EE_365_200708FA_homework_01_solution

EE_365_200708FA_homework_01_solution - 1 Homework 01...

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Unformatted text preview: 1 Homework 01 Jeffrey Mayer 2005-2007. All rights reserved. EE 365 Fall 2007 1. + 1 ( ) 2 120cos( ) V v t t = 1 ( ) v t 0.5 2.5 mH 2 ( ) v t 20 40 mH 50 F + ( ) i t + 1 120 0 V = 0.5 0.9425 j 20 15.08 j 18.85 mS j + I 2 V 2 3 1 1 18.85 10 20 15.08 30.56 4.972 Z j j j = + + = + 120 0 (0.5 0.9425) (30.56 4.972) 3.795 10.8 A I j j = + + + = 2 120 (3.7952 10.8 )(0.5 0.9425) 117.5 1.54 V V j = + = ( ) 2 3.795cos( 10.8 ) A i t t = ( ) 2117.5cos( 1.54 ) V v t t = 2 Homework 01 Jeffrey Mayer 2005-2007. All rights reserved. EE 365 Fall 2007 ( ) 2 240cos( 15 ) V v t t = + ( ) 218cos( 15 ) A i t t = 2. * (240 15 )(18 15 ) 4.320 30 kVA 3.741 2.160 kVA S j = = = + 4.320 kVA S = 1 cos (30 ) 0.866 lagging pf = = 3.741kW P = 2.160 kVAR Q = 2.160 kVAR Q = 4.320 kVA S = 3.741kW P = 3 Homework 01 Jeffrey Mayer 2005-2007. Jeffrey Mayer 2005-2007....
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This note was uploaded on 10/08/2009 for the course EE 365 at Pennsylvania State University, University Park.

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EE_365_200708FA_homework_01_solution - 1 Homework 01...

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