Problem 14-2    Source:    Problem 4.12 - page 96 -    NILSSON & RIEDEL -    Book: Circuitos Elétricos - 8ª edição - Ed. Pearson Education do Brasil - 2010.

In the circuit show in Figure 14-02.1, calculate:

a) The voltage of the nodes.

b) The currents i₁ , i₂ , i₃ and i₄.

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

Solution of the problem using Nodal Voltage   Click here!

Solution of the problem using Thévenin/Norton   Click here!

Solution of the Problem 14-2   -    Superposition Method

Item a

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. We will call i_1a , i_2a , i_3a and i_4a the currents calculated with the voltage source of 144 volts . We will call i_1b , i_2b , i_3b and i_4b the currents calculated with the current source of 3 A.

So let's keep the voltage source at 144 volts and eliminate the current source from 3 A. Remember that when we eliminate a current source the effect is of an open circuit, as we can see in the Figure 14-02.2.

Note that, in this case, i_3a = i_4a. Initially, let's calculate i_1a. To do so, we must calculate the series of 80 ohms and 5 ohms resistors. Then calculate the parallel of this with the 10 ohm resistor . After the calculations, we find the value of 8.95 ohms . In addition to the 4 ohms resistor, we find the total resistance of the circuit that is 12.95 ohms . From this information we can calculate the value of i_1a.

Knowing the value of i_1a and applying a current divider, we will find the values of i_2a , i_3a and i_4a, that is:

Another way to calculate i_3a and i_4a would be to apply Kirchhoff's law. Try to do !!!!

Now, we can calculate the currents using the other source, that is, we will use the current source of 3 A . For this, we must eliminate the voltage source of 144 volts and we must not forget that for this we must shorten the source. See how the circuit is in the Figure 14-02.3.

Let us first calculate the current i_3b. For this, we must calculate the parallel of the resistors of 10 ohms e 4 ohms, resulting in an equivalent resistance of 2.86 ohms . Adding this value to the 80 ohms resistor, which is in series with the 2.86 ohms equivalent resistance, we find 82.86 ohms . Now we can apply a current divider and calculate i_3b, that is:

Note that the value of i_3b is negative. With the value of i_3b, we can now, by applying a current divider, calculate i_1b and i_2b.

Note that the value of i_1b also is negative. The value of i_2b, can be calculated by applying a current divider or Kirchhoff's law. Using the current divider, we have:

To calculate the value of i_4b we can use Kirchhoff's law and verify that i_4b = 3 + i_3b, then:

Well, so far we have calculated the currents for two different situations. One when we use only the voltage source of 144 volts and the other when we use only the current source of 3 A . Now, we need to put together this information to find the end result. To do so, let's add algebraically the currents with equal indices, or:

See that using the Superposition method, we have been able to solve the problem by simplifying the circuit. Of course, this requires some work . This problem was purposely resolved step by step, showing in detail the solution of the problem.

Item b

After all these calculations, it is still necessary to calculate the voltages of the nodes e₁ and e₂. However, this is quite simple because you just have to use the law of Ohm , or:

If we want proof that these results are correct, we can take stock of the circuit's powers.

Let's reproduce the original circuit ( see Figure 14-02.4 ) and analyze its solution by BCM (Basic Circuit Method).

Notice that in this case we do not yet have the configuration of the basic circuit . But, we easily realize that doing a source transformation on the right side of the circuit, we will find the desired setting. See in the Figure 14-02.5 the new configuration of the circuit.

It was enough to multiply the value of the current source by the resistor that was in parallel and to add the values of the resistors. And ready. We have the configuration we wanted. Now let's calculate R_sp, or:

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 find the value of i₂, or:

Exactly the value calculated by the traditional method . And possessing this value, we quietly calculate the other values. Note that in this circuit, the fundamental is determine the value of i₂. The calculation of the values of the other currents and voltages in the circuit becomes obvious. Calculating V_b we found the others currents. As already done, we leave it as an exercise.

And so, once again we demonstrate the ease and speed with which we arrive at the final result. It is enough to look at the circuit, to make the necessary changes and the problem is solved.