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A circuit problem.

  1. May 10, 2007 #1
    Hi everyone.

    Now I ve got this problem, but not a homework. I hope some of you can help. There is a battery say B, and n small bulbs A1, A2...An. I can make a circuit as : B-> A1-> A2....An . I can also change it to B->A2-A1...An . Of course there are many ways to make a circuit so that all the bulbs will be lit.
    The question here is: how many possible ways are there to make a circuit.
  2. jcsd
  3. May 10, 2007 #2


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    do u just wanna connecting in one loop or multiple loops etc... ? the number will be drastically different
  4. May 11, 2007 #3
    Thanks for your replying. There's no loop. It can be a tree or a series, but not a loop.
  5. May 11, 2007 #4
    Being from the old school I am having a bit of trouble trying to decide what you are describing, must be new math. But from what I can glean from your post then I would say that the answer would be close to infinity. You have not given the battery capacity, lamp watts, etc.
    Last edited: May 11, 2007
  6. May 11, 2007 #5
    In fact this is a math problem, not a electrical. The capacity of the battery is consider to light all the bulbs given. Here is my drawing for the case that there are only two lamps A1 and A2. In this case, there are 3 possible ways to make the circuits. If there are 3 bulbs, a tree circuit can be drawn.

    Attached Files:

  7. May 11, 2007 #6
    I hope this thread could be moved back to the maths box.
  8. May 11, 2007 #7
    Since I have to take my shoes off to count over ten then this is out of my experience in regards to math. But by looking at your drawing then it looks like you have two circuits in series and one in parallel. Since lamps are a resistive load then this changes the properties of the circuit.
  9. May 11, 2007 #8
    As I mentioned above, this one is not electrical. We have to consider the battery is infinitely strong !! and also all the bulbs are circuited parallelly meaning they all have one potentials without any voltage drop even if the number of bulbs could be very high.
  10. May 11, 2007 #9
    If you have a infinitely strong battery and a n number of lamps then I still don't fathom the question other than my first response to your post. The battery could be made of half of the material in the universe and the lamps made out of the other half. Then the brightness of the lamps would still be decided if the circuit was series or parallel.
  11. May 11, 2007 #10
    I wasn't trying to be funny but you posted it as an electrical problem and I thought that maybe the math was leading you to think that a electrical solution would find the answer. This is an electrical forum and you will receive electrical answers.

    With an infinite power souce and infinite lamps then the answer would be infinite. In physics they say that when infinities are encountered then there is a problem with the math.

    In the answer box I would draw a figure 8 sideways.
    Last edited: May 11, 2007
  12. May 11, 2007 #11
    Thanks any way. I will try to post it again in the maths box.
  13. May 11, 2007 #12


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    mmm... sounds like you wanna see how many ways (topologically inequivalent) you can build a tree with N distinct objects. A graph theory type problem i think...
    but firstly, what is the definition of the bulb being lit? could there be a situation when battery is not connected to any bulbs? or are you really saying that all N distinct objects must be included in the tree graph?

    anyway, it seems that a systematic way to do this is to first work out how many topological inequivalent graph can you have (assuming all items are the same first). then for each inequivalent graph work out the number of ways you can permute the N items. mm...not gonna be easy, perhaps graph theory may help
  14. May 11, 2007 #13
    You seem to make the question more clear. The problem here is that each bulb is different from any other bulb. That's why I have numbered it A1, A2..An. For this reason, just when you make it a circuit in series, meaning B-A1-A2...An and you permute the knots, there are (n+1)j!/2 ways. But you can also make it trees and the number rises rapidly, I can not find the final solution. Hope any one can help this.
  15. May 11, 2007 #14
    pixel01>> do we assume : B-A1-A2 is the same as A2-A1-B?
  16. May 11, 2007 #15


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    Looks like (n+1)!/2 to me. The 2 would come from B-A1-A2... being the same as ...A2-A1-B. Or am I wrong?
  17. May 11, 2007 #16
    Yes, that's right.
  18. May 11, 2007 #17
    (n+1)!/2 is true only for the case of series: B-A1-A2..An. If n>2, we can make it trees and the final number must add up.
  19. May 11, 2007 #18
    Hello, I am new to this forum so Hi everyone. I did not read all the replys to your question yet but the answer is 3. Series, Parallel and a combination thereof. I hope that is what you were looking for.
  20. May 11, 2007 #19
    If the drawing given was expected to work, it won't. You must have a loop of some sort or at least a ground for electron flow. Your diagrams you gave are not good. Sorry.
  21. May 12, 2007 #20


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    So that's why I suggested to first work out all inequivalent topology with same items for all nodes first, then work out how many permutation you can have for each topology,..eg: for 5 items





    then fill in the boxes... the first one 0-0-0-0-0, there are (n+1)!/2
    the second one you probably get 5 x n!/2 (you will have to check these)

    by the way, can you do a three or four way connection?


    do a few for n small first then you may be able to find a pattern and extrapolate
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