Problem 78-4    Source:   Problem developed by the author of the site.

Determine the power dissipated by the 20 ohm resistor in the circuit shown in Figure 78-04.1.

Solution of the Problem 78-4   

Initially we must realize that the 45 ohm resistor in series with the -j5 capacitor, are in parallel with the voltage source. Therefore, they can be removed from the circuit without prejudice for the solution of the same. On the other hand, by making the parallel between the 10 ohm resistor and the -j10 capacitor we will obtain the value of 5 - j5. Then, by solving the circuit to the left of the coils, we obtain the Figura 78-04.2.

The next step is to determine the Thévenin equivalent of the circuit shown in Figure 78-04.2 . We must calculate the open circuit voltage, where we find the voltage value equal to V_th = 13.72 ∠ -59° V. And to determine the impedance value of Thévenin, we short-circuit the voltage source. With that, the resistor 15 ohm is in series with the j10 inductor. And this set forms a parallel with 5 - j5. Performing the calculation, we find 5.59 - j2.65 Ω. Thus, the Thévenin equivalent of the circuit is shown on the right in Figura 78-04.2.

Having established the circuit above, we can determine the equations that will solve the problem. For the mesh I₁ we have:

And for the mesh I₂ we have:

So, we got a system of two equations with two unknowns and that can be rearranged, such that:

So, using programs like Octave, Matlab, scientific calculator, etc ..., let's get:

So the power that the 20 ohm resistor dissipates is given by: