222CHAPTER 21 (c) It is impossible to deliver the needed power with an input voltage of 4.50 kV. The maximum line current with an input voltage of 4.50 kV occurs when the line is shorted out at the customer’s end, and this current is ()rmsrmsmax4 500 V15.5 A290 lineVIR∆===Ω()()()rmsrmsmaxmax4.5010 V15.5 AinputVI=∆=×PThe maximum input power is then This is far short of meeting the customer’s request, and all of this power is lost in the transmission line. 346.9810 W6.98 kW=× =21.43From vfλ=, the wavelength is 863.0010 m s4.0010 m4 000 km75 Hz×=×=vfThe required length of the antenna is then,
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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.