| Size of wire B. & S. gauge | Area in circular mils. | .80 power factor | .70 power factor | ||||||||
| Spacing of conductors | Spacing of conductors | ||||||||||
| 1" | 3" | 6" | 12" | 24" | 1" | 3" | 6" | 12" | 24" | ||
| 500,000 | 500,000 | 1.00 | 1.15 | 1.30 | 1.47 | 1.62 | 1.00 | 1.00 | 1.16 | 1.33 | 1.49 |
| 300,000 | 300,000 | 1.00 | 1.00 | 1.09 | 1.16 | 1.25 | 1.00 | 1.00 | 1.00 | 1.02 | 1.12 |
| 0,000 | 211,600 | 1.00 | 1.00 | 1.00 | 1.03 | 1.10 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 |
| 000 | 167,800 | 1.00 | 1.00 | 1.00 | 1.00 | 1.01 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 |
| 00 | 133,100 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 |
| 0 | 105,500 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 |
| 1 | 83,690 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 |
| 2 | 66,370 | ||||||||||
| 3 | 52,630 | ||||||||||
| 4} | 41,740 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 |
| 5} | 33,100 | ||||||||||
| 6} | 26,250 | ||||||||||
| 7} | 20,820 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 | 1.00 |
| 8} | 16,510 | ||||||||||
| 9} | 13,090 | ||||||||||
| 10} | 10,380 | ||||||||||
EXAMPLE.—A circuit supplying current at 440 volts, 60 frequency, with 5% loss and .8 power factor is composed of No. 2 B. & S. gauge wires spaced one foot apart. What is the drop in the line?
According to the formula
| % loss × volts | |
| drop = | —————— × S |
| 100 |
Substituting the given values, and value of S as obtained from the table for frequency 60
| 5 × 440 | |
| drop = | ————— × 1 = 22 volts |
| 100 |
Current.—As has been stated, the effect of power factor less than unity, is to increase the current; hence, in inductive circuit calculations, the first step is to determine the current flowing in a circuit. This is done as follows:
| apparent load | |
| current = | ——————— (1) |
| volts |
and
| watts | |
| apparent load = | ————— (2) |
| power factor |