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Unformatted text preview: P Load = I 3, RMS ( ) 2 " R Load Thus I 3, RMS ( ) 2 = P Load R Load = 10 6 W 10 " 2 # = 10 8 A 2 I 3, RMS = 10 4 A Continued on the next page Practice Problem 3 Solution, continued From the given numbers of turns, obtain: I 2, RMS = I 3, RMS " n 4 n 3 # $ % & ’ ( = 10 4 A ( ) " 1,000 200,000 # $ % & ’ ( = 50 A Thus P Line = I 2, RMS ( ) 2 " R Line = 50 A ( ) " 1 # ( ) = 2,500 W = 2.5 kW Answer: C...
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This note was uploaded on 12/06/2010 for the course EECS 314 taught by Professor Ganago during the Spring '07 term at University of Michigan.
 Spring '07
 Ganago

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