EE_387_200809FA_homework_01_solution

EE_387_200809FA_homework_01_solution - 1. cos(t ) sin(t )...

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1 Homework 01 Solution © Jeffrey Mayer 2005-2008. All rights reserved. EE 387 Fall 2008 1. cos( ) cos( )cos( ) sin( )sin( ) cos( ) sin( ) sin( ) sin( ) cos( ) sin( ) cos( ) cos( ( )) sin( ( )) cos(( ) ) sin(( ee e e e e e tt t t t t t t t ω ϕω ϕ ωω ωωϕ ++ + + ⎡⎤ = ⎢⎥ +− + + + ⎣⎦ −+ = −− + = )) e t ωϕ
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2 Homework 01 Solution © Jeffrey Mayer 2005-2008. All rights reserved. EE 387 Fall 2008 + 1 ( ) 2 120cos( ) V vt t ω =⋅ 1 () 0.5 Ω 2.5 mH 2 20 Ω 40 mH 50 F μ + it + 1 120 0 V =∠ ° ± 0.5 Ω 0.9425 j Ω 20 Ω 15.08 j Ω 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 = + =+ Ω 120 0 (0.5 0.9425) (30.56 4.972) 3.795 10.8 A I jj ∠ ° = ++ + ° ± 2 120 (3.7952 10.8 )(0.5 0.9425) 117.5 1.54 V Vj =− −° + ° ± ( ) 2 3.795cos( 10.8 ) A t = ( ) 2117.5cos( 1.54 ) V t = 2.
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3 Homework 01 Solution © Jeffrey Mayer 2005-2008. All rights reserved. EE 387 Fall 2008 ( ) 2 240cos( 15 ) V vt t ω =+ ° ( ) 218cos( 15 )A it t =− ° * (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 = P = 3.
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4 Homework 01 Solution © Jeffrey Mayer 2005-2008. All rights reserved.
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EE_387_200809FA_homework_01_solution - 1. cos(t ) sin(t )...

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