Ans. Because the product of such a current multiplied by the pressure does not represent any watts expended.
A man lifting a weight, and then allowing it to descend the same distance to its initial position, as shown in figs. 1,355 to 1,357, presents a mechanical analogy of wattless current.
Let the movement of the weight represent the current and the weight the pressure. Then calling the weight 10 pounds (volts), and the distance two feet (amperes). The work done by the man (alternator) on the weight in lifting it is
10 pounds × 2 feet = 20 foot pounds (1)
(10 volts × 2 amperes = 20 watts.)
The work done on the man by the weight in forcing his hand down as his muscles relax is
10 pounds × 2 feet = 20 foot pounds (2)
(10 volts × 2 amperes = 20 watts.)
From (1) and (2) it is seen that the work done by the man on the weight is equal to the work done by the weight on the man, hence no useful work has been accomplished; that is, the potential energy of the weight which it originally possessed has not been increased.
Figs. 1,355 to 1,357.—Mechanical analogy of wattless current. If a man lift a weight any distance, as from the position of fig. 1,355 to position of fig. 1,356, he does a certain amount of work on the weight giving it potential energy. When he lowers it to its original position, as in fig. 1,357, the weight loses the potential energy previously acquired, that is, it is given back to the man, the "system" (man and weight) having returned to its original condition as in fig. 1,355. During such a cycle, the work done by the man on the weight is equal to the work done by the weight on the man and no useful external work has been accomplished.
Why the Power Factor is equal to Cos φ.—In the preceding figures showing power curves for various phase relations between current and pressure, the curves show the instantaneous values of the fluctuating power, but what is of more importance, is to determine the average power developed.