The method followed in calculating the size of conductors for the three-wire system is based on the same laws as for the two-wire system. When the two sides of the three-wire system are balanced, with no current in the neutral wire, the E. M. F. of the circuit is double that of a two-wire system, and since the current for one lamp also passes through another lamp in series, the total current required is only one-half of that necessary for the same number of lamps on the two-wire system. For an equal percentage drop of potential, therefore, the size of wire need be only one-fourth of that necessary in the two-wire system. If, however, the drop is to remain the same in each case, the size of wire in the three-wire system will be one-half that required for the two-wire system; for, although the resistance of the conductor for the three-wire system would thereby be doubled, the current is of only one-half the value, so that in the one case E = C R, and in the other case E = 1/2 C x 2R, which values are identical.

For the three-wire system, therefore,

Rf = 2 E

N F

(12.)

The resistance per foot of the conductor in a 110-220 volt three-wire system is equal to twice the drop in volts divided by the lamp feet.

For large conductors, the area in circular mils

A = 10.8 N F = 5.4 N F .

(13.)

## Rule

In determining the safe carrying capacity of conductors for the three-wire system, remember that, with the same number and kind of lamps, the conductor carries only one-half the current of a conductor installed on the multiple-arc system.

## Example

What size of conductor should be used with the three-wire system with 5 per cent. loss, when 52 lamps of 16 candlepower, 110 volts, are 540 feet from the dynamo ?

## Solution

According to formula 10, the initial voltage is

V ' = 100 x 110 = 115.8 volts , and the drop is, by formula 11, E = 115.8 - 110 = 5.8 volts. The lamp feet = 52 X 540 = 28,080 = N F. Then, by formula 12, the resistance per foot of the wire will be

Rf = 2 x 5.8 = .000413 ohm.

28.080

= .413 ohm per 1,000 feet.

By reference to Table 1, it will be seen that this corresponds to No. 6 wire. The current to be carried is one-half of that taken by the same number of lamps on a 110-volt two-wire circuit, or and since the wire may carry 35 amperes, it is safe. Ans.

52 x .5 = 13 amperes.

12. Amount of Copper in Three-Wire System.

When the three-wire system is adopted in place of the two-wire system, the total E. M. F. is doubled, as we have seen, and the current is reduced by one-half. But the same amount of energy is delivered, and the drop will be the same if the conductor area is decreased in proportion to the decrease of current. Then, since the third, or neutral, wire is usually made equal to the outer wires, the total amount of copper is reduced by one-fourth. When the percentage loss is the same, however, the outer wires may be in cross-section only one-fourth that of the conductors necessary for the two-wire system. But an additional conductor, the neutral wire, of the same cross-section, must be supplied, having, however, only one-half the length of the total circuit, or one-eighth the metal of the multiple-arc system.

The total metal of the three-wire system is, therefore, 1/4 + 1/8 = 3/8 of the metal required in a multiple-arc system for the same number and kind of lamps, and for an equal percentage drop of potential.

The neutral wire could be made considerably smaller than the outside conductors, because it normally carries only a small fraction of the current. However, for practical reasons, it is usually made of the same size as the other conductors.