In a balanced three-phase system, the three-phase apparent power S equals

In a balanced three-phase system, the three-phase apparent power S equals

• A. S = √3 V_L I_L
• B. S = 3 V_L I_L
• C. S = V_L I_L
• D. S = √3 V_Ph I_Ph

Answer: (A)

In a balanced three-phase system, the total apparent power comes from combining the three phase powers and taking the vector magnitude of real and reactive power. The standard relation uses line values: S = √3 times the line voltage times the line current. This comes from how the three phase powers add up: P = 3 V_Ph I_Ph cos φ and Q = 3 V_Ph I_Ph sin φ lead to S = √(P^2 + Q^2) = 3 V_Ph I_Ph, and with V_L = √3 V_Ph and I_L = I_Ph (for a Y connection) or V_L = V_Ph and I_L = √3 I_Ph (for Δ), both yield S = √3 V_L I_L. Equivalently, S can also be written as 3 V_Ph I_Ph, which is the same value. So the correct form is S = √3 V_L I_L. The other expressions don’t match the standard relationship: missing the √3 factor, or using phase quantities incorrectly, or multiplying by 3 in the line-voltage form.

In a balanced three-phase system, the total apparent power comes from combining the three phase powers and taking the vector magnitude of real and reactive power. The standard relation uses line values: S = √3 times the line voltage times the line current. This comes from how the three phase powers add up: P = 3 V_Ph I_Ph cos φ and Q = 3 V_Ph I_Ph sin φ lead to S = √(P^2 + Q^2) = 3 V_Ph I_Ph, and with V_L = √3 V_Ph and I_L = I_Ph (for a Y connection) or V_L = V_Ph and I_L = √3 I_Ph (for Δ), both yield S = √3 V_L I_L. Equivalently, S can also be written as 3 V_Ph I_Ph, which is the same value.

So the correct form is S = √3 V_L I_L. The other expressions don’t match the standard relationship: missing the √3 factor, or using phase quantities incorrectly, or multiplying by 3 in the line-voltage form.