DIPOLES_LINEAIRES_correction - Correction du T.D 8 Exercice...

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Correction du T.D 8 Exercice 8.1 6 . 17 Ω Exercice 8.2 1 - Loi des mailles : U AB = E 1 - r 1 i 1 = E 3 - r 3 i 3 = E 2 + U K Loi des noeuds : i 1 = - i 3 Donc : i 1 = E 1 - E 3 r 1 + r 3 et U K = U AB - E 2 = E 1 - r 1 r 1 + r 3 ( E 1 - E 3 ) - E 2 A.N U K = 0 . 22 V 2 - Loi des mailles : U AB = E 1 - r 1 i 1 = E 2 - r 2 i 2 = E 3 - r 3 i 3 Donc : i 1 = E 1 - E 2 + r 2 i 2 r 1 et i 3 = E 3 - E 2 + r 2 i 2 r 3 Loi des noeuds : i 1 + i 2 + i 3 = 0 d’o`u on tire : i 1 = E 1 r 1 + E 1 r 3 - E 2 r 2 - E 3 r 3 1 + r 1 r 2 + r 1 r 3 ; i 2 = E 2 r 1 + E 2 r 3 - E 1 r 1 - E 3 r 3 1 + r 2 r 1 + r 2 r 3 ; i 3 = E 3 r 1 + E 3 r 2 - E 1 r 1 - E 2 r 2 1 + r 3 r 1 + r 3 r 2 Soit i 1 = 4 . 64 10 - 1 A ; i 2 = - 0 . 35 10 - 1 A ; i 3 = - 4 . 28 10 - 1 A Exercice 8.3 Il suffit d’´ ecrire la loi des mailles dans les diff´ erentes mailles, apr` es simplification des intensit´ es en utilisant la loi des noeuds. On obtient ainsi un syst` eme de 3 inconnues ` a 3 ´ equations dont la solution est : I = E 2( R + r ) A.N : I = 1 A Exercice 8.4 Il suffit de faire des simplifications au niveau des r´ esistances pour trouver : U AB = 1 . 18 V ; U BC = 0 . 788 V ; U DE = 1 . 01 V ; U EF = 0 . 338 V ; U DF = 1 . 35 V Exercice 8.5 1 -
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