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How to use the calculator
- Set the frequency (50 or 60 Hz) and motor unit (kW or hp) at the top; they apply to every tab.
- Power: pick the system, star or delta, and whether you know the current or the power. You get P, Q, S, line and phase values, impedance per phase and a power triangle.
- PF correction: enter kW, present and target power factor and line voltage. Add the transformer's kVA and %Z for the resonance check, or leave it blank.
- Motor current: enter the nameplate output, voltage, efficiency and power factor.
Results update as you type. Copy link saves the problem in the address, for example ?tab=pfc&p=100&pf1=0.75&pf2=0.95&tr=1000&z=6 after the page URL.
| Link parameter | Meaning |
|---|---|
tab, hz, pu | power, pfc or motor; 50 or 60; kW or hp |
ph, conn, mode | 3 or 1; star or delta; fwd (from current) or rev (from power) |
vl, il, pf, pft, eff | Line voltage, line current, power factor, lag or lead, efficiency in % |
kn, pv | Known power: kW, kVA or out (motor output), and its value |
p, pf1, pf2, tr, z, mva, st | Correction: kW, present and target PF, transformer kVA, %Z, upstream fault level in MVA, bank steps |
out | Motor rated output in the chosen unit |
Why √3 appears in three-phase power
A balanced supply has three equal voltages 120° apart. The voltage between two lines is the difference of two phase voltages, which is 2 × cos 30° = √3 ≈ 1.732 times the phase voltage, so 230 V per phase gives about 400 V between lines.
Total power is three times the power of one phase, P = 3 × Vph × Iph × cos φ. In line values, for star and delta alike:
P = √3 × VL × IL × PF, S = √3 × VL × IL, Q = S × sin φ
Here φ is the angle between voltage and current in each phase, and the power factor PF = cos φ. A single-phase load uses the same formulas without the √3.
Star vs delta: line and phase values
| Connection | Phase voltage | Phase current |
|---|---|---|
| Star (wye) | VL ÷ √3 | IL |
| Delta | VL | IL ÷ √3 |
The tool also gives the impedance per phase, Z = Vph ÷ Iph, with R = Z cos φ and X = Z sin φ, the usual next exam step. At 400 V, 10 A and PF 0.8 a star load has 23.09 Ω per phase; a delta load with the same line current has 69.28 Ω.
The power triangle
Real power P (kW) does the work. Reactive power Q (kVAR) builds the magnetic fields of motors and transformers and flows back and forth. Apparent power S (kVA) is what the cables and transformer must carry. They form a right triangle: S² = P² + Q² and PF = P ÷ S. At PF 0.8 the sides are in the ratio 4 : 3 : 5. A lagging (inductive) load has positive Q; a leading (capacitive) load has negative Q, and the triangle points down.
Correcting power factor with capacitors
A capacitor supplies kVAR next to the load, so less reactive current flows from the supply. For a load of P kW:
Qc = P × (tan φ₁ − tan φ₂), where φ₁ = cos⁻¹(present PF) and φ₂ = cos⁻¹(target PF).
The capacitance depends on the connection. In delta, each capacitor sees the full line voltage; in star, each sees VL ÷ √3, so star needs three times the capacitance for the same kVAR:
Cdelta = Qc ÷ (3 × 2πf × VL²) Cstar = Qc ÷ (2πf × VL²)
Single-phase: C = Qc ÷ (2πf × V²). At 60 Hz the same kVAR needs 5/6 of the 50 Hz capacitance. The current falls in the ratio PF₁ ÷ PF₂ and cable losses by 1 − (PF₁ ÷ PF₂)²: correcting 100 kW from 0.75 to 0.95 at 400 V takes 55.32 kVAR, cuts the current from 192.5 A to 151.9 A and the losses by 37.7%.
Harmonic resonance: the check most tools skip
A capacitor bank and the inductance of the supply transformer form a parallel resonant circuit. Its harmonic order is estimated from the short-circuit power at the bus:
Ssc = transformer kVA ÷ (%Z ÷ 100) h = √(Ssc ÷ Qc)
Drives, rectifiers and UPS units inject 5th, 7th, 11th and 13th harmonic currents. If h lands near one of them, those currents are amplified, and capacitor fuses blow or cans fail. In Eaton's example, a 1,500 kVA, 5% transformer gives about 30 MVA, and a 250 kVAR bank resonates near the 11th (h = 10.95). Eaton also warns that a 5.3rd resonance with 5th-harmonic current present makes problems likely.
The tool flags h within 6% of a characteristic harmonic as a risk and within 10% as a caution, checks every switching step (each step moves the resonance) and lists the bank sizes to avoid. An optional upstream fault level is added in series with the transformer, which lowers h. The usual fix is a detuned bank: about 7% series reactors tune each step to h = 1 ÷ √0.07 = 3.78, which is 189 Hz on 50 Hz or 227 Hz on 60 Hz, below the 5th.
Worked examples
1. A star-connected load
A star load draws 10 A at 400 V and PF 0.8 lagging. S = √3 × 400 × 10 = 6.928 kVA, P = 6.928 × 0.8 = 5.543 kW and Q = 6.928 × 0.6 = 4.157 kVAR (φ = 36.87°). Each phase sees 400 ÷ √3 = 230.9 V at 10 A, so Z = 23.09 Ω, with R = 18.48 Ω and X = 13.86 Ω.
2. Capacitors for a 100 hp motor (a textbook problem)
This problem is from V.K. Mehta's Principles of Power System (chapter 6): a 400 V, 50 Hz motor delivers 100 hp (74.6 kW) at 93% efficiency and PF 0.75, and a delta bank raises the PF to 0.95. Input: 74.6 ÷ 0.93 = 80.22 kW. Qc = 80.22 × (0.8819 − 0.3287) = 44.38 kVAR, or 14.79 kVAR per phase, so each delta capacitor is 294.3 µF at 400 V. The book builds each phase from four 100 V capacitors in series, each 4 × 294.3 = 1,177 µF; it prints 1,173.8 µF because it rounds its steps. The current falls from 154.4 A to 121.9 A.
3. A bank that resonates near the 7th harmonic
A plant takes 600 kW at PF 0.72 from a 1,000 kVA, 5.75% transformer at 400 V, 50 Hz. Correcting to 0.95 needs 381.1 kVAR and drops the current from 1,203 A to 911.6 A. Ssc = 1,000 ÷ 0.0575 = 17.39 MVA, so h = √(17,391 ÷ 381.1) = 6.755 (337.8 Hz), 3.5% below the 7th. In 4 steps, the first (95.28 kVAR) sits at h = 13.5, near the 13th. Banks of 293.3 to 438.2 kVAR land within 10% of the 7th here. With drives on the bus, use a detuned bank.
4. Motor full-load current, and why the code table differs
A 10 hp, 460 V, 60 Hz motor at 90% efficiency and PF 0.85 takes 10 × 745.7 ÷ 0.9 = 8.286 kW, so I = 8,286 ÷ (√3 × 460 × 0.85) = 12.23 A. For US conductor sizing, the NEC's Table 430.250 lists 14 A for this motor. Use the formula to understand a motor, and the code table or nameplate to size its circuit.
Reading the result and common mistakes
- Phase voltage in the √3 formula.
√3 × V × Ineeds the line voltage. With the phase voltage, use3 × Vph × Iph. - kW vs kVA. Transformers are rated in kVA: 500 kVA cannot supply 500 kW at PF 0.8.
- Star capacitors sized as delta. Star needs three times the µF, at a lower voltage rating.
- The wrong horsepower. The tool uses 745.7 W per hp; some textbooks use the metric 735.5 W.
- Correcting to exactly 1. The last step from 0.95 to 1.0 on a 100 kW load costs another 32.87 kVAR, and at light load the bank can make the power factor leading.
- Skipping the resonance check. A bank safe at full size can resonate at a smaller step.
Embed the Three-Phase Power Calculator on your website
Teachers, bloggers and course sites can add this tool to any web page for free. Paste the code below into an HTML or embed block (WordPress, Google Sites, Canvas, Moodle, Blogger, Wix). It loads the same tool you see above, without our menus or ads. Please keep the credit line under the frame. Sizes, a live preview and step-by-step guides for each platform are on the calculator widget page.
Frequently asked questions
What is the formula for three-phase power?
Real power is P = √3 × VL × IL × PF, using line voltage and line current. Apparent power is S = √3 × VL × IL and reactive power is Q = S × sin φ. For 400 V, 10 A and PF 0.8, P = 5.543 kW.
How do I calculate the kVAR needed for power factor correction?
Multiply the real power by the difference of the tangents: Qc = P × (tan φ1 − tan φ2), with φ = cos⁻¹(PF). Raising 100 kW from PF 0.75 to 0.95 needs 55.32 kVAR.
Why does a star-connected capacitor bank need three times the capacitance?
A capacitor's kVAR grows with the square of its voltage. In star each one sees VL ÷ √3, so it gives a third of the kVAR it would give in delta and needs three times the capacitance.
How many amps does a three-phase motor draw?
I = output ÷ (√3 × V × efficiency × PF). An 11 kW motor at 400 V with 89% efficiency and PF 0.84 draws 21.24 A. For circuit sizing follow your code; the US NEC uses its Table 430.250.
What is harmonic resonance in a capacitor bank?
The bank and the supply inductance resonate at h = √(Ssc ÷ Qc) times the supply frequency. If h is near a 5th, 7th, 11th or 13th harmonic that your loads produce, that current is amplified and can overheat the capacitors. A detuned bank avoids it.
Should I correct power factor all the way to 1?
Usually not. From 0.95 to 1.0 on a 100 kW load needs another 32.87 kVAR for little extra gain, and the bank can make the power factor leading at light load. Check what your utility requires.