Solution to Homework 9

Solution to Homework 9 - Homework 9 Solutions P10.3-5 18 V...

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Unformatted text preview: Homework 9 Solutions P10.3-5 18 V A = 18 2 16 ms T =- = 2 2 393 rad/s 0.016 T π π ϖ = = = 16 18 cos ( ) 27 θ θ = ⇒ = ° ( 29 ( 29 18 cos 393 27 V v t t = + ° P10.4-1 120 400cos300 i s di di L R v i t dt dt + = - ⇒ + = - Try cos300 sin300 then 300 sin300 300 cos300 f f di i A t B t A t B t dt = + = - + . Substituting and equating coefficients gives 0.46 300 120 1.15 300 120 400 A A B B B A = -- + = = - + =- Then ( ) 0.46cos300 1.15sin300 1.24cos(300 68 ) A i t t t t = -- =- ° P10.4-2 500 500cos1000 2 s v dv dv i C v t dt dt- + + = ⇒ + = Try cos1000 sin1000 then 1000 cos1000 1000 cos1000 f f dv v A t B t A t B t dt = + = - + . Substituting and equating coefficients gives 1000 500 0.2 0.4 1000 500 500 A B A B B A- + = = ⇒ = + = Then ( ) 0.2cos1000 0.4sin1000 0.447cos(1000 63 ) V v t t t t = + =- ° P10.5-1 (5 36.9 ) (10 53.1 ) 50 16.2 10 16.2 2 5 10.36 (4 j3)(6 j8) 10 j5 5 26.56 ° ° ° ° ° ° ∠ ∠- ∠- ∠- = = = ∠ +-- ∠- P10.5-2 3 2 45 3 5 81.87 4 3 5 81.87 [4 3 36.87 ] 5 5 2 8.13 5 81.87 (4.48 3.36) 5 81.87 (5.6 36.87 ) 28 45 14 2...
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This note was uploaded on 09/30/2010 for the course ECE 2025 taught by Professor Juang during the Spring '08 term at Georgia Tech.

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Solution to Homework 9 - Homework 9 Solutions P10.3-5 18 V...

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