Solution d'Exercice 1 : Théorème de Millman - Circuits électriques
Ici, E 1 = 1 V, E 2 = 2 V, E 3 = 3 V
Z 1 = 1 Ω, Z 2 = 2 Ω, Z 3 = 3 Ω
∴ Y 1 = 1
, Y 2 = 0.5
, Y 3 = Y = 3 
Par le théorème de Millman, le circuit équivalent est représenté.
∴ ......
et
∴ ......
Appliquons la loi des noeuds au noeud A:

Z 1 = 1 Ω, Z 2 = 2 Ω, Z 3 = 3 Ω
∴ Y 1 = 1




Par le théorème de Millman, le circuit équivalent est représenté.
∴ ......

et
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∴ ......

Rappel :
Ce théorème très pratique permet de déterminer la différence de potentiel aux bornes de plusieurs branches en parallèle.
Soit un circuit linéaire en régime permanent.
I1+ I2 + I3 + I = 0
U = E1 – R1I1 = E2 – R2I2 = – R3I3
I1 = (E1 – U) / R1 , I2 = (E2 – U) / R2 , I3 = -U / R3
I1+ I2 + I3 + I = 0 → (E1 – U) / R1 + (E2 – U) / R2 – U / R3 + I = 0
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