Problem 15-6    Source:   Test 1 - Problem 3 - Electrical Circuits I - UFRGS - 1975.

In the circuit shown in Figure 15-06.1, calculate current I_B when the S key is closed, knowing that circuit B is represented by the graph of the Figure 15-06.2

Solution of the Problem 15-6   -    Thévenin/Norton Method

When we analyze the graph provided by the problem, we conclude that it represents a resistor of value equal to 1 ohm , because the arctan of 45 degrees is equal to 1. The Thévenin equivalent of the circuit must be computed within the dotted rectangle identified by letter A .

Initially, we calculate the value of V_th, which is the open circuit voltage between the points P-R, as shown in the Figure 15-06.3. Looking closely at the circuit, we notice that the potential difference between points a-b is "e" volts. As the value of the resistor that lies between these two points is 1 ohm, then by Ohm's law , we conclude that an electric current of "e" ampère will circulate through it. Obviously, by the 2 ohms resistor , which is in parallel with the source of 26 A (left side of the circuit), a current of 26 - e ampère will circulate.

The figure above shows the circuit with the currents indicated in its various components. Notice that the 1 ohm resistor between the points c and P has been removed, because this does not interfere with the calculations. You can easily calculate the value of e by making the loop in anti-clockwise formed by 1 Ω, 2 Ω and 3 Ω resistors, including the 40V source, thay is:

Doing the calculation, we find the value of e, or:

As the voltage "e = 3 volts" , then the current e flowing through the resistor of 1 ohm between the points a-b value "e = 3 A" .

Thus, knowing the value of the current e, one can calculate V_th, because the resistor of 4 ohms circulates the current e - 1 = 2 A. Soon:

In this way, the value of K can be calculated. To do so, making the right-hand loop of the Figure 15-06.3, clockwise, and remembering that e = 3 A, we get:

By performing the calculation, the value of K is:

At this point, only need to calculate the value ofR_th. You must delete the independent sources. The current source becomes an open circuit and the voltage source a short circuit. Note that the dependent source can not be deleted. Then, in the Figure 15-06.4 the modified circuit is shown.

Note that to calculate R_th we must know the value of I. For this purpose we have introduced a 4 volt power supply in parallel with the 4 ohms resistor. Therefore, a current of 1 A is circulated by this resistor. Then by the dependent source "11 e" will pass a current I - 1 . On the left side of the circuit we have two resistors of equal value ( 3 ohms each). Then, the current I - 1 will be divided into two equal parts or (I - 1) / 2 . Now we can calculate the value of e for this new situation.

Substituting this value of "e" into the dependent source and making the loop through the points c and b we have the equation:

Performing the calculation:

With this result you can calculate the value of R_th:

The Figure 15-06.5 shows the final circuit that will allow the calculation of I_B.

Therefore, by directly applying the Ohm law to the circuit, we find I_B:

You can check if the results are correct by doing a power balance. After closing of the key S, we have e = 3.2885 volts and we notice that the current of 1 A that circulates at 3 ohms resistor, which is in series with the voltage source of 40 volts, drops to a value equal to 0.7143 A. This voltage source absorbs circuit power as well as the dependent source. Therefore, the only source that supplies power to the circuit is the current source of 26 A . After the calculations we will find a power of 1181.1434 watts provided by this source. The rest of the circuit consumes this power. Check !!!! .