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Power Factor Correction

Electronics & PhysicsPower and Energy
Load
Supply
System
Frequency
Capacitor bank needed (kvar) —
Capacitance (µF, per phase in delta)
—
Apparent power before (kVA)
—
Apparent power after (kVA)
—
Line current before (A)
—
Line current after (A)
—
Current reduction
—

Power factor correction capacitors supply Qc = P × (tan φ1 − tan φ2) kvar. For a 100 kW load at 0.70 corrected to 0.95, tan φ falls from 1.020 to 0.329, so the bank needs about 69.2 kvar, and apparent power drops from 142.9 kVA to 105.3 kVA.

About this tool

Electricians, plant engineers and students use this to size a capacitor bank before quoting it. Enter the load's real power in kW, its present power factor and the value you want to reach, then pick single- or three-phase, the supply voltage and 50 or 60 Hz. The result gives the reactive power the capacitors must supply in kvar, the capacitance in microfarads (per phase for a delta-connected bank on line voltage), and how far apparent power and line current fall. One limit: it treats the load as steady and sinusoidal. Loads with variable-speed drives or rectifiers carry harmonics, and plain capacitors there can resonate with the transformer, so detuned banks or filters are sized from a harmonic study instead.

How to use it

  1. Describe the load

    Type the real power in kW and the measured power factor, for example from a utility bill or a power analyser.

  2. Set the target

    Move the target power factor slider; 0.95 is a common tariff threshold.

  3. Describe the supply

    Choose single- or three-phase, enter the voltage (line-to-line for three-phase) and pick 50 or 60 Hz.

  4. Read the bank size

    Use the kvar figure to pick a standard capacitor rating and the µF value to check individual units.

Examples

100 kW plant, 0.70 to 0.95

Real power
100
Present power factor
0.7
Target power factor
0.95
System
Three-phase
Voltage (line-to-line for three-phase)
400
Frequency
50 Hz

Result Capacitor bank needed (kvar): 69.2
Capacitance (µF, per phase in delta): 458.6
Apparent power before (kVA): 142.9
Apparent power after (kVA): 105.3
Line current before (A): 206.2
Line current after (A): 151.9
Current reduction: 26.3%

  1. φ1 = arccos(0.7) → tan φ1 = 1.0202
  2. φ2 = arccos(0.95) → tan φ2 = 0.3287
  3. Qc = 100 kW × (1.0202 − 0.3287) = 69.15 kvar
  4. C per phase (delta) = Qc ÷ (3 × 2π × 50 × 400²) = 458.6 µF
  5. Current falls from 206.2 A to 151.9 A

A motor-heavy 400 V load at 0.70 lagging needs about 69 kvar of capacitors; apparent power drops from 142.9 kVA to 105.3 kVA.

10 kW at 230 V, 0.80 to 1.00

Real power
10
Present power factor
0.8
Target power factor
1
System
Single-phase
Voltage (line-to-line for three-phase)
230
Frequency
50 Hz

Result Capacitor bank needed (kvar): 7.5
Capacitance (µF, per phase in delta): 451.3
Apparent power before (kVA): 12.5
Apparent power after (kVA): 10
Line current before (A): 54.3
Line current after (A): 43.5
Current reduction: 20%

  1. φ1 = arccos(0.8) → tan φ1 = 0.75
  2. φ2 = arccos(1) → tan φ2 = 0
  3. Qc = 10 kW × (0.75 − 0) = 7.5 kvar
  4. C = Qc ÷ (2π × 50 × 230²) = 451.3 µF
  5. Current falls from 54.3 A to 43.5 A

tan(arccos 0.8) is 0.75, so 7.5 kvar cancels all reactive power, which takes about 451 µF at 50 Hz.

How it is calculated

Qc = P × (tan(arccos PF1) − tan(arccos PF2)); C = Qc ÷ (2πf V²) single-phase, C = Qc ÷ (3 × 2πf V²) per phase in delta

P
real power of the load, kW
PF1, PF2
present and target power factor (cos φ)
Qc
reactive power the capacitors supply, kvar
V, f
supply voltage and frequency
C
capacitance in farads, shown in µF

Real power stays the same; only the reactive part changes. Reactive power is P × tan φ, so the bank must cover the difference between the reactive power now and at the target angle. Apparent power is P ÷ PF, and current is S ÷ V single-phase or S ÷ (√3 × V) three-phase. Capacitor reactive output rises with the square of voltage and in direct proportion to frequency, which is why the µF value depends on both.

When not to use it

  • Loads with heavy harmonic content need a detuned bank sized from a harmonic study.
  • Rapidly varying loads such as welders call for automatic or thyristor-switched banks, not one fixed size.
  • It does not size the capacitor's fuses, contactors or cables.

Common mistakes

  • Entering phase-to-neutral voltage for a three-phase bank that is connected in delta.
  • Using kVA instead of kW as the real power input.
  • Correcting a lightly loaded motor to unity, which turns the power factor leading when load drops.
  • Ordering a 400 V capacitor for a 440 V network, which shortens its life.

Frequently asked questions

Why does the target power factor matter so much near 1.0?

tan φ changes steeply near unity. Going from 0.95 to 1.00 for 100 kW needs about 33 kvar, nearly half of what it took to go from 0.70 to 0.95, yet it trims apparent power by only about 5 kVA. That is why most installations stop at 0.95 to 0.98.

Is the capacitance shown per capacitor or for the whole bank?

For single-phase it is the total capacitance. For three-phase it is the value of each of the three capacitors in a delta-connected bank, each seeing the full line voltage. A star-connected bank on the same supply needs three times that per phase, because each unit sees only V ÷ √3.

Does correction reduce my energy (kWh) use?

Not directly. The load still draws the same real power. Lower current does cut I²R losses in cables and transformers, and it removes reactive-power penalties where the tariff charges for them, but the kWh at the equipment stays the same.

How do I find the present power factor?

Read it from a power analyser or a multifunction meter at the main board, or work it out from a bill as kWh ÷ √(kWh² + kvarh²) over the same period. A bill-based figure is an average, so check the peak-load value too before sizing a fixed bank.

Can I use this for a single motor?

Yes, using the motor's input power and its power factor at typical load. Keep a motor-mounted capacitor below the motor's no-load reactive power, usually about 90 % of it, or self-excitation can cause overvoltage when the motor is switched off.

Why does 60 Hz give a smaller capacitance?

A capacitor's reactive output is V² × 2πf × C. At 60 Hz the same capacitor delivers 20 % more kvar than at 50 Hz, so fewer microfarads cover the same kvar. The kvar requirement itself is the same at either frequency.

Cite this tool

Using this tool in an assignment, article or report? Copy a citation, or a link for your website.

APA

Tools Directory Editorial Team. (2026, October 12). Power Factor Correction. Tools Directory. Retrieved October 27, 2026, from https://tools.milunir.com/electronics-physics/power-and-energy/power-factor-correction/

MLA

Tools Directory Editorial Team. “Power Factor Correction.” Tools Directory, 12 Oct. 2026, https://tools.milunir.com/electronics-physics/power-and-energy/power-factor-correction/. Accessed 27 Oct. 2026.

Chicago

Tools Directory Editorial Team. “Power Factor Correction.” Tools Directory. Last modified October 12, 2026. Accessed October 27, 2026. https://tools.milunir.com/electronics-physics/power-and-energy/power-factor-correction/.

Link<a href="https://tools.milunir.com/electronics-physics/power-and-energy/power-factor-correction/">Power Factor Correction – Tools Directory</a>