This section is from the book "A Treatise On Architecture And Building Construction Vol3: Stair Building, Ornamental Ironwork, Roofing, Sheet-Metal Work, Electric-Light Wiring And Bellwork", by The Colliery Engineer Co.. Also available from Amazon: A Treatise On Architecture And Building Construction.

60. The size of wire required to supply current to any group of lamps depends on the number of lamps, each of which takes a certain average current. The following table gives the value of this current for lamps of different candle-power and with a line E. M. F. of 110 and 55 volts:

110 Volts. | Lamp. | 55 Volts. |

.5 Ampere. | 16 Candlepower. | 1 Ampere. |

1.0 Ampere. | 32 Candlepower. | 2 Amperes. |

1.5 Amperes. | 50 Candlepower. | 3 Amperes. |

3.0 Amperes. | 100 Candlepower. | 6 Amperes. |

Since the lamps are connected across the leads in multiple arc (Art. 3), whether on the two-wire or the three-wire system, it is evident that the main conductors must be able to carry current for each lamp on the circuit. If the average current is .5 ampere, and there are 30 lamps on the circuit, the main conductors will have to carry 30 X .5 = 15 amperes.

61. In nearly all cases a very simple formula may be used in calculating the size of wire to be furnished for any particular circuit. According to Ohm's law, the loss in volts divided by the current flowing gives the resistance of the line (Art. 14); if this result be divided by the length of the line in feet, the resistance in ohms per foot is obtained, and the size of wire corresponding to this resistance can be found directly in a wire table. Since resistances are usually given in ohms per thousand feet owing to the low resistance of copper, it will be necessary to multiply by 1,000 to find the corresponding gauge of wire.

We may express these operations in the shape of a formula as follows:

Rf = E

CL'

(4.) where Rf is the resistance per foot of the wire, E the drop in volts, C the current, and L the length of the wire in feet.

We have seen, however (Art. 60), that a 16-candlepower lamp on the usual 110-volt circuit takes .5 ampere. Then for C we may substitute the number of lamps N multiplied by the current required for one, or N x .5. The total length of line L is also equal to 2 F, when F is the distance in feet as measured along the double line between the point of supply and the lamps. We may, therefore, write formula 4 as follows:

Rf = E/.5N x 2F = E/NF.

(5.)

The term N F is the number of lamps multiplied by the distance from the source of current to those lamps; in other words, this expression gives the lamp feet.

From the preceding we may derive the following rule:

In order to find the resistance per foot of the conductor to use in any 110-volt circuit with 16-candlepower lamps, divide the loss in volts, as predetermined, by the lamp feet.

The corresponding wire may then be found from Table 1.

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