Series/Parallel of 10 resistors question

[John S]

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

[TTman]

"John S" <sop...@invalid.org> wrote in message
news:j7ndtl$8ud$1...@dont-email.me...

I'm too old too. Go to bed and think about it... guaranteed to keep you
awake all night :)

[Tim Wescott]

You're just too lazy to list out all the possibilities.

Just keep in mind when you're sitting in front of a piece of paper,
trying to find the pencil's "on" switch (those things are so obscure
these days!) that you also want to pay attention to power dissipation --
one resistor is 18 ohms and 25 watts, but nine resistors arranged as
three series groups of three parallel resistors is 18 ohms and 225 watts.

--
www.wescottdesign.com

[Jim Thompson]

On Wed, 19 Oct 2011 16:42:13 -0500, Tim Wescott <t...@seemywebsite.com>
wrote:

Yes.

Any SYMMETRICAL arrangement of N resistors will be capable of N x 25
Watts.

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice:(480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

[Martin Brown]

And some of the more asymmetric combination networks will leave
individual resistors under rather more stress than the rest. The OP
would do well to bootstrap up from 2,3,4 by hand until he gets bored.

Given N identical 2 port devices and a pair of nodes to connect using
some or all of them how many distinct network paths are possible? (gives
an upper bound)

A more sensible approach would be to decide what load(s) and power
ratings you actually require the unit to provide.

--
Regards,
Martin Brown

[John S]

In retrospect, I probably should not have mentioned the wattage rating
as that is not the burning question. The wattage can be adjusted so as
not to exceed the wattage rating of any one resistor. I think I can
calculate the greatest wattage dissipated by any one resistor for a
given configuration. But, I need to have the configuration.

The question of resistance remains. And I am still too old to do this alone.

Thanks.

[John S]

On 10/19/2011 5:00 PM, Martin Brown wrote:

> On 19/10/2011 22:42, Tim Wescott wrote:
>> On Wed, 19 Oct 2011 16:02:42 -0500, John S wrote:
>>
>>> 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?
>>
>> You're just too lazy to list out all the possibilities.
>>
>> Just keep in mind when you're sitting in front of a piece of paper,
>> trying to find the pencil's "on" switch (those things are so obscure
>> these days!) that you also want to pay attention to power dissipation --
>> one resistor is 18 ohms and 25 watts, but nine resistors arranged as
>> three series groups of three parallel resistors is 18 ohms and 225 watts.
>
> And some of the more asymmetric combination networks will leave
> individual resistors under rather more stress than the rest. The OP
> would do well to bootstrap up from 2,3,4 by hand until he gets bored.

I don't grok bootstrapping resistors.

> Given N identical 2 port devices and a pair of nodes to connect using
> some or all of them how many distinct network paths are possible? (gives
> an upper bound)

Okay, please tell me. This may be what I am seeking.

[Timo Schneider]

On 2011-10-19, John S <sop...@invalid.org> wrote:

Hi John.

This is an interesting problem, my solution would be based on the principal how
you calculate the resistance of a circuit:

- If two resistors R_1, R_2 are in series you replace them with a new resistor
R_n1,2 with the value R_1+R_2

- If two resistors are in parallel you replace them with a new resitor with the
value (R1+R2)/(R1*R2)

With these two simple rules we can analyze all resitor circuits. So we can also
apply these rules "in reverse":

If we have a circuit with n resistors we can replace any of those with either a
parallel or a serial two-resistor-subcircuit. This gives as 2n new circuits (of
which some can be equivalent to each other!). If we have only 1 resistor we can
build exactly 1 circuit. So the formula for the number of circuits in
dependence on the number of available resistors is:

Circuits(1 Resistor) -> 1
Circuits(n + 1 Resistors) -> R(n) + 2n

If we solve this we get

R(n) = n^2 - n + 1

Now, since in your case all of the resistors have the same value, the real
number of "different" circuits will be lower. Unfortunately I can not come up
with a simple solution to this. So if you want to know, just write a program
that starts with one resistor and uses the "expansion rules" described above
until n resistors are used up and then backtracks.

Since the number of states is qudratic this should give you an answer in
seconds for small resistor numbers (<1000).

Regards,
Timo

PS: I am a computer scientist, not an electronics engineer, so sorry if i got
the resistor formulas wrong - that's just what i remember from highschool.

[Timo Schneider]

On 2011-10-19, John S <sop...@invalid.org> wrote:

Hi John.

This is an interesting problem, my solution would be based on the principal how
you calculate the resistance of a circuit:

- If two resistors R_1, R_2 are in series you replace them with a new resistor
R_n1,2 with the value R_1+R_2

- If two resistors are in parallel you replace them with a new resitor with the
value (R1+R2)/(R1*R2)

With these two simple rules we can analyze all resitor circuits. So we can also
apply these rules "in reverse":

If we have a circuit with n resistors we can replace any of those with either a
parallel or a serial two-resistor-subcircuit. This gives as 2n new circuits (of
which some can be equivalent to each other!). If we have only 1 resistor we can
build exactly 1 circuit. So the formula for the number of circuits in
dependence on the number of available resistors is:

Circuits(1 Resistor) -> 1

Circuits(n + 1 Resistors) -> Circuits(n) + 2n

If we solve this we get

Circuits(n) = n^2 - n + 1

[John S]

Thank you, Timo. That may be a good start to get to my answer.

Cheers,
John S

[John S]

On 10/19/2011 4:42 PM, Tim Wescott wrote:
> On Wed, 19 Oct 2011 16:02:42 -0500, John S wrote:
>
>> 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?
>
> You're just too lazy to list out all the possibilities.

So, you're saying my ancient mind is up to the task? Although I doubt
you are qualified to judge that, I thank you, Tim.

I did list some of the possibilities on paper:
Two resistors have only two possibilities.
Three resistors have four possibilities.
Four resistors have eight possibilities.

Then my brain broke. If the trend continues, would ten resistors have
512 possibilities?

I pass the ball to you, Tim.

John S

[Timo Schneider]

On 2011-10-19, John S <sop...@invalid.org> wrote:

> On 10/19/2011 4:42 PM, Tim Wescott wrote:
>> On Wed, 19 Oct 2011 16:02:42 -0500, John S wrote:

Hi,

>>> 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?

> I did list some of the possibilities on paper:
> Two resistors have only two possibilities.

Mm... why?

With two resistors you can build those three circuits:

---R----R---

----R----
--| |--
----R----

-----R------

[John Larkin]

On Wed, 19 Oct 2011 16:02:42 -0500, John S <sop...@invalid.org>
wrote:

I took an Electrical Machinery course once (mandatory, two semisters)
and we had some cool load banks.

Imagine two busses running vertically side by side, being the two
terminals of a composite resistor. Between them is a series string of
power resistors. At each resistor-resistor junction is a knife switch
that can go to the left bus, to the right bus, or open. There are all
sorts of series and parallel combinations possible.

Actually, I learned a lot in that course. I got an A with about 60%
grades on tests, because the class average was about 15. The labs were
fun, too... lots of sparks and smoke.

John

[George Herold]

> John S- Hide quoted text -
>
> - Show quoted text -

Woah... two resistors have more than two possibilities, unless I mis-
understood the question.

George H.

[George Herold]

Oh it's something like (N!)(N-M!)/M! - 1... :^) You can always start
with one and work your way up.... by the time you get to 5 or 6 maybe
you can guess the rest.

George H.

[k...@att.bizzzzzzzzzzzz]

I get three:

Parallel
Series
Single (one shorted)

Four, if you count two shorted (zero ohms)

Five, if you count no-connects.

[j...@myplace.com]

On Wed, 19 Oct 2011 16:02:42 -0500, John S <sop...@invalid.org>
wrote:

If this was the 1960s, and I had a bag of good weed, and had a 20 year
old brain again, I bet I'd figure out the answer after smoking that
bag of weed, and even tell you the meaning of life based on
resistance, tolerance, and power.

As soon as I figure out how to go back to the 1960's, regain my 20
year old brain, and find a bag of really good weed, I'll give you the
answer.


Until then, I can honestly tell you the answer is somewhere between
zero and infinite, and I know that is 100% correct and accurate, based
on my PHD.
You can view my PHD here: http://tinyurl.com/3cxkx75

[Martin Brown]

ROFL. I had not intended a double meaning. Perhaps I should have said by
induction starting from one resistor and working up.

>
>> Given N identical 2 port devices and a pair of nodes to connect using
>> some or all of them how many distinct network paths are possible? (gives
>> an upper bound)
>
> Okay, please tell me. This may be what I am seeking.

I think it might be slightly more obvious to cast the problem as using
precisely M resistors how many choices are possible. Unfortunately
determining which of them have distinct values is somewhat harder.

M p(M) Values
1 1 R
2 2 2R, R/2
3 4 3R, 3R/2, 2R/3, R/3
4 8 4R, 5R/2, 5R/3, 4R/3, 3R/4, 3R/5, 2R/5, R/4

You can derive the next line from the previous by adding an R in series
or in parallel connected to one external node and every other internal
node. There might be some others come out of the woodwork for larger N.

2^10 is still a bit tedious though it would only take a few seconds to
compute in any of the algebra packages or in a spreadsheet.

And you should think really carefully about 9R||R load capacity.

--
Regards,
Martin Brown

[John S]

Because the single R falls into the category of the number of ways of
connecting one resistor (one way).

[John S]

Your last two possibilities fall outside the scope of the question since
they require no load bank at all.

[k...@att.bizzzzzzzzzzzz]

Maybe, depending on your definition of "load bank". That's why I separated
them. Specification of the problem, matter.

[John S]

I came up with the same p(M) for M -> 1 to 4. I wonder if the p(M)
sequence continues in the same fashion up to M = 10. If so, then p(M)
would be 512 for M = 10. Also, wouldn't that mean that the total number
if ways would be 512 + the sum of all the ways less than ten?

So, would it be 2^9 + 2^8 + 2^7 +....2^1 + 2^0 = 2^10 ??

If so, then the total number of possibilities would be 1024.

> And you should think really carefully about 9R||R load capacity.

I have. I will limit total power to 25W regardless of the configuration.

[John S]

> ^^^^
(2^10)-1

> If so, then the total number of possibilities would be 1024.

> ^^^^
1023

[ehsjr]

There are 55 different value combinations with 10 equal value R's, not
including shorted or open resistors. 9 possible purely parallel, and 9
possible purely series. Then, the combination of N series with P
parallel, where P+N < 11, and P>1. Note that P must be equal to at
least 2, else it doesn't represent a parallel configuration and yields
unwanted duplicates. For example, 8 in series added to P in parallel
where P = 1 would be the same as 9 in series.

It's easier to program than to figure out.

Ed

[Martin Brown]

That seems a bit silly. You could easily choose to use only those
configurations that will handle close to maximum power and end up with
something a lot more useful and a heck of a lot easier to build.

eg. 1R = 3 x R||R||R (gets you 9x25W of dissipation)

>>
>
> There are 55 different value combinations with 10 equal value R's, not
> including shorted or open resistors. 9 possible purely parallel, and 9
> possible purely series. Then, the combination of N series with P
> parallel, where P+N < 11, and P>1. Note that P must be equal to at
> least 2, else it doesn't represent a parallel configuration and yields
> unwanted duplicates. For example, 8 in series added to P in parallel
> where P = 1 would be the same as 9 in series.
>
> It's easier to program than to figure out.
>
> Ed

It is also incorrect. The parallel resistor can go to any node in the
chain of series resistors and different numbers of resistors already in
series can be placed in parallel. I couldn't convince myself that the
construction I suggested above would get every possible network but
according to Wolfram MathWorld my p(m) = 2^(m-1) is complete.

http://mathworld.wolfram.com/ResistorNetwork.html

(they do it for 1 ohm resistors)


--
Regards,
Martin Brown

[John S]

Yes, it is silly for the moment. However, I am trying to simplify the
problem so as to come up with the answer for resistances before I look
at the power situation. Once I have a complete solution somewhere, I can
go back and do the power thing, I hope.

>>
>> There are 55 different value combinations with 10 equal value R's, not
>> including shorted or open resistors. 9 possible purely parallel, and 9
>> possible purely series. Then, the combination of N series with P
>> parallel, where P+N < 11, and P>1. Note that P must be equal to at
>> least 2, else it doesn't represent a parallel configuration and yields
>> unwanted duplicates. For example, 8 in series added to P in parallel
>> where P = 1 would be the same as 9 in series.
>>
>> It's easier to program than to figure out.
>>
>> Ed
>
> It is also incorrect. The parallel resistor can go to any node in the
> chain of series resistors and different numbers of resistors already in
> series can be placed in parallel. I couldn't convince myself that the
> construction I suggested above would get every possible network but
> according to Wolfram MathWorld my p(m) = 2^(m-1) is complete.
>
> http://mathworld.wolfram.com/ResistorNetwork.html
>
> (they do it for 1 ohm resistors)
>

Thanks to your link, I now see that the answer to my OP is 1023
different resistance values. Now, if I can figure out how to continue
your sequence above, I'll put it into a spreadsheet that will tell me
the resistance of the combinations.

Many thanks.

[Jon Kirwan]

On Thu, 20 Oct 2011 18:37:08 +0100, Martin Brown
<|||newspam|||@nezumi.demon.co.uk> wrote:

><snip>
>http://mathworld.wolfram.com/ResistorNetwork.html

Martin, that limits itself to parallel and series
combinations. At some point, starting with 5, this isn't the
only way. There are delta/y configurations. For example,
take 7 resistors of the same value but place them in an
unbalanced wheatstone bridge configuration:

|
+
/ \
/ R
R \
/ R
/ \
+-----R-----+
\ /
R /
\ R
R /
\ /
+
|

This is irreducible by parallel-series analysis.

I'm not convinced yet that the Wolfram page you mentioned
addresses itself fully to these additional configuration
possibilities -- most particularly as N grows large.

Or maybe I'm not visualizing this as well as I should.

Jon

[John S]

Hi, Jon -

Martin did say that other possibilities may come out of the woodwork.
You seem to have found one.

However, I've decided that just knowing that there are over 1000 ways to
connect them to get a value between 1.8 and 180 ohms is sufficient. I
think I will not chase all the combinations and record their possible
values as the number of possibilities seem to exceed the accuracy of any
value I am likely to need between the extremes.

But, I don't mean to suggest you curtail your conversation with Martin
by any means. This is highly interesting to me.

John S

[Jim Thompson]

Do a delta-to-wye conversion, then series/parallel o:-)

>
>I'm not convinced yet that the Wolfram page you mentioned
>addresses itself fully to these additional configuration
>possibilities -- most particularly as N grows large.
>
>Or maybe I'm not visualizing this as well as I should.
>
>Jon

Not exactly a real-world problem.

[John S]

On 10/20/2011 1:50 PM, Jon Kirwan wrote:

> On Thu, 20 Oct 2011 18:37:08 +0100, Martin Brown
> <|||newspam|||@nezumi.demon.co.uk> wrote:
>
>> <snip>
>> http://mathworld.wolfram.com/ResistorNetwork.html
>
> Martin, that limits itself to parallel and series
> combinations. At some point, starting with 5, this isn't the
> only way. There are delta/y configurations. For example,
> take 7 resistors of the same value but place them in an
> unbalanced wheatstone bridge configuration:
>
> |
> +
> / \
> / R
> R \
> / R
> / \
> +-----R-----+
> \ /
> R /
> \ R
> R /
> \ /
> +
> |
>
> This is irreducible by parallel-series analysis.

Jon, can you show that statement to be true?

[Jon Kirwan]

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.

>>I'm not convinced yet that the Wolfram page you mentioned
>>addresses itself fully to these additional configuration
>>possibilities -- most particularly as N grows large.
>>
>>Or maybe I'm not visualizing this as well as I should.
>>
>>Jon
>
>Not exactly a real-world problem.
>
> ...Jim Thompson

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

Jon

[Jon Kirwan]

Actually, it would be better (a conclusive proof) if you'd
show me how you'd do it with series-parallel as it relates to
the Wolfram's approach in designing its counting method. My
purpose was simply to show one example that doesn't appear to
be included in their approach for counting orientations. I
could be wrong, as I said. But it looks like they weren't
including the above configuration, to me.

I am getting a glimmer of how to approach the problem --
hypercubes and Hamiltonion walks and generating functions are
in mind, right now. Probably I'll change my thinking. But I
need to let it rest for a few days, as I'm on other things
right now.

Jon

[Jon Kirwan]

On Thu, 20 Oct 2011 14:07:49 -0500, John S
<sop...@invalid.org> wrote:

By the way, I did manage a quick google using my word of
'irreducible' and unbalanced wheatstone bridge and shock of
all shocks I found this page:

http://www.allaboutcircuits.com/vol_1/chpt_10/3.html

I don't mean to argue by authority and feel free to take that
author for what you will. But at least someone else talks
like me about it. So there are at least two of us in the
world, for what it is worth.

Jon

[Bitrex]

I found this link you might be interested in:

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

[John S]

Oh, I didn't mean to suggest that you were wrong. I simply wanted to see
how you arrived at your answer. FWIW, I couldn't find a solution in
series/parallel for your example either.

[John S]

Yes. Thank you.

[John S]

It didn't look like it to me, either. But, who am I? I'm having a hard
time these days visualizing these things.

> I am getting a glimmer of how to approach the problem --
> hypercubes and Hamiltonion walks and generating functions are
> in mind, right now. Probably I'll change my thinking. But I
> need to let it rest for a few days, as I'm on other things
> right now.
>
> Jon

Well, thanks for your input. I always value it.

John S

[Jim Thompson]

The sort of thing that gives PhD's a bad name... WTF ?:-)

[Jon Kirwan]

On Thu, 20 Oct 2011 15:56:50 -0400, Bitrex
<bit...@de.lete.earthlink.net> wrote:

Damn! More to read. But then, maybe I don't have to think.
Not sure whether that is good or bad. ;)

Jon

[John S]

Maybe? What's YOUR definition of a load bank? Ahhh - Never mind, this is
another request for you to ESAD.

[ehsjr]

Martin Brown wrote:

<snip>

>
>>>
>>
>> There are 55 different value combinations with 10 equal value R's, not
>> including shorted or open resistors. 9 possible purely parallel, and 9
>> possible purely series. Then, the combination of N series with P
>> parallel, where P+N < 11, and P>1. Note that P must be equal to at
>> least 2, else it doesn't represent a parallel configuration and yields
>> unwanted duplicates. For example, 8 in series added to P in parallel
>> where P = 1 would be the same as 9 in series.
>>
>> It's easier to program than to figure out.
>>
>> Ed
>
>
> It is also incorrect.

Yes, you're right. I failed to include "stacking" R's made
in parallel into series strings, then combining the result.
So a number of values were not included. eg: -5p-4p-1s- or -2p-2p-2p-
etc. I realized that before reading your post *but* you
also pointed out that series strings could be paralleled to
create different values, like -3s-||-2s- or -3s-||-4s- or
-2s||-3s-||-4s- etc. Good point. Thanks!

Ed

[John S]

The total resistance is 7/5 (if R = 1). But, I don't know how to get
there using on series/parallel connections.

You win.

John S

[John S]

> ^^
only

> You win.
>
> John S

[Jon Kirwan]

On Thu, 20 Oct 2011 17:02:18 -0500, John S

Yup. Same value I got, 1.4 ohms.

>But, I don't know how to get

>there using on[ly] series/parallel connections.
>
>You win.

:) I wasn't trying to win, just point out that it takes a
little more analysis that was shown by Martin on the Wolfram
website.

These problems are indeed fun, even if as Jim says
impractical. Stretches the mind. No idea what practical
thing can happen. But John Conway hoped nothing practical
would ever be done with his work on 26 dimensions -- until
someone found something despite his hopes. So you never know
where something like this takes you until you go for it.

Jon

[DonMack]

There is a mapping from n to R that gives the result.

F_i(A,B) = i*(A + B) + (1 - i)*A*B/(A + B)

is one such mapping,

F_a_k(....F_a_0(R,R),...,R)

where a_k is the binary sequence for an integer m <= n.

e.g., for m = 1, a = {1,0,0,0,...}, m = 5, a = {1, 0, 1, 0,...}

One can compute F for each n.

In fact F/R is independent of R

I'm not sure if F has any generating function but I imagine it would.

There are other possible mappings such as

G_i(A,B) = (A+B)^(1 - 2*i)*(A*B)^i

F_0(A,B) = G_0(A,B) = A + B
F_1(A,B) = G_1(A,B) = A*B/(A + B)

So you can think of F and G as "super" combinations that incorporate both
parallel and series. I is the indicator or a parameter that gives the
amount. F is the linear interpolation and G is the geometric interpolation.

Since you are simply composing different combinations we can do this but
require a parameter for each "branch" of the combination. Since we are using
0 and 1 to signify the combinations we can map these to the binary numbers.
What this means is that any topology of series and parallel combinations you
use can be thought of as a simple number.

so

-R-R-

can be thought of as 0

and

-R-
-R-


note that this produces idempotent in the expansion of F and possibly G
which can significantly reduce the complexity.

It would be nice to know the generating function for F and/or G. I'm sure it
exists but it will be the function that maps N(or R) to R.

[k...@att.bizzzzzzzzzzzz]

Zero and open are useful values. If they're free, why not?

>Ahhh - Never mind, this is another request for you to ESAD.

You're no better at keeping technical discussions technical than Slowman.
Gotta be a prick.

[Jon Kirwan]

On Thu, 20 Oct 2011 15:13:14 -0500, John S

I had arrived at it from: (1) hard work as a teenager when I
saw the problem and didn't know there was an easy approach
such as Norton and Thevenin eqivalents and hadn't come up
with branch-current, mesh, and nodal analyses yet on my own
(I _much_ prefer nodal analysis, by the way, as it 'sings in
my mind' like a beautiful song whereas the others are tinny
and clunky by comparison.) (2) Reading others' works saying
so. (3) Trying my imagination at it from time to time
afterwards, just to make sure.

But it is provable, I think. Lay out all possible nodes that
reach from the starting point to the ending point without
visiting any node twice. For example, in the above
unbalanced wheatstone bridge, with 0 the starting point and 5
the ending point:

0
/ \
/ R
R 1
/ R
/ \
2-----R-----3
\ /
R /
4 R
R /
\ /
5

We get:

0 2 4 5
0 2 3 5
0 1 3 5
0 1 3 2 4 5

You might notice above that there is a "2 3" in one path and
a "3 2" in another, which implies a reversal. Not sure if
that means anything yet. But there it is.

Let's take another 7 resistor version, but only with series
parallel connections:

0--R--1--R--2--R--3--R--4--R--5
| | |
| '-----R-----+
'---R-------------'

In this case, we get:

0 1 2 3 4 5
0 1 4 5
0 1 2 4 5

No reversals. (Remember that we aren't allowed to visit a
node more than once.)

Let's try this one:

,---R-------,
| |
0--R--1--R--2--R--3--R--4--R--5
| |
'-----R-----+

0 1 3 4 5
0 1 2 4 5
0 1 2 3 4 5
0 1 3 2 4 5

Hmm. There is that pesky reversal, again. A "2 3" and then
a "3 2" appears.

Might be a clue here somewhere about defining what is series
parallel only and what might be excluded from that privileged
set.

But I'm just shooting in the dark.

Jon

[Timo Schneider]

On 2011-10-19, Timo Schneider <ti...@tesla.lan> wrote:
> On 2011-10-19, John S <sop...@invalid.org> wrote:

Hi,

>> 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.

> This is an interesting problem, my solution would be based on the principal how
> you calculate the resistance of a circuit:
>
> - If two resistors R_1, R_2 are in series you replace them with a new resistor
> R_n1,2 with the value R_1+R_2
>
> - If two resistors are in parallel you replace them with a new resitor with the
> value (R1+R2)/(R1*R2)
>
> With these two simple rules we can analyze all resitor circuits.

The last statement is wrong. I just realized that there are circuits that are
valid in the sense of the problem definition which can not be reduced to a
single resistor by those two rules, such as:

|---R----R---|
| | |
+ --| R |-- -
| | |
|---R----R---|

None of the resistors are in parallel or in series - so none of the above rules
can be applied.

Regards,
Timo

[Jim Thompson]

Be sure to wear steel-toed shoes ;-)

[ehsjr]

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

[Martin Brown]

On 20/10/2011 20:07, John S wrote:
> On 10/20/2011 1:50 PM, Jon Kirwan wrote:
>> On Thu, 20 Oct 2011 18:37:08 +0100, Martin Brown
>> <|||newspam|||@nezumi.demon.co.uk> wrote:
>>
>>> <snip>
>>> http://mathworld.wolfram.com/ResistorNetwork.html
>>
>> Martin, that limits itself to parallel and series

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.

>> combinations. At some point, starting with 5, this isn't the
>> only way. There are delta/y configurations. For example,
>> take 7 resistors of the same value but place them in an
>> unbalanced wheatstone bridge configuration:
>>
>> |
>> +
>> / \
>> / R
>> R \
>> / R
>> / \
>> +-----R-----+
>> \ /
>> R /
>> \ R
>> R /
>> \ /
>> +
>> |
>>
>> This is irreducible by parallel-series analysis.
>
> 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.

--
Regards,
Martin Brown

[Jon Kirwan]

On Fri, 21 Oct 2011 09:24:57 +0100, Martin Brown

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

[Martin Brown]

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

--
Regards,
Martin Brown

[John Fields]

On Fri, 21 Oct 2011 09:24:57 +0100, Martin Brown

---
Here's a tabulation of all of the 55 possible combinations of from one
to ten simple series, parallel and series-parallel connected equal
value resistors normalized to one ohm:

news:n7u2a75jqn978kb45...@4ax.com

--
JF

[Jon Kirwan]

On Fri, 21 Oct 2011 13:13:56 +0100, Martin Brown
<|||newspam|||@nezumi.demon.co.uk> wrote:

>On 21/10/2011 11:01, Jon Kirwan wrote:

>><snip>

>>
>> 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.
>
>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

[mike]

John S wrote:
> 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.

[Tauno Voipio]

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.

--

Tauno Voipio

[John S]

Read the thread and concentrate on those posts by Martin Brown. Perhaps
they will answer your question.

[j...@myplace.com]

On Wed, 19 Oct 2011 22:28:55 +0100, "TTman" <pcw1...@ntlworld.com>
wrote:

>
>"John S" <sop...@invalid.org> wrote in message
>news:j7ndtl$8ud$1...@dont-email.me...

>>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
>

>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

[Martin Brown]

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.

--
Regards,
Martin Brown