Phasor Calculation: Solve -43.62+j20.52

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Discussion Overview

The discussion revolves around calculating phasor notation for complex numbers, specifically focusing on the conversion of Cartesian coordinates to polar form and the implications of quadrant placement on angle determination. Participants explore the calculation of angles and magnitudes in phasor representation, as well as the nuances of dividing phasors with zero phase angles.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Conceptual clarification

Main Points Raised

  • One participant calculates the angle for the phasor -43.62+j20.52 using tan-1(20.52/-43.62) and arrives at -25.19 degrees, questioning why the answer should be adjusted by 180 degrees.
  • Another participant suggests that drawing a sketch of the phasor could clarify the angle determination and emphasizes the importance of considering the quadrant when using arctan.
  • A participant reflects on the limitations of calculators in computing arctan and presents a related problem involving the division of phasors, expressing confusion about the phase angle of 0 and its implications for the calculation.
  • Further clarification is provided regarding the representation of -2∠0° and the correct approach to dividing phasors, noting that the magnitude should always be positive in phasor notation.

Areas of Agreement / Disagreement

Participants express various viewpoints on the calculation of angles and the handling of phase angles, indicating that multiple competing views remain regarding the correct approach to phasor calculations.

Contextual Notes

Participants discuss the limitations of using calculators for angle calculations and the need to consider quadrant placement, which may not be fully resolved in their explanations. The nuances of representing negative resistances and phase angles are also highlighted but not definitively settled.

Who May Find This Useful

This discussion may be useful for students and practitioners in electrical engineering or physics who are learning about phasor notation and complex number calculations.

Steve13579
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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?
 
Last edited:
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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.
 
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!
 
Steve13579 said:
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°
 
I forgot about that.. Thanks so much!
 

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