Re: A lost treasure (Series within Parallel resistor combinations.)

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On Wed, 18 Jul 2007 18:22:50 +0200, "Philippe 92"
<nospam@xxxxxxxxxxxx> wrote:

Quentin Grady wrote :
G'day G'day Folks,
...
Connecting R1 and R2 in series gives a total resistance of R1 + R2
Connecting R3 and R4 in series gives a total resistance of R3 + R4
Putting these series combination in parallel gives a combined
resistance of (R1 + R2)(R3 + R4)/(R1 + R2 + R3 + R4) = Rx (say)
If I connect the node between R1 & R2 to the node between R3 & R4
the circuit is transformed to a parallel within series connection.
The resistance is now Ry = (R1 x R3)/(R1 + R3) + (R2 x R4)/(R2 + R4)
The maths challenge this time is that I'd like, R1, R2, R3, R4, Rx,
and Ry to all be positive integers, preferably under 100.

I obtained an incredible answer from someone, most probably ksbrown in
which it was pointed out the advisability of adding some further
conditions

1. The solutions when the nodes were connected and left unconnected
should be different ie R1 x R4 should not equal R2 x R3.

2. Not only should Rx and Ry be integers but so should their
components. (R1R3)/(R1+R3) and (R2R4)/(R2+R4)

What I'd like is a table containing perhaps a dozen solutions.

I'd be quite happy if someone produced the answers using a brute force
computer method. While pure mathematical methods are more elegant I'm
no longer sure I could follow them.

Best wishes,

Hi,

may be this recreation from my Web site will give some usefull ideas :
http://chephip.free.fr/pba_en/pb036.html
It gives all possible integer combinations of two resistors in parallel
to get 30 Ohms, and the method to get them.

G'day G'day Phillipe,

Thank you. I was fascinated by the possibility of getting the
fourteen integer answers to getting 30 Ohms and seeing the method of
solution you used.

However, when I went to the URL provided I was unable to get the
solutions to appear. I'm not sure why this is. It could be inbuilt
protection against popups, Active X etc. and be peculiar to my
browser.

How do others find it?

The method can be adapted easily to any value, and repeating it, could
give the set of values your are searching.

a little program with that strategy gives many many solutions, for
instance :

r1=12, r2=21, r3=60, r4=28 : rx=24, ry=10+12=22
r1=12, r2=24, r3=60, r4=48 : rx=27, ry=10+16=26
r1=15, r2=20, r3=30, r4=180 : rx=30, ry=10+18=28
r1=12, r2=30, r3=60, r4=45 : rx=30, ry=10+18=28
r1=12, r2=36, r3=60, r4=36 : rx=32, ry=10+18=28
r1=12, r2=24, r3=60, r4=552 : rx=34, ry=10+23=33
r1=11, r2=33, r3=110, r4=88 : rx=36, ry=10+24=34
r1=15, r2=30, r3=30, r4=150 : rx=36, ry=10+25=35
r1=14, r2=28, r3=35, r4=364 : rx=38, ry=10+26=36
r1=20, r2=40, r3=20, r4=120 : rx=42, ry=10+30=40
r1=12, r2=48, r3=60, r4=80 : rx=42, ry=10+30=40
r1=11, r2=55, r3=110, r4=66 : rx=48, ry=10+30=40
r1=12, r2=60, r3=60, r4=60 : rx=45, ry=10+30=40
r1=20, r2=40, r3=20, r4=160 : rx=45, ry=10+32=42
r1=20, r2=40, r3=20, r4=280 : rx=50, ry=10+35=45
r1=12, r2=60, r3=60, r4=84 : rx=48, ry=10+35=45
r1=15, r2=45, r3=30, r4=360 : rx=52, ry=10+40=50
r1=15, r2=45, r3=30, r4=630 : rx=55, ry=10+42=52
r1=11, r2=55, r3=110, r4=220 : rx=55, ry=10+44=54
r1=11, r2=88, r3=110, r4=88 : rx=66, ry=10+44=54
r1=15, r2=90, r3=30, r4=90 : rx=56, ry=10+45=55
r1=20, r2=60, r3=20, r4=300 : rx=64, ry=10+50=60
r1=20, r2=60, r3=20, r4=540 : rx=70, ry=10+54=64
r1=12, r2=108, r3=60, r4=108 : rx=70, ry=10+54=64
r1=11, r2=88, r3=110, r4=154 : rx=72, ry=10+56=66
r1=12, r2=102, r3=60, r4=187 : rx=78, ry=10+66=76
r1=11, r2=99, r3=110, r4=264 : rx=85, ry=10+72=82
r1=20, r2=120, r3=20, r4=240 : rx=91, ry=10+80=90

I had to interrupt my program after about 70 pages of solutions...

Getting too many answers appears to be more the problem when one uses
a computing approach. Quite different from working without one.

In the 70 pages of solutions were there more that had all resistors
less than 100.


Regards.

Thank you once again. Over the years I've put together integer
solutions of resistors in parallel and with the help of others, three
resistors in parallel. Three resistors in parallel are extremely
useful for problems involving two meshes.

I find it fascinating that so many of the answers to today's problem
have an absolute difference between rx and ry of 1 or 2.

Aesthetically I prefer solutions that have none of the resistors equal
in value and rx and ry far apart.

Best wishes,

--
Quentin Grady ^ ^ /
New Zealand, >#,#< [
/ \ /\
"... and the blind dog was leading."

http://homepages.paradise.net.nz/quentin
.

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