CHAP. VIL] ELECTRICITY AND MAGNETISM. 341 



as "efficient" as steam-engines of equal power, the 

 necessary zinc is still 24 times as dear as the equivalent 

 amount of coal. This calculation does not take into 

 account the cost of acids of the batteries. In fact, where 

 strong currents are wanted, batteries are abandoned in 

 favour of dynamo-electric machines, worked by steam or 

 water power, or by gas engines. 



In the case of transmission of power, as in the pre- 

 ceding paragraph, the expense may be far smaller if the 

 original water-power costs little. The dynamo-machine 

 may turn 8 5 per cent of the mechanical power into the 

 energy of electric currents, and the electromotor may 

 convert back 85 per cent of the current-energy (or 72 

 per cent of the original power) into work. 



The mechanical -work of a current may be calculated as follows : A 

 current whose strength is C conveys through the circuit in t seconds a quan- 

 tity of electricity = Ct. \ But the number of ergs of work W, done by a 

 current is equal to the product of the quantity of electricity into the differ- 

 ence of potentials through which it is transferred (Art. 367), provided these 

 latter are expressed in "absolute" C.G.S. units ; or 



av = w. 



Now if W ergs of work are done in t seconds, the rate of working is got by 

 dividing W by t; whence 



CV = ^. 



If C and V are expressed in webers and volts respectively, and it is desired 

 to give the rate of working in horse-power, it must be remembered that 

 i iveber= lo- 1 C.G.S. units of current ; that i volt = io 8 C.G.S. units of 

 E.M.F. ; and that i horse-power (as defined by Watt) =; 550 foot-pounds 

 per second = 76 kilogramme-metres per second = 76 X io 5 gramme-centi- 

 metres per second = 745 X io 7 ergs per second, whence 



C*feyrxVwgj = rate of doing work in H> _ R 



745 



For example, to find the rate at which actual work is consumed in an 

 electric lamp : measure the whole current in webers ; measure the difference 

 of potential between the terminals of the lamp in volts; multiply them 

 together and divide by 745 ; the result will be the number of horse-power 

 used up in the lamp. 



