120 Current and ResistanceAdditional ProblemsP27.54(a)IVR=∆soP==IVVR∆∆af2RV=∆Ωafaf2212025 0576 VW.andRV=∆Ωafaf120100144VW(b)IVQtt===∆∆∆25 00208100...W120 VAC∆t480..C0.208 AsThe bulb takes in charge at high potential and puts out the same amount of charge at lowpotential.(c)=25 0..WJ∆U∆t00400..J25.0 WsThe bulb takes in energy by electrical transmission and puts out the same amount of energyby heat and light.(d)Ut= ×25 086 40030 064 8 108...JssddJbgbgafThe electric company sells energy .CostJ$0.070 0kWhk1000WsJh3600 sCost per joulekWhkWh3.60 10 JJ6=×FHGIKJFHGIKJ⋅FHGIKJFHGIKJ==×FHGIKJ−64 8 1026070 094 1068.$1.$0.$1.*P27.55The original stored energy is UQVQCii12122∆.(a)When the switch is closed, charge Qdistributes itself over the plates of Cand 3Cin parallel,presenting equivalent capacitance 4C. Then the final potential difference is
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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 .