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Series/Parallel of 10 resistors question

E

ehsjr

Jan 1, 1970
0
Jon said:
But the strength of my example was built on the approach used
by Wolfram's web page. It makes an assumption about
structure that is then used to evolve the math. The
assumption is false, as it relates to the OP's question, so
the conclusions don't necessarily apply.

I wasn't arguing that it series parallel isn't a part of some
solution approach as a practical matter. I was addressing
myself to the theory applied on the web page for counting
combinations.

Different thing.




Well, there is that. But the OP made it clear, I think, that
the question was theoretic, not practical.

Jon

Oh crap. That's 2 _more_ things I missed. I thought the op wanted
simple series/parallel and practical solutions, not complex arrays
& theoretical solutions.
Grumble. Displeased with myself, but glad you posted.

Ed
 
M

Martin Brown

Jan 1, 1970
0
Yes. I agree. I was slightly pleased to see it and the way I read it
confirmed that my construction was sufficient. I did say that I wasn't
sure if some networks of deltas might generate some new values.
Jon, can you show that statement to be true?

I agree with Jon (this was the sort of net that I was worried about).

But this particular one is too symmetrical and does not contribute any
new values to the sequence. A basic symmetry argument shows that the two
nodes are at V/3 and 2V/3 and the overall impedance is R. If you draw it
right the equipotential surfaces become clear.

I think the first one that could introduce a novel value requires one
extra resistor so one leg has 3R in it. Can anyone find or derive the
full sequence for every possible network from identical resistors?

I'd also be interested in the number of unique values.

The two that my construction does not generate for N=4 are

R||R+R||R and (R+R)||(R+R)

I think that is all of them

I haven't so far been able to sketch all the missing N=5 networks.
It is obviously at least 4.

It seems clear that for any composite number PQ less than the total
number of resistors N you can take them in groups of P or Q and apply
the series parallel rule to the clumps which definitely generates new
networks. 6 is the first composite with distinct factors.
 
J

Jon Kirwan

Jan 1, 1970
0
Yes. I agree. I was slightly pleased to see it and the way I read it
confirmed that my construction was sufficient. I did say that I wasn't
sure if some networks of deltas might generate some new values.


I agree with Jon (this was the sort of net that I was worried about).

But this particular one is too symmetrical and does not contribute any
new values to the sequence. A basic symmetry argument shows that the two
nodes are at V/3 and 2V/3 and the overall impedance is R. If you draw it
right the equipotential surfaces become clear.

I think the first one that could introduce a novel value requires one
extra resistor so one leg has 3R in it. Can anyone find or derive the
full sequence for every possible network from identical resistors?

I'd also be interested in the number of unique values.

The two that my construction does not generate for N=4 are

R||R+R||R and (R+R)||(R+R)

I think that is all of them

I haven't so far been able to sketch all the missing N=5 networks.
It is obviously at least 4.

It seems clear that for any composite number PQ less than the total
number of resistors N you can take them in groups of P or Q and apply
the series parallel rule to the clumps which definitely generates new
networks. 6 is the first composite with distinct factors.

Bitrex nailed a good paper here:

http://arxiv.org/ftp/arxiv/papers/1004/1004.3346.pdf

Take a look. It's all you could hope to want, I think, if
you include the references as well.

Jon
 
J

Jon Kirwan

Jan 1, 1970
0
Thanks! Not only that but some of the results are applicable to software
testability (a subject I am very interested in).

Please share! What parts relate to that (what do you _see_
in your mind about this?) I'm very curious.

Jon
 
M

mike

Jan 1, 1970
0
John said:
I have 10 resistors of 18 ohms/25 watts each. I will use them for a load
bank I am building.

In trying to achieve flexibility, I began asking myself what maximum
number of different resistances are possible with any combination of
them. It is obvious that two in parallel is 9 ohms, three in parallel is
6 ohms, etc, down to 1.8 ohms. It is also obvious that two in series is
36 ohms, three is 54 ohms, etc up to 180 ohms.

Then I began thinking of combinations of series/parallel like two in
series (or parallel) with one in parallel (or series), and so on.

I think you get my drift. My brain too old and out of practice now to
set up the problem and I was never good at that Binomial Theorem thing
to begin with. Can somebody enlighten me?

Many thanks,
John S

How many ways can you hook up one resistor?
ONE.
How many unique ways can you add a resistor to make a composite resistor?
TWO, series or parallel.
For each of the two composite resistors, how can you add a resistor
to make a new composite resistor?
TWO, series or parallel.
For each of the 4 possible composite resistors, how can you add a resistor
to make a new composite resistor?
repeat 'till you run out of resistors...add 'em up.

Is it not that simple?

For unequal resistors, gets much more complicated,
but the same concept should work.
 
T

Tauno Voipio

Jan 1, 1970
0
How many ways can you hook up one resistor?
ONE.
How many unique ways can you add a resistor to make a composite resistor?
TWO, series or parallel.
For each of the two composite resistors, how can you add a resistor
to make a new composite resistor?
TWO, series or parallel.
For each of the 4 possible composite resistors, how can you add a resistor
to make a new composite resistor?
repeat 'till you run out of resistors...add 'em up.

Is it not that simple?

For unequal resistors, gets much more complicated,
but the same concept should work.


This method leaves out the combinations where the new
composite is built of composite blocks, as you're adding
one resistor at a time. For instance, it does not generate
a series connection of two parallel-connected blocks.
 
J

John S

Jan 1, 1970
0
How many ways can you hook up one resistor?
ONE.
How many unique ways can you add a resistor to make a composite resistor?
TWO, series or parallel.
For each of the two composite resistors, how can you add a resistor
to make a new composite resistor?
TWO, series or parallel.
For each of the 4 possible composite resistors, how can you add a resistor
to make a new composite resistor?
repeat 'till you run out of resistors...add 'em up.

Is it not that simple?

Read the thread and concentrate on those posts by Martin Brown. Perhaps
they will answer your question.
 
I'm too old too. Go to bed and think about it... guaranteed to keep you
awake all night :)

The correct answer is:

18362390002030400342502466214352424101003520520206203414135725725624524
13011204205245248562476247624924924682346724875913948275247646824359245
12342385245684287524858456745672713841344756456252345922592472456467285
8272772752757257278572572752475723457257275177312347246534

Which is:

One novemoctogintillion,
eight hundred thirty-six octooctogintillion,
two hundred thirty-nine septoctogintillion,
two hundred three quinoctogintillion,
forty quattuoroctogintillion,
thirty-four treoctogintillion,
two hundred fifty duooctogintillion,
two hundred forty-six unoctogintillion,
six hundred twenty-one octogintillion,
four hundred thirty-five novemseptuagintillion,
two hundred forty-two octoseptuagintillion,
four hundred ten septseptuagintillion,
one hundred sexseptuagintillion,
three hundred fifty-two quinseptuagintillion,
fifty-two quattuorseptuagintillion,
twenty treseptuagintillion,
six hundred twenty duoseptuagintillion,
three hundred forty-one unseptuagintillion,
four hundred thirteen septuagintillion,
five hundred seventy-two novemsexagintillion,
five hundred seventy-two octosexagintillion,
five hundred sixty-two septsexagintillion,
four hundred fifty-two sexsexagintillion,
four hundred thirteen quinsexagintillion,
eleven quattuorsexagintillion,
two hundred four tresexagintillion,
two hundred five duosexagintillion,
two hundred forty-five unsexagintillion,
two hundred forty-eight sexagintillion,
five hundred sixty-two novemquinquagintillion,
four hundred seventy-six octoquinquagintillion,
two hundred forty-seven septenquinquagintillion,
six hundred twenty-four sexquinquagintillion,
nine hundred twenty-four quinquinquagintillion,
nine hundred twenty-four quattuorquinquagintillion,
six hundred eighty-two trequinquagintillion,
three hundred forty-six duoquinquagintillion,
seven hundred twenty-four unquinquagintillion,
eight hundred seventy-five quinquagintillion,
nine hundred thirteen novemquadragintillion,
nine hundred forty-eight octoquadragintillion,
two hundred seventy-five septenquadragintillion,
two hundred forty-seven sexquadragintillion,
six hundred forty-six quinquadragintillion,
eight hundred twenty-four quattuorquadragintillion,
three hundred fifty-nine trequadragintillion,
two hundred forty-five duoquadragintillion,
one hundred twenty-three unquadragintillion,
four hundred twenty-three quadragintillion,
eight hundred fifty-two novemtrigintillion,
four hundred fifty-six octotrigintillion,
eight hundred forty-two septentrigintillion,
eight hundred seventy-five sextrigintillion,
two hundred forty-eight quintrigintillion,
five hundred eighty-four quattuortrigintillion,
five hundred sixty-seven tretrigintillion,
four hundred fifty-six duotrigintillion,
seven hundred twenty-seven untrigintillion,
one hundred thirty-eight trigintillion,
four hundred thirteen novemvigintillion,
four hundred forty-seven octovigintillion,
five hundred sixty-four septenvigintillion,
five hundred sixty-two sexvigintillion,
five hundred twenty-three quinvigintillion,
four hundred fifty-nine quattuorvigintillion,
two hundred twenty-five trevigintillion,
nine hundred twenty-four duovigintillion,
seven hundred twenty-four unvigintillion,
five hundred sixty-four vigintillion,
six hundred seventy-two novemdecillion,
eight hundred fifty-eight octodecillion,
two hundred seventy-two septendecillion,
seven hundred seventy-two sexdecillion,
seven hundred fifty-two quindecillion,
seven hundred fifty-seven quattuordecillion,
two hundred fifty-seven tredecillion,
two hundred seventy-eight duodecillion,
five hundred seventy-two undecillion,
five hundred seventy-two decillion,
seven hundred fifty-two nonillion,
four hundred seventy-five octillion,
seven hundred twenty-three septillion,
four hundred fifty-seven sextillion,
two hundred fifty-seven quintillion,
two hundred seventy-five quadrillion,
one hundred seventy-seven trillion,
three hundred twelve billion,
three hundred forty-seven million,
two hundred forty-six thousand,
five hundred thirty-four
 
M

Martin Brown

Jan 1, 1970
0
Please share! What parts relate to that (what do you _see_
in your mind about this?) I'm very curious.

Jon

The permutations in some of the equations derived there are closely
related to the path testing coverage problem for branch decision points
in software if you do a little bit of relabelling of the entities.

The traditional metric is cyclomatic complexity index first derived by
McCabe - it sort of works (it is correct for what it sets out to do).
What it is also good for is spotting latent maintenance traps in
inherited code. If the complexity is beyond a certain limit there is a
very good chance that it will contain bugs.
 
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