Problem + Hard 54-1    Source:  Problem 11.16 - page 232 -    EDMINISTER, Joseph A. -   Book: Electric Circuits - 2nd Edition - Ed. McGraw Hill - 1971.

Three impedances, Z₁ = 10 ∠+30°, Z₂ = 20 ∠0° and Z₃ = 5 - j5, must be connected successively to terminals A-B of the circuit shown in the figure below. Determine the power dissipated in each of them.

Solution of the Problem + Hard 54-1   

To solve this problem, the best method is to use Thevenin's theorem. Like this, we must determine the Thevenin impedance that the circuit presents for the voltage source. And to determine the Thevenin voltage just calculate the voltage an open circuit between points A-B.

Initially we are going to determine the Thevenin impedance and for this we put the voltage source short-circuited.

See in the figure to the side how the circuit was when we put the voltage source in short circuit.

Note that the part of the circuit highlighted by the rectangle in green forms a configuration delta. In this way we can transform it to a star configuration or Y.

To carry out this transformation, we will use the equations studied in the chapter 5. We can remember them Here!

After the calculations, in the star configuration, we find two branches with values ​​equal to 2.5 + j 2.5 Ω and another equal to 1.25 - j 1.25 Ω.

Above, in the left figure, we can see how the circuit was after the transformation delta-star. The figure on the right shows the same circuit, but we perform the sum of the components that were in series and we rearranged the topology. But the circuit, electrically, is the same. This circuit will allow you to calculate Z_th. Therefore, we must calculate the parallel of the two upper branches and add to the impedance 2.5 + j2.5 that is in series with point A.

Now just add the impedance 2.5 + j2.5 and we find Z_th. This way:

To find the value of the Thevenin voltage we need to calculate the voltage that appears between the A-B terminals to open circuit. For this, we define the currents I, I₁ and I₂ that circulate through the circuit as it is shown in the adjacent figure.

To determine I, you must find the equivalent impedance that the circuit offers to the voltage source. Then we must calculate the parallel of the impedances formed by the resistors in series with the inductors, which we will call Z', and later, we add the resistance of 5 ohms which is in series with Z'. Like this:

With the value of Z_eq we can easily find the value of I, because applying Ohm's law, we have:

We need to calculate the values ​​of I₁ and I₂. For such, we can calculate the voltage value between the points x - z. For that, just apply the Ohm's law, that is, V_xz = I Z'. Dividing this tension by impedance of the branch corresponding to each current, we have:

For I₂, we apply the same principle.

Now that we know the values ​​of I₁ and I₂, we can calculate V_a and V_b.

And as we know, V_ab = V_a - V_b. Therefore:

We managed to establish the circuit of Thèvenin and now we pass the second step of the problem.

In the figure on the side we can see the model that we will use to calculate the items requested in the problem statement. Z_L represents the load that will be placed between the terminals A-B. Keep in mind that this circuit simulates the original circuit perfectly.

Item a   ⇒   Z_L = 10 ∠+30° = 8.66 + j 5

We need to calculate the electric current flowing through the circuit. Before we go calculate the equivalent impedance that the circuit offers to the voltage source.

Applying Ohm's law we find I. Like this:

As we want to calculate the average power that the load dissipates, we must take into account consideration only the real part of the impedance Z_L, that is, the value of R_L = 8.66 Ω. Soon:

Item b   ⇒   Z_L = 20 ∠0°   

Since the load angle is zero, then Z_L is a resistor of value equal to 20 ohms. Following the same steps as above, we obtain:

Applying Ohm's law we find I. Like this:

In this case, R_L = 20 Ω. Soon:

Item c   ⇒   Z_L = 5√2 ∠-45° = 5 - j 5

As in the previous items, we have:

Applying Ohm's law we find I. Like this:

As we want to calculate the average power that the load dissipates, we must take into account consideration only the real part of the impedance Z_L, that is, the value of R_L = 5 Ω. Soon: