Solve AC Voltage Sources Series Problem: Find U_AB Mag. & Phase Angle

[Korisnik]

Homework Statement

If we have two AC sinusoidal voltage sources in series with +ive poles in the same direction (A... +u1- ... +u2-...B),
Find voltage phasor $\underline{U}_{AB}=?$ (underlined means phasor, complex) - magnitude and phase angle, and find momentary value of voltage $u_{ab}$ at moment t.$u_1=U_{1m}\sin(\omega\cdot t+\alpha_1)\\ u_2=U_{2m}\sin(\omega\cdot t+\alpha_2)$,

$U_{1m}=3V\\ U_{2m}=9V\\ f=52Hz\\ t=4\cdot 10^{-3} s\\ \alpha_1=1,3 rad\\ \alpha_2=0,1 rad$

The Attempt at a Solution

Now, I've found momentary value, (I don't know how to translate this well, it's the value of voltage in that moment t = 4ms), just by plugging in then adding both voltages and the answer is correct (10, 41V). Then I transformed both voltages in complex form, then added real and imaginary parts, got the value of the phasor at that moment (the magnitude, 7,4V (which is basically U_ab/sqrt(2)), but when I divide Im(Uab) by Re(Uab) then make tangent^-1 of that, I get the wrong angle. I get the angle of 6°... and the right is 21,22°. can someone explain the procedure after I've gotten the phasor in complex form. (x + yj, where j = i = sqrt(-1)).

Ill write my attempt:
This is U1: http://m.wolframalpha.com/input/?i=3(sin(104*pi*4*10^-3+1.3)+i*cos (104*pi*4*10^-3+1.3))&x=0&y=0
And U2
http://m.wolframalpha.com/input/?i=9(sin(104*pi*4*10^-3+0.1)+i*cos+(104*pi*4*10^-3+0.1))&x=0&y=0

Now adding these 2 together:
Uab=10.408 - j*1.113 V
Magnitude is square root of real and imaginary squares: Uab=10.4673
Effective: 10.4673/sqrt2=7.4015

But now when I am looking for the angle: arctg (1.113/10.408)=6.1º.

What should I do?

 

[NascentOxygen]

You represented sin(wt+ɸ) as sin( ) + i.cos( )
but this means that at t=0 the angle would be arctan(cos / sin)
when you know it must be arctan(sin / cos)

So, try representing sin(wt+ɸ) as cos( ) + i.sin( )

I'm a bit vague on this myself.

 

[Korisnik]

You represented sin(wt+ɸ) as sin( ) + i.cos( )
but this means that at t=0 the angle would be arctan(cos / sin)
when you know it must be arctan(sin / cos)

So, try representing sin(wt+ɸ) as cos( ) + i.sin( )

I'm a bit vague on this myself.

True, I wasn't even looking at this, I was just looking at the numbers.

Again, I get correct magnitude (r) of phasor but not the angle (I get 96° and not 21°):
http://www.wolframalpha.com/input/?...104pi*4*10^-3++0.1)+i*sin(104pi*4*10^-3+0.1))http://i.imgur.com/GwkylxI.png?1

This is the picture, so if there's something I've missed...

 

[NascentOxygen]

Have you checked that 21° is the right answer?

Jump to the foot of this article on the maths behind the addition of two sinewaves: http://2000clicks.com/mathhelp/GeometryTrigEquivPhaseShift.aspx

 

[Korisnik]

Have you checked that 21° is the right answer?

Jump to the foot of this article on the maths behind the addition of two sinewaves: http://2000clicks.com/mathhelp/GeometryTrigEquivPhaseShift.aspx

I've figured it out, it has to be when t=0ms, because it's a phasor. In order to fully describe it, you need to specify maximum (or effective, RMS value, as was specified here) voltage and starting, phase angle alpha. So I got 21,22°.