V₂ = + E₂ + E3 = +9V +7V = 16 V However, note also that point a is connected to point b so V₂ = V₁ = 16 V c. The voltage
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V₂ = + E₂ + E3 = +9V +7V = 16 V However, note also that point a is connected to point b so V₂ = V₁ = 16 V c. The voltage
V₂ = + E₂ + E3 = +9V +7V = 16 V However, note also that point a is connected to point b so V₂ = V₁ = 16 V c. The voltage Vab can be determined directly from Vab= V₁ - V₂ = 16V - 16 V = OV or applying Kirchhoff's voltage law around the closed loop we have +Vab-E₁ + E₂ + E3 = 0 Vab = E₁ - E3 - E₂ = 16 V - 7V-9V = OV and EXAMPLE 5.30 For the series network of Fig. 5.66, determine a The voltage V₁. b. The voltages V, and Vc. The voltage Vab- Solutions: & Let us first determine the current: E 72 V 72 V = 4 A + E = 72 V R₁ on v R₂ 80 V₂ R₁40 V;