778_Physics ProblemsTechnical Physics

778_Physics ProblemsTechnical Physics - 118*P27.47 Current...

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118 Current and Resistance *P27.47 The energy taken in by electric transmission for the fluorescent lamp is P t = F H G I K J F H G I K J F H G I K J F H G I K J F H G I K J = 11 100 3600 396 10 08 088 6 6 Js h s 1 h J cost J kWh k 1000 Ws J h 3600 s af . . $0. $0. For the incandescent bulb, t = F H G I K J × F H G I K J = =− = 40 100 144 10 08 32 32 088 232 7 7 W h s 1 h J cost J 3.6 10 J saving 6 . . $0. $0. $0. $0. $0. P27.48 The total clock power is 270 10 2 50 243 10 61 2 × F H G I K J F H G I K J clocks clock s 1 h Jh ej .. . From e W Q = out in , the power input to the generating plants must be: Q t Wt e in out ±Jh == × 0250 972 10 12 12 . . . and the rate of coal consumption is Rate × F H G I K J = 100 295 10 295 12 5 . . . kg coal 33.0 10 J kg coal h metric ton h 6 . P27.49
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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