Electrical: Reduce the expression to the form V_mcos(wt+theta)

Click For Summary

Homework Help Overview

The problem involves reducing the expression 15sin(wt-45°) + 5cos(wt-30°) + 10cos(wt-120°) to the form Vmcos(wt+θ), focusing on the conversion of sine and cosine terms into phasor form and subsequent calculations of magnitude and angle.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting sine terms to cosine form and the implications of using phasor representation. Questions arise regarding the correct quadrant for the resulting phasor based on the signs of the components.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the calculated values and questioning the correctness of the angle derived from the phasor's position. Guidance is offered regarding quadrant determination and the use of specific functions for angle calculation.

Contextual Notes

Participants are navigating potential discrepancies in their calculations and the expected results, indicating a need for careful consideration of phasor representation and angle conventions.

Northbysouth
Messages
241
Reaction score
2

Homework Statement




Reduce the Expression:

15sin(wt-45°) + 5cos(wt-30°) + 10cos(wt-120°)

to the form Vmcos(wt+θ)

Homework Equations





The Attempt at a Solution





My theta value at the end isn't coming out right.

My first step was to put make sure each term was in terms of cosine, hence:

15cos(wt-45°-90°) + 5cos(wt-30°) + 10cos(wt-120°)

Which in phasor form is:

15∠-135 + 5∠-30 + 10∠-120

Then I used cos and sin to find the real values and the j values

15cos(-135) + j15sin(-135) + 5cos(-30) + j5sin(-30) + 10cos(-120) + j10sin(-120)

= -11.28 -j21.77

Which in a phasor format is:

24.52∠62.6

24.53 = (11.282+ 21.772)1/2

62.6 = arctan(21.77/11.28)

But apparently the solution is:

24.52∠-117.39°

What am I missing?

Any help would be appreciated. Thank you
 
Physics news on Phys.org
Note the signs of the complex components of your phasor sum. In what quadrant will the phasor lie?
 
Are you saying that because I have -11.28 - j21.77 as negatives that the phasor will be in the third quadrant and consequently the angle made with the positive x-axis will be 180-62.6?
 
Northbysouth said:
Are you saying that because I have -11.28 - j21.77 as negatives that the phasor will be in the third quadrant and consequently the angle made with the positive x-axis will be 180-62.6?

The phasor will be in the third quadrant, yes. You can either add 180 or subtract 180, but generally the convention is to choose the result with a magnitude less than or equal to 180. Alternatively, employ the atan2(y,x) function rather than the atan function (if your calculator has it); it automatically sorts out the quadrant from the given y,x data.