Phasor Addition Help

Thread Starter

ctishman

Joined Oct 12, 2011
9
So I'm trying to work my way through some of the exercises in my ac course, but I just can't seem to make it make sense. Anyhow, I was hoping someone would be able to help me out with it.

The example I'm given, and I'm quoting from this (terrible) textbook:

Introductory Circuit Analysis said:
It can be shown using the vector algebra described in section 14.10* that 1V\(\angle\)0° + 2V\(\angle\)90° = 2.236V\(\angle\)63.43°.

In other words, if we convert v\(_{1}\) and v\(_{2}\) to phasor form using

v = V\(_{m}\)sin(ωt ± θ) \(\Rightarrow\) V\(_{m}\)\(\angle\)±θ

and add them using complex number algebra, we can find the phasor form for V\(_{T}\) with very little difficulty.
Anyhow, the entire book reads like it was written from one engineer to another engineer, rather than to a student, but I won't get into that too much. I'm just trying to figure out what the heck they're doing and how it's supposed to work to add phasors together.

*note that section 14.10 says nothing about adding vectors in polar form, and everything I've seen seems to indicate that it's impossible unless the angles are multiples of 180°.
 

MrChips

Joined Oct 2, 2009
30,621
Firstly, know how to convert from polar coordinates to orthogonal coordinates (same as Cartesian or X-Y coordinates) and vice versa.

For example,

1V
0° + 2V
90° = 2.236V
63.43°

work backwards:

2.236V
63.43°

is

x = r cos(θ), i.e. x = 1

y = r sin(θ), i.e. y = 2

Thus, to add two vectors, convert to X-Y coordinates and add the X-values and the Y-values. Then convert to polar coordinates:

r = √(x^2 + y^2)

θ = atan(y/x)
 

Thread Starter

ctishman

Joined Oct 12, 2011
9
Okay, NOW it makes sense! I'd been banging my head about it for hours and had lost perspective. I really appreciate the help with that!
 
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