For a balanced three-phase load with line voltage V_L, line current I_L, and power factor cosφ, how is real power P calculated?

For a balanced three-phase load with line voltage V_L, line current I_L, and power factor cosφ, how is real power P calculated?

• A. P = √3 V_L I_L cosφ
• B. P = V_L I_L cosφ
• C. P = 3 V_L I_L cosφ
• D. P = √3 V_Phase I_Phase cosφ

Answer: (A)

In a balanced three-phase system, real power comes from the combined contribution of all three phases, and the line quantities relate to the phase quantities in a specific way. The correct expression is P = √3 V_L I_L cosφ. This comes from adding the power of each phase: P = 3 V_Ph I_Ph cosφ, and using V_Ph = V_L/√3 and I_Ph = I_L gives P = 3 (V_L/√3) I_L cosφ = √3 V_L I_L cosφ. This also shows the equivalent form using phase quantities: P = 3 V_Ph I_Ph cosφ, which is the same total power as the line-quantity form. The other options don’t fit: using a single-phase formula ignores the three-phase nature; using 3 V_L I_L cosφ overcounts by a factor of three; and using √3 with phase quantities would misplace the factor (the correct phase-quantity form uses 3, not √3).

In a balanced three-phase system, real power comes from the combined contribution of all three phases, and the line quantities relate to the phase quantities in a specific way. The correct expression is P = √3 V_L I_L cosφ. This comes from adding the power of each phase: P = 3 V_Ph I_Ph cosφ, and using V_Ph = V_L/√3 and I_Ph = I_L gives P = 3 (V_L/√3) I_L cosφ = √3 V_L I_L cosφ.

This also shows the equivalent form using phase quantities: P = 3 V_Ph I_Ph cosφ, which is the same total power as the line-quantity form.

The other options don’t fit: using a single-phase formula ignores the three-phase nature; using 3 V_L I_L cosφ overcounts by a factor of three; and using √3 with phase quantities would misplace the factor (the correct phase-quantity form uses 3, not √3).