FIGURE 65. PEAK AND PEAKTOPEAK VALUES.

between those two limits. You could determine the average value

by adding together a series of instantaneous values of the

alternation (between 0 and 180), and then dividing the sum by

the number of instantaneous values used. The computation would

show that one alternation of a sine wave has an average value

equal to 0.636 times the peak value. The formula for average

voltage is:

Eavg = 0.636 x Emax

where Eavg is the average voltage of one alternation, and Emax is

the maximum or peak voltage. Similarly, the formula for average

current is:

Iavg = 0.636 x Imax

where Iavg is the average current of one alternation, and Imax is

the maximum or peak current.

Do not confuse the above definition of an average, value with

that of the average value of a complete cycle. Because the

voltage is positive during one alternation and negative during

the other

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