Problema 32-4    Source:    Problem 11 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 32-04.1, when E₂ = 0 results I₁= 4 and the ammeter reads 2 A. Determine the parameters "Y".

Solution of the do Problem 32-4

The Y parameters are given by the following equations:

The problem statement says that E₂ = 0. This means that we can short-circuit the output. In this case, I₂ = - 2 and observing the circuit, we conclude that i = - I₂ = 2. Thus, the source of current is   4 i = 4 x 2 = 8 A. Then the current through the resistance R it's equal to   4 i + I₁ = 8 + 4 = 12 A.

Since we know the value of i, then we can calculate the value of the voltage source V /3. To do so, making the mesh indicated by the arrow orange, and as the output is shorted, then - V/3 + 2 i = 0. We know that i = 2 A , we conclude that V = 12 volts. Obviously, V / 3 = 4 volts. See the Figure 32-04.3, the circuit situation.

Note that E₁ = V, then we have all the values we need to calculate Y₁₁ and Y₂₁. However, just to complement the problem, note that after finding the value of E₁ and knowing the current that pass through the R resistor, we can calculate its value, or   R = 12 / 12 = 1 ohm.

In the same way, to calculate Y₂₂ and Y₂₁, we must have E₁ = 0, that is, a short-circuit at the input (port 1). If E₁ = 0, then V = 0. Thus, on the right side of the circuit, instead of the voltage V / 3, we have a short circuit.

Let us place a voltage source at the output (port 2) with a convenient value, for example 2 volts. In this way, will pass a current of 1 A for each resistance of 2 ohms. Therefore, it is concluded that i = - 1 A and I₂ = 2 A. Thus, on the left side of the circuit, the current source has a value of - 4 A. See the Figure 32-04.3, that the direction of the current source has been inverted due to the negative signal of the current.

As we have all the necessary information, we can calculate Y₂₂ and Y₁₂.