20_Ch 21 College Physics ProblemCH21 Alternating Current Circuits and Electromagnetic Waves

20_Ch 21 College Physics ProblemCH21 Alternating Current Circuits and Electromagnetic Waves

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
220 CHAPTER 21 (b) For an ideal transformer, () () ( ) av av 2, rms 2, rms input ouput VI == PP () ( ) Thus, av 9.0 input P V 0.400 A 3.6 W 21.39 The power input to the transformer is ( ) 5 1, rms 1, rms 3 600 V 50 A 1.8 10 W input =∆ = = × P For an ideal transformer, ( ) av 2, rms 2, rms av ouput input = P P so the current in the long- distance power line is 5 av 2,rms 2, rms 1.8 1.8 A 100 000 V input I V × = P 2 22 1.8 A 100 3.2 lost line IR = × P The 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 2 5 3.2 0% 100% 0.18% 1.8  × = ×=  ×  % Lost 10 lost input P P 21.40 (a) Since the transformer is to step the voltage down from 120 volts to 6.0 volts, the secondary must have fewer turns than the primary. (b) For an ideal transformer,
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/09/2011 for the course PHYS 2020 taught by Professor Staff during the Fall '10 term at FIU.

Ask a homework question - tutors are online