Apparent Power

From our analysis of dc networks (and resistive elements above), it would seem apparent that the power delivered to the load of Fig. 1 is simply determined by the product of the applied voltage and current, with no concern for the components of the load; that is, $P = VI$.
Defining the apparent power to a load.
Fig. 1: Defining the apparent power to a load.
However, we found in Chapter 13 that the power factor ($\cos \theta$) of the load will have a pronounced effect on the power dissipated, less pronounced for more reactive loads. Although the product of the voltage and current is not always the power delivered, it is a power rating of significant usefulness in the description and analysis of sinusoidal ac networks and in the maximum rating of a number of electrical components and systems. It is called the apparent power and is represented symbolically by S. * Since it is simply the product of voltage and current, its units are voltamperes, for which the abbreviation is VA. Its magnitude is determined by $$\bbox[10px,border:1px solid grey]{ S = VI} \, \text{(volt-amperes, VA)} \tag{1}$$ or, since $ V=IZ$ and $ I = {V \over Z}$ then $$ \bbox[10px,border:1px solid grey]{S = I^2Z} \, \text{(VA)}$$ and $$ \bbox[10px,border:1px solid grey]{S = {V^2 \over Z} \, \text{(VA)}$$ The average power to the load of Fig. 1 is $$ P = VI \cos \theta$$ However, $$S=VI$$ Therefore, $$ \bbox[10px,border:1px solid grey]{P = S \cos \theta}$$ and the power factor of a system $F_p$ is $$F_p = \cos \theta = {P \over S}$$ The power factor of a circuit, therefore, is the ratio of the average power to the apparent power. For a purely resistive circuit, we have $$P = VI = S$$ and $$F_p = \cos \theta = {P \over S} = 1$$ In general, power equipment is rated in volt-amperes (VA) or in kilovolt-amperes (kVA) and not in watts. By knowing the volt-ampere rating and the rated voltage of a device, we can readily determine the maximum current rating. For example, a device rated at 10 kVA at 200 V has a maximum current rating of $I = 10,000 VA/200 V = 50 A$ when operated under rated conditions. The volt-ampere rating of a piece of equipment is equal to the wattage rating only when the $F_p$ is 1. It is therefore a maximum power dissipation rating. This condition exists only when the total impedance of a system $Z \angle \theta$ is such that $\theta = 0$.
The exact current demand of a device, when used under normal operating conditions, could be determined if the wattage rating and power factor were given instead of the volt-ampere rating. However, the power factor is sometimes not available, or it may vary with the load.
Demonstrating the reason for rating a load in kVA rather than kW
Fig. 2: Demonstrating the reason for rating a load in kVA rather than kW.
The reason for rating some electrical equipment in kilovolt-amperes rather than in kilowatts can be described using the configuration of Fig. 2. The load has an apparent power rating of 10 kVA and a current rating of 50 A at the applied voltage, 200 V. As indicated, the current demand of 70 A is above the rated value and could damage the load element, yet the reading on the wattmeter is relatively low since the load is highly reactive. In other words, the wattmeter reading is an indication of the watts dissipated and may not reflect the magnitude of the current drawn. Theoretically, if the load were purely reactive, the wattmeter reading would be zero even if the load was being damaged by a high current level.