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Finding resistance R1

  1. Apr 10, 2016 #1
    1. The problem statement, all variables and given/known data
    In the network of sinusoidal current , R2 = 10 ohm , X2 = -30*sqrt(3) ohm. Find the resistance R1 so that voltage U2 is in phase delay behind voltage U for angle alpha = - pi / 6.


    2. Relevant equations


    3. The attempt at a solution
    Z2 = sqrt(R^2 + X2^2) = 20*sqrt(7) ohms
    fi2 = arctg(X2/R2) = -79. 107 degrees
    fi = 30 + fi2 = -49.107 degrees

    the rest is in the picture below.

    The problem is that this last equation can only give complex solutions and R1 should be in real domain.
     

    Attached Files:

  2. jcsd
  3. Apr 10, 2016 #2

    cnh1995

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    Homework Helper

    See if this works:
    Take U as reference(∠0°). You already have angle of Z2 w.r.t. U. You also have angle of U2 i.e I*Z2. Write an expression for current I which includes R1 and from the known angles, calculate R1.
     
  4. Apr 10, 2016 #3

    gneill

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    Staff: Mentor

    Basically you want the phase angle of ##10 - j30\sqrt{3} + R1## to be -49.107°.
     
  5. Apr 10, 2016 #4
    You mean something like this?
    image.png
    uploading pictures
    I put I at angle + 19 degrees so that U2 is at phase delay 30 degrees behind U.But still I am stuck with complex numbers.

    Yes if I put I at zero degrees Z = U / I = Z * exp(j*(-49)) but can't figure out how to use that information.
     
  6. Apr 10, 2016 #5

    cnh1995

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    This gives exactly what gneill said in #3.
    Z2 is at an angle -79.107°. I*Z2 is at angle -30°. So, I must be at an angle 49.107°.
    Since,
    10-j30√3+R1=U∠0°/I∠49.107°,
    angle of (10+R1)-(j30√3) is -49.107°.
    How do you compute angle of a complex number a±ib?
     
    Last edited: Apr 10, 2016
  7. Apr 10, 2016 #6
    arctg(-X2/(R1+R2)) = -49.107
    -30sqrt(3) = -1.155R1 - 11.547
    R1 = 35 ohms.

    I am courious what was wrong with my approach?
     
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