Problema 32-3    Source:    Problem 9 of the Quadripoles section of the list of Electrical Circuits I, of the School of Engineering of UFRGS, 2011, Prof. Dr. Valner Brusamarello.

In the circuit shown in Figure 32-03.1, determine the parameters "Y"

Solution of the Problem 32-3

The Y parameters are given by the following equations:

We calculate Y₁₁ and Y₂₁ making E₂ = 0, that is, a short circuit at the output. Connect a voltage source (or current) E₁ on the input port, as shown in the Figure 32-03.2.

We adopted E₁ = 10 volts (just a convenient value, it could be any other value), in order to find a relationship between I₁ , I₂ and i. Note that the resistor of 4/19 ohms can be eliminated because the short-circuit is in parallel with it. See the Figure 32-03.3.

By making the mesh of the circuit highlighted in green, in the clockwise direction, as indicated by the blue arrow, the following relation is found:

Therefore, solving algebrically, we obtain:

On the other hand, observing the circuit, we conclude that:

Using the Kirchhoff's Law for the currents that arrive and leave the node a, we have:

Relating the last two equations and working algebraically the same, we find:

See in the Figure 32-03.4, the circuit drawn with the values found. Note that we replaced the font 3 K by a font 3 i , inverting its polarity, because K = - i . Making the mesh marked by the arrow in red, in the clockwise direction, in the highlighted part in green, we find:

Solving algebrically, we have:

But we know that I₁ = I₂ and that i = 2 I₂ then replacing this information in equation above, and solving it, we obtain:

Now, with these informations, we easily calculate the values of Y₁₁ and Y₂₁. Then:

In the same way, to calculate Y₂₂ and Y₂₁, we must have V₁ = 0, that is, a short-circuit at the input (port 1). Let us place a voltage source at the output (port 2) with a convenient value, for example E₂ = 4 V. In this way, a current of 19 A will circulate by the resistor of 4/19 ohms. The modified circuit and the values of the variables can be seen in the Figure 32-03.5.

In the highlighted part of yellow and making the mesh indicated by the arrow red clockwise, we have:

Solving algebrically, we have:

From the circuit, it is known that by the resistor of 2 ohms passes a current I₁ - i, generating the voltage V. Using these last two informations, we can write:

In the Figure 32-03.6, see how the circuit was. Making the mesh in the highlighted part in green counterclockwise, as indicated by the arrow in red, we have:

Solving algebraically we find the value of I₁, or:

Now, we easily calculate I₂, because it is known from the circuit that:

Now, we calculate the values of Y₂₂ and Y₁₂, that is: