Why power factor matters
A load with a power factor of 0.85 draws 18 percent more current than the work needs (1 ÷ 0.85 = 1.18) (see AC Power Basics). That extra current heats the cables and transformers, drops voltage, uses up supply capacity, and, in many tariffs, is billed as a kVA demand or a power-factor penalty. Most of the reactive current is the magnetizing current of motors and transformers, and it can be supplied locally by a capacitor, which relieves the whole network upstream of it.
Correcting power factor
Place a capacitor bank across the load. The capacitor’s leading reactive power cancels the motor’s lagging reactive power, and the supply sees only the real power and the remaining small reactive part. The reactive power needed to improve the power factor from cos φ₁ to cos φ₂ is:
Qc = P (tan φ₁ − tan φ₂) C (per phase, delta) = Qc ÷ (3 ω V²)
Methods: fixed capacitors at a motor (switched with it), a central automatic bank with several steps controlled by a power-factor relay, or active filters that also remove harmonics. Do not correct beyond about 0.95 to 0.98 lagging: a leading power factor can raise the voltage and cause instability when the load is light.
Worked example: the press motor and the whole workshop
Worked example: from 0.85 to 0.95
- The motor: P = 12.09 kW; tan φ₁ = 0.620 (cos φ 0.85), tan φ₂ = 0.329 (0.95). Qc = 12.09 × (0.620 − 0.329) = 3.52 kvar: a standard 3.5 or 4 kvar capacitor.
- Capacitance (delta): C = 3 518 ÷ (3 × 400² × 314.2) = 23.3 µF per phase (70 µF if star-connected); capacitor current 5.1 A in each line.
- The line current falls from 20.5 A to 18.4 A (11 percent less) and the apparent power from 14.22 to 12.72 kVA. Since losses go as I², the cable and transformer losses fall to 80 percent of the old value, 20 percent less.
- The workshop: a plant of 80 kW at 0.78 needs Qc = 80 × (0.802 − 0.329) = 37.9 kvar: a 12.5 kvar × 4 automatic bank (50.0 kvar).
- The bill: with a kVA demand charge of 12 per kVA a month, the motor’s demand falls from 14.22 to 12.72 kVA, saving 18.0 a month (216 a year) against a capacitor costing about 88: a payback of 0.4 years for this one motor, and much faster across a larger plant with a power-factor penalty.
Harmonics and non-linear loads
A linear load (a resistor, a motor) draws a current of the same sinusoidal shape as the voltage. A non-linear load (a rectifier in a drive, a switch-mode supply, an LED driver, a welder) draws current in pulses, and that distorted current is the sum of the fundamental and harmonics: currents at multiples of 50 Hz. A six-pulse drive produces mainly the 5th (250 Hz), 7th, 11th and 13th harmonics; single-phase electronics produce triplen (3rd, 9th) harmonics that add in the neutral. The total harmonic distortion is THD = √(Σ Ih²) ÷ I₁.
| Harmonic | Current (percent of the fundamental, typical 6-pulse drive with a line reactor) | In the 20.5 A line |
|---|---|---|
| 5th (250 Hz) | 25 % | 5.13 A |
| 7th (350 Hz) | 12 % | 2.46 A |
| 11th (550 Hz) | 7 % | 1.44 A |
| 13th (650 Hz) | 5 % | 1.03 A |
| 17th (850 Hz) | 3 % | 0.62 A |
| 19th (950 Hz) | 2 % | 0.41 A |
The THD is 29 percent, so the RMS current is I₁ × √(1 + THD²) = 20.5 × 1.042 = 21.4 A, slightly more than the fundamental. The harmonics heat transformers (eddy-current losses rise as the square of the frequency), overheat neutrals and capacitors, distort the voltage for other users and make meters read wrongly. The remedies are a line reactor or DC choke (which brings the 80 to 100 percent THD of a bare six-pulse drive down to about 30 percent, as in the table), a 12-pulse or active front end for large drives, passive or active filters, and keeping the neutral oversized. Standards: IEC 61000-3-2 and -3-12 for equipment, IEEE 519 and EN 50160 for the network.
Resonance and detuned capacitors
A capacitor bank and the supply’s inductance form a resonant circuit; at the resonant harmonic order hr = √(Ssc ÷ Qc) the harmonic currents are greatly amplified. With the transformer’s short-circuit power of Ssc = 100 ÷ 0.04 = 2 500 kVA and the 50.0 kvar bank, hr = √(2 500 ÷ 50.0) = 7.1, close to the 9th and between the 7th and 11th harmonic currents the drives produce: a recipe for capacitor failure. The solution is a detuned bank with a series reactor of 7 percent: the resonance moves to f = 50 ÷ √0.07 = 189 Hz (order 3.78), below the 5th harmonic, where there is no current to excite it. Use detuned banks wherever drives or other non-linear loads exceed 20 to 25 percent of the load.
Demand, energy and the bill
A commercial bill has: energy (kWh), demand (the peak kW or kVA in the month, often over 15 or 30 minutes), a fixed charge, reactive-power or power-factor charges, and taxes. The load factor is the average load divided by the peak: the press motor runs 48 352 kWh a year at 12.1 kW, a load factor of 46 percent (4 000 hours of 8 760). A low load factor means expensive demand charges for the energy used; shifting or smoothing the peaks (staggered starts, soft starters, batteries) lowers the bill without using less energy.
Saving energy
- Motors: premium-efficiency (IE3 or IE4), correct sizing, drives on fans and pumps (see Induction Motors and Drives).
- Lighting: LED with controls; typically 50 to 80 percent less than fluorescent.
- Compressed air and pumping: fix leaks, reduce the pressure, run only when needed (see Pneumatic Systems).
- Standby and idle: measure the overnight and weekend load; it is often 30 to 50 percent of the day load.
- Metering and monitoring: sub-meters on the big loads; trend the kW, kvar and power factor; compare with the production.
- Transformers: switch off lightly loaded units, choose low-loss designs, balance the phases.
At 0.12 per kWh the press motor costs 5 802 a year to run; a 10 percent energy saving is worth 580 a year from this one machine.
Related
- Previous: Protection, Earthing and RCDs. Next: Solar PV, Batteries and Backup Power.
- Three-Phase Power and Power Factor Calculator.
- Capacitors and the Reactance and Resonance Calculator.
- Electrical Power: the whole guide.
FAQ
Do I pay for kvar?
Often indirectly: as a power-factor penalty, a kVA demand charge, or a fee for reactive energy (kvarh) beyond a threshold, depending on the tariff.
Can I over-correct?
Yes. A leading power factor is penalized by some utilities and can cause overvoltage at light load. Keep it between 0.95 and unity.
Why do capacitors fail?
Over-voltage, harmonic currents, resonance, and heat. Use detuned banks and keep them cool and ventilated.
Does a drive correct the power factor?
A drive’s input has a high displacement power factor (close to 1) but a poor distortion power factor from the harmonics. True power factor = displacement factor × distortion factor.
References
- IEEE Std 519 (harmonic control in power systems), IEC 61000-3-2 / -3-12 (harmonic limits for equipment), IEC 60831 / 61048 (shunt capacitors for power systems).
- Dugan, R. C., McGranaghan, M. F., Santoso, S., Beaty, H. W., Electrical Power Systems Quality, McGraw-Hill; Wildi, T., Electrical Machines, Drives and Power Systems.
- US DOE, Improving Motor and Drive System Performance; utility tariff documents for the real charges.
Last reviewed: October 2026. Tariffs, costs and harmonic spectra are typical examples.