# Solving for y parameters using a system of equations

simulate this circuit – Schematic created using CircuitLab

Question: Find the y parameters for the circuit shown below.

My attempt:

KCL

$$\I_1 = I_3 + I_6 \$$

$$\I_4 = I_3 + I_5\$$

$$\I_5 = I_2 + I_6\$$

EFC:

$$\I_3 = \frac{V_1 - V_a}{20}\$$

$$\I_4 = \frac{V_a-0}{10}\$$

$$\I_5 = \frac{V_2 - V_a}{4}\$$

$$\I_6 = \frac{V_1 - V_2}{8}\$$

Solving this systems for $$\ V_1 \$$ and $$\ V_2\$$ yields:

$$\ V_1 = \frac{35}{2} + \frac{25}{2} \$$

$$\ V_2 = \frac{25 }{2} + \frac{27}{2} \$$

Therefore our Z parameters are as follows:

$$\begin{bmatrix} \frac{35}{2} & \frac{25}{2} \\ \frac{25}{2} & \frac{27}{2} \end{bmatrix}$$

I know how to find the y parameters using delta Z, but I thought there was a way to solve for these items directly in the system of equations. Can I solve for the y parameters directly?

• y-parameters is just admittance instead of impedance, so ofcourse it is possible. But why do it if its easier for you to find the z parameters first, and then convert it? Commented Nov 30, 2018 at 20:38
• @Linkyyy I solved it this way as it was the only way I knew how. I would rather solve for y instead of z in the system of equations Commented Nov 30, 2018 at 20:40
• @Linkyyy are you saying, in general, it is easier to do it the way I'm doing it now? Commented Nov 30, 2018 at 21:03
• Look up the definition of y parameters, and you will see that it is not that dificult when you already understand what z parameters is. The difference is you short the opposite port, instead of having it open circuited. Then you calculate the admittance instead of the impedance (which is really just the reciprocal of each other) Commented Nov 30, 2018 at 21:08

hint: $$\ Y_{11} = \frac{I_1}{V_1} \$$ when you short-circuit $$\ V_2 \$$.
This arises from $$\ I_1 = V_1 Y_{11} + V_2 Y_{12} \$$. The short-circuited $$\ V_2 \$$ forces the $$\V_2 Y_{12} \$$ term to zero.
The other equation that applies is $$\ I_2 = V_1 Y_{21} + V_2 Y_{22} \$$ .