Annealing.-Resistance is lessened by annealing. Matthiessen gives the following relative conductivities for copper and silver, the comparison being made with pure silver at 100° C.: Metal. Copper... Silver........ Dr. Siemens compared the conductivities of copper, silver, and brass with the following results. Ratio of hard to annealed: Copper....... .... 1 to 1.058 Silver....... 1 to 1.145 Brass....... 1 to 1.180 Standard of Resistance of Copper Wire. (Trans. A. I. E. E., Sept. and Nov. 1890.)-Matthiessen's standard is: A hard-drawn copper wire 1 metre long, weighing 1 gramme, has a resistance of 0.1469 B. A. unit at 0° C. Relative conducting power (Matthiessen): silver, 100; hard or unannealed copper, 99.95; soft or annealed copper, 102.21. Conductivity of copper at other temperatures than 0° C., Ct = Co(1 − .003877+.00000900972). The resistance is the reciprocal of the conductivity, and is Rt Ro(1.00387t+.0000059712). The shorter formula R = Ro(1+.00406t) is commonly used. A committee of the Am. Inst. Electrical Engineers recommend the following as the most correct form of the Matthiessen standard, taking 8.89 as the sp. gr. of pure copper: A soft copper wire 1 metre long and 1 mm. diam. has an electrical resistance of .02057 B.A. unit at 0° C. From this the resistance of a soft copper wire 1 foot long and .001 in. diam. (mil-foot) is 9.720 B.A. units at 0° C. Internat. Legal Ohms. Ohms. 61 "23°.9 66 75° F... 10.53 For tables of the resistance of copper wire, see pages 218 to 220, also pp. 1034, 1035. Taking Matthiessen's standard of pure copper as 100%, some refined metal has exhibited an electrical conductivity equivalent to 103%. Matthiessen found that impurities in copper sufficient to decrease its density from 8,94 to 8.90 produced a marked increase of electrical resistance. DIRECT ELECTRIC CURRENTS. Ohm's Law.-This law expresses the relation between the three fundamental units of resistance, electrical pressure, and current. It is: electrical pressure, resistance I= E Current = ; ;; whence E = IR, and R In terms of the units of the three quantities, volts -; volts = amperes X ohms; ohms = ohms volts amperes Amperes = EXAMPLES: Simple Circuits.-1. If the source has an effective electrical pressure of 100 volts, and the resistance is two ohms, what is the current ? E 100 50 amperes. 2. What pressure will give a current of 50 amperes through a resistance of 2 ohms? E IR 50 X 2 = 100 volts. 3. What resistance is required to obtain a current of 50 amperes when the pressure is 100 volts? RE÷1 100 50 2 ohms. Ohm's law applies equally to a complete electrical circuit and to any part thereof. Series Circuits.-If conductors are arranged one after the othe are said to be in series, and the total resistance of the circuit is the sum of a T2 R1 R2 A Rg the resistances of its several parts. Let A, Fig. 170, be a source of current, such as a battery or generator, producing a difference of potential or E. M. F. of 120 volts, measured across ab, and let the circuit contain four conductors whose resistances, 71, 72, 73, 74, are 1 ohm each, and three other resistances, R1, R2, R3, each 2 ohms. The total resistance is 10 ohms, and by Ohm's law the current I= E÷R = 120 ÷ 10 = 12 amperes. This current is constant throughout the circuit, and a series circuit is therefore one of constant current. The, drop of potential in the whole circuit from a around to b is 120 volts, or E RI. The drop in any portion depends on the resistance of that portion; thus from a to R1 the resistance is 1 ohm, the constant current 12 amperes, and the drop 1 x 12 = 12 volts. The drop in passing through each of the resistances R1, R2, R3 is 2 x 12 24 volts. FIG. 170. Parallel, Divided, or Multiple Circuits.-Let B, Fig. 171, be a generator producing an E. M. F. of 220 volts across the terminals ab. The a с e B R1 OR2 d g h FIG. 171. current is divided, so that part flows through the main wires ac and part through the "shunt" s, having a resistance of 0.5 ohm. Also the current has three paths between c and d, viz, through the three resistances in parallel R1, R2, R3, of 2 ohms each. Consider that the resist ance of the wires is so small that it may be neglected. Let the conductances of the four paths be represented by Cs, C1. C2, C3. The total conductance is Cs+ C1 + C2+ C3 = C and the total resistance R=1+ C. The conductance of each path is the reciprocal of its resistance, the total conductance is the sum of the separate conductances, and the resistance of the combined or allel" paths is the reciprocal of the total conductance. The current I= E÷R = 770 amperes. Conductors in Series and Parallel.-Let the resistances in parallel be the same as in Fig. 171, with the additional resistance of 0.1 ohm in each of the six sections of the main wires, ac, bd, etc., in series. The voltage across ab being 220 volts, determine the drop in voltage at the several points, the total current, and the current through each path. The problem is somewhat complicated. It may be solved as follows: Consider first the points eg; here there are two paths for the current, efgh and eg. Find the resistance and the conductance of each and the total resistance (the reciprocal of the joint conductance) of the parallel paths. Next consider the points cd; here there are two paths-one through e and the other through cd. Find the total resistance as before. Finally consider the points ab here there are two paths-one through c, the other through s. Find the conductances of each and their sum. The product of this sum and the voltage at ab will be the total amperes of current, and the current through any path will be proportional to the conductance of that path. The resistances, R, and conductances, C, of the several paths are as follows • Total current 220 x 3.0332 667.3 amperes. "eRag = 139.96 × 0.5 The drop in voltage in any section of the line is found by the formula E R1, R being the resistance of that section and I the current in it. As the R of each section is 0.1 ohm we find E for ac and bd each = 22.7 volts, for ce and dg each 14.0 volts, and for ef and gh each 6.67 volts. The voltage across cd is 220-2 × 22.7 = 174.6 volts; across eg, 174.62 x 14.0 146.6, and across fh 146.62 x 667 = 133.3 volts. Taking these voltages and the resistances R R2R3, each 2 ohms, we find from I ER the current through each of these resistances 87.3, 73.3, and 66.65 amperes, as before. Internal Resistance.-In a simple circuit we have two resistances, that of the circuit R and that of the internal parts of the source of electromotive force, called internal resistance, r. The formula of Ohm's law when the internal resistance is considered is I E÷ (R + 1). Power of the Circuit.-The power, or rate of work, in watts = current in amperes X electro-motive force in volts = IX E. Since I = E÷R, watts E2R = electro-motive force2 resistance. EXAMPLE.-What H.P. is required to supply 100 lamps of 40 ohms resistance each, requiring an electro-motive force of 60 volts? The number of volt-amperes for each lamp is E2 60° Ꭱ = 40 1 volt-ampere = 12 H.P. (electrical) very nearly. Electrical, Brake, and Indicated Horse-power. The power given out by a dynamo volts x amperes 1000 Kilowatts, kw. Volts x amperes 746 electrical horse-power, E.H.P. The power put into a dynamo shaft by a direct-connected engine or other prime mover is called the shaft or brake horse-power, B.H.P. If e, is the efficiency of the dynamo, B.H P. = E.H.P. ÷ €1. If e, is the mechanical efficiency of the engine, the indicated horse-power, I.H.P. brake H.P. ÷ e2 = E.H.P. + (e1 x e2). If e1 and e2 each 91%, I.H.P. E.H.P. 1.194 = kw. x 1.60. In directconnected units of 250 kw. or less the rated H.P. of the engine is commonly taken as 1.6 x the rated kw. of the generator. Electric motors are rated at the H.P. given out at the pulley or belt. H.P. of inotor = E.H P. supplied efficiency of motor. Heat Generated by a Current.-Joule's law shows that the heat developed in a conductor is directly proportional, 1st, to its resistance; 2d, to the square of the current strength; and 3d, to the time during which the current flows, or H = I2Rt. Since I E÷R, Or, heat = current2 × resistance X time electro-motive force X current X time electro-motive force2 X time resistance. = quantity of electricity flowing EQ; or heat electro-motive force x quantity. The electro-motive force here is that causing the flow, or the difference in potential between the ends of the conductor. = The electrical unit of heat, or "joule " = 107 ergs heat generated in one second by a current of 1 ampere flowing through a resistance of one ohm = 239 gramme of water raised 1° C. H = I2Rt X .239 gramme calories = I2Rt x .0009478 Br tish thermal units. In electric lighting the energy of the current is converted into heat in the lamps. The resistance of the lamp is made great so that the required quantity of heat may be developed, while in the wire leading to and from the lamp the resistance is made as small as is commercially practicable, so that as little energy as possible may be wasted in heating the wire. Heating of Conductors. (From Kapp's Electrical Transmission of Energy.)-It becomes a matter of great importance to determine before hand what rise in temperature is to be expected in each given case, and if that rise should be found to be greater than appears safe, provision must be made to increase the rate at which heat is carried off. This can generally be done by increasing the superficial area of the conductor. Say we have one circular conductor of 1 square inch area, and find that with 1000 amperes flowing it would become too hot. Now by splitting up this conductor into 10 separate wires each one tenth of a square inch cross-sectional area, we have not altered the total amount of energy transformed into heat, but we have increased the surface exposed to the cooling action of the surrounding air in the ratio of 1: /10, and therefore the ten thin wires can dissipate more than three times the heat, as compared with the single thick wire. Prof. Forbes states that an insulated wire carries a greater current without overheating than a bare wire if the diameter be not too great. Assuming the diameter of the cable to be twice the diam. of the conductor, a greater current can be carried in insulated wires than in bare wires up to 1.9 inch diam. of conductor. If diam. of cable = 4 times diam. of conductor, this is the case up to 1.1 inch diam. of conductor. Heating of Bare Wires.-The following formulæ are given by Kennelly: temperature of the wire and t that of the air, in Fahrenheit degrees; current in amperes, d diameter of the wire in mils. This latter formula gives for the carrying capacity in amperes of bare wires almost exactly the figures given for weather-proof wires in the Fire Underwriters' table except in the case of Nos. 18 and 16, B. & S. gauge, for which the formula gives 8 and 11 ampere, respectively, instead of 5 and 8 amperes, given in the table. Heating of Coils. -The rise of temperature in magnet coils due to the passage of current through the wire is approximately proportional to the watts lost in the coil per unit of effective radiating surface, thus: t being the temperature rise in degrees Fahr.; S, the effective radiating surface; and k a coefficient which varies widely, according to conditions. In electromagnet coils of small size and power, k may be as large as 0.015. Ordinarily it ranges from 0.012 down to 0.005; a fair average is 0.007. The more exposed the coil is to air circulation, the larger is the value of k; the larger the proportion of iron to copper, by weight, in the core and winding, the thinner the winding with relation to its dimension parallel with the magnet core, and the larger the space factor" of the winding, the larger will be the value of k. The space factor is the ratio of the actual copper cross-section of the whole coil to the gross cross-section of copper, insulation, and interstices. See also the discussion of magnet windings under Electromagnets, p. 1050. Fusion of Wires.-W. H. Preece gives a formula for the current required to fuse wires of different metals, viz., I=ad, in which d is the diameter in inches and a a coefficient whose value for different metals is as follows: Copper, 10244; aluminum, 7585; platinum, 5172; German silver, 5230; platinoid, 4750; iron, 3148; tin, 1462; lead, 1379; alloy of 2 lead and 1 tin, 1318. Allowable Carrying Capacity of Copper Wires. For insulated aluminum wire the safe-carrying capacity is 84 per cent of that of copper wire with the same insulation. Underwriters' Insulation. The thickness of insulation required by the rules of the National Board of Fire Underwriters varies with the size of the wire, the character of the insulation. and the voltage. The thickness of insulation on rubber-covered wires carrying voltages up to 600 varies from inch for a No. 18 B. & S. gauge wire to inch for a wire of 1 000 000 circular mils. Weather-proof insulation is required to be slightly thicker. For voltages of over 600 the insulation is required to be at least 1/16 inch thick for all sizes of wire under No. 8 B. & S. gauge, and to be at least 3/32 inch thick for all sizes greater than No. 0000 B. & S. gauge. Copper-wire Table.-The table on pages 1034 and 1035 is abridged from one computed by the Committee on Units and Standards of the American Institute of Electrical Engineers (Trans. Oct. 1893). ELECTRIC TRANSMISSION, DIRECT CURRENTS. Cross-section of Wire Required for a Given Current.resistance of a given line of copper wire, in ohms: Let R 1 mil-foot of copper; length of wire, in feet; drop in voltage between the two ends; = current, in amperes; sectional area of wire, in circular mils; e = Ꭱ e L -; R = T "A; whence A rIL = e. The value of r for soft copper wire at 75° F. is 10.505 international ohms. For ordinary drawn copper wire the value of 10.8 is commonly taken, corresponding to a conductivity of 97.2 per cent. For a circuit, going and return, the total length is 2L, and the formula becomes A = 21.6IL e, L here being the distance from the point of supply to the point of delivery. If E is the voltage at the generator and a the per cent of drop in the line, Ea 100, and A then e = = 2160IL P and A E' If P = |