hw9key - t > 200 ( ) s 4 200 2 10 s s-= ( ) 4 4 4 4 1 10...

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1 HW9Key First, 0 < t < 200 , we have two deferential equations: ( I1 is the current go through the inductor and I2 is the current go through resistor. s μ 1 2 1 2 10 dI L I R dt I I = + = Solve these equations( with initial condition: I1=0 when t = 0), we will get: 4 4 1 10 1 1 10 2 10 10 10 10 10 10 R t t L R t t L I e e I e e - - × - - × = - = - = =
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2 HW9Key Second,
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Unformatted text preview: t > 200 ( ) s 4 200 2 10 s s-= ( ) 4 4 4 4 1 10 2 10 1 10 1 2 1@(200 ) 1 10 10 10 9 R t t L s t I I I e e e e--- - - = = =- = 3 HW9Key Energy conserve 2 6 2 2 2 3 1 1 1 10 40 0.4( ) 2 2 10 10 CV CV LI I A L-- = = = =...
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This note was uploaded on 05/06/2008 for the course PHYS 1304 taught by Professor Ye during the Spring '08 term at SMU.

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hw9key - t > 200 ( ) s 4 200 2 10 s s-= ( ) 4 4 4 4 1 10...

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