For a three-phase system with line voltage V_L = 480 V and line current I_L = 60 A at unity power factor, what is the real power P?

For a three-phase system with line voltage V_L = 480 V and line current I_L = 60 A at unity power factor, what is the real power P?

• A. 12.0 kW
• B. 25.0 kW
• C. 49.9 kW
• D. 60.0 kW

Answer: (C)

In a balanced three-phase system, the real power using line quantities is P = √3 × V_L × I_L × cosφ. When the power factor is unity, cosφ = 1, so P = √3 × V_L × I_L. With V_L = 480 V and I_L = 60 A, P = 1.732 × 480 × 60 = 1.732 × 28,800 ≈ 49,881 W, about 49.9 kW. This directly gives the correct value. If you tried using P = V_L × I_L without the √3 factor, you’d get 28.8 kW, which is not correct for three-phase line quantities. The √3 factor is essential when working with line voltage and line current in a three-phase system.

In a balanced three-phase system, the real power using line quantities is P = √3 × V_L × I_L × cosφ. When the power factor is unity, cosφ = 1, so P = √3 × V_L × I_L. With V_L = 480 V and I_L = 60 A, P = 1.732 × 480 × 60 = 1.732 × 28,800 ≈ 49,881 W, about 49.9 kW. This directly gives the correct value.

If you tried using P = V_L × I_L without the √3 factor, you’d get 28.8 kW, which is not correct for three-phase line quantities. The √3 factor is essential when working with line voltage and line current in a three-phase system.