RLC Locus Diagrams Homework: Solve for Z,Y Variables

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The discussion focuses on solving for the port impedance (Z) and port admittance (Y) in a series circuit with resistance (R) and capacitance (C) at 50Hz. The user initially struggles with mapping the equations correctly, particularly in separating the real and complex parts. After some attempts, they clarify that Z should be expressed as R - jX and provide the correct formulas for G and B. The final solution involves completing the square to show that the loci represent a circle centered at (0, -1/2x) with a radius of -1/2x. The user successfully concludes their homework with this understanding.
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Homework Statement

In a series circuit V = V<0 @50hz with an R and C, show that the graph of the loci or the port impedance, Z, and port admittance, Y, as the resistance R is varied from 0 to inf ohms are as shown.
[URL]http://ivila.net/prob.png[/URL][URL]http://ivila.net/loci.png[/URL]

Homework Equations


The Attempt at a Solution



Z = R + j*X;
Y = Z-1;

Map to Y = G + jB (Separate Real and Complex)
G = R/(R2 + X2);
B = -X/(R2 + X2);

G2 + B2 = 1/(R2+X2) = B/-X

Then I get lost,

I know I need to arrange into the format below with the condition that the R term (varying from 0 to inf) remains in the LHS, ie, out of the radius. however I am stuck .
(G-x)2 + (B-y)2 = radius2

Any help would be much appreciated.
 
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I found the solution!

Firstly.

Z = R - jX (incorrectly stated in my first post)
Y = G + jB
G = R/(R^2 + X^2)
B = X/(R^2 + X^2)
G^2 + B^2 = 1/(R^2 + X^2) this is very close to B, infact B/X = G^2 + B^2

using this we can say.

G^2 + B^2 = 1/X * B
G^2 + (B^2 - B/X) = 0

Now Complete the Square

G^2 + (B^2 - B/X + a^2) = a^2
2a = -1/x, a = -1/2x

Therefore

(G^2) + (B + -1/2x)^2 = (-1/2x)^2

or, a circle, origin at 0,-1/2x radius -1/2x

Thanks!
Alex
 

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