Lessons in practical electricity; principles, experiments, and arithmetical problems, an elementary text-book . on short circuit by Formula E 12 X 12(81) W = -= —0— =48 watts, or four times the energy required by the lamp. The resistance of the lamp is3 ohms, Formula (72), and when connected to the 6 cells inseries will receive 2 amperes, Formula r6Voitr (37). The pressure sending the current rVT through the cells isCXr = 2x3 = 6 volts, or one-half the E. M. F., and 6 voltsare also maintained across the lamp termi-nals. The external power is thus 2x6= 12 watts and the watts lost in the cell 12
Lessons in practical electricity; principles, experiments, and arithmetical problems, an elementary text-book . on short circuit by Formula E 12 X 12(81) W = -= —0— =48 watts, or four times the energy required by the lamp. The resistance of the lamp is3 ohms, Formula (72), and when connected to the 6 cells inseries will receive 2 amperes, Formula r6Voitr (37). The pressure sending the current rVT through the cells isCXr = 2x3 = 6 volts, or one-half the E. M. F., and 6 voltsare also maintained across the lamp termi-nals. The external power is thus 2x6= 12 watts and the watts lost in the cell 12watts, or the total watts = 24, or one-halfthe power the cells could deliver on shortcircuit. 227. Efficiency of a Battery.—Byefficiency is meant the relation of the usefulwork done to the total energy perfect battery, or dynamo (that is, onewith no internal resistance), would deliverall of the energy to the external circuit, butas some portion of it is lost in the internalresistance the useful energy is always less than the totalenergy expended. If the total energy expended is represented. r = 3p>* evoiu I*— evoltspA/WVWWV-J C SLmparaa t tm I Li*. 3 6VoJt8 Fig. 201.—Lost on therial Resistance VoltsInter- ELECTRICAL WORK AND POWER. 225 by 100, and one-half of this amount is unavailable for usefulwork, the efficiency would be 50 per centum or 50 per no machine, then, can the efficiency be 100 per cent. To Find the Efficiency of a Battery : Divide the resistance of the external circuit by the resistance ofthe external circuit plus the resistance of the battery. Eft=RT-r <83)- Prob. 90 : What is the efficiency of the battery in Prob. 89 ? Thetotal resistance of the battery is 3 ohms and the resistance of the lampis 3 ohms. By Formula (85) Eff. = ^-^ = ^~] = -50 or 50%. The total watts expended in ^| 226 are 24, and the usefulwatts in the lamp 12, so that the efficiency is the ratio of Useful watts 12 = _ 5Q = ^ ag before_ Total watts expende
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