Solve 4-Mesh Network for Rx & Lx w/ R, C & Source Freq.

[sandy.bridge]

Homework Statement

The four-mesh network has the mesh currents
I_1, I_2, I_3, I_4
in the indicated regions 1, 2, 3 and 4 respectively. In this circuit, the resistor R and the identical capacitors (Zc) are adjusted such that
I_4=0
For this condition, the unknowns Rx and Lx can be expressed in terms of R, C and the source frequency (rad/s). Find the expressions for Rx and Lx.

What I did was I set up loop equations for loop 2, 3 and 4. I neglected loop 1.

Loop 2:
-Z_CI_1+(R+2Z_C)I_2-Z_CI_4=0\rightarrow{-Z_CI_1+(R+2Z_C)I_2=0}\rightarrow{I_1=\frac{(R+2Z_C)I_2}{Z_C}}

Since the potential across Zd is zero, we have:
(-j\omega{L_x}I_3)/Z_C=I_2
applying substitution we have,
I_1=\frac{(R+2Z_C)((-j\omega{L_x}I_3)/Z_C)}{Z_C}

Next, around loop 3:
-R_xI_1+(R_x+j\omega{L_x})I_3=0\rightarrow{-R_x(\frac{(R+2Z_C)((-j\omega{L_x}I_3)/Z_C)}{Z_C})+(R_x+j\omega{L_x})I_3=0}
The current I3 cancels, and algebraic manipulation results in:
L_x=\frac{1}{j\omega{^3}CR+2\omega{^2}-\frac{j\omega}{R_xC}}
No matter what I do, I cannot seem to get Lx in terms of simply C, R and the angular frequency; that is, the expression always has Rx when solving for Lx, and Lx when solving for Rx.

Any suggestions?

 

[The Electrician]

Replace Rx and Lx with a single impedance Zx and eliminate loop 4. Solve for Zx in terms of the other components. That result will be a complex impedance; convert it to an admittance Yx by taking the complex reciprocal. Rx will be the inverse of the real part of Yx and the imaginary part can be converted to an inductance Lx by use of the radian frequency.

 

[special]

the electrician,

can you please show the working out. I've have got the answers but am unsure of my working out.

 

[The Electrician]

Go have a look at this thread:

http://forum.allaboutcircuits.com/showthread.php?t=84745