Phasor Calculation: Solve -43.62+j20.52

[Steve13579]

Homework Statement

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.

Homework Equations

tan^-1(X/R)

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?

 

[SteamKing]

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.

 

[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!

 

[SteamKing]

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!

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°

 

[Steve13579]

I forgot about that.. Thanks so much!