Problem 11-5    Source:    Problem 3.58 - page 140 -    IRWIN, J. David -    Book: Análise de Circuitos em Engenharia - 4ª edição - Ed. Pearson - 2013.

In the circuit show in Figure 11-05.1, calculate:

The voltages of the essencial nodes, i₁ , i₂ and the voltage V_o.

Solution of the problem using Node Voltage Method   click here!

Solution of the problem using Thévenin/Norton Method   clique aqui!

Solution of the Problem 11-5 -  Method of Transforming Sources

Analyzing the circuit in a first moment, it is realized that when calculating the voltage of the e₂ node, we can compute V₀. And also the current through resistance of 2 ohms. From the circuit, we easily conclude that the current i₂ is the sum of the two current sources, that is, 6 + 8 = 14 A . Knowing the value of these two currents, the current across the 4 ohms resistance can be calculated. In the Figure 11-05.2, the explosion of the current source of 6 A is shown in the circuit.

As shown in the figure above, with the current source explosion, two circuits can be obtained: a highlighted in yellow and another highlighted in green, which can be transformed into something else simple. Initially we are going to do source transformations on the circuit highlighted in yellow. See the Figure 11-05.3 for the circuit.

We transform the current source in parallel with the 3 ohms resistor in a source 18 volts (3 x 6 = 18) in series with the 3 ohms resistor. Now we have two voltage sources in series with opposite polarities. Therefore we must subtract its values and the positive pole will point at sense of the source of higher voltage, in this case, that of 18 volts. We can continue the transformation until we get a single source and a resistance as can be seen in the Figure 11-05.4.

Pay close attention to all transformations that have been made step by step. Firstly, we subtract the voltage sources by finding a voltage source of 6 volts (left circuit of the figure above). Subsequently, we transform this voltage source into a current source of 2 A in parallel with a resistance of 2 ohms which is the resistance equivalent of the parallel of the 3 ohms and 6 ohms resistors, as shown in the figure above. Note the arrow on the current source pointing downward as well as the positive pole of the 6 volt voltage source.

And finally we come to our goal of having only one source associated with a resistor as shown in the Figure 11-05.5.

On the other hand, in the highlighted green circuit, all that remains is to add the current sources and eliminate the resistor of 12 ohms, since we know that we can remove resistors in series with current sources. As for the voltage source of 64 volts, in series with the 2 ohms resistor , we can transform it into a current source in parallel with the 2 ohms resistor . See the Figure 11-05.6 how the circuit was after the transformations.

Notice that we have two sources of currents pointing up, so we can add their values. And the source of 4/6 points down, so we must subtract its value from the previous result. The two resistors are in parallel. Calculating the equivalent resistance we arrive at the circuit shown in the Figure 11-05.7.

And finally, one can calculate e₂ by applying the law of Ohm, that is:

With value of the e₂, you can calculate V₀, or:

Also, notice that the two sources of current in the circuit, 6 A and that of 8 A together, produce a current of 14 A. This current must be bypassed by 12 ohms. This current is i₂. As a consequence, we have a voltage drop on this resistor of   12 x 14 = 168 volts.   Therefore, the relationship between e₂ and e₄ is:

Substituting for the numerical values we find:

Knowing the value of V₀, we can calculate the current through the resistor of 2 ohms . We find the value of 2 A (4/2 = 2 A). In this way, it is easy to calculate the current value i₃.

Knowing i₃ let's calculate the value of e₁.

Knowing e₁, you can calculate i₁, or:

And e₃ is:

Therefore, we can summarize the values found.