Parallel and Series Circuit

a1234

Homework Statement

I'm asked to find a combination of resistors (parallel and/or series) that uses resistors of 25 Ω, 100 Ω, 50 Ω, and 50 Ω. They should add up to give a total resistance of 62.5 Ω.

Homework Equations

Req for parallel = 1/R1 + 1/R2 + ...
Req for series = R1 + R2 + ...

The Attempt at a Solution

This seems like a relatively simple problem, but I've been unable to find a combination that gives the desired resistance. I noticed, however, that if the 50 Ω, 50 Ω, and 25 Ω resistors are placed in parallel, we get an equivalent resistance of 12.5 Ω, which may help in getting to the total of 62.5 Ω. I'm also required to use all the resistors.

Homework Helper
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Can you think of a way to make the numbers smaller?
Since 100>62.5, what will you need to do with that one?

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I hate this problem. I also failed to find a combination that works. I can find it if the 100 was a 50 or if there was another 100 (so the two 100s could be combined in parallel for 50). Maybe there is one. @haruspex ,without giving away too much, do you see a combination that works?

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Gold Member
do you see a combination that works?
Yes, it has a solution.
There is a methodical approach, but it is a bit tedious.

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Gold Member
Yes, it has a solution.
There is a methodical approach, but it is a bit tedious.
Thanks.
@a1234 , has your class discussed any procedure to solving these?

I finally found it. After a1234 has solved it, I would like to know about a more methodical way, tedious or not, unless it is just trying all combinations. Even "branch and bound" would be some help.

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Homework Helper
Gold Member
do you see a combination that works?
Yes, there is a solution.

But I don't know what hint I should give without accidently giving away the answer. Will post later when I have the right words for that hint.

Edit: I see Factchecker has also found it.

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fishspawned
the best way i found to working out the solution was first assuming that the 100 is in a parallel circuit.then it was a matter of calculating as i went to see if i was getting higher or lower than 62.5, thus determining what had to be arranged to get higher or lower.

honestly tho, this question to me was really a furious game of mix and match with little in the way of logically determining your steps beyond that. there is a question, but it really isn't satisfying when you get there, just the relief that you're done.

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Homework Helper
Gold Member
Pick one resistor and call it R.

Find the resistance Rs that needs to be connected in series with R which gives Req=62.5 ohms.
Try all the combinations of the remaining three resistors to find which one gives Rs.

If that doesn't work,
find the resistance Rp that needs to be connected in parallel with R which gives Req=62.5 ohms. (Here, Rp can only be 100 ohms).
Try all the combinations of the remaining three resistors to find which one gives Rs.

Using some common sense, you can mentally work things out when you are dealing with the remaining three resistors.

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a1234
I finally solved it. I think the most helpful thing is to notice that the equivalent resistance in a parallel circuit is less than each of the individual resistors.

Thanks for all the help!

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Homework Helper
Gold Member
the most helpful thing is to notice that the equivalent resistance in a parallel circuit is less than each of the individual resistors.
Yes, that is what I used, and conversely that the max resistance that can be made from a set is all in series.

My first step was to factor out to reduce the problem to creating a 5 from 2, 4, 4, 8.
It also helps to develop a notation.
I write _ for series and | for parallel, and [ ... ] to denote an unspecified circuit built from the resistors listed.
The outermost structure is two paths in parallel or two paths in series.

We can immediately rule out 2|[4,4,8] and 4|[2,4,8] as being <5.
8|[2,4,4]≤8|10<5.
[8,2]|[4,4]≤10|8<5.
Etc.
So that rules out outermost parallel.
Working through the "outermost in series" cases is tougher.

I suspect there is some Diophantine approach, but it is too subtle for me.

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