Series/Parallel of 10 resistors question

[John S]

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)

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.

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]

<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]

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

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

 

[John S]

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]

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

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.

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]

Jon, can you show that statement to be true?

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]

Jon, can you show that statement to be true?

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

 

[John S]

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

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]

I found this link you might be interested in:

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

Yes. Thank you.

 

[John S]

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.

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

 

[Jon Kirwan]

I found this link you might be interested in:

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

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

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

 

[ehsjr]

Martin Brown wrote:

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]

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

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]

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

You win.

John S

 

[Jon Kirwan]

The total resistance is 7/5 (if R = 1).

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.

 

[[email protected]]

Maybe? What's YOUR definition of a load bank?

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]

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.

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]

Hi,

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