Problem 14-4    Source:   Adapted from the Exercise 4.8 - page 161 -    CARLSON, A. Bruce -    Book: Circuits - Preliminary Edition - Ed. John Wiley - 1996.

In the circuit show in Figure 14-04.1, calculate i₁ and i₂.

Solution of the Problem 14-4   -    Superposition Method

Remember : a voltage source is eliminated by short-circuiting it. To differentiate the calculation of currents with different sources, let us assume the index a for the calculation with the first source and the b index for the calculation with the second source . Let's assume the first source as the source of the voltage. Short circuit it obtains the circuit below:

Looking at the circuit in the Figure 14-04.2, we realize that by the resistorof 4 ohms , which is between the points b-c , circulates a current which value is i_1a + 3. And by the 2 ohms resistor, which lies between the points c-d , circulates a current which value is i_2a + 3. Taking this into consideration and applying Kirchhoff to the voltage across the two resistors of 2 ohms and the two of 4 ohms , we have the following equation:

Grouping similar terms and rearranging the equation we have:

Dividing the two members by 6 , we get:

Since we have an equation and two unknowns, we must find another equation that relates these two variables. Thus, by making another mesh, from point a, passing through the points b-c and returning by the resistor of 12 ohms , because of this resistor circulates a current equal to i_1a - i_2a, we can write the equation:

Dividing the two members by 6 , we get:

Now we have a system of two equations with two unknowns of easy solution. Thus, we find the values of:

These calculated currents are due only to the current source . Let's go now eliminate the current source and calculate the currents due to the voltage source . In the Figure 14-04.3 we see the circuit.

First we will calculate the parallel of the resistor of 12 ohms with the two resistors that are in series, between the point c and the negative pole of the source , passing through the point d . Doing the calculation we find the value of 4 ohms . With this, we can add this result with the other two resistors, since all are in series, finding the total resistance that the circuit offers to the voltage source, that is, R_total = 10 ohms. Therefore, to calculate i_1b just apply the Ohm law , or:

Note that this current arrives at the node c and follows two different paths: one by the resistor of 12 ohms and another by the resistors that are in series (2 + 4 = 6 ohms) . Thus, a current divider can be applied to calculate i_2b. Then:

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:

Let's reproduce the original circuit ( see Figure 14-04.4 ) and analyze its solution by BCM (Basic Circuit Method).

Note that the current source of 3 A is connected between the points b-d . One of the transformations we can make is "explode" this source, as shown in the Figure 14-04.5.

In this way, we can obtain two current sources in parallel with two resistors. Then, as we learned in the previous chapters, we can turn these circuits into voltage sources in series with their respective resistors. See, in the Figure 14-04.6, the new configuration of the circuit after these transformations.

Note that on the left side of the circuit, we have two voltage sources in series with polarities contraries. Then it is possible to add them algebraically, resulting in a single voltage source equal to 24 volts , with the positive polarity facing up.

And that's it. After these transformations we can get to our basic circuit . Just apply the practical method and solve the problem very quickly.

Notice that we quickly compute the denominator of the equation, represented by R_sp, and given by:

Thus, replacing the numerical values of the resistors that make up the circuit, where   we have the values of R₁ = 6, R₂ = 12 and R₃ = 6, we find:

Now, just find the numerator by doing the cross multiplication between the sources voltage of 24 and 6 volts , representing V₁ and V₂, respectively, by the resistances R₃ and R₁, or:

And finally, dividing this value by R_sp, we find the value of i₁₂.

To find e₂, just calculate:

Notice that the important thing in this circuit is to find the value of i₁₂, because finding the values of the other currents in the circuit is trivial.

This is a very quick and direct way to calculate the currents in the circuit.