Free · Physics models
Balanced Delta and Star Power
Relate RMS phase and line values with apparent, active and reactive power for balanced three-phase loads.
- Formula & worked example
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Calculator inputs
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How to use this calculator
- Enter the known values in the units shown. Results update as you type.
- Where results are editable, change one to solve backwards. Lock a value to hold it fixed.
- Use the worked example to check the method. Reset restores the starting fields.
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Formula and method
A delta phase current of 2 amperes at 250 volts gives 1.5 kVA apparent power and 1.2 kW active power at power factor 0.8.
Balanced apparent power =3Vphase Iphase; P=S cosφ; reactive magnitude=S sin(arccos(power factor)).
Worked example
Enter these known values and leave the other values blank.
- Phase current (I==ph==)
- 2 A
- Phase voltage (V==ph==)
- 250 V
- Power factor
- .8
- Line current (I==line==)
- 3.46 A
- Line voltage (V==line==)
- 250 V
- Phase angle (φ)
- 37 deg
- Apparent power (S)
- 1.5 kVA
- Active power (P)
- 1.2 kW
- Reactive power (Q)
- 0.9 kvar
Assumptions and limitations
- The three phases are balanced, sinusoidal and measured as RMS values. Delta and star branches preserve their distinct line-to-phase relations.
- Reactive output is a nonnegative magnitude obtained from the principal phase angle; it does not assign capacitive versus inductive sign.
- Unbalanced loads, harmonics, cable losses and neutral currents are omitted.
- Clear the unknown numerical field before using a supported inverse; selection controls retain their captured branches.
Common questions
What assumptions matter for this model?
The three phases are balanced, sinusoidal and measured as RMS values. Delta and star branches preserve their distinct line-to-phase relations.
What does this model leave out?
Unbalanced loads, harmonics, cable losses and neutral currents are omitted.
References
Model requirements: Power factor magnitude must lie between zero and one. Zero active power is permitted for a purely reactive balanced load.