EE_365_200708FA_homework_04_solution

EE_365_200708FA_homework_04_solution - 1. I1 + ~ V1 - ideal...

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1 Homework 04 © Jeffrey Mayer 2005-2007. All rights reserved. EE 365 Fall 2007 1. 2 230 0 V V =∠ ° ± 1 ~ V + + 0.3 j 0.65 0.003 j 0.0065 2400:240 ideal 2 I ± 1 I ± j 1 k 4.5 k 75 kW 0.85 lagging 22 tan( ) 75tan(31.7 ) 46.48 kvar QP φ = = ° = * 2 2 2 * 75 46.48 kVA 230 326.1 202.1 A 383.6 31.78 A S I V j j ⎛⎞ = ⎜⎟ ⎝⎠ + = =− −° ± ± 2 75 46.48 kVA SPj Q j =+ P 2 =75 kW 1 1 cos ( ) cos (0.85) 31.7 pf = = 2 1 240 383.6 31.78 A 2400 38.36 31.78 A N II N ′ = = ∠− ° ±± 10 230 0 2300 0 V Va V ′ = = ⋅∠ ° = ∠° 1 ~ V + + 0.3 j 0.65 0.015 j 0.024 1 I ± j 1 k 4.5 k 2 38.36 31.78 A I ′ = ° ± 2 2300 0 V V ′ = ± 2 1 2 2 2400 (0.003) 240 0.3 N rr N ′ = = 2 1 2 2 2400 (0.0065) 240 0.65 ll N XX N ′ = = 1 2 3 12 2 2 2 () (2300 V) (0.3 0.65 )(38.36 31.78 A) 2323 15.13 V 2323 0.37 V ml EVr j X I j j ′′ =++ = ++ Ω ° ° ±
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2 Homework 04 © Jeffrey Mayer 2005-2007. All rights reserved. EE 365 Fall 2007 1 1 1 || 2323 0.37 V 1000 || 4500 0.5313 2.32 A 2.38 77.1 A m m mc E I jX R j j = ∠° = =− =∠ −° ± ± 12 1 (38.36 31.78 A) (0.5313 2.32 A) 33.14 22.53 A 40.07 34.21 A m III j j =+ ° + ±±± 11 1 1 1 () (2323 15.13 V) (0.3 0.65)(40.07 34.21 A) 2347 29.91 V 2348 0.730 V ml VE rj X I jj j + + + ° =∠° ±± ± * 1 * (2348 0.730 V)(40.07 34.21 A) 94.08 34.94 kVA 77.12 53.88 kVA SV I j = ° ° ° 1. (cont.)
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3 Homework 04 © Jeffrey Mayer 2005-2007.
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EE_365_200708FA_homework_04_solution - 1. I1 + ~ V1 - ideal...

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