Problem 10-11    Source:   Question 24 of the access test Professor of Electrical of the IF of Santa Catarina (Brazil) Edital 42/2014.

In the circuit show in Figure 10-11.1, calculate the value of the currents i₂ , i₃ and i₄ about resistors R₂ , R₃ and R₄, respectively.

Solution of the problem 10-11

Since there are two series resistors between the e-f points, one can add their values and replace them with a single resistor with a value equal to 5 ohms. Note in the Figure 10-11.2 how the circuit was.

From the figure above it is verified that after the identification of the points, the three resistors are in parallel. Therefore, one can redesign the circuit including the other components that were left out in the previous figure. Note that the resistors of 1 and 0.75 Ω are in series. Thus, adding its values gives the result as shown in the Figure 10-11.3.

Calculating the equivalent resistance of the three resistors in parallel, we find 1.25 Ω. As this equivalent resistance is in series with the 1.75 Ω resistor, the total resistance that the circuit offers to the voltage source is 3 ohms. It is now easy to calculate the total current flowing through the circuit, or:

With the value of i we can calculate the currents requested by the problem by two different paths.

As we know the current flowing through the resistance of 1.25 Ω then we can calculate the potential difference between points c-f, or:

Therefore, the currents are:

The other way is to apply a current divider because the current flowing through the equivalent resistance of the resistors in parallel is known. The current value is 1 A and the equivalent resistance value is 1.25 Ω. Soon:

Therefore, two different methods were presented that lead to identical results.