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Series/Parallel of 10 resistors question

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

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  1. John S

    John S Guest

    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.
  2. Jon Kirwan

    Jon Kirwan Guest

    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 /
    \ /

    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.

  3. John S

    John S Guest

    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
  4. John S

    John S Guest

    Jon, can you show that statement to be true?
  5. Jon Kirwan

    Jon Kirwan Guest

    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

    Different thing.
    Well, there is that. But the OP made it clear, I think, that
    the question was theoretic, not practical.

  6. Jon Kirwan

    Jon Kirwan Guest

    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.

  7. Jon Kirwan

    Jon Kirwan Guest

    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.

  8. John S

    John S Guest

    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.
  9. John S

    John S Guest

  10. John S

    John S Guest

    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
  11. Jon Kirwan

    Jon Kirwan Guest

  12. John S

    John S Guest

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

    ehsjr Guest

    Martin Brown wrote:

    Yes, you're right. I failed to include "stacking" R's made
    in parallel into series strings, then combining the result.
    So a number of values were not included. eg: -5p-4p-1s- or -2p-2p-2p-
    etc. I realized that before reading your post *but* you
    also pointed out that series strings could be paralleled to
    create different values, like -3s-||-2s- or -3s-||-4s- or
    -2s||-3s-||-4s- etc. Good point. Thanks!

  14. John S

    John S Guest

    The total resistance is 7/5 (if R = 1). But, I don't know how to get
    there using on series/parallel connections.

    You win.

    John S
  15. John S

    John S Guest

  16. Jon Kirwan

    Jon Kirwan Guest

    Yup. Same value I got, 1.4 ohms.
    :) I wasn't trying to win, just point out that it takes a
    little more analysis that was shown by Martin on the Wolfram

    These problems are indeed fun, even if as Jim says
    impractical. Stretches the mind. No idea what practical
    thing can happen. But John Conway hoped nothing practical
    would ever be done with his work on 26 dimensions -- until
    someone found something despite his hopes. So you never know
    where something like this takes you until you go for it.

  17. DonMack

    DonMack Guest

    There is a mapping from n to R that gives the result.

    F_i(A,B) = i*(A + B) + (1 - i)*A*B/(A + B)

    is one such mapping,


    where a_k is the binary sequence for an integer m <= n.

    e.g., for m = 1, a = {1,0,0,0,...}, m = 5, a = {1, 0, 1, 0,...}

    One can compute F for each n.

    In fact F/R is independent of R

    I'm not sure if F has any generating function but I imagine it would.

    There are other possible mappings such as

    G_i(A,B) = (A+B)^(1 - 2*i)*(A*B)^i

    F_0(A,B) = G_0(A,B) = A + B
    F_1(A,B) = G_1(A,B) = A*B/(A + B)

    So you can think of F and G as "super" combinations that incorporate both
    parallel and series. I is the indicator or a parameter that gives the
    amount. F is the linear interpolation and G is the geometric interpolation.

    Since you are simply composing different combinations we can do this but
    require a parameter for each "branch" of the combination. Since we are using
    0 and 1 to signify the combinations we can map these to the binary numbers.
    What this means is that any topology of series and parallel combinations you
    use can be thought of as a simple number.



    can be thought of as 0



    note that this produces idempotent in the expansion of F and possibly G
    which can significantly reduce the complexity.

    It would be nice to know the generating function for F and/or G. I'm sure it
    exists but it will be the function that maps N(or R) to R.
  18. Guest

    Zero and open are useful values. If they're free, why not?
    You're no better at keeping technical discussions technical than Slowman.
    Gotta be a prick.
  19. Jon Kirwan

    Jon Kirwan Guest

    I had arrived at it from: (1) hard work as a teenager when I
    saw the problem and didn't know there was an easy approach
    such as Norton and Thevenin eqivalents and hadn't come up
    with branch-current, mesh, and nodal analyses yet on my own
    (I _much_ prefer nodal analysis, by the way, as it 'sings in
    my mind' like a beautiful song whereas the others are tinny
    and clunky by comparison.) (2) Reading others' works saying
    so. (3) Trying my imagination at it from time to time
    afterwards, just to make sure.

    But it is provable, I think. Lay out all possible nodes that
    reach from the starting point to the ending point without
    visiting any node twice. For example, in the above
    unbalanced wheatstone bridge, with 0 the starting point and 5
    the ending point:

    / \
    / R
    R 1
    / R
    / \
    \ /
    R /
    4 R
    R /
    \ /

    We get:

    0 2 4 5
    0 2 3 5
    0 1 3 5
    0 1 3 2 4 5

    You might notice above that there is a "2 3" in one path and
    a "3 2" in another, which implies a reversal. Not sure if
    that means anything yet. But there it is.

    Let's take another 7 resistor version, but only with series
    parallel connections:

    | | |
    | '-----R-----+

    In this case, we get:

    0 1 2 3 4 5
    0 1 4 5
    0 1 2 4 5

    No reversals. (Remember that we aren't allowed to visit a
    node more than once.)

    Let's try this one:

    | |
    | |

    0 1 3 4 5
    0 1 2 4 5
    0 1 2 3 4 5
    0 1 3 2 4 5

    Hmm. There is that pesky reversal, again. A "2 3" and then
    a "3 2" appears.

    Might be a clue here somewhere about defining what is series
    parallel only and what might be excluded from that privileged

    But I'm just shooting in the dark.

  20. The last statement is wrong. I just realized that there are circuits that are
    valid in the sense of the problem definition which can not be reduced to a
    single resistor by those two rules, such as:

    | | |
    + --| R |-- -
    | | |

    None of the resistors are in parallel or in series - so none of the above rules
    can be applied.

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