nagle_differential_equations_ISM_Part54

Nagle_differential_e - Exercises 7.9 Therefore we have the following system for the currents I 1 I 2 and I 3 005 I 1 10 I 2 = 50 01 I 3 10 I 2 2 I

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Unformatted text preview: Exercises 7.9 Therefore, we have the following system for the currents I 1 , I 2 , and I 3 : . 005 I 1 + 10 I 2 = 50 . 01 I 3 + 10 (- I 2 + 2 I 3 ) = 0 I 1- I 2- I 3 = 0 (7.29) with initial conditions I 1 (0) = I 2 (0) = I 3 (0) = 0. Let I 1 ( s ) := L{ I 1 } ( s ), I 2 ( s ) := L{ I 2 } ( s ), and I 3 ( s ) := L{ I 3 } ( s ). Using the initial conditions, we conclude that L{ I 1 } ( s ) = s I 1 ( s )- I 1 (0) = s I 1 ( s ) , L{ I 3 } ( s ) = s I 3 ( s )- I 3 (0) = s I 3 ( s ) . Using these equations and taking the Laplace transform of the equations in (7.29), we come up with . 005 s I 1 ( s ) + 10 I 2 ( s ) = 50 s- 10 I 2 ( s ) + (0 . 01 s + 20) I 3 ( s ) = 0 I 1 ( s )- I 2 ( s )- I 3 ( s ) = 0 . Expressing I 2 ( s ) = I 1 ( s )- I 3 ( s ) from the last equation and substituting this into the the first two equations, we get (0 . 005 s + 10) I 1 ( s )- 10 I 3 ( s ) = 50 s- 10 I 1 ( s ) + (0 . 01 s + 30) I 3 ( s ) = 0 ....
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This note was uploaded on 02/12/2011 for the course MA 221 taught by Professor Mazmani during the Spring '08 term at Stevens.

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Nagle_differential_e - Exercises 7.9 Therefore we have the following system for the currents I 1 I 2 and I 3 005 I 1 10 I 2 = 50 01 I 3 10 I 2 2 I

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