Problem 78-6    Source:  Problem developed by the author of the site.

Calculate the current I and the power dissipated by the 10 ohm resistor in the circuit shown in Figure 78-06.1 when the circuit is powered by V₁ = 182 cos (10t + 60°).

Solution of the Problem 78-6   

In this problem, the current I is already entering the points of the coils, as shown in Figure 78-06.1. It is clear that this is the Case 1 studied in item 3. Equivalence of Inductances. So, we can replace the circuit formed by the coils with their equivalent inductance, given by eq. 78-03.

Therefore, the circuit equation can be written as:

As we have only one unknown, the value of I is:

Note that this value is the peak current value, as we are using the voltage value of a trigonometric function, that is, this is the peak voltage value. To find the real power dissipated by the resistor, we must find the effective or rms value of the current. Like this:

Therefore, the real power dissipated by the 10 ohm resistor is: