ART. 222.-Unit of Electromotive Force; Volt. The electromotive force of a circuit is the amount of work done on a unit of positive electricity in passing once round the circuit. It is expressed in terms of W per Q moved round; hence in the C.G.S. system, by erg per Qe.g... The Congress adopted as a practical unit the unit which had been defined and adopted by the British Association, namely, 108 erg per Qo.g.s., denominated the volt. Hence 1 volt = 108 erg per Qc.g.s., The volt, as will be seen from the following short table, is nearly equal to the electromotive force of a Daniell's cell. A customary abbreviation for the term electromotive force is e.m.f. ELECTROMOTIVE FORCE OF VOLTAIC CELLS. Name of Cell. Daniell. Grove. volts Bunsen. Latimer-Clark. Leclanché. 1.42 ART. 223.-Unit of Capacity; Farad. We have seen that the idea of capacity is expressed in terms of Q per (W per Q). The ordinary C.G.S. unit is Qo.g.s. per (erg per Qo.g.). The Congress adopted the practical unit of the British Association, namely, the farad. It is defined by or 1 farad coulomb per volt, - = 10-9 Qc.g.s. per (erg per Qc.g.s.). The microfarad, which is the one millionth part of the farad, is the most convenient unit for actual work. A cable is an infinitely long cylindrical condenser. For a cable having a metallic core of a L radius, an insulating sheath of b L radius, and a specific inductive capacity k, the capacity per unit of length is ART. 224.-Unit of Resistance; Ohm. When a steady current of electricity flows round a circuit, the amount of the current is the same at every cross-section of the circuit. When the electromotive force is varied, the circuit being kept the same and at the same temperature, the amount of the current is found to be proportional to the electromotive force. Hence we have the law, discovered by Ohm, k W per Q = Q per T. This gives us the idea of electric resistance. The C.G.S. unit is erg per Qc.g. per (Qc.g.s. per sec.). The practical unit, originated by the British Association and adopted by the Electrical Congress, is volt per ampere. The single equivalent term is ohm ; so that 1 ohm = volt per ampere, = - 109 C.G.S. unit of resistance. ART. 225.-The Standard Ohm. The British Association after defining the ohm appointed a committee to construct a standard which should realize the definition. The result of their measurements was that the ohm is represented by the resistance of a column of pure mercury at 0° C., one square millimetre in section and 105 centimetres long. In accordance with this result, standard coils were constructed of an alloy of two parts of silver to one of platinum, and issued to experimenters. Subsequent measurements, made by various experimenters, agreed in showing that the standard ohm was slightly less than the ohm of the definition. The Paris Congress appointed a committee of electricians to make a fresh determination; and on their report the standard ohm has been authoritatively defined as the resistance of a column of mercury at 0° C., having a section of one square millimetre and a length of 106 centimetres. The Siemens unit of resistance was defined as the resistance of a column of mercury at 0° C., having one sq. mm. in section, and 1 metre long. S ART. 226.-Watt; Joule. We have 1 W per T = (W per Q moved round) × Q passing round per T. A special unit of activity, called the watt, is obtained from this equivalence by making W per Q the volt, and Q per T the ampere. Thus 1 watt volt by ampere, = The joule is the corresponding unit of energy, 1 joule = volt by coulomb. These two denominations were proposed by Sir W. Siemens in his address to the British Association, 1882; they appear likely to be adopted, but they have not as yet the authoritative stamp of the other denominations defined in this section. Also since 1 W per T = {(W per Q) / (Q per T)} × (Q per T) × (Q per T), = Rx (Q per T)2; The relation of the watt to the horse-power is Ex. 1. Find the multiplier for changing the electrostatic unit of potential from the centimetre, gramme, and second, to the metre, kilogramme, and second. The old unit of potential is expressed by Now and But and erg per Qc.g.s. 1 erg=gm. by cm. per sec. per sec. by cm.; 1 Q2.g... = dyne by cm.2, =gm. by cm. per sec. per sec. by cm. hence, when we substitute instead of cm. and gm., we obtain = X 1 erg 001 × 01 × 01 kgm. by metre per sec. per sec. by metre, = '0000001 Also 1 Q2.g.s. = '001 x '01 × ('01)2 kgm. by metre per sec. per sec. by metre2, = •000000001 1 Qo.g..=10000/10 1 √kgm. by metre per sec. per sec. by metre. The kilogrammetre here meant is the absolute not the gravitational unit. Ex. 2. Find the current in a circuit of 50 ohms, generated by a dynamo machine having an internal resistance of 5 ohms, when the electromotive force of the dynamo is 450 volts. 1. Find the multipliers in the electrostatic system for changing the units of Quantity, Capacity, and Current from centimetre, gramme, and second, to metre, kilogramme, and second. Find the corresponding multipliers when the units mentioned above belong to the electromagnetic system. 3. Find the multipliers for changing the electromagnetic units of Electromotive Force, Current, and Resistance, from the C.G.S. units to the F.P.S. units. 4. A table of electromotive forces is expressed in terms of the millimetre, milligramme, and second; find the factor for changing to the C.G.S. unit. 5. Compare the millimetre-milligramme-second unit of current with the ampere. 6. Find the multiplier for changing cheval-vapeur to watt. 7. Compare the kilogrammetre with the joule. 8. If an electromotive force of 90 volts is maintained between the terminals of an incandescent lamp, and a current of 15 amperes flows through the lamp; what is the rate at which energy is supplied to the lamp? 9. A dozen incandescent lamps, each having a resistance of 2.75 ohms, are joined in a single circuit, and the resistance of the wires connecting the terminals with the terminals of the dynamo machine is 1.2 ohms. If the maximum electromotive force of the dynamo is 250 volts, what is the maximum current which can be sent through the lamps? 10. A single Grove's cell is employed to send a current through an external resistance of 100 ohms. What is the strength of the current taking the internal resistance of the cell at 25 ohm ? 11. Calculate, in terms of the watt, the activity of the above circuit. 12. When the poles of a battery were connected with the terminals of a tangent-galvanometer, a current of 24 amperes was produced; and when the resistance of the circuit was increased by 1 ohm, all else remaining as before, the strength of the current was 11 amperes. Find the electromotive force of the battery. 13. A circuit is formed containing galvanometer, battery, and connecting wires, the total resistance of the circuit being 4.85 ohms; the galvanometer shows a deflection of 48. When a piece of platinum wire is introduced into the circuit, the deflection falls to 29°. Calculate the resistance of the platinum wire, given tan 48° 1.121, and tan 29° = 0.554. = 14. The ratio of the electrostatic to the electromagnetic unit of quantity is 3 × 1010 in the C.G.S. system; what is it in the F.P.S. system? 15. A battery of 50 Grove cells, having a total internal resistance of 13.5 ohms, is joined by a short-circuit; find the current which will be given. 16. The Board of Trade, acting under the Electric Lighting Act, have adopted a unit of energy which is defined as "the energy contained in a current of 1,000 amperes flowing under an electromotive force of one volt during one hour.” Compare this unit with the joule. SECTION XLIX.-RESISTANCE. ART. 227.-Resistance of a Substance. When a steady current of electricity flows along a wire of uniform material, having a uniform cross-section, the strength of the current is directly |