy to see that we 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. This is the "2 Ohm Problem", i.e., find a combination of distinct
E12 resistors that give exactly 2 ohms when connected in parallel.
One approach to finding such an expression is to first consider each individual
basic resistor, multiplied by powers of 10, focusing on eliminating the large
primes from the denominators of these individual sums. For the basic resistors
22, 33, 39, 47, 58, 68, and 82 the smallest sums that eliminate the primes
greater than 5 from the denominators are
1 1 1
-- + --- = --
22 220 20
1 1 1
-- + --- = --
33 330 30
1 1 77
-- + + ----- = ----
39 39000 3000
1 1 1 213
-- + ----- + ------ = -----
47 47000 470000 10000
1 1 143
-- + ----- = ----
56 56000 8000
1 1 1 1 653
-- + --- + ---- + ------ = -----
68 680 6800 680000 40000
1 1 1 1 1 271
-- + --- + ---- + ----- + ------ = -----
82 820 8200 82000 820000 20000
The sum of the seven series listed above is 3561/20000, which differs from 1/2 by
6439/20000. Thus we seek sums k1 to k5 of reciprocal powers of 10 such that
/ 1 \ / 1 \ / 1 \ / 1 \ / 1 \ 6439
( --- )k1 + ( --- )k2 + ( --- )k3 + ( --- )k4 + ( --- )k5 = -----
\ 10/ \ 12/ \ 18/ \ 15/ \ 27/ 20000
Since there are only 32 distinct sums of reciprocal powers of 10 up to 10^4, it
is feasible to make an exhaustive search. The first solution we find is
1 1 1 1 1 1
k1 = 1 + ---- + ---- + ---- k2 = k3 = 1 + ---- + ---- k4 = 1 + ---- k5 = 0
10^2 10^3 10^4 10^1 10^2 10^4
Thus we can produce 2 ohms by combining the following thirty-two integer resistances
from the E12 series:
10 1000 10000 100000
12 120 1200
15 150000
18 180 1800
22 220
33 330
39 39000
47 47000 470000
56 56000
68 680 6800 680000
82 820 8200 82000 820000
This solution makes no use of any resistors from the "27" sequence. There are, of
course, many other solutions. Consider, for example, the expression
1.1111 1.1101 1.001 1.0011 1.001 1.1 1.1 1.0111 1.0111 114653
------ + ------ + ----- + ------ + ----- + --- + --- + ------ + ------ = ------
82 68 56 47 39 33 22 18 12 360000
To complete a "2 ohm" solution based on this, we need to find series using the
resistances 10, 15, and 27 that sum to Q = 65347/360000. Each of the following
expressions equals this quantity, so each of them (along with the expression above)
gives a distinct solution of the 2 ohm problem:
Q = 1.10001/10 + 1.01111/15 + 0.11100/27
= 1.10101/10 + 1.01011/15 + 0.11010/27
= 1.10111/10 + 1.01001/15 + 0.11001/27
= 1.11101/10 + 1.00011/15 + 0.10110/27
= 1.11111/10 + 1.00001/15 + 0.10101/27
One peculiar feature of these solutions can be seen if we interpret the numerators
of these terms as binary numbers. For example, the numerator of 10 in the first of
these expressions is 1.10001, which we interpret as the binary number 110001 = 49.
Similarly we determine the base 2 corresponding to each of the other numerators,
adding a zero to each numerator of 27 so that it has the same number of digits
as the other numerators. This gives the results
49 47 56
53 43 52
55 41 50
61 35 44
63 33 42
The differences between consecutive numbers in the first column are 4, 2, 6, 2,
and these are the negatives of the differences of consecutive numbers in the
second and third columns. If we reverse the digits of the numerators, but still
interpret them as binary numbers, we have the values
35 61 14
43 53 22
59 37 38
47 49 26
63 33 42
Now the differences between consecutive numbers in the first column are 8, 16, -12,
16, and these are the same as the differences of the third column, and the negatives
of the differences of the second column.
We can simplify the situation by removing all the terms from the numerators
that are common to all rows, leading to the denominators for five equal
expressions (swapping the third and fourth sets to put them in numerical order)
.0000 .1110 .1100
.0100 .1010 .1010
.0110 .1000 .1001
.1100 .0010 .0110
.1110 .0000 .0101
Taking these as binary digits in forward order or reverse order, we get the
sequences
forward reverse
0 14 12 0 7 3
4 10 10 2 5 5
6 8 9 6 1 9
12 2 6 3 4 6
14 0 5 7 0 10
In the forward case, the differences between consequtive numbers in the first
column are equal and opposite the differences in the second, which are twice
the differences of the third column, whereas in the reverse case the differences
in the first and third columns are equal and opposite the differences in the
second column.
The forward case occurs because the component-wise difference between any two
sets of numerators, interpreted as decimal numbers, is proportional to 1, -1,
-0.9, which implies that each set gives the same quantity when divided by their
respective denominators, because of the basic equality
1 1 / 1 1 \ 1 1 0.9
--- - --- - ( --- - --- ) = --- - --- - --- = 0
10 15 \ 27 270/ 10 15 27
The decimal differences between the numerators of consecutive sets (multiplied
by 10000) are shown below.
.0000 .1110 .1100
100 -100 -90
.0100 .1010 .1010
10 -10 -9
.0110 .1000 .1001
990 -990 -891
.1100 .0010 .0110
10 -10 -9
.1110 .0000 .0101
On the other hand, when we interpret the numbers in binary, the decimal
difference given by 1 - .1 = .9 becomes 1 - .1 = .1 (which is 1/2), so the
differences are proportional to 1, -1, -0.5 instead of 1, -1, -0.9. It would
be interesting to know if there exist larger sets of numbers with this same
property, i.e., the component-wise differences are in fixed proportions in
two different bases.
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