Problem + Hard 15-3    Source:    Problem 2 of P1 - 1st semester - 2016 - Discipline: Electrical Circuit Engineering Analysis - ULBRA - Prof. doctor João Carlos Vernetti dos Santos.

In the circuit of the Figure 15-03.1, when the switch S is in position 1, the current I = 1.2 A. Determine the value of the voltage V_ab, when the key S is in position 2.

Solution of the Problem + Hard 15-3

When the key "S" is in position 1, the points a and b are interconnected by a resistor equal to 1 ohm and through which a electrical current of 1.2 A. This information makes it possible to calculate the Thevenin equivalent for the circuit that is highlighted by the dashed line in blue.

To do so, eliminate the independent sources and calculate the Thevenin resistance. Note that when we eliminate the voltage sources E₁ and E₂ (they are short-circuited), the three 12 ohms resistors are in parallel with each other. Thus, we get a single equivalent resistance equal to 4 ohms. On the other hand, when we eliminate the current sources I₁ and I₂, they behave like an open circuit and therefore we can eliminate them from the circuit.

In the Figure 15-03.2 we can see how the simplified circuit that will allow you to calculate the Thevenin resistance. Thus, the two 4 ohms resistors that are in series, can be replaced by a single one with a value equal to 8 ohms.

This, in turn, will be in parallel with the 24 ohms resistor. As a result of parallel, we obtain an equivalent resistor of 6 ohms. And finally, adding the value of this resistor with the resistor of 2 ohms which is linked to the point a, we find the value of R_th1. Let's call it this way, because we must calculate R_th2 referring to the circuit highlighted by the dashed lines in green in Figure 15-03.1. Soon:

With the value of R_th1 and considering the key S in position 1, we can now calculate the value of V_th1. For that, follow the circuit shown in the Figure 15-03.3.

We know that when we insert the 1 ohm resistor into the Thevenin equivalent, a current of 1.2 A through the circuit. Therefore, applying Ohm's law to the circuit we easily calculate the value of V_th1. The result will be:

Let's pay attention to the other circuit (on the right at points a-b), and calculate its Thevenin equivalent.

First, let's find out what the corresponding graph means to circuit A. From the graph, we see that when i = 0 the voltage V₁ is equal to 15 volts. However, we know that when i = 0 this represents an open circuit. On the other hand, when V₁ = 0 (ie a short circuit) then the electric current takes on the value of 3 A.

Notice the circuit shown in the Figure 15-03.4. When the circuit behaves as an open circuit, we have i = 0 and therefore V₁ = 15 V. However, if we apply a short circuit to its output (V₁ = 0 V), for a current to flow i = 3 A, we must have a resistor in series with a voltage source equal to R = 15/3 = 5 ohms.

Thus, the circuit shown above satisfies the conditions required by the graph. In this way, we can redraw the circuit highlighted with a dashed line in green, and replace the "box" that represents circuit A, shown in dashed red, as shown in the Figure 15-03.5.

In the figure to the side we see the redrawn circuit. Note that in addition to introducing the circuit from "box A", we made a explosion voltage source of 7.2 volts (Right side of circuit).

Now it is easy to see that we can transform the entire circuit that is highlighted by the green dashed line, across a single voltage source in series with a resistor. Follow along in Figure 15-03.6 below, the transformations made in the circuit.

Note that the circuit shown on the right in the figure above is exactly the same. Thevenin equivalent of the circuit we are analyzing. Note that R_th2 = 4 Ω and V_th2 = 12.27 V. We could have used another method to find the result. However, note that doing font transformations quickly we get to the final result.

As we have the two Thevenin equivalents it only remains to connect them and calculate the value of V_ab. See how it looks in the Figure 15-03.7 the final circuit.

Note how the circuit in the figure on the right made it easy to determine the value of V_ab, as it is enough to apply Ohm's law to the circuit. Note that this case is when the key S is in position 2. Therefore, we can calculate the current I.

With the value of the current flowing through the circuit, we can calculate the voltage V_ab, or:

For this calculation we use the circuit to the left of the points ab. We would arrive at the same result using the circuit on the right of points a - b.