Find R1: Resistance for Sinusoidal Current Network

  • Thread starter Thread starter Ivan Antunovic
  • Start date Start date
  • Tags Tags
    Resistance
Ivan Antunovic
Messages
109
Reaction score
4

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

  • Capture2.JPG
    Capture2.JPG
    3.9 KB · Views: 422
  • Untitled.png
    Untitled.png
    28.1 KB · Views: 482
Physics news on Phys.org
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?
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
6K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 26 ·
Replies
26
Views
4K