Problem 15-5    Source:    Problem 5.18 - page 115 -    JOHNSON, David E. , HILBURN, John L. , JOHNSON, Johnny R. -    Book: Fundamentos de Análise de Circuitos Elétricos - 4ª edição - Ed. Prentice Hall do Brasil - 1994.

Calculate v by replacing everything except the 4 ohms resistor, by its equivalent of Thévenin in the circuit shown in Figure 15-05.1

Solution of the Problem 15-5   -    Thévenin/Norton Method

Since we can not use the 4 ohms resistor, let's remove it from the circuit. Removing the resistor, the terminals a-b appear. From these terminals we must calculate the Thévenin equivalent for the circuit. In this way, we can make a star-delta transformation between points a - b - d . With the delta circuit, all resistors will be in parallel with the current sources. Then, after performing these transformations, we arrive to the circuit shown in the Figure 15-05.2. Note the new resistor values after the star-delta transformation. This transformation is necessary since we are interested in calculating the open circuit voltage at the terminals a-b . This voltage will be the Thévenin voltage .

Paying attention to the new configuration, we realize that we can carry out transformation of sources at all current sources, as they have a resistor in parallel. Therefore, after the transformations we have the circuit shown in the Figure 15-05.3.

Note, in the circuit of the figure above, that all voltage sources are in series and their polarities point in the same direction. So we can sum the values of all of them. We will call this value of V_total = 152 + 190 + 57 = 399 volts. In the same way we will do for the resistors, because all are in series. Then, by adding them up, we find R_eq = 12.67 + 38 + 9.5 = 60.17 ohms. Since we are looking for the Voltage of Thévenin for the points a-b , we can calculate the current I (as indicated by the green arrow) that flows through the circuit, that is:

Solving the equation, we find the value of I, or:

Now, by making the mesh equation for the voltage source of 57 volts , the resistor of 9.5 ohms and V_th, counterclockwise, we have:

Solving this equation, we find the value of V_th. So:

Now we need to calculate the value of R_th. To do this, simply eliminate all the voltage sources that appear in the circuit of the figure above. Putting all voltage sources in short circuit , we realize that the resistor of 9.5 ohms is in parallel with the other two which are in series. Therefore, R_th is:

We are already able to assemble the equivalent of Thévenin and calculate the value of v. See Figure 15-05.4 for the final circuit.

To find the value of v, calculate the value of I.

Then, the value of v is: