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tro-motive force in "volts" and R the resistance in "ohms" then C = *

But the number of volts to overcome the resistance of the circuit must be small as compared with the electro-motive force used for doing commercial work. Depending on the cost of coal and copper, it is common to allow 5 per cent, 10 per cent, or 15 per cent loss. In no case is the safe carrying capacity (without heating) of the wire used as a measure of the amount of current to be carried. This carrying capacity of a wire is far in excess of the amount that it would be commercially profitable to allow, for the loss of energy to force such amounts of current as a wire might carry would be at least 75 per cent of the whole voltage of the dynamo.

Arc and incandescent lighting and motors require different systems of wiring, and under each head peculiarities will be described and illustrations of calculating given.


Arc lamps are run in series; that is, the same current that lights one lamp lights the next one, and so on through the entire circuit operated by one dynamo. As the voltage must overcome the sum of the resistance of all the lamps, it is impracticable and unsafe to supply more than 50 lamps from one machine. It is therefore usual to have as many different circuits as there are dynamos.

Arc lamps are commonly rated as "full arcs" or "half arcs," estimated roughly by the amount of current they use or by the light they furnish. The former use 9.6 amperes and furnish 2000 candle-power lamps, the later 6.8 amperes and 1200 candle-power.

Let it be required to determine the size of wire necessary to use on a circuit of 50 full arc lamps, with a loss of but 10 per cent, allowing 45 volts for each lamp. Fifty lamps would require 2250 volts. But only go per cent is to be available for lighting purposes, so -^tHP- volts = 2500 volts,—the amount the dynamo must supply: 2500—2250 = 250 the number of volts to be lost in overcoming line resistance.

C-l since C = g.6 and E=250 and R = * = = 26 + ohms. 8 miles equals 42,240 feet, hence the resistance of 1000 feet would be

T4far--6i55 ohms-
Looking at the table on pages 40 and 41 in

the column of resistances per 1000 feet, .62849 is the nearest to this number. The size of wire corresponding in the first column is seen to be No. 8 (.128 inches diameter).

It is usual to allow No. 6 wire for such arc circuits, thus bringing the loss as low as 5 per cent, while No. 8 is well adapted to half arc circuits. Arc wiring is the simplest to calculate; indeed, often there is no calculation done at all, as one set of figures answers for the entire circuit, but in


a series of similar figuring must be done as the current passes successively through "bus" bars, mains, feeders, service wires, branches, and individual lamps. An incandescent lamp system is spoken of as having a uniform pressure (in volts) everywhere. This is only approximate, however, and depends upon the amount of current flowing in any particular wire and its distance from the dynamos.

It is common to arrange this system after the style of a gas or water plant, with a center of distribution, from which the main supply wires emerge in all directions. Extras or feeder wires are usually necessary. Supply wires tapped to the mains enter every house or block to be lighted. Finally these individual wires supply each lamp or group of lamps. The wires are always in duplicate, one to bring the current to the lamp, the other to carry it away. Each lamp in circuit exists independent of all the others, can trace its path back to the dynamo without touching another lamp. In order that all lamps may burn at a uniform brilliancy the sizes of wire should be calculated separately for each installation.

The calculation will depend primarily on the kind of apparatus to be employed, whether direct current, the ordinary two-wire system, the three-wire system, or the alternating system with transformers. Secondarily it will depend on the particular make and candlepower of the lamps. Incandescent lamps have been run on arc circuits, but their use is dangerous, and is discouraged by insurance companies.

Let it be required to determine the size of wires to use in an installation of 100 lamps drawing their current from a center of distribution of a capacity of iooo lamps one-half mile from the station. The distance of the house in question, from this center is 120 feet. The lamps are to be located on four floors, 25 lamps on each. The center of the room to which the current is brought is 50 feet from the main junction boxes for each floor. Each lamp is wired separately to the supply wires, there being no groups or chandeliers. The lamps are no volt .6 ampere each, and it is decided to allow different percentages of loss for each of these three stages into which the wiring will be divided.

The twelve or thirteen lamps on each side of the junction point to the supply wires can get their current from one size of wire large enough to carry the entire current to the furthest lamp with the f per cent loss. This will be 12x.6=7.2 amperes, and the distance 15 feet, I per cent loss will 99} available 110 + 991 = 110.8 + , 110.8—uo=.8 volts lost. R = c =5^=0.111 ohms for 30 feet—, twice the fifteen feet to allow for the return. Hence for 1000 feet (in order to be found in the table of resistances) it would be 3.7 ohms. The nearest to this in the table is 3.1 and the size corresponding is No. 15 (.057).

To determine the size of the 50-foot lengths that connect with the main supply wires, and

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