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Kirchoff laws for phasors

  1. Nov 23, 2004 #1


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    Something's bugging me. Suppose we take kvl around a loop in a circuit, we have:


    Suppose v1, v2, v3(t) are all sinusoidal (they can be written as Acos(wt+s)).

    So we have

    Suppose we replace all of them by their phasors, this should also equal zero but why? I'll write it out here (without suppressing e^jwt, but just adding the imaginary parts)

    (A1cos(wt+s1)+A2cos(wt+s2)+....) + j(A1sin(wt+s1) + A2sin(wt+s2)+....)

    If I know the group of real terms add to zero, does that necessary imply that the group of imaginary terms add to zero? Is there a proof of this?
  2. jcsd
  3. Nov 23, 2004 #2


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    http://www.csupomona.edu/~zaliyazici/f2001/ece209/ece209-02.pdf [Broken]

    I just saw a very unsatisfying derivation of kvl for phasors here. How do you just remove the "R", and assume kvl holds for the entire phasor (not just the real part). I saw a similar technique elsewhere. They write kvl for the reals, and then R {V1+V2+...}=0 (where V1, V2 are phasors), then just remove the R. This removing of the R is bothering me. I don't see how it is justified.
    Last edited by a moderator: May 1, 2017
  4. Nov 24, 2004 #3


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    Figured out the answer. It's simple. I wish they just put it in the proof in my book:

    If we know

    then take the derivative of both sides


    divide by -w on both sides:


    so the imaginary parts in phasor notation add to zero.
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