220CHAPTER 21 (b) For an ideal transformer, () ()( )avav2, rms2, rmsinputouputVI==∆PP()( )Thus, av9.0inputPV0.400 A3.6 W21.39The power input to the transformer is ( )51, rms1, rms3 600 V50 A1.810 Winput=∆==×PFor an ideal transformer, ( )av2, rms2, rmsavouputinput=PPso the current in the long-distance power line is 5av2,rms2, rms1.81.8 A100 000 VinputIV×=∆P2221.8 A100 3.2lostlineIRΩ=×PThe power dissipated as heat in the line is then The percentage of the power delivered by the generator that is lost in the line is 253.20%100%0.18%1.8×=×=×% Lost10lostinput=×PP21.40(a) Since the transformer is to step the voltage downfrom 120 volts to 6.0 volts, the secondary must have fewer turns than the primary. (b) For an ideal transformer,
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This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.