Problem + Hard 31-2    Source:    Question 1 - Electrical Circuits test - Engineering School - Ufrgs - 1975.

In the circuit shown in the figure below, it is known that the quadripole "B" has:

Furthermore, the quadripole "A" is passive and it is known that when the key "S" is in:

position 1 ⇒    I = 1 A   and   V_o = 10 volts

position 2 ⇒    V_o = 25/ 3 volts

position 3 ⇒    I₁ = 5 A

Using this information, obtain the Z parameters of the quadripole "A" and the constant K.

Solution of the Problem + Hard 31-2

Key "S" in position 1

With switch "S" in position 1 we have the circuit shown in the figure below. Depending on the value of I = 1 A, we show in the circuit the values ​​of the currents possible to be calculated by inspection.

In the resistor that is in parallel with the current source of 16 A, a current of 15 A will flow and this causes a voltage drop of 75 volts across this resistor. From the problem data, V_o = 10 volts. This allows calculating the value of V₁ by making the mesh indicated by the line dashed in orange in the figure above. Then:

Calculating the mesh with the source of 125 volts and V₁, you can calculate the value of I_{x< /sub>}. Like this:

So the value of I₁ will be:

The value of V_ae is:

Consequently:

So:

Now you can easily calculate the value of I₂, because:

And to calculate the value of V₂:

Now we have all the necessary values ​​to write the equations of the quadripole "A", remembering that it is passive. So, Z₁₂ = Z₂₁. To simplify the notation, we adopted Z = Z₁₂ = Z₂₁. Therefore:

Key "S" in position 2

With the switch "S" in position 1, the quadripole equation "A" was determined. Passing the key "S" to the position 2, you must find the value of the parameters Z. Therefore, in the circuit shown in the figure below, the values ​​of some calculated currents in the circuit are indicated simply by inspection.

The data provided by the problem statement is   V_o = 25/3 volts. Thus, the value of I will be equal to   I = (25/3) / 5 = 5/3 A. Then the current source value will be   3. I = 3 x (5/3) = 5 A. For another side, we have that V_ab = 25/3 volts. So V_ae = 50/3 volts. As a result, the current flowing across 5 ohms resistance is equal to 10/3 A. So the current between points m and a will be 5 A, which is exactly the current supplied by the current source. So, it is concluded that the current I₂ = 0. Furthermore, from the circuit V_ma = 25 volts and since V_ae = 50/3 volts, it is easily calculates the value of V₂, because:

From eq. 32-2a (shown above), we can write that:

Looking to the left of the quadrupole circuit "A", we get the following relation:

It is known that the quadripole is passive and Z = Z₁₂ = Z₂₁ was adopted. Furthermore, do not forget that I₂ = 0 and V₂ = 125/3 . Then, using the two previous equations and replacing by the values ​​found, simplify the equations of the quadripole parameters to:

We found a system of two equations with two unknowns. To solve it, divide the first equation by the second and get deletes the term I₁. Solving, the value of Z₁₁ is calculated, or:

Substituting this value in the Z equation, we find:

From eq. 31-2b, calculate the value of Z₂₂, or:

Therefore, the parameters Z of the quadripole "A" can be written as:

Key "S" in position 3

With switch "S" in position 3 the circuit is summarized as shown in the figure below. Note that on the right side of the quadripolo "B", a font transformation was performed. As stated in the problem I₁ = 5 A and so, V₁ is calculated by meshing the left side of the quadripole "A".

Knowing I₁ and V₁ it is possible to calculate the value of I₂ using the eq. 31-2c. Then:

You can calculate the value of V₂ using eq. 31-2d. Soon:

Making the mesh indicated by the arrow highlighted in green, the value of I is calculated.

So, I = I₃ = 9 A. V₃ is calculated by doing:

To find the values ​​of I₄ and V₄, meshing on the right side of the "B" quadripole:

Substituting the value of V₄ in the second quadripole equation "B" (provided in the problem statement), by the equation above, and substituting I₃ = 9 A, the value of I₄ is calculated, or:

Entering the values ​​of  I₃ = 9 A,   I₄ = 9 A   It is   V₃ = -5 V   in the first equation from the quadripole "B" (provided in the problem statement), the value of K can be calculated. Soon

Performing the calculate: