118 Current and Resistance*P27.47The energy taken in by electric transmission for the fluorescent lamp isP∆t=FHGIKJ=×=×FHGIKJFHGIKJ⋅FHGIKJFHGIKJ=111003 6003 96103 96100808866Jshs1 hJcostJkWhk1 000W sJh3 600 saf..$0.$0.For the incandescent bulb,P∆t=FHGIKJ=×=××FHGIKJ==−=401003 6001 44101 441008323208823277Whs1 hJcostJ3.610Jsaving6af..$0.$0.$0.$0.$0.P27.48The total clock power is270102 503 6002 4310612×FHGIKJFHGIKJ=×clocksJ sclocks1 hJ hej...From eWQ=outin, the power input to the generating plants must be:QtWteinoutJ hJ h∆∆==×=×2 43100 2509 72101212...and the rate of coal consumption isRate =××FHGIKJ=×=9 72101 002 9510295125.
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