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TD #1 - Exercice 4.65 page 210 1 = R C 1 Z q = R si L =0 C...

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( ) ( ) 1 1 1 0 1 0 31623 rad s 15 éq éq éq Z R jL j R j L C C Z R si L C avec LC Z ω ω ω ω ω ω ω ω ω = + = + = = = > = = Ω Exercice 4.65, page 210 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 1 2 2 1 2 rad s 2 0 V 4 4 4 1 1 1 2 2 4 Transformation de la source de tension en source de courant 2 0 0,5 0 A 4 Diviseur de courant 1 S L C S S S L L j Z R Z R Z j L j Z j C j j j j R j Z j ω ω ω ω ω ω ω ω = = = = = = = = = − = − = − = = = × = D D D Exercice 4.72, page 212 V V I I I ( ) ( ) ( ) ( ) ( ) ( ) ( ) 1 2 1 0,5 0 4 1 1 1 1 1 1 4 4 2 4 0,26 71,6 A 0,26cos 2 71,6 A L C L L j Z Z Z Z j j j i t t ω × = + + + + + + = ∠ − = D D D I
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( ) ( ) a) Impédance de Thévenin Le schéma est déjà dans le domaine des phaseurs Élimination de la source de tension court-circuit 1 11,3 45 1 1 8 8 8 a) Source de Thévenin Diviseur de Th Z j j = = ∠ − + + D Exercice 4.78, page 213 ( ) ( ) ( ) ( ) ( ) tension 8 8 5 30 8 8 7,07 15 V 8 8 8 8 8 8 S Th j j j j j j + ∠ − + = = = + + + + D D V V
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( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 3 6 1 1 2 2 2 2 1000 rad s 4 30 V 120 60 10 1000 60 1 1 12,5 10 1000 80 Méthode des noeuds: 0 7,16 33,4 V 7,16cos 1000 33,4 V i o L L C i L C o o o j Z R Z j L j j Z j C j j Z Z Z v t t ω ω ω ω = = = = = = × = = − = − × = − = = = = ∠ − = D D D Exercice 4.84, page 213 V V V V V V V V V Z C =-j(1/C V S (j ) + - V 1 Z L =jL Z o =R L V 2 Z C =-j(1/C + V o -
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( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) 377 rad s 230 0 V 25 26,53 36,45 46,7 ) 1 25 26,53 37,15
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