Problem 14-3    Source:    Example 5.10 - page 188 -    IRWIN, J. David -    Book: Análise de Circuitos em Engenharia - 4ª edição - Ed. Pearson Education do Brasil - 2013.

In the circuit show in Figure 14-03.1, calculate:

a) The voltage of all the essential nodes.

b) The voltage at the terminals a-b.

c) The currents i_k , i_x and i_y.

Solution of the problem using Method of Transforming Sources   Click here!

Solution of the problem using Nodal Voltage   Click here!

Solution of the Problem 14-3   -    Superposition Method

To use the Superposition method we must keep in the circuit a single source at a time. As we will have two calculations for the circuit currents, one for each source, let's differentiate these currents with an additional index.

Initially, let's eliminate two of the sources. Remembering : we eliminate a voltage source by short-circuiting its terminals, and a current source becomes an open circuit. To differentiate the calculation of the currents with different sources, let us assume the index "a" for the calculation with the first source, the index "b" for the calculation with the second source and for the third source, the "c" index.

In the Figure 14-03.2 we see the circuit with the elimination of two sources. Like this, e₁ and e₂ they are the same, as well as i_xa and i_ya. Making the parallel of the resistors 3 and 6 ohms, we find the equivalent resistance of 2 ohms . Therefore, by applying a current divider we can calculate the currents and voltages we want, that is:

With the values of the currents we can calculate the value of the voltages, or:

Now it is possible to calculate the value of e_3a e V_ab1, or:

We then compute the requested variables in the problem using only the current source of 16 A . Now we must recalculate the variables, using only the voltage source of 12 volts. See in the Figure 14-03.3, how the circuit was.

Well, the process is the same. We calculate the equivalent resistance of the whole circuit, which is equal to 6.43 ohms ( 6||8 + 3) and from this value we find the value of i_total of the circuit.

Note that the calculated current (i_total) has negative value, meaning that its direction is contrary to what appears in the Figure 14-03.3. To calculate i_xb = i_yb, we apply a current divider , emphasizing that we must obey the direction of the currents in the circuit.

Now we can calculate the requested voltages in the problem.

To calculate the last step, we will use the current source of 2 amp and eliminate the other two. In the Figure 14-03.4 we see the circuit.

Analyzing the figure above, we notice that in the part of the circuit highlighted in yellow we have the resistors of 3 and 6 ohms in parallel and this set is in series with that of 2 ohms . Therefore, we have an equivalent resistance of 4 ohms . In the highlighted part in green , we have two resistors in series, forming a equivalent resistance of 6 ohms . Therefore, by applying a current divider, we will calculate the values of the currents indicated in the circuit.

Note that i_xc has negative value, indicating that its direction is contrary to that indicated in the Figure 14-03.4. To calculate i_kc, we must realize that this is a portion of i_xc. Then:

From this moment, we can calculate the voltages of the circuit, or:

With all the partial values of the currents, we can now, adding algebraically the same, find the values of the currents in the circuit, that is:

We will do the same for the voltages of the circuit, therefore:

As you can see, in a circuit that has many sources, the Superposition method can become quite laborious and time-consuming.

Let's reproduce the original circuit and analyze its solution by BCM (Basic Circuit Method).

On the left side of the circuit shown in the Figure 14-03.5, the 16 A source can be transformed with the 3 ohms resistor which is in parallel with the source. This results in a voltage source of 48 volts in series with the 3 ohms resistor. Thus, there are two sources of voltage in series, but with opposite polarities. Then one must subtract his values and the positive pole will be in the sense of the source of greater value, that is, the value will be 36 volts with the positive pole facing up.

Looking at the right side of the circuit, we realize that we can add the two resistors of 3 ohms each, totaling 6 ohms . And this resistor is in parallel with the current source of 2 A . Therefore, by doing the transformation we will find a voltage source equal to 12 volts with the positive pole facing top. And adding the resistances, we find a value of 8 ohms . Okay, we already have the circuit in the configuration of the basic circuit. See in the Figure 14-03.6, as the circuit was.

Stay tuned for the fact that we kept the point intact e₂. So what we are going to calculate is the current i_k. The methodology is always the same. Let's calculate R_sp as we have done in previous problems. Then:

Making the cross product of the sources with the resistors, adding them (because the positive pole of the two sources are facing up) and dividing by R_sp, we found the value of i_k, or:

With the value of i_k, easily calculate the value of e₂ by applying the Ohm law , or:

To calculate the value of i_x, we subtract the 12 volt source of the value of e₂ and we divide by the 8 ohms resistor. Soon:

Returning to look at the original circuit we see that:

The other values have already been calculated and will not be repeated here. Notice how much less "operations" we accomplished and we quickly came to the results of the problem.