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  1. Required Capacitance (µF)

    Required Capacitance (µF): Power Factor Correction Calculator

    C in farads (×10^6 for µF); Q_C in VAR = 1000·P·(tan theta1 − tan theta2)

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Results

Required Capacitor Bank
69.15
kVAR of correction needed
Capacitance required 1,375.74 µF
tan(φ₁) at current PF 1.0202
tan(φ₂) at target PF 0.3287

What is Power Factor Correction?

Power factor (PF) measures how effectively electrical power is converted into useful work. A low power factor means more reactive power is drawn, increasing current, losses, and often utility penalties. Power factor correction adds capacitors that supply reactive power locally, raising PF toward unity and reducing the apparent power demand.

Power triangle showing real, reactive and apparent power with capacitor reducing reactive power
Power factor correction reduces reactive power Q, shrinking apparent power S toward real power P.

How to Use the Calculator

Enter your real power (kW), the current (measured) power factor, the target power factor you want to reach, and the supply voltage and frequency. The tool returns the reactive power (kVAR) the capacitor bank must supply and the equivalent capacitance in microfarads.

The Formula Explained

The required reactive compensation is:

\(Q_C = P \cdot (\tan(\varphi_1) - \tan(\varphi_2))\), where \(\varphi = \arccos(\text{PF})\).

Once \(Q_C\) (in VAR) is known, single-phase capacitance is:

$$C = \frac{Q_C}{2\pi \cdot f \cdot V^2}$$

Here \(f\) is the line frequency in Hz and \(V\) is the line voltage in volts.

Two angle arrows comparing original and corrected power factor angles
The required capacitor kVAR equals the difference between the original and target reactive power.

Worked Example

A 100 kW load at PF 0.70 to be corrected to 0.95 on a 400 V, 50 Hz supply:

\(\tan(\arccos 0.70) = 1.0202\), \(\tan(\arccos 0.95) = 0.3287\). $$Q_C = 100 \times (1.0202 - 0.3287) = 69.15 \text{ kVAR}$$ $$C = \frac{69{,}152}{2\pi \times 50 \times 400^2} = 0.001376 \text{ F} \approx 1375.7\ \mu\text{F}$$

FAQ

Why not correct all the way to 1.0? Over-correction can cause leading power factor and voltage rise; utilities usually require around 0.95.

Is this single-phase or three-phase? The capacitance formula shown is the single-phase equivalent for the given V. For three-phase, divide \(Q_C\) by 3 per phase and use phase voltage.

What units does it use? Power in kW, voltage in volts, frequency in Hz; results in kVAR and µF.

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