Question 1

The power factor of a load is equal to the&#10;A) sine of its phase angle.&#10;B) ratio of true power to apparent power.&#10;C) ratio of its resistance to reactance.&#10;D) each of the above.

Accepted Answer

The power factor of a load is calculated as the ratio of true power to apparent power. It indicates how much of the apparent power is being utilized as useful or true power. The sine of the phase angle is a measure of the phase difference between voltage and current, which can be used to calculate the reactive power of the load but not the power factor. The ratio of resistance to reactance is known as the impedance angle and can also be used to calculate the power factor, but it is not as accurate as the true power to apparent power ratio.

Question 2

Most industrial loads are __________ and, therefore, have ___________ power factors.&#10;A) resistive-inductive, lagging&#10;B) resistive-inductive, leading&#10;C) resisitive-capacitive, lagging&#10;D) resistive-capacitive, leading

Accepted Answer

Industrial loads typically have a combination of resistive and inductive components, which leads to a lagging power factor. Capacitive loads would result in a leading power factor, but they are not as common in industrial settings.

Question 3

Improving; or 'correcting' power factor means to change an existing power factor so that it&#10;A) approaches unity.&#10;B) moves away from unity.&#10;C) increases beyond unity.

Accepted Answer

Improving or correcting power factor means to bring the existing power factor closer to unity (1). A power factor of 1 is considered to be the ideal power factor, indicating that all the supplied power is being used efficiently without any wastage. Therefore, the best choice is A, as it correctly states that improving or correcting power factor means approaching unity.

Question 4

The effect of a 'poor' or low power factor is causes the load to&#10;A) draw more energy than is necessary.&#10;B) operate below its power rating.&#10;C) draw more load current than necessary.&#10;D) operate less efficiently.

Accepted Answer

A low power factor causes the load to draw more current than necessary, which can lead to increased losses in the electrical system and higher energy costs.

Question 5

The greatest beneficiary of power-factor improvement is the&#10;A) residential consumer.&#10;B) industrial consumer.&#10;C) the electricity network company.&#10;D) each of the above.

Accepted Answer

The electricity network company benefits the most from power-factor improvement as it reduces losses in the system, improves voltage regulation, and increases the capacity of the network.

Question 6

The __________ derives absolutely no benefit whatsoever from power-factor improvement.&#10;A) residential consumer.&#10;B) industrial consumer.&#10;C) the electricity network company.

Accepted Answer

Power-factor improvement benefits the electricity network company and industrial consumers by reducing the amount of reactive power that needs to be generated and distributed. However, for residential consumers, it does not make a significant difference as their power usage is lower and their electricity bills are usually calculated based on the total energy consumption.

Question 7

The most common method of power-factor improvement is by connecting a suitable capacitor&#10;A) in series with the load.&#10;B) in parallel with the load.&#10;C) in series or in parallel with the load.

Accepted Answer

Connecting a capacitor in parallel with the load is the most efficient and popular way of improving power factor. This is because capacitors are reactive devices that generate reactive power, which can offset the reactive power drawn by the inductive load, hence increasing the power factor of the system.

Question 8

When a capacitor is connected to improve power factor, the current drawn by the load __________, while the current drawn from the supply __________.&#10;A) increases, reduces&#10;B) reduces, remains the same&#10;C) increases, remains the same.&#10;D) remains the same, reduces

Accepted Answer

When a capacitor is connected to improve power factor, it reduces the reactive power in the circuit, which in turn reduces the current drawn from the supply. However, the current drawn by the load remains the same. Therefore, the correct answer is D - remains the same, reduces.

Question 9

Power-factor correction capacitors are usually rated in&#10;A) microfarads.&#10;B) watts.&#10;C) reactive volt amperes.&#10;D) volt amperes.

Accepted Answer

Power-factor correction capacitors are rated in reactive volt amperes (VARs) because they are used to provide reactive power to improve the power factor in AC electrical systems.

Question 10

The main incentive for industrial consumers to improve the power factor of their loads is&#10;A) a reduction in the energy they consume.&#10;B) any motors will run more efficiently.&#10;C) they will not be subject to penalty surcharges.&#10;D) their loads draw less current.

Accepted Answer

Improving the power factor of their loads helps industrial consumers avoid penalty surcharges that may be imposed by utility companies for low power factor usage. It does not necessarily lead to a reduction of energy consumption or more efficient motor operation, although it may result in their loads drawing less current.

Question 11

It is rarely economical for an industrial consumer to improve the power factor of their load to unity because&#10;A) of the capital cost of the necessary power-factor correction equipment.&#10;B) their loads are not constant.&#10;C) the cost of energy varies over time.

Accepted Answer

The answer of It is rarely economical for an industrial...

Deck 21: Power Factor Improvement