Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

If you have $N$ 1ohm resistors, how many distinct equivalent resistances can you create? Assume that only parallel and series and mixture of them is allowed and no bridging between two parallel connections is allowed.

share|cite|improve this question
What do you mean by "bridging between two parallel connections?" – George G Mar 28 '14 at 11:39
@George: my assumption was that you couldn't create a ladder or mesh. – Carl Witthoft Mar 28 '14 at 11:50
Oh, that makes sense now. – George G Mar 28 '14 at 11:51
This is a combinatorics problem. Since all your N resistors are indistinguishable, you can start w/ all combinations (not permutations) of spanning sets. Really, this is purely a math problem, with a little work at the end to see how many spanning collections produce the same net resistance (if any). – Carl Witthoft Mar 28 '14 at 11:52

There are basically two kinds of addition here (ordinary addition and inverse-sum-inverse), representing series and parallel arrangements.

You can represent the thing as a tree with alternating nodes of addition and ISI layers. The thing resolves pretty much down to a tree with N leaves.

The magic is dealt with here . It talks about leaves on trees, but the various branch-points represent a set of resistors in series or parallel, depending on the parity of steps between the level and the ground. Because this presuposes say, a series layer, you have to double the number.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.