Power factor describes how effectively an AC electrical system turns supplied electrical current into useful work. In factories, large motors often draw current that lags behind voltage, which increases the current needed for the same real power output. A low power factor wastes capacity in wires, transformers, and generators, even when the useful power in kilowatts has not changed.
Engineers monitor and correct power factor to reduce losses, avoid utility penalties, and improve system efficiency.
The power triangle connects real power P, reactive power Q, and apparent power S using right triangle geometry. Motor loads are usually inductive, so they consume positive reactive power measured in kVAR and create a lagging power factor. Capacitor banks supply leading reactive power that cancels part of the motor's inductive reactive demand.
This reduces apparent power and line current while keeping the real power delivered to the load nearly the same.
Understanding Engineering: Power Factor and Correction
In an alternating current circuit, voltage and current continually reverse direction. With a resistor, their peaks occur at the same time. Energy flows from the supply into the resistor and becomes heat.
An induction motor behaves differently because its windings must create a magnetic field before the rotor can turn. Building that field requires current.
Some energy enters the magnetic field during one part of each cycle, then returns toward the supply during another part. This back and forth exchange does not produce useful shaft output, but the cables and supply equipment still have to carry the current.
Current creates heating in every conductor because of electrical resistance. The heating rises with the square of current. This means a modest current increase can cause a much larger increase in cable loss.
Consider two factory loads that deliver the same mechanical output. The load with poorer power factor needs more current from the same voltage supply. It therefore runs cables, switchgear, transformers, and generators closer to their current limits.
A transformer may have enough capacity in kilovolt amperes but not enough room for another machine, even when the total useful load in kilowatts seems low. Voltage drop can become worse on long cable runs, which can make motors run hotter or struggle during starting.
Capacitors are placed near inductive loads because they store energy in an electric field. Their current leads the voltage, which has the opposite timing effect to the lagging current of an inductor. A properly chosen capacitor bank supplies much of the magnetic field current locally.
The upstream supply then delivers less reactive current. The motor still needs nearly the same real input power to turn its shaft, so correction does not create free energy or make a heavily loaded motor more powerful.
Its main benefit is releasing capacity and reducing current related losses in the parts of the system before the capacitor location. Banks may be fixed for a steady motor load or switched in steps as the plant load changes.
Correction must be controlled carefully. Too much capacitance can produce a leading power factor. This may raise voltage and cause problems with lightly loaded transformers or generators.
Capacitors can also interact with harmonic currents from variable speed drives, welding equipment, computer power supplies, and LED lighting. Harmonics can overheat capacitors or create resonance, where certain currents become unusually large. Engineers often use detuned reactors with capacitor banks to reduce this risk.
Students should separate three ideas when solving problems. Real power tells how much useful energy per second the equipment needs. Apparent power determines equipment current ratings.
Reactive power explains the timing difference that links them. Check whether a load is inductive or capacitive, keep units consistent, and remember that correction changes the reactive part rather than the useful work requirement.
Key Facts
- Real power P is the useful power converted to work or heat, measured in watts or kilowatts.
- Reactive power Q is the power exchanged with magnetic or electric fields, measured in VAR or kVAR.
- Apparent power S is the total AC power capacity required, measured in VA or kVA.
- Power triangle relation: S^2 = P^2 + Q^2.
- Power factor: PF = P / S = cos(theta), where theta is the phase angle between voltage and current.
- Capacitor correction reduces inductive reactive power: Q_corrected = Q_load - Q_capacitor.
Vocabulary
- Real Power
- Real power is the average power that performs useful work in a circuit, such as turning a motor shaft or producing heat.
- Reactive Power
- Reactive power is power that moves back and forth between the source and reactive components without being permanently consumed.
- Apparent Power
- Apparent power is the product of RMS voltage and RMS current in an AC circuit and represents the total power capacity the source must provide.
- Lagging Power Factor
- A lagging power factor occurs when current lags voltage, usually because the load is inductive, such as a motor or transformer.
- Capacitor Bank
- A capacitor bank is a group of capacitors used to supply leading reactive power and improve the power factor of an AC system.
Common Mistakes to Avoid
- Confusing kW with kVA is wrong because kW measures useful real power while kVA measures total apparent power drawn from the supply.
- Assuming reactive power is wasted energy is wrong because reactive power is not consumed like heat, but it still increases current and equipment loading.
- Adding a capacitor that overcorrects the load is wrong because too much capacitive reactive power can create a leading power factor and cause voltage or resonance problems.
- Using PF = S / P is wrong because power factor is the ratio of useful real power to apparent power, so the correct formula is PF = P / S.
Practice Questions
- 1 A motor load uses P = 80 kW with a power factor of 0.80 lagging. Find the apparent power S and the reactive power Q.
- 2 An industrial load has P = 120 kW and Q = 90 kVAR lagging. A capacitor bank supplies 50 kVAR. Find the corrected Q, corrected S, and corrected power factor.
- 3 Explain why adding capacitors to an inductive motor load can reduce the current in the supply wires even though the motor's real power output stays approximately the same.