Phasors are vectors.

This issue has been discussed (argued?) at length in this very forum: http://forum.allaboutcircuits.com/showthread.php?t=15741

I'm not sure if there was a final resolution.

Vectors with standard angular velocity that rotate in the cartesian plane.

All my classes confirmed that and wikipedia seems to agree: http://en.wikipedia.org/wiki/Phasor
In fact, it reads "In physics and engineering, a phase vector ("phasor") is a representation of a sine wave whose amplitude (A), phase (θ), and frequency (ω) are time-invariant."

In another wiki-type page: http://en.wikibooks.org/wiki/Circuit_Theory/Phasors

we find this quote:

"Phasors will always be written out either with a large bold letter (as above), or will be written out with a vector notation, such as the letter \vec{V} This wikibook prefers the former notation for the simple reason that phasors are not vectors (or else presumably, we would call them "vectors")."

Actually, in EE textbooks written before, say, 1950 (they can still be found in your local university library), what we now call phasors were still given the name "vectors". The word phasor (and for a short period, sinor) hadn't yet been coined, and there was considerable discussion as to whether the concept was needed, and whether it differed from the vector concept.

 

If this is the case, I won't argue about it anymore. Let the mad scientists with the fuzzy hair sort it out themselvers. All I want to do is build things, not discover if the chicken hatched from the egg, or the egg was laid by the chicken (Greek proverb whose english match I do not know).

 

If I were to take the original problem and swap the inductor for a capacitor. Would I also swap out...
R+j\(\omega\)L
for
R-j\(\omega\)C

or would addition still be necessary

 

If I were to take the original problem and swap the inductor for a capacitor. Would I also swap out...
R+j\(\omega\)L
for
R-j\(\omega\)C

or would addition still be necessary

Instead of the term jωL you'd want a term 1/jωC like this:

\(I1 = \frac{V1}{R+\frac{1}{j \omega C}}\)

The j in the denominator will cause the sign of that term to change when you move the j to the numerator.

Continue as before.

 

so if I had \(\frac{1}{j0.0188}\)

that would be equivalent to -j0.0188


edit: or would it be equivalent to -j53.19

 

so if I had \(\frac{1}{j0.0188}\)

that would be equivalent to -j0.0188


edit: or would it be equivalent to -j53.19

The latter.