IRWIN 9e 11_58 - Irwin Basic Engineering Circuit Analysis...

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Unformatted text preview: Irwin, Basic Engineering Circuit Analysis, 9/E 1 11.53 .-‘i. balanced thi'ee-phuae source server; Ihe i'nJlam-‘ing lnada: Ll‘Jlel l: 60 EVA. all 0.3 pf Jaggng Lile 2: 3E1 M’s-1. m 0.7'5 111' lugging The line wit-age at the Juml 1'5 Eliii ‘v' mm at H) HR. Determine the line eurrent and the eemhjneLl power meter at the lead. SOLUTION:LOG\0L it 60 KVA at 0'8 1430012: sokvA a): one; PF 49%;»? 1QabI: 208v Arm m: 60H; Van; IzoLo" V/‘ng ’3_¢ = 60k LCOJ’I(O'8) 15%,, = €0L368¥D /<\/A ' § SI]. " '3’¢ :3 523,¢ -'— 30k L601-) (0175—) 20 (56. 81° km Chapter 11: Polyphase Circuits Problem 11.58 Irwin, Basic Engineering Circuit Analysis, 9/E 6; = 30 6.41.41” mi 3‘15 51’ ¢ = :50 (.4-I- 41° km E s 3— + 5 72—19 ’1—16 21-15 51¢ = 20k £56.81" +Ioi< mum 3 Tug; = 29. 08 [58.36 Km 120 L0” \7 an = ($29.98 k (38. 38') ’6 “A = 2W. 83 é3asa° A my FF: v611438.389 == 0'789 _______________._.____._..———————-—-— Problem 11.58 Chapter 11: Polyphase Circuits ...
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This note was uploaded on 10/04/2010 for the course ECE ECE212 taught by Professor Micahstickel during the Fall '10 term at University of Toronto.

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IRWIN 9e 11_58 - Irwin Basic Engineering Circuit Analysis...

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