The line loss is 3 volts = E; the number of lamps is 24 of 16 candlepower, and 3 of 32 candlepower, equivalent to a total of 24 + 6 = 30 of 16 candlepower = N ; the distance is 75 feet = F. Then, by formula 6, the resistance per foot

Rf = 3 = .00067 ohm.

2 x 30 x 75

= .67 ohm per 1,000 feet

The corresponding wire is No. 8. The current in the circuit is 30 X 1 = 30 amperes ; but, as the limiting current for this size of wire is 25 amperes, we must take the next size larger, and use No. 7. Ans.

65. In such a case as the above, we might wish to know what the resulting drop will be, since we have to use a different size of wire. This is easily obtained by a reversal of formula 6, for a 55-volt circuit:

E = Rf x 2 NF = 2 Rf NF. (7.)

No. 7 wire has a resistance per 1,000 feet of .519 ohm, or .000519 ohm per foot. As before, N = 30, and F = 75. Then

E = 2 X .000519 X 30 X 75 = 2.33 volts, the required drop.

In the same manner, the drop may be found for a 110-volt circuit, when the size of wire, the number of lamps, and the distance are known, by a modification of formula 5,

E = Rf NF. (8.)


What will be the drop in a 110-volt circuit, in which is located a group of 15 lamps of 16 candlepower, the distance from the point of supply being 270 feet, and the wire No. 10 B. & S. ?


The resistance per 1,000 feet of No. 10 wire = 1.04 ohms, or per foot = .00104 ohm = Rf. The number of 16-candlepower lamps is 15 = N, and the distance in feet is 270 = F. Then, by formula 8, E = .00104 X 15 X 270 = 4.2 volts, the drop. Ans.

66. When, in calculating the wire for a lighting system, it is necessary to use a size of wire larger than that required for the given drop, the effect may be balanced by using a smaller wire than that called for in another part of the same circuit. Suppose, for instance, a total drop of 6 volts were allowed in a circuit composed of two different lines, and it was decided to provide for a drop of 2 volts in one part, and 4 volts in the other. If, then, it should be found that the wire to give 4 volts drop was too small to carry the current safely, and a larger wire had to be used, giving a drop of only 3 1/2 volts, the second wire could be designed to allow a drop of 2 1/2 instead of 2 volts, provided, of course, that the safe carrying capacity were not exceeded.

When the calculations call for a wire smaller than No. 14, it should not be used, as this size is the smallest allowed by the underwriters for interior wiring.

67. In calculating the wire for a loop circuit such as shown in Fig. 50, the total length of conductor is evidently (4 X 75) + (4 X 20) = 380 feet, but the actual length traversed by the current for any one lamp, such as a, is 10+ 75 + 20 + 75 + 10 = 190 feet for the double distance. If there are 40 lamps on the loop, the lamp feet (Art. 61) will be 40 times one-half this double distance, or 40 X 95 = 3,800 lamp feet, giving the same result as if the lamps were all bunched at the further end.

Solution 500

Fig. 50.


In determining the size of wire for a loop circuit, multiply the number of lamps by one-half the distance, in feet, around the loop. This will give the lamp feet N F, to be used in the formulas already given.


What size of wire is required for a loop circuit similar to Fig. 50, having 26 lamps, 110-volt, 16-candlepower, the length of the room being 80 feet, and the width 30 feet ? Allow for 3 volts drop.


The total length of circuit through any one lamp is (2 X 80) + (2 X 30) = 220 feet, and the single distance is 220/2 = 110 feet = F. The number of lamps is 26 = N, and the allowable drop is 3 volts = E. Then, since 16-candlepower lamps are installed, and the E. M. F. of the circuit is 110 volts, the resistance per foot of the conductor will be, by formula 5,

Rf = 3 = .00105 ohm.

26 x 110

= 1.05 ohms per 1,000 feet.

By reference to the wire table, it will be seen that this corresponds almost exactly to No. 10 wire. The current is 26 X .5 = 13 amperes, which is within the safe limit, and No. 10 wire will be used. Ans.