PS3 W08 e2 p6s

# PS3 W08 e2 p6s - P Load = I 3, RMS ( ) 2 " R Load Thus...

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Chapter 16 Practice Problem 3 In the circuit shown on this diagram, the numbers of the turns in the two transformers are: n 1 = 1,000 n 2 = 200,000 n 3 = 200,000 n 4 = 1,000 The resistances are: R Line = 1 Ω R Load = 0.01 Ω The load resistor R Load absorbs 1 MW of average power. How much average power absorbs the line resistor R Line ? A. 250 kW B. 4 kW C. 2.5 kW D. Not enough information E. None of the above Solution The average AC power absorbed by the load equals

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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.

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PS3 W08 e2 p6s - P Load = I 3, RMS ( ) 2 " R Load Thus...

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