# Series/Parallel of 10 resistors question

• posted

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

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

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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.

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Yes.

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

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| James E.Thompson, CTO                            |    mens     |
• posted

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.

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Martin Brown```
• posted

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.

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I don't grok bootstrapping resistors.

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

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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 (

• posted

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

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 (

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

Cheers, John S

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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

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Mm... why?

With two resistors you can build those three circuits:

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

----R----

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

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

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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

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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.

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is

s

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

George H.

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I get three:

Parallel Series Single (one shorted)

Four, if you count two shorted (zero ohms)

Five, if you count no-connects.

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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:

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

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.

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