EE_387_200809SP_homework_01_solution

EE_387_200809SP_homework_01_solution - 1 cos(t sin(t cos(et...

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1 Homework 01 Solution © Jeffrey Mayer 2005-2009. All rights reserved. EE 387 Spring 2009 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 ++ + + ⎡⎤ = ⎢⎥ +− + + + ⎣⎦ −+ = −− + = )) cos(( ) ) sin(( ) ) e e e t t t ωϕ ωω ϕ =
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2 Homework 01 Solution © Jeffrey Mayer 2005-2009. All rights reserved. EE 387 Spring 2009 + 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 e it t ω = ( ) 2117.5cos( 1.54 ) V e vt t = 2. + 1 ( ) 2120cos( ) V e t = 1 () 0.5 Ω 2.5 mH 2 20 Ω 40 mH 50 F μ +
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3 Homework 01 Solution © Jeffrey Mayer 2005-2009. All rights reserved. EE 387 Spring 2009 ( ) 2 240cos( 15 ) V e vt t ω =+ ° ( ) 218cos( 15 )A e 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 2005-2009. All rights reserved.
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This note was uploaded on 10/08/2009 for the course EE 387 taught by Professor Jefferymayer during the Spring '08 term at Penn State.

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EE_387_200809SP_homework_01_solution - 1 cos(t sin(t cos(et...

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