Find R1: Resistance for Sinusoidal Current Network

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The discussion revolves around calculating the resistance R1 in a sinusoidal current network where R2 is 10 ohms and X2 is -30√3 ohms, ensuring that voltage U2 lags voltage U by -π/6 radians. Participants explore the phase angles of the impedances and currents, aiming to express the relationship between R1 and the known angles. The correct approach involves setting the angle of the complex expression (10 - j30√3 + R1) to -49.107°, leading to the calculation of R1 as 35 ohms. The original poster expresses confusion about their initial method, prompting clarification on how to compute the angle of a complex number. Overall, the focus is on deriving R1 while addressing the complexities of phase angles in AC circuits.
Ivan Antunovic
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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.
 

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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?
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.

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?
 

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