(watts/volts) × feet × 2,160
circular mils = ————————————
% loss × volts


watts × feet × 2,160
= ————————————  (5)
% loss × volts2

This formula (5) applies to a direct current two wire circuit, and to adapt it to any alternating current circuit it is only necessary to use the letter M instead of the number 2,160, thus

watts × feet × M
circular mils =  ——————————  (6)
% loss × volts2

in which M is a coefficient which has various values according to the kind of circuit and value of the power factor. These values are given in the following table:

VALUES OF M

SYSTEM		POWER FACTOR
1.00 .98.95 .90.85 .80.75 .70.65 .60
Single phase 2,160 2,249 2,400 2,660 3,000 3,380 3,840 4,400 5,112 6,000
Two phase
(4 wire)		1,080 1,1251,200 1,3301,500 1,6901,920 2,2002,556 3,000
Three phase
(3 wire)		1,080 1,1251,200 1,3301,500 1,6901,920 2,2002,556 3,000

NOTE.—The above table is calculated as follows: For single phase M = 2,160 ÷ power factor2 × 100; for two phase four wire, or three phase three wire, M = ½ (2,160 ÷ power factor2)× 100. Thus the value of M for a single phase line with power factor .95 = 2,160 ÷ .952 × 100 = 2,400.

It must be evident that when 2,160 is taken as the value of M, formula (6) applies to a two wire direct current circuit and also to a single phase alternating current circuit when the power factor is unity.

In the table the value of M for any particular power factor is found by dividing 2,160 by the square of that power factor for single phase and twice the square of the power factor for two phase and three phase.

Ques. For a given load and voltage how do the wires of a single and two phase system compare in size and weight, the power factor being the same in each case?

Ans. Since the two phase system is virtually two single phase systems, the four wires of the two phase systems are half the size of the two wires of the single phase system, and accordingly, the weight is the same for either system.