Kirchoff laws for phasors

1. Nov 23, 2004

learningphysics

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

v1(t)+v2(t)+....=0

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

So we have
A1cost(wt+s1)+A2cost(wt+s2)+....=0

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. Nov 23, 2004

learningphysics

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
3. Nov 24, 2004

learningphysics

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

If we know
A1cos(wt+s1)+A2cos(wt+s2)+....=0

then take the derivative of both sides

-A1wsin(wt+s1)-A2wsin(wt+s2)+...=0

divide by -w on both sides:

A1sin(wt+s1)+A2sin(wt+s2)+....=0

so the imaginary parts in phasor notation add to zero.