# Series/Parallel of 10 resistors question

Discussion in 'Electronic Design' started by John S, Oct 19, 2011.

1. ### John SGuest

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

2. ### TTmanGuest

I'm too old too. Go to bed and think about it... guaranteed to keep you
awake all night 3. ### Martin BrownGuest

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.

4. ### John SGuest

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.

5. ### John SGuest

I don't grok bootstrapping resistors.

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

6. ### Timo SchneiderGuest

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.

7. ### Timo SchneiderGuest

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 (<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.

8. ### John SGuest

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

Cheers,
John S

9. ### John SGuest

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

10. ### Timo SchneiderGuest

Mm... why?

With two resistors you can build those three circuits:

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

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

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

11. ### Guest

I get three:

Parallel
Series
Single (one shorted)

Four, if you count two shorted (zero ohms)

Five, if you count no-connects.

12. ### Guest

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

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

13. ### Martin BrownGuest

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.

14. ### John SGuest

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

15. ### John SGuest

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

16. ### Guest

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

17. ### John SGuest

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.

19. ### ehsjrGuest

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

20. ### Martin BrownGuest

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.

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

(they do it for 1 ohm resistors)  