Solved Problems about AC Powers

Problem 52-7 Source: Adapted of Problem 12.17 - page 262 - EDMINISTER, Joseph A. - Book: Circuitos Elétricos - Ed. McGraw Hill - 1971.

In the circuit shown in the Figure 52-07, the load connected to the terminals A-B consists of a variable resistance R_L and a capacitive reactance X_C, variable between 2 and 8 ohms. Determine the value of R_L and X_C which carry the maximum power transfer to the load. Determine the value of that maximum power supplied to the load.

Solution of the Problem 52-7

Initially, we must compute the Thévenin equivalent of the circuit.

Note that the impedance Z = 2 + j10 is connected in parallel with the resistance of 8 ohms when the voltage source is shorted in order to calculate the Thévenin equivalent. Like this:

Z_th = 8 (2 + j10) / (8 + 2 + j10)

By performing the calculation, we obtain:

Z_th = 5.76975 ∠ 33.69° = 4.8 + j3.2 Ω

In order to calculate the Thévenin voltage, we need to calculate the open-circuit voltage at points A-B. Using a voltage divider:

V_th = 50∠45° x (2 + j10) / (8 + 2 + j10)

By performing the calculation, we find:

V_th = 36.06 ∠78.69° = 7.072 + j35.36 V

By calculating the Thévenin equivalent, there are conditions to address the problem requests. See the figure below, the circuit that will be used to solve the problem.

In the Figure 52-07.a, we call the load composed of R_L and X_C by Z_L, that is, load impedance. This situation refers to case 2 when we studied the maximum power transfer theorem. Immediately, it can be deduced that the value of R_L must be equal to the value of R_th, or R_L = 4.8 Ω. In addition, the following condition must be satisfied:

Z_L = Z_th^*

That is, the load must be the conjugated complex of Z_th. Then, the value of the capacitive reactance must have a value that cancels the inductive reactance of Z_th. Thus, the entire circuit will be represented by a purely resistive impedance. For this to happen, we should have:

-jX_C = -j3.2 Ω

As the problem statement says that X_C can range from 2 to 8 ohms, so the calculated value is within specification.

Therefore, we can write the value of Z_L as:

Z_L = 4.8 - j3.2 Ω

In order to calculate the maximum power transferred to the load, we must calculate the electric current that flows through the circuit, or:

I = V / (R_L + R_th) = 36.06 ∠ 33.69° / (4.8 + 4.8)

Note that the imaginary parts of the impedances have been ignored because they cancel out. Performing the calculation, we find:

I = 3.76 ∠ 33.69° Ω

Thus, the power will be:

P = R_L |I|² = 4.8 x (3.76)² = 67.86 W

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