Problem 10-9    Source:   Problem elaborated by the author of the site.

In the circuit show in Figure 10-09.1, calculate:

a) The currents i₁ and i₂.

b) What is the power dissipated in the 20 ohms resistor?

Solution of the problem 10-9

Item a

It is possible to combine the two voltage sources that are in series. Since they have opposite polarities, you must subtract your values by finding 45 volts. The two resistors, 4 and 7 ohms, are also in series and can be replaced by one of equivalent value equal to 11 ohms. See in the Figure 10-09.2 the circuit.

Note that this circuit is similar to problem 10-4. The difference is that there is a resistor of 20 ohms in parallel with the set of three resistors. In the Figure 10-09.3 we can see the redesigned circuit.

Making the necessary transformation, it is verified that all resistors are in parallel. In the circuit to the right of the figure above, this is evident. Therefore, the equivalent resistance of the set of resistors can be calculated. This results in R_eq = 4 ohms. The Figure 10-09.4 shows the circuit after replacing the parallel by R_eq.

Using the Ohm's law makes it very easy to calculate the value of i₂. However, it should be noted that the positive polarity of the voltage source is below (at point d), which means that the value of i₂ will be negative. Soon:

Therefore, we conclude that the direction of current i₂ is opposite to that shown in the circuit. From this moment on, there are two different ways to calculate the i₁.

Way 1

With the value of i₂ you can calculate the value of V_da and from this result, applying the Ohm's law, we calculate i₁.

Note that the positive (and not negative) value of i₂ has been used because you want to calculate the voltage of the point d with respect to the point a. This value is positive because the current i₂ has opposite direction to that represented in the circuit. As i₁ crosses the resistor of 20 ohms, applying more once the Ohm's law, we find the value of i₁, that is:

Way 2

Another possible way is to apply a current divider. The total current flowing through the four resistors in parallel, represented by i₂, is known. The value is also known of the equivalent resistance of the parallel, equal to 4 ohms. Then:

Therefore, by an alternative method we find the same value.

Item b

To calculate the electrical power dissipated in the resistor of 20 ohms it is necessary to know the electric current that crosses the resistor. But, looking at the electric circuit, this current is i₁, whose value is 0.6 A, already calculated previously. Applying the equation for power: