Working with phasors (Circuits, such as complex power)

[Cocoleia]

Homework Statement

I am going over examples in my textbook and I came across this:

I don't understand how they converted 18.265 at angle of 39.9 to 14.02+j11.71

Homework Equations

I know how to convert from the imaginary numbers into the angle form, usually I use:

Is there another equation when going in the other direction, or do I use the same ones. I will have two equations and two unknowns, one with tan and one with the square root? It seems a bit complicated and I feel like I am missing out on something, but I can't find it on my formula sheet or in my notes so I am a bit confused. Thanks !

 

[gneill]

It's akin to converting a vector in polar form to rectangular form. Use cos and sin to extract the real and imaginary component magnitudes.

 

[Cocoleia]

It's akin to converting a vector in polar form to rectangular form. Use cos and sin to extract the real and imaginary component magnitudes.

I will use something like this

even if I don't have an exponential ?

 

[Cocoleia]

It's akin to converting a vector in polar form to rectangular form. Use cos and sin to extract the real and imaginary component magnitudes.

Ok, Ok. I take the cos of the angle and multiply by the coefficient that becomes the real part, and then I take sin and multiply which is the imaginary part?

 

[gneill]

I will use something like this
View attachment 195299
even if I don't have an exponential ?

Yes. For phasors the exponential is implied, and we use the magnitude and phase angle for its shorthand notation.

The full form of the phasor is ##P = A e^{j(ω t + Φ)}##. The "ωt" part represents the rotating motion of the phasor. Splitting it: ##P = A e^{jωt} e^{j Φ}##. When the angular frequency ω is the same for all phasors in a system we just drop the rotating component from the notation and take it as implied. That leaves ##A e^{j Φ}## as the unique part of the phasor, and that can be represented by a complex number (rectangular form) or magnitude and angle (polar form) in phasor "shorthand".

Ok, Ok. I take the cos of the angle and multiply by the coefficient that becomes the real part, and then I take sin and multiply which is the imaginary part?

Yes.