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AC Voltage sources

  1. Nov 14, 2014 #1
    1. The problem statement, all variables and given/known data
    If we have two AC sinusoidal voltage sources in series with +ive poles in the same direction (A... +u1- ... +u2-...B),
    Find voltage phasor [itex]\underline{U}_{AB}=?[/itex] (underlined means phasor, complex) - magnitude and phase angle, and find momentary value of voltage [itex]u_{ab}[/itex] at moment t.

    [itex]u_1=U_{1m}\sin(\omega\cdot t+\alpha_1)\\ u_2=U_{2m}\sin(\omega\cdot t+\alpha_2)[/itex],

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

    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 Im looking for the angle: arctg (1.113/10.408)=6.1º.

    What should I do?
  2. jcsd
  3. Nov 14, 2014 #2


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    Staff: Mentor

    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.
  4. Nov 15, 2014 #3
    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°):


    This is the picture, so if there's something I've missed...
    Last edited: Nov 15, 2014
  5. Nov 15, 2014 #4


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    Staff: Mentor

  6. Nov 15, 2014 #5
    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°.
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