22_Ch 21 College Physics ProblemCH21 Alternating Current Circuits and Electromagnetic Waves

22_Ch 21 College Physics ProblemCH21 Alternating Current Circuits and Electromagnetic Waves

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222 CHAPTER 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 () rms rms max 4 500 V 15.5 A 290 line V I R == = ()() ( ) rms rms max max 4.50 10 V 15.5 A input VI =∆ P The 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. 34 6.98 10 W 6.98 kW =× = 21.43 From vf λ = , the wavelength is 8 6 3.00 10 m s 4.00 10 m 4 000 km 75 Hz × = × = v f The 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.

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