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 3 600 3 96 10 3 96 10 08 088 6 6 J s h s 1 h J cost J kWh k 1 000 W s J h 3 600 s a f . . \$0. \$0. For the incandescent bulb, P t = F H G I K J = × = × × F H G I K J = = = 40 100 3 600 1 44 10 1 44 10 08 32 32 088 232 7 7 W h s 1 h J cost J 3.6 10 J saving 6 a f . . \$0. \$0. \$0. \$0. \$0. P27.48 The total clock power is 270 10 2 50 3 600 2 43 10 6 12 × F H G I K J F H G I K J = × clocks J s clock s 1 h J h e j . . . From e W Q = out in , the power input to the generating plants must be: Q t W t e in out J h J h = = × = × 2 43 10 0 250 9 72 10 12 12 . . . and the rate of coal consumption is Rate = × × F H G I K J = × = 9 72 10 1 00 2 95 10 295 12 5 .
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