Power Factor Correction

Power factor correction is one of the most practically important topics in AC circuit analysis. In any AC system that supplies inductive loads, the current drawn lags behind the voltage, resulting in a power factor less than unity. This increases the current that the source must supply for the same useful power delivered, wasting conductor capacity and increasing losses. Correcting the power factor by adding reactive compensation brings the system closer to a purely resistive condition, which is both economically and technically desirable.

Core Concept Explanation

In an AC circuit supplying an inductive load, the current phasor lags the voltage phasor by an angle φ. The cosine of this angle, cos φ, is defined as the power factor (PF). Unity power factor (PF = 1) means all supplied power is converted into useful work. A lagging power factor means a portion of the current is circulating reactive energy between the source and the inductance, doing no real work but still occupying conductor and transformer capacity.

The three quantities in AC power analysis are: real power P measured in watts (W), reactive power Q measured in volt-ampere reactive (VAR), and apparent power S measured in volt-ampere (VA). These are related by the power triangle where S² = P² + Q². The power factor equals P divided by S.

When a capacitor is connected in shunt (parallel) with the inductive load, the capacitor supplies a leading reactive current that partially or fully cancels the lagging reactive current demanded by the inductor. This reduces the net reactive power Q drawn from the source, which shrinks the apparent power S and the total current, even though the real power P remains unchanged.

The process of adding a capacitor to reduce the lagging reactive power demand is called power factor correction or power factor improvement. In industrial installations this is done using capacitor banks, which are switched in or out depending on the reactive demand of the load at any given time.

Mathematical Expression

If an inductive load draws real power P at an original power factor cos φ1, the reactive power demanded is Q1 = P tan φ1. To improve the power factor to a new value cos φ2, the new reactive power required is Q2 = P tan φ2. The capacitor must supply the difference in reactive power.

The required capacitive reactive power is Qc = Q1 - Q2 = P (tan φ1 - tan φ2). Since the capacitor connected across a voltage V at frequency f carries reactive power Qc = V² / Xc = V² ω C, the required capacitance is:

C = Qc / (ω V²) = P(tan φ1 - tan φ2) / (2π f V²)

This formula directly gives the capacitor bank rating in farads needed to shift the power factor from cos φ1 to cos φ2. All quantities must be in consistent SI units. Note that this is purely a shunt connection; the load parameters do not change.

Practical Understanding

Industries pay electricity bills based on both kWh consumed and kVA demand. A poor power factor increases kVA without increasing kWh of useful output, resulting in higher tariff penalties from utilities. Correcting the power factor reduces the kVA demand charge and also reduces I²R losses in the distribution cables, since the current magnitude is lowered.

Over-correction must be avoided. If excessive capacitance is added, the net reactive power becomes leading, the system now operates at a leading power factor, and similar current increase problems resurface. The ideal correction targets exactly unity power factor or a slightly lagging value as specified by the utility, typically 0.95 lagging.

Given:
Load: P = 50 kW, original power factor = 0.7 lagging (cos φ1 = 0.7)
Target power factor = 0.95 lagging (cos φ2 = 0.95)
Supply voltage V = 400 V (line), frequency f = 50 Hz

Why this formula applies:
Shunt capacitor must supply reactive power equal to Q1 - Q2 to bring PF from 0.7 to 0.95

Formula:
C = P(tan φ1 - tan φ2) / (2π f V²)

Substitution:
φ1 = cos⁻¹(0.7) = 45.57° → tan φ1 = 1.0202
φ2 = cos⁻¹(0.95) = 18.19° → tan φ2 = 0.3287
Qc = 50000 × (1.0202 - 0.3287) = 50000 × 0.6915 = 34575 VAR
C = 34575 / (2π × 50 × 400²)

Calculation:
C = 34575 / (2π × 50 × 160000)
C = 34575 / 50265482
C ≈ 687.9 µF

Final Answer: Required capacitance = 687.9 µF (approximately 688 µF capacitor bank)

Exam Tip: In GATE, power factor correction questions always give PF in lagging form. Compute tan of both angles from cos values, find Qc = P(tan φ1 - tan φ2), and derive C. Never substitute reactive power for real power in the formula. Also remember Q is positive for lagging (inductive) loads.

Mechanism of Correction

• The inductive load draws lagging current, creating positive reactive power Q absorbed from the source.
• A shunt capacitor generates leading current, supplying reactive power Qc back toward the source.
• The net reactive power seen by the source is Q_net = Q_load - Qc, which is smaller in magnitude.
• The apparent power S = sqrt(P² + Q_net²) decreases, so the source current decreases.
• Real power P delivered to the load is unchanged since the capacitor is lossless and purely reactive.
• Unity power factor is achieved when Qc = Q_load, making Q_net = 0 and S = P.

Quick Revision

• Power factor = cos φ = P/S; unity means all apparent power is real power.
• Inductive loads cause lagging PF; capacitors are added in shunt to correct this.
• Required capacitor reactive power: Qc = P(tan φ1 - tan φ2).
• Required capacitance: C = P(tan φ1 - tan φ2) / (2π f V²).
• Real power P stays unchanged; only reactive and apparent power are reduced.
• Over-correction leads to leading PF, which is equally undesirable.
• GATE trap: Always use RMS voltage in the capacitance formula, not peak voltage.