Answer

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**Hint**

We know that the circuit that is excited using an alternating source is called an AC Circuit. The alternating current (AC) is used for domestic and industrial purposes. In an AC circuit, the value of the magnitude and the direction of current and voltages is not constant, it changes at a regular interval of time. Alternating current (AC) is the type of electric current generated by the vast majority of power plants and used by most power distribution systems. Alternating current is cheaper to generate and has fewer energy losses than direct current when transmitting electricity over long distances. Based on this concept we have to solve this question.

**Complete step by step answer**

We know that,

$ \text{Power Factor = }\dfrac{\text{ watts }}{\text{ volt-amperes }} $

$ =\dfrac{\text{P}}{\text{S}}=\dfrac{\text{VIcos}\phi }{\text{VI}}=\cos \phi $

We said previously that in a pure resistive circuit, the current and voltage waveforms are in-phase with each other so the real power consumed is the same as the apparent power as the phase difference is zero degrees $ \left(0^{\circ}\right) $ . So, the power factor will be:

Power Factor, $ \text{pf}=\cos {{0}^{{}^\circ }}=1.0 $

That is the number of watts consumed is the same as the number of volt-amperes consumed producing a power factor of 1.0, or $ 100 \% $ . In this case it is referred to a unity power factor.

We also said above that in a purely reactive circuit, the current and voltage waveforms are out-of-phase with each other by $ 90^{\circ} . $ As the phase difference is ninety degrees $ \left(90^{\circ}\right), $ the power factor will be:

Power Factor, $ \text{pf }=\cos {{90}^{{}^\circ }}=0 $

That is the number of watts consumed is zero but there is still a voltage and current supplying the reactive load. Then reducing the reactive VAr component of the power triangle will cause $ \theta $ to reduce improving the power factor towards one, unity. It is also desirable to have a high-power factor as this makes the most efficient use of the circuit delivering current to a load.

Then we can write the relationship between the real power, the apparent power and the circuits power factor as:

Real Power $ (\mathrm{P})= $ Apparent Power $ (\text{S}) $ $ (\mathrm{S}) \times $ Power Factor $ (\mathrm{pf}) $

Power Factor $ (\text{pf})=\dfrac{\text{ Real Power }(\text{P})\text{ in Watts }}{\text{ Apparent Power }(\text{S})\text{ in volt-amps }} $

An inductive circuit where the current "lags" the voltage is said to have a lagging power factor, and a capacitive circuit where the current "leads" the voltage is said to have a leading power factor. The value of the power factor is zero in both inductive and capacitive circuits. Here, capacitive circuit is not an option.

**Therefore, only option (A) is the correct answer.**

**Note**

We know that the power factor (PF) is the ratio of working power, measured in kilowatts (kW), to apparent power, measured in kilovolt amperes (kVA). Apparent power, also known as demand, is the measure of the amount of power used to run machinery and equipment during a certain period. The ideal power factor is unity, or one. Anything less than one means that extra power is required to achieve the actual task at hand. All current flow causes losses both in the supply and distribution system.

It should also be known that since power factor is a cosine function of the relative phase angle between current and voltage there are no possible values greater than one. Cosine of any angle that is not zero is less than 1.

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