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Kirchoff's Law

  1. Jun 6, 2004 #1
    I'm having a problem with Kirchoff's law because It seems it can be interpreted 2 ways. Perhaps they both arrive at the same answer, but if you could help me out as to which way to do it or if it matters.
    1. the Integral[E*dl]=0, in a closed loop
    2. the algebraic sum of the voltages in a closed loop equals zero.

    The difference arrises in the sign. For instance, for the Integral[E*dl] going across a voltage source is -V because you are going against the electric field. But for the algebraic sum of voltages you are going to a higher potential so it is +V. Similarly for resitors the electric field (Integral[E*dl]) gives a +V, but for algebraic sum of voltages it is -V because there is a loss in potential. I don't understand I have never heard a explanation for this in my EE circuits class or physics class. What do you think?
     
  2. jcsd
  3. Jun 6, 2004 #2
    It doesn't matter because in both case the answer is zero. If you want you could write instead:
    [tex]-\oint \underline E . \underline dl = 0[/tex]
    which is exactly the same thing and corresponds to the second form you have there.

    Matt
     
  4. Jun 6, 2004 #3

    turin

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    Homework Helper

    I concur with baffledMatt.

    I do want to mention, though, that the first form that you have mentioned will lead to a non-trivial violation for high speed circuits. There are various generalizations that will allow it to show up in the second form, but it is better explained by a generalization of the first (called Faraday's Law). If you are an EE undergrad, you may never run into it. I think they give you the criterion for being able to use the circuit theory that you are learning as applying to circuits smaller than about a tenth of a wavelength. This takes care of the low speed approximation.
     
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