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# Problem with phasors?

Hi.

See following:

124,13@0° + (298,37@-25,8°)(0,62+ j35,15) =

124,13@0°+ (298,37@-25,8°)(35,15@90°) =

(128,69 + j9,44) =

129,04@4,2°

In line 2 I know how they get 35,15 but not how they get the degree.

In line 3 and 4 I have no idea how they get what they get except I know how they get the degree in the last line.

Could someone please explain.

Thank you.

### 2 Answers

- godzyxLv 68 years agoFavorite Answer
Complex number conversion

Rectangular to polar:

x + jy = √(x² + y²) ∠ arctan (y / x)

Polar to rectangular:

A ∠ θ = A cos θ + j A sin θ

(0,62+ j35,15) = √((0,62)² + (35,15)²) ∠ arctan ((35,15) / (0,62)) = 34,16 ∠ 88.99°

For multiplication and division, convert to polar. For addition and subtraction, convert to rectangular.

As for your numbers, the formula in 2 does not equal the answer in 3, even if the last operation is division not multiplication. So you are on your own here.

- ?Lv 78 years ago
Here are my recommendations.

1. If you want to add phasors, convert them to rectangular coordinates and add the real parts together and the imaginary parts together. If you want to you can convert the result back to polar coordinates.

2. If you want to multiply 2 phasors, convert them to polar coordinates and multiply the magnitudes and add the phase angles. If you want to you can convert the result back to rectangular coordinates.

3. If you want to divide 2 phasors, convert them to polar coordinates and divide the magnitude of the numerator by the magnitude of the denominator. Then subtract the phase of the denominator from the phase of the numerator.