# Four-mesh network

1. Jan 15, 2012

### sandy.bridge

1. The problem statement, all variables and given/known data

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?

2. Jan 16, 2012

### 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.

3. May 13, 2013

### special

the electrician,

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

4. May 13, 2013