Problem + Hard 11-4    Source:    Adapted Question 2 - Test of Electrical Circuits - Computer Engineering - UFRGS - 2016.

In the circuit in the Figure 11-04.1, find the values ​​of I_y, V_x and V_o.

Solution of the Problem + Hard 11-4   -    Transf. of Sources

Initially, we can carry out two transformations: the first, highlighted by the red rectangle, transforming the circuit delta in a star circuit; the second, highlighted by the brown rectangle, combine the fonts into a single one, remembering that the 18 ohm resistor, for now, can be eliminated, as it is in series with a current source. In this way, using the eq. 5-1 for the Delta → Estrela and combining the sources we have the circuit in the Figure 11-04.2.

Note that the star circuit, highlighted in the red rectangle, is equivalent to the delta circuit of the initial circuit. The part of circuit highlighted in the brown rectangle, is equivalent to the sources of the initial circuit, also highlighted in brown. To find the value of the dependent source on the right, we must make the branch highlighted by the green arrow in the circuit above and determine the relationship that exists between I_y and V_x. Therefore:

Then the source of 120 V_x, after replacing the value found above, turns into 720 I_y - 720. When we transform to a current source in parallel with the resistor, the value of the current source is equal to 120 I_y - 120 (since we divide by the value of the resistor). Of this amount, we subtract the source 3 I_y, as it points in the opposite direction. We found the value 117 I_y - 120. After this step, we should have a voltage source in series with a resistor. To do this, we multiply the value of current source by the value of the 6 ohms resistor, finding the value that appears in the circuit above, this value being equal to 702 I_y - 720.

Now, in the circuit above, notice that the two 6 ohms resistors are in series. Results a resistor of 12 ohms. We are back to a current source of value 58.5 I_y - 60. And the two 12 ohms resistors were in parallel, resulting in an equivalent resistor of 6 ohms. Doing one more source transformation, we get the circuit in the Figure 11-04.3.

Based on the above equation, we can calculate the current flowing through the 3 ohms resistor. This current depends on I_y, whose value is 2 I_y - 2, as indicated in the circuit above. Therefore, the current flowing through the 6 ohms resistors is 3 I_y - 2. So, we can write the equation mesh that includes the dependent source. Thus, we have:

Rearranging the equation, we find:

So, we easily find the value of I_y, or:

As we already know the relationship between I_y and V_x, then:

And from the initial circuit, we easily relate V_o and I_y, that is: