Series/Parallel of 10 resistors question

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.

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

(2^10)-1

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

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)

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.

(they do it for 1 ohm resistors)

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.

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.

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

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

Not exactly a real-world problem. ...Jim Thompson

Jon, can you show that statement to be true?

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

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

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:

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

I found this link you might be interested in:

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.

Yes. Thank you.

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

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

John S

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

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

Jon

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