Problem 14-1    Source:   Problem elaborated by the author of the site.

In the circuit show in Figure 14-01.1, calculate:

a) The currents i₁ , i₂ and i₃.

b) The voltage of the node e₁.

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 Kirchhoff Method   Click here!

Solution of the problem using Thévenin/Norton   Click here!

Solution of the Problem 14-1   -    Superposition Method

Item a

As we know, the Superposition method allows to calculate the currents of the circuit using one source at a time. In this problem we will delete one of the sources.

Remember that: we eliminate a voltage source by short-circuiting it, as shown in the Figure 14-01.2. To differentiate the calculation of currents with different sources, we will assume the a index for the calculation with the first source and the index b for the calculation with the second source .

By calculating the equivalent resistance of the circuit, one can find the value of i_1a. For this, we calculate the parallel of the resistors of 6 ohms and 2 ohms , which results in 1.5 ohms and add up to that of 4 ohms , since they are in series, obtaining 5.5 ohms . Then, the current i_1a is:

With the value of i_1a, by applying a current divider , we find i_2a, or:

Thus, we discover the value of i_3a:

Then, we calculate the three currents of the circuit using only the voltage source of 16 volts . Now we must recalculate the currents using only the voltage source of 20 volts , shorting the 16 volt source. See in the Figure 14-01.3 for the circuit.

Well, the process is the same. We calculate the equivalent resistance, which is equal to 7.33 ohms and, from this value, we find the value of i_3b.

Note that the calculated current (i_3b) has a negative value, meaning that the direction its is contrary to what appears in the Figure 14-01.3. To calculate i_2b and i_1b we apply a current divider , emphasizing that we must obey the direction of the currents chosen in the circuit. For the calculation of i_2b we use -i_3b, because they are pointing in opposite directions. It is like i_3b has a negative value, it results that i_2b will have a positive value. For the calculation of i_1b, we use i_3b, because they point in the same direction. Then, i_1b will have a negative value as can be seen in the calculations below.

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

Item b

Since we have the value of i₂ it is easy to calculate the value of e₁, because it is enough to apply the law of Ohm , or:

Thus, we obtained the same results that were found using the other methods of circuit resolution.

If we want proof that these results are correct we can take stock circuit power.

Addendum

As the Superposition method is used for circuits that have as main characteristic the LINEARITY , it is possible to develop an equation that relates a variable to the two sources, whether voltage or current. In our case, we will relate i₂ with the voltage sources of 16 and 20 volts, which we will call, respectively, V₁ and V₂. Then we can write:

We should calculate K₁ and K₂. For this, canceling V₂, that is, equating to ZERO , as we did at the beginning of the problem, we can calculate K₁. In this way, we have:

Making V₁ equal to zero, we can calculate K₂:

After we calculate K₁ and K₂, we are able to write the equation that governs the circuit, that is:

Note that with this equation, we can calculate the current i₂ for any value of V₁ and V₂. We can do this for any variable in our interest. See the Table 14-01.1 with several values of V₁ and V₂ (in volts) and the respective value of i₂ (ampère).

Let's reproduce the original circuit (see Figure 14-01.4 ) and analyze its solution by BCM (Basic Circuit Method).

Notice that we can quickly calculate the denominator of the equation, represented by R_sp, which is given by:

Thus, replacing the numerical values of the resistors that compose the circuit, where we have the values of R₁ = 4, R₂ = 2 and R₁ = 6, we find:

Now, just find the numerator by doing the cross multiplication between the voltage sources of 16 and 20 volts, representing V₁ and V₂ respectively, by the resistors R₃ and R₁, or:

And finally, dividing this value by R_sp, we find the value of i₂.

Exactly the value found when using the traditional method. With the value of i₂, to find e₁ multiply by the value of R₂ (law of Ohm), resulting in:

Note that the important thing in this circuit is to find the value of i₂, because finding the other values is trivial.

This is a very quick and direct way to calculate the currents in the circuit. In the next problems we will see that through one or two source transformations we will be able to arrive at the configuration of this circuit and solve it easily.

To conclude, see how easy it is to calculate K₁ and K₂.

Ratifying the values previously found. In this way, we can write the equation that governs this circuit, or: