Problem 11-8    Source:   Problem elaborated by the author of the site.

In the circuit show in Figure 11-08.1, calculate:

a) The current I₁.

b) The power supplied or received by the 40 volt voltage source.

Solution of the Problem 11-8 -  Method of Transforming Sources

Item a

One technique we will use in solving this problem is the possibility of transforming a current source into two or more current sources.

See the circuit shown in the Figure 11-08.2 where the current source that interconnected the points b - c has been broken down into two current sources. One leaves the point b and reaches the point g. The other one leaves the point g and reaches the point c. Note that the value of the current source does not change. It remains 12 I ₁. In this way, the law of the nodes for the node g also does not change, since the 12 I₁ that arrives at node g leaves the other source causing the algebraic sum of these currents to be zero. Therefore there was no change in the equations of the circuit. The same is true for node b. Before modification was output 12 I₁ and, after the modification, it continues to exit 12 I₁. The same reasoning holds for the node c. We also grouped the two resistors in series with the 90 volt source in a single 20 ohms resistor.

From this modified circuit it is possible to transform sources and find the circuit shown in the Figure 11-08.3. With this, one can formulate the equations that will allow to find the solution of the same.

For the node a we can establish the following relation:

Since we have three variables we need two more equations to solve the system. So let's go apply the law of Kirchhoff to tension, in the mesh indicated by arrow brown in the figure above.

Doing the same for the mesh indicated by the green arrow we have:

Substituting I₃ in this last equation by the relation of the node a and rearranging the terms of the last two equations, is found the following system:

Solving this system we find the value of I₁, or:

Item b

To calculate the power at the source of 40 volts we must pay attention to the initial circuit and calculate the currents in the circuit. For a better understanding, we show in the Figure 11-08.4 the currents in the circuit. Let's see the steps necessary to calculate them. Notice that we can easily calculate the voltage between points a and d by subtracting the voltage drop across the two resistors that are in series with the source, from the source value. Thus, we find V_ad = 90 - 1.09 x 20 = 68.2 volts .

With the value of V_ad we can compute V_cd = V_ad + 40 = 108.2 volts . Then the current flowing through the resistance between points c and d is I_cd = 108.2 / 30 = 3.62 A . Then, using the law of Kirchhoff for the node c, we find the current that flows through the source of 40 volts, so I₄₀ = 13.1 - 3.62 = 9.48 A . This current enters the positive pole, therefore it is absorbing power of the circuit and its value is positive, that is: