Find R1: Resistance for Sinusoidal Current Network

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Homework Statement


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.

Homework Equations

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.
 
Attachments
  • Circuit schematic of a series AC network: resistor R1, resistor R2 and reactance X2 in series between terminals marked + and U, with voltage U2 measured across R2 and X2.
    Capture2.JPG
    3.9 KB · Views: 437
  • Handwritten derivation of a sinusoidal circuit with phasor equations for U, U1, U2 and I, a phasor diagram in the upper left, and a final quadratic in R1 with complex coefficients.
    Untitled.png
    28.1 KB · Views: 503
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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.
 
Basically you want the phase angle of ##10 - j30\sqrt{3} + R1## to be -49.107°.
 
cnh1995 said:
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.
You mean something like this?
Handwritten derivation in complex phasor form: I = U / sqrt((R1+R2)^2 + X2^2) e^(j19), then U2 = I Z2 and U2 = U sqrt(R2^2 + X2^2) / sqrt((R1+R2)^2 + X2^2) e^(-j30), with U1 = I R1 and U = U1 + U2. A small red phasor diagram appears in the top left corner.

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.

gneill said:
Basically you want the phase angle of ##10 - j30\sqrt{3} + R1## to be -49.107°.
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.
 
cnh1995 said:
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.
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:
arctg(-X2/(R1+R2)) = -49.107
-30sqrt(3) = -1.155R1 - 11.547
R1 = 35 ohms.

I am courious what was wrong with my approach?