- #1

sandy.bridge

- 798

- 1

## Homework Statement

The four-mesh network has the mesh currents

[tex]I_1, I_2, I_3, I_4[/tex]

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

[tex]I_4=0[/tex]

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:

[tex]-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}}[/tex]

Since the potential across Zd is zero, we have:

[tex](-j\omega{L_x}I_3)/Z_C=I_2[/tex]

applying substitution we have,

[tex]I_1=\frac{(R+2Z_C)((-j\omega{L_x}I_3)/Z_C)}{Z_C}[/tex]

Next, around loop 3:

[tex]-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}[/tex]

The current I3 cancels, and algebraic manipulation results in:

[tex]L_x=\frac{1}{j\omega{^3}CR+2\omega{^2}-\frac{j\omega}{R_xC}}[/tex]

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?