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Question about resistances please

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data

    just a questiion aren't the 1-ohm resistance(upper left) and the 3-ohm resistance in series? . The definition for resitance in series is "Two or more elements are in series if they are cascaded or connected sequentially " ,aren't they sharing the same wire? so why aren't they in series?the wire i'm talking about doesn't separate in two branches too , i'm confused
    look: View attachment 54025
    thanksssss
     
  2. jcsd
  3. Dec 17, 2012 #2

    lewando

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    Re: question about resistancesss please

    Yes, they are in series. The 1Ω and 5Ω connected at "b" are also in series.
     
  4. Dec 17, 2012 #3
    Re: question about resistancesss please

    Hi tahnks for your reply well i don't think they are in series because in the solution they don't use the rules of resistances in seires or parallel to solve the problem .check it out sdfgf.jpg . but i need to know why . if someone knows please explain
     
  5. Dec 17, 2012 #4

    lewando

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    Thanks for adjusting the problem statement :smile:. When you apply a voltage across a and b or a current through a and b, then there is a branch at a and at b, so they are not in series.
     
  6. Dec 17, 2012 #5
    actually the original figure doesn't show that the branch at point "a" bifurcates. it only appears in the solution , but i guess i finally get it , i think it's because black dots (nodes) mean that there is always a bifurcation . now everything makes sense , thanks !!!!!
     
  7. Dec 17, 2012 #6

    lewando

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    I read your original post literally. In my mind, a dot is a dot. To be more clear, show the attached voltage/current source. Then the branches will stand out. Glad you are clear!
     
  8. Dec 17, 2012 #7
    haha there is no a voltage source in this problem, the problem just say that we have to prove that equivalent resistance between "a" and "b" is 27/17 ohm.That's it. :smile:
     
  9. Dec 17, 2012 #8

    lewando

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    Well something was applied to inject a current, I, into a and outfrom b. Again, resulting in branches, therefore no series resistance.
     
    Last edited: Dec 17, 2012
  10. Dec 17, 2012 #9

    SammyS

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    If you consider what is meant by "equivalent resistance", then it is implied that you are finding what resistor could be placed between a & b and provide the same resistance to current flow as the given set of resistors. In other words, it's implied that there is a potential difference applied across a & b and an equivalent resistor would allow the same current to flow as given set of resistors.

    This set of resistors cannot be analyzed on the basis of series/parallel analysis, unless you first do Y-Delta transformation.

    The other common way to solve this is with Kirchhoff's circuit laws .
     
    Last edited: Dec 17, 2012
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