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EE_365_200708FA_homework_04_solution

# EE_365_200708FA_homework_04_solution - 1 I1 ~ V1 ideal 0.3...

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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 2 2 tan( ) 75tan(31.7 ) 46.48 kvar Q P φ = = ° = * 2 2 2 * 75 46.48 kVA 230 326.1 202.1A 383.6 31.78 A S I V j j = + = = = ∠− ° ± ± 2 2 2 75 46.48 kVA S P jQ j = + = + P 2 =75 kW 1 2 2 1 cos ( ) cos (0.85) 31.7 pf φ = = = ° 2 2 2 1 240 383.6 31.78 A 2400 38.36 31.78 A N I I N = = ∠ − ° = ∠ − ° ± ± 2 2 10 230 0 2300 0 V V aV = = ∠ ° = ∠ ° ± ± 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 2 2 2400 (0.003) 240 0.3 N r r N = = = Ω 2 1 2 2 2 2 2400 (0.0065) 240 0.65 l l N X X N = = = Ω 1 2 3 1 2 2 2 2 ( ) (2300 V) (0.3 0.65 )(38.36 31.78 A) 2323 15.13 V 2323 0.37 V m l E V r jX I j j = + + = + + Ω ∠ − ° = + = ° ± ± ±

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