Solved Problems about AC Powers

Problem + Hard 52-3 Source: Adapted from Problem 3 of the 1st Exam of 2019 - Discipline Electric Circuits of Engineering School - PUCRS - 2019.

For the circuit shown in the figure below, it is known that:

Load A consumes 100 VA with unity power factor.
The load B has an inductive power factor and is formed by a resistor of 50 ohms and a reactance of 50 ohms.
The load C consumes 200 W and has an inductive power factor equal to 0.7071.
The effective value of the power supply is equal to 100∠0°.

That said, remembering the principle of conservation of complex power, calculate:

a) The total power factor of the loads.

b) The effective current consumed by the loads.

c) What reactive power must a capacitor bank connected in parallel generate to make the power factor of the unit circuit.

d) John, a cautious engineer designed this capacitor bank with twice the reactive power calculated in item previous. Explain the consequences of this error.

Solution of the Problem + Hard 52-3

Initial Considerations

If load A has unity power factor, it follows that S_A = P_A and therefore P_A = 100 W.
The load B, in turn, has an impedance equal to Z_B = 50 + j50 = 50√2∠45° . So there is a factor of power equal to FP = 0.7071.
If the load C has a power factor equal to FP = 0.7071, then the impedance angle Z_C is equal at 45°. And from there, it follows that R_C = X_C. Furthermore, this charge consumes 200 W.

Based on this last information, as R_C = X_C, so P_C = Q_C. Of that Thus, the complex power of charge C can be written as:

S_C = 200 + j200 = 200√2∠45° VA

On the other hand, the load A is resistive and consumes 100 W. Therefore, one can write the complex power consumed by the loads A and C, or:

S_A + S_C = 100 + 200 + j200 = 300 + j200 VA

To calculate S_B, you must calculate I₂. Like this:

I₂ = V / Z_B = 100∠0° / 50√2∠45°

Carrying out the calculation:

I₂ = √2∠-45° A

With this data, S_B can be calculated, remembering that to calculate this power it is necessary to use the complex current I₂, or:

S_B = V I^*₂ = 100√2∠45° = 100 + j100 VA

It is now possible to calculate the total complex power consumed by the circuit. Then:

S_T = S_A + S_B + S_C = 400 + j300 = 500∠36.87° VA

Item a

Based on the angle value of S_T, the power factor of the circuit is obtained, or:

FP_total = cos (36.87°) = 0.800

Item b

To calculate the value of I_T, use the fact that S_T = V I^*_T. Then:

I^*_T = 500∠36.87° / 100∠0°

Carrying out the calculation:

I^*_T = 5∠36.87° A

So the value of I_T is:

I_T = 5∠-36.87° A

Item c

To make the power factor unity, the reactive power must be null. As already calculated, S_T = 400 + j300 . So the power of the capacitors must nullify the j300 portion of S_T, that is:

Q_C = - 300 VAr

Item d

As engineer John calculated twice the required power, that is, Q_C = - 600 VAr, then the power factor goes from 0.800 inductive to 0.800 capacitive.

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