# Homework Help: Finding resistance R1

1. Apr 10, 2016

### Ivan Antunovic

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.

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2. Apr 10, 2016

### cnh1995

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.

3. Apr 10, 2016

### Staff: Mentor

Basically you want the phase angle of $10 - j30\sqrt{3} + R1$ to be -49.107°.

4. Apr 10, 2016

### Ivan Antunovic

You mean something like this?

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.

5. Apr 10, 2016

### cnh1995

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
6. Apr 10, 2016

### Ivan Antunovic

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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