# Homework Help: Phasor calulations

1. Apr 29, 2014

### Steve13579

1. The problem statement, all variables and given/known data
Calculate the phasor notation for -43.62+j20.52
My answer is 180 degrees off and I don't know why you add it in this case. I just want to know how to calculate angle, the magnitude I found fine.

2. Relevant equations
tan-1(X/R)

3. The attempt at a solution
tan-1(20.52/-43.62) = -25.19 degrees
That's the answer I get but the answer is 180 degrees plus my result above resulting in 154.8 degrees

edit: I'm guessing it may be because it's in quadrant 2 or 3?

Last edited: Apr 29, 2014
2. Apr 29, 2014

### SteamKing

Staff Emeritus
If you will draw a simple sketch of your phasor, the answer should appear immediately.

Remember, doing a simple arctan calculation on a calculator returns only the principal angle θ such that -π/2 ≤ θ ≤ π/2. You must examine the components of a particular phasor to determine the proper quadrant.

3. Apr 29, 2014

### Steve13579

Got it, makes sense! It helped to think of the limitations of my calculator computing arctan with only one value input rather than two if that makes sense...

If you wouldn't mind I came across something that is probably a similar situation.
I have -2∠0°/(0.45-j0.15) which I turned into -2∠0°/0.474∠-18.435° and solved resulting in: -4.22∠18.435°
But apparently you can not do that... the answer is 4.22∠-161.75°
I can get that answer too by writing the polar notation of -2∠0° as -2 and than dividing by 0.45-j0.15. Can I not solve the way I initially tried because of a phase angle of 0? It has no reluctance and only a real resistance of -2, well not really a negative resistance but ya. Is that the reason I can not try what I did? Thanks!

4. Apr 30, 2014

### SteamKing

Staff Emeritus
Remember, -2∠0° = 2∠180°

You always want the first number in phasor notation to represent the magnitude of the phasor, hence it is always positive.

Your division problem would then be

(2/0.474)∠(180°-(-18.435°)) = 4.22∠198.435° = 4.22∠-161.55°

5. Apr 30, 2014

### Steve13579

I forgot about that.. Thanks so much!