The formula
S = √3 × V_L × I_L ; P = S × cos φ ; Star: V_ph = V_L/√3, I_ph = I_L ; Delta: V_ph = V_L, I_ph = I_L/√3
- V_L, I_L
- line voltage and line current (volts, amperes)
- V_ph, I_ph
- voltage across and current through one phase of the load
- S, P, Q
- apparent, real and reactive power
- cos φ
- power factor of the load
Worked example
A star-connected load on a 415 V line drawing 10 A at 0.85 power factor.
- S = √3 × 415 × 10 = 1.732 × 4150 = 7188 VA
- P = 7188 × 0.85 = 6110 W
- Q = √(7188² − 6110²) = 3786 VAr
- Star connection: V_ph = 415 / √3 = 239.6 V, I_ph = I_L = 10 A
7.19 kVA total, delivering 6.11 kW with 3.79 kVAr reactive; each phase sees 239.6 V.
Where you'll use it
Industrial load calculations, motor ratings and distribution design. The 415 V / 240 V pair in the example is the standard Indian LT supply — 415 V between lines, 240 V from any line to neutral.