118 Current and Resistance*P27.47The energy taken in by electric transmission for the fluorescent lamp isP∆t=FHGIKJ=×FHGIKJFHGIKJ⋅FHGIKJFHGIKJ=111003600396 100808866Jshs1 hJcostJkWhk1000WsJh3600 saf..$0.$0.For the incandescent bulb,∆t=FHGIKJ×FHGIKJ==−=40100144 1008323208823277W hs1 hJcostJ3.6 10 Jsaving6..$0.$0.$0.$0.$0.P27.48The total clock power is270 102 50243 10612×FHGIKJFHGIKJclocksclocks1 hJhej...From eWQ=outin, the power input to the generating plants must be:QtWteinout±Jh∆∆==×0250972 101212...and the rate of coal consumption isRate ×FHGIKJ=100295 10295125...kg coal33.0 10 Jkg coal hmetric ton h6.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 .