Poor power factor means the distribution system carries more current than the load actually needs. Utilities charge for this reactive burden.
A correctly sized capacitor bank eliminates the excess reactive current at source and reduces both the utility bill and heat in cables.
This guide covers the kVAR sizing formula, the power triangle, a full worked example for a 500 kW industrial plant, and an interactive calculator with demand charge savings estimate.
The required capacitor bank size in kVAR equals the active load in kW multiplied by the difference between tan(φ₁) and tan(φ₂).
φ₁ is the existing power factor angle, φ₂ is the target. The result is exactly how much reactive power the capacitors must supply.
Why Low Power Factor Costs Money
Every inductive load (a motor, transformer, or fluorescent fitting) draws two types of current from the supply.
Active current does actual work. Reactive current creates and collapses the magnetic field in the load, contributing nothing to useful work but adding to the total current the supply cable and transformer must carry.
When total current is higher than the active component alone requires, the utility bills for larger apparent power demand than the plant actually uses productively.
In India, DISCOM regulations typically impose surcharges when pf falls below 0.90, and offer rebates when it exceeds 0.95. The reactive component also increases I²R losses in cables and heats the transformer.
Adding capacitors in parallel with the inductive load supplies the reactive current locally. The supply sees only the residual reactive current and a higher power factor. No additional equipment changes -- just the addition of capacitors at the panel or at individual large motors.
The Power Triangle: kW, kVAR and kVA
The relationship between active power, reactive power, and apparent power is a right triangle. kW is the horizontal leg, kVAR is the vertical leg, and kVA is the hypotenuse.
Power factor is the cosine of the angle between kVA and kW.
Correction shifts the reactive leg downward by supplying capacitive kVAR to cancel the inductive kVAR. The kW leg stays unchanged.
Only kVAR and kVA reduce, giving a smaller kVA demand and higher power factor. See the kW, kVA and kVAR guide for the full power triangle.
Q = Reactive power (kVAR) -- reduced by capacitor bank
S = Apparent power (kVA) -- what the utility bills for demand charges
pf = cos φ = P / S
tan φ = Q / P = √(1/pf² − 1)
The kVAR Sizing Formula
P = active load power (kW) -- from energy meter or nameplate sum
φ₁ = existing power factor angle = cos⁻¹(pf_existing)
φ₂ = target power factor angle = cos⁻¹(pf_target)
tan φ = √(1/pf² − 1) -- use a calculator or the table below
Alternative form using pf directly:
Q_c = P × (√(1/pf₁² − 1) − √(1/pf₂² − 1))
tan φ Quick Reference Table
| Power Factor | Angle φ (°) | tan φ | Common application |
|---|---|---|---|
| 0.70 | 45.6° | 1.0202 | Old induction motor drives, poor condition |
| 0.75 | 41.4° | 0.8819 | Mixed motor and lighting loads, uncorrected |
| 0.80 | 36.9° | 0.7500 | Typical uncorrected industrial plant |
| 0.85 | 31.8° | 0.6197 | Minimum for most utility penalty avoidance |
| 0.90 | 25.8° | 0.4843 | DISCOM minimum standard in most Indian states |
| 0.95 | 18.2° | 0.3287 | Standard target for new industrial installations |
| 0.97 | 14.1° | 0.2511 | High efficiency target, VFD dominated plants |
| 1.00 | 0° | 0.0000 | Unity -- avoid over correcting to this point |
Worked Example: 500 kW Plant at 0.72 Power Factor
An industrial plant draws 500 kW at a measured power factor of 0.72. The utility applies a demand surcharge on kVA above the contracted level. Target power factor is 0.95. Calculate the required capacitor bank rating and the resulting kVA demand reduction.
tan φ₁ = √(1/0.72² − 1) = √(1.929 − 1) = √0.929 = 0.9639tan φ₂ = √(1/0.95² − 1) = √(1.108 − 1) = √0.108 = 0.3287Q_c = 500 × (0.9639 − 0.3287) = 500 × 0.6352 = 317.6 kVARRound up to 320 kVAR (e.g. 8 × 40 kVAR steps in an APFC panel)S₁ = P / pf₁ = 500 / 0.72 = 694.4 kVAS₂ = P / pf₂ = 500 / 0.95 = 526.3 kVAΔS = 694.4 − 526.3 = 168.1 kVA (24.2% reduction)Capacitor Bank Sizing Calculator
Fixed Bank vs Automatic APFC Panel
| Type | How It Works | Best For | Risk if Wrong |
|---|---|---|---|
| Fixed capacitor bank | Permanently connected kVAR. No switching. Operates whenever the main supply is on. | Constant, stable loads -- single large motor, small factory with uniform shift pattern | Over correction during light load periods (nights, weekends) -- voltage rise, leading pf penalty |
| Automatic APFC panel | Controller measures pf continuously and switches capacitor steps in and out to maintain a set target pf. Typically 6 to 12 steps. | Variable loads -- process plants, hospitals, commercial buildings with wide daily load swings | Switching surges if reactor free and harmonics are present -- fit detuned reactors for VFD loads |
| Motor mounted capacitor | Individual capacitor fitted directly at each motor terminal. Corrects at the point of reactive demand. Switched with the motor. | Large motors (above 22 kW) as a supplement to a central bank, or where cable losses between panel and motor are high | Self excitation on direct online motors if capacitor is too large relative to motor no load magnetising current |
Reduced Utility Bill
Lower kVA demand reduces demand charges and avoids reactive power surcharges. In plants with high demand billing, this is typically the largest saving. The electrical energy consumption guide covers how to read and interpret your demand billing.
Lower Cable and Transformer Losses
Reducing total current by 20 to 25% cuts I²R losses by 36 to 44% in the same cable. This also reduces transformer operating temperature and extends insulation life.
Released System Capacity
A transformer already loaded to 80% apparent power can accept additional load after correction without replacement. The same kVA capacity now delivers more productive kW. This defers capital expenditure on transformer upgrades.
Improved Voltage Regulation
Reactive current in feeders causes voltage drop. Supplying reactive power locally reduces the voltage drop between the supply transformer and the load, improving voltage regulation for sensitive equipment.
Watch: Capacitor Bank Sizing Calculation Explained
Power Factor Improvement Questions
External References
- Power Factor Correction -- A Plant Engineer's Guide -- Eaton
- Inside the Capacitor Bank Panel -- Electrical Engineering Portal
What We Learn Today
- Q_c = P × (tan φ₁ − tan φ₂) gives the capacitor bank rating in kVAR
- tan φ = √(1/pf² − 1) -- convert power factor to angle before subtracting
- kVA demand reduces by 10 to 30% when pf improves from 0.72 to 0.95
- Never correct beyond 0.98 -- over correction causes voltage rise and leading pf penalties
- Variable loads need an APFC panel; fixed loads can use a simple switched bank
- VFD dominated plants need detuned reactors in series with capacitors to avoid harmonic resonance