Problem 31-2    Source:    Exercicie 17-3 - page 699   HAYT, William H. Jr. , KEMMERLY, Jack E. , DURBIN, Steven M. - Book:    Análise de Circuitos em Engenharia - Ed. McGraw Hill - 7ª Edição - 2008.

In the circuit shown in Figure 31-02.1, determine the parameters "Z".

Solution of the Problem 31-2

The Z parameters are given by the following equations:

We calculate Z₁₁ when we do I₂ = 0, that is, an open circuit in port 2. Therefore, if there is no current flowing at the exit of the quadripole, then Z₁₁ will be the serial-parallel association of the resistors that make up the circuit. Note that in this case, resistor of 20 ohms is in series with that of 5 ohms, having 25 ohms as a result of the serial association. This association, in turn, is in parallel with the resistor of 10 ohms. These three resistors result in an equivalent resistance of:

The resistor of 40 ohms is in series with R_eq. Soon:

To calculate Z₂₂, we should have I₁ = 0. Then, looking at the output of the quadripole, it is calculated the resistance that the circuit presents when I₁ = 0. Notice that the reasoning is the same. In this case, the resistor of 20 ohms is in series with that of 10 ohms and this set in parallel with that of 5 ohms . As a result, we obtain an equivalent resistance of:

Likewise, the 40 ohms resistor is in series with R_eq, then:

It's still missing to calculate Z₁₂ and Z₂₁. To calculate Z₁₂ we should have I₁ = 0 and we calculate the relation V₁ /I₂.So, setting a value for V₂ and calculating V₁ and I₂, as in the circuit shown ion the Figure 31-02.2, we get:

Note that we arbitrate a convenient value for V₂ of 44.286 volts, because we have already calculated the impedance Z₂₂ = 44.286 Ω. Thus, we easily calculate the value of I₂, whose result is1 A. Knowing that I₁ = 0 and I₂ = 1 A, we can find the value of V₁, because this will be the sum of the voltage drops in the resistors of 10 and 40 ohms, resulting the value of 41.43 volts . To calculate V₁, the current is calculated through the 20 ohms and 5 ohms resistors using a current divider. In the circuit shown in the figure above, all the values are found. Now we can calculate Z₁₂, or:

To calculate Z₂₁ let's take the same procedure. For that, let's adopt V₁ = 47.143 volts. This value is convenient because we get I₁ = 1 A. Note that at 40 ohms resistor , it will also pass 1 A, causing a drop of 40 volts on this resistor. Therefore, it is easy to calculate the other variables. In the Figure 31-02.3, we see the values found.

Knowing that I₁ = 1 A e V₂ = 41.43 volts, then:

Note that Z₁₂ = Z₂₁ and with this, it is concluded that the circuit is PASSIVE or RECIPROCAL.

Alternative Method

Another way to solve this problem is to use a Delta-Star transformation in the set of resistors of 20 Ω , 10 Ω and 5 Ω that form a Delta circuit. So, starting from the circuit Delta shown in the Figure 31-02.4 and arriving at the circuit Star shown in the Figure 31-02.5.

Replacing this transformation in the original circuit, we have the circuit shown in the Figure 31-02.6

Note that the value 41.43 is the result of the sum of 40 + 1.43 . Let's compare this circuit with our model for reciprocal circuits that can be seen in Theory. This way we can obtain the following relations:

When the calculation is made, we find the value of Z₁₁, or:

On the other hand, it is known that:

When the calculation is made, we find the value of Z₂₂.

Note that these are the same values as previously found, unless numerical rounding.