Three-phase power calculator
Work out kW, kVA and kVAr for a three-phase load, and the capacitor needed to improve its power factor.
A clear answer. Here’s how.
For a balanced three-phase load the apparent power is S = √3 × V × I ÷ 1000 kVA with V the line-to-line voltage, the real power is P = S × power factor, and the reactive power is Q = √(S² − P²). The phase angle is arccos of the power factor. Raising the power factor from pf₁ to pf₂ at the same real power needs a capacitor bank of P × (tan φ₁ − tan φ₂) kVAr, after which the apparent power is P ÷ pf₂ and the line current falls in the same proportion. Harmonics, unbalance and capacitor switching are not modelled.
How this is checked
3 reference checks passed These 3 cases were run when this page was built, on 20 October 2026, and all reproduce within the stated tolerance. The same cases are part of the automated tests that run on every change to Toolmoz. Engine version 1.
See the checks and where each expected answer comes from
Case 1
- Inputs
- You know ampsLine-to-line voltage (V) 400Line current (A) 100Power factor 0.8Target power factor (0 to skip) 0.95
- Expected, and what Toolmoz gave
- Apparent power (kVA) expected 69.282, Toolmoz 69.28203230275508Real power (kW) expected 55.4256, Toolmoz 55.42562584220407Reactive power (kVAr) expected 41.5692, Toolmoz 41.56921938165305Phase angle (°) expected 36.87, Toolmoz 36.86989764584401Capacitor needed for the target power factor (kVAr) expected 23.3517, Toolmoz 23.35169714772972Line current after correction (A) expected 84.21, Toolmoz 84.21052631578947
- Where the expected answer comes from
- √3 × 400 V × 100 A = 69.282 kVA; × 0.8 = 55.4256 kW; Q = √(69.282² − 55.4256²) = 41.5692 kVAr; 55.4256 × (tan 36.87° − tan 18.19°) = 23.3517 kVAr; 55.4256 ÷ 0.95 kVA ÷ (√3 × 400 V) = 84.21 A (hand calculation)
- Allowed difference
- ±0.005 absolute
Case 2
- Inputs
- You know kwLine-to-line voltage (V) 400Real power (kW) 50Power factor 0.8Target power factor (0 to skip) 0
- Expected, and what Toolmoz gave
- Line current (A) expected 90.21, Toolmoz 90.21097956087901Apparent power (kVA) expected 62.5, Toolmoz 62.499999999999986
- Where the expected answer comes from
- I = 50 000 W ÷ (√3 × 400 V × 0.8) = 90.21 A and S = 50 ÷ 0.8 = 62.5 kVA
- Allowed difference
- ±0.005 absolute
Case 3
- Inputs
- You know ampsLine-to-line voltage (V) 230Line current (A) 10Power factor 1Target power factor (0 to skip) 0
- Expected, and what Toolmoz gave
- Real power (kW) expected 3.9837, Toolmoz 3.9837168574084174Reactive power (kVAr) expected 0, Toolmoz 0
- Where the expected answer comes from
- A unity power factor has no reactive power, and √3 × 230 V × 10 A = 3.9837 kW
- Allowed difference
- ±0.0001 absolute
What this shows: the engine reproduces independently worked values for these cases. It does not prove every possible input is right. To test an answer you were given, see how to check a number; to see what is and is not checked across Toolmoz, see the accuracy ledger.
See a worked example
Worked example
A fixed demonstration, separate from the values you enter above.
- You know
- The line current
- Line-to-line voltage (V)
- 400
- Line current (A)
- 100
- Power factor
- 0.8
- Target power factor (0 to skip)
- 0.95
Example result
- Apparent power (kVA)
- 69.282
- Real power (kW)
- 55.426
- Reactive power (kVAr)
- 41.569
- Phase angle (°)
- 36.87
- Line current (A)
- 100.00
- Capacitor needed for the target power factor (kVAr)
- 23.352
- Apparent power after correction (kVA)
- 58.343
- Line current after correction (A)
- 84.21
- Current saved (%)
- 15.8
Balanced three-phase: S = √3 × V × I, P = S × power factor and Q = √(S² − P²), with V the line-to-line voltage. Correction adds capacitance to cancel reactive power: kVAr = P × (tan φ₁ − tan φ₂). Harmonics, unbalance and a leading power factor are not modelled; size real capacitor banks with a qualified engineer.