Power systems / load analysis

Three-phase power

Resolve a balanced three-phase load into real, reactive and apparent power, then switch between star and delta to see the winding-level voltage and current. The capacitor size needed for 0.95 power factor is included.

ConnectionStarPhase voltage230.9 VPF statusAcceptable

Real power

38.410 kW

Useful work delivered

Apparent power

43.648 kVA

Sizing basis for cables and transformers

Reactive power

20.731 kVAr

Exchanged with inductive loads

Power factor

0.880

28.4° phase displacement

Power triangle and phase quantities

Apparent, real and reactive power with the per-phase values for the selected connection.

Star (Y)

Power triangle and phase quantities
Quantity	Value	Basis
Apparent power S	43.648 kVA	√3 · V_L · I_L
Real power P	38.410 kW	S · cos φ
Reactive power Q	20.731 kVAr	S · sin φ
Power factor	0.880	Phase angle 28.36°
Phase voltage	230.94 V	V_L / √3
Phase current	63.00 A	I_L (star)
Real power per phase	12.803 kW	P / 3 for a balanced load
kVAr to reach 0.95 pf	8.107 kVAr	Capacitor correction requirement

Line quantities are identical in both connections; only the winding-level voltage and current change by √3.

Power components

The power triangle: real, reactive and apparent.

Real power vs power factor

Holding apparent power fixed, real power is a straight line in cos φ.

Canonical equations

The exact relationships implemented by this engine.

S = √3·V_L·I_L · balanced load only

P = S·cos φ · the component that does work

Q = S·sin φ · the component that magnetises

V_ph = V_L/√3 or I_ph = I_L/√3 · line quantities are unchanged

Show model notation and computationOpen the symbolic form, the variable definitions, and the step-by-step derivation when you need the detail.

A clear path from inputs to results.

A balanced three-phase load is fully described by line voltage, line current and power factor. This engine resolves those into real, reactive and apparent power, converts between star and delta phase quantities, and sizes the capacitor needed to correct the power factor.

Step-by-step methodWorked exampleAssumptions explained

How to use

Enter the line-to-line voltage.

Enter the line current.

Set the power factor — the cosine of the phase angle between voltage and current.

Choose star or delta to see the winding-level quantities.

Read real, reactive and apparent power.

Check the correction figure if the power factor is below the utility’s threshold.

The formula

S = \sqrt{3}\,V_L I_L, \qquad P = S\cos\varphi, \qquad Q = S\sin\varphi, \qquad V_{ph} = \frac{V_L}{\sqrt{3}}\ \text{(star)}, \quad I_{ph} = \frac{I_L}{\sqrt{3}}\ \text{(delta)}

Reference: IEC 60038 (standard voltages); balanced-load power relations per IEEE 141.

Worked example

400 V, 63 A, 0.88 power factor (default inputs)

Inputs

Line voltage 400 V.
Line current 63 A.
Power factor 0.88 lagging.
Star connection.

Calculation path

1
Apparent power: √3 × 400 × 63 = 43.65 kVA.
2
Real power: 43.65 × 0.88 = 38.41 kW.
3
Phase angle: arccos(0.88) = 28.36°, so Q = 43.65 × sin 28.36° = 20.73 kVAr.
4
Phase voltage in star: 400 / √3 = 230.94 V.
5
Real power per phase: 38.41 / 3 = 12.80 kW.

Result

43.65 kVA · 38.41 kW · 20.73 kVAr · phase voltage 230.9 V · 12.80 kW per phase.

Interpreting the result

The utility bills for apparent power (kVA) but you consume real power (kW). The 20.7 kVAr of reactive power is doing no useful work — it is magnetising the load — but it still occupies transformer and cable capacity. Correcting to 0.95 would recover about 8 kVAr of that capacity, which is why large sites install capacitor banks.

Assumptions & limitations

The load is balanced across all three phases. Unbalanced loads need per-phase analysis.

Voltages and currents are sinusoidal. Harmonic-rich loads need true-RMS measurement and distortion power.

Steady state; no transient or starting conditions.

Power factor is displacement power factor, not true power factor including distortion.

Line impedance is neglected, so the voltage at the load equals the voltage at the measurement point.

Frequently asked questions

Why is there a √3 in the formula?

Because the three phase powers add as phasors 120° apart. The vector sum of three equal phasors at 120° gives √3 times the per-phase value when expressed in line quantities.

What is a good power factor?

Above 0.95 is usually the utility threshold; below 0.9 often attracts a penalty. Unity is theoretically ideal but not economic to chase beyond about 0.98.

Does this work for single-phase loads?

No. For single phase, S = V × I and the √3 disappears. A three-phase system with a mix of single-phase loads needs a per-phase balance calculation.

Do line quantities change between star and delta?

No. The line voltage and line current are the same at the terminals. What changes is how they divide inside the machine: star divides voltage by √3, delta divides current by √3.

How do I size the correction capacitor?

Q_c = P(tan φ₁ − tan φ₂). At 38.41 kW going from 0.88 to 0.95, that is roughly 8 kVAr. The engine computes it for the entered load.

What does lagging versus leading mean?

Lagging means current lags voltage — the normal case for motors and transformers. Leading means capacitive dominance, usually from over-correction. The arithmetic here is identical; only the sign of Q changes.