Bonjour,
en suivant l'énoncé détaillé :
1) U(CD) = R₁ x I₁ = 1,0.10³ x 6,0.10⁻³ = 6,0 V
2) U(DE) = U(BF) - U(CD) = U(AF) - U(CD) = 12,0 - 6,0 = 6,0 V
3) I₂ = U(DE)/R₂ = 6,0/1,5.10³ = 4.10⁻³ A = 4,0 mA
I₃ = I₁ - I₂ = 6,0 - 4,0 = 2,0 mA
R₃ = U(DE)/I₃ = 6,0/2,0.10⁻³ = 3,0.10³ Ω = 3 kΩ
