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

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() 11 1 0 1 0 31623 rad s 15 éq éq éq ZR j Lj R j L CC s i L C avec LC Z ωω ω ⎛⎞ =+ ⎜⎟ ⎝⎠ =− = => = Exercice 4.65, page 210 ( ) ( ) ( ) 22 1 2r a d s 20 V 4 4 4 1 2 2 4 Transformation de la source de tension en source de courant 0,5 0 A 4 Diviseur de courant 1 S L C S S S L L j Zj L j C j j j j R j Z j = =∠ == = × = D D D Exercice 4.72, page 212 V V I I I () ( ) 12 1 0,5 0 4 1 1 1 1 44 2 4 0,26 71,6 A 0,26cos 2 71,6 A LC L L j ZZ j j j it t ∠× = ++ +− + 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 11 888 a) Source de Thévenin Diviseur de Th Z jj == + −+ D Exercice 4.78, page 213 ( ) tension 88 5 3 0 7,07 15 V 88 8 S Th jjjj +∠ + = −++ D D V V
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() ( ) 3 6 1 12 2 2 2 1000 rad s 43 0 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 oL L C i LC o o o j ZR Zj L j j C j j ZZ Z vt t ω = =∠ == × = =− × = −−= D D D Exercice 4.84, page 213 V VV VV 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 45,6 cos 45,6 0,7 ) 230 0 6,19 45,6 A 37,15 45,6 ) 230 0 46 S L éq S S S éq SL L éq j Zj a fp a j j Z b jZ j Z ω = =∠ =− = =+ − = = == = ∠−
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This note was uploaded on 04/21/2009 for the course ELE ELE1403 taught by Professor Khaled during the Spring '09 term at NYU Poly.

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

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