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Home / Technical Articles / Reactive power and power factor correction essentials to ALWAYS keep on mind

Why reactive power and PF correction?

Why reactive power and power factor correction are so important? Well, for experienced engineers, this is not the question but the fact. For young students who meet this matter and terms for the first time, it’s important to explain the matter very carefully, because much future knowledge will be built on its basis.

Reactive power and power factor correction essentials to ALWAYS keep on mind
Reactive power and power factor correction essentials to ALWAYS keep on mind (on photo: Low voltage capacitors installed in an enclosure to improve power factor)

Introduction to terms

In AC circuits, the current absorbed by a load can be represented by two components: active component IR and reactive component IQ.

The active component IR, in phase with the supply voltage, is directly related to the output (and therefore to the part of electric energy converted into energy of different types: mechanical energy, light energy, thermal energy…);

The reactive component IQ, in quadrature to the voltage, is used to generate the flow necessary for the conversion of powers through the electric or magnetic field and it is index of the transfer of energy between supply and load. Without this, there could be no net transfer of power, for example, thanks to the magnetic coupling in the core of a transformer or in the air gap of a motor.

In the most common case, in the presence of ohmic-inductive type loads, the total current I lags with respect to the active component IR. Therefore, in an electrical installation, it is necessary to generate and transmit, in addition to the active power P, a certain reactive power Q, which is essential for the conversion of the electrical energy but is not available to the load because exchanged with the network.

The complex of the power generated and transmitted constitutes the apparent power S. Power factor cosϕ is defined as the ratio between the active component IR and the total value of the current I; Φ is the phase angle between the voltage and the current.

Improving the power factor means taking the necessary steps to increase the power factor in a defined section of the installation by locally delivering the necessary reactive power so that the value of the current and consequently of the power flowing through the upstream network can be reduced, at the same required output power.

In this way, the lines, the generators and the transformers can be sized for a lower apparent power, as better explained in the following paragraphs.


  1. Reactive power and its effects
  2. The effects of low power factor
    1. Voltage Drop
    2. Efficiency and cost
  3. Power Factor Correction
    1. Series Connection
    2. Shunt Connection of Capacitance
  4. Power Factor improvement at the load
    1. Correction at constant real power
    2. Correction at constant S
  5. Capacitor Installations
  6. Energy Saving Costs using PFC
    1. Energy Metering and charging methods
    2. Loss Reduction
    3. Inrush current

1. Reactive power and its effects

The necessary presence of reactive elements in AC circuits and the associated requirements of reactive power or VARs to service these elements cause significant problems in the operation of such AC supply systems.

The net reactive power requirements must be met by the source and this can have significant impact on the operation if some steps are not taken to alleviate the problem.

The complex power associated with a voltage V and current I is defined as:
S = P + jQ = V × I* (where I* is the complex conjugate of I)

We define the reactive power to be positive when it is absorbed (as in a lagging power factor circuit).

a. Pure capacitance element – For a pure capacitance element, P=0 and I leads V by 90° so that complex power is:

S = jQ = (V ∠0°) (I ∠90°)
S = V×I ∠−90°
S = −jV×I

Thus the capacitance element generates reactive power.

b. Inductive element – Similarly, for an inductive element, P = 0 and I lags V by 90° so that:

S = jQ = (V ∠0°) (I ∠−90°)
S = V×I ∠+90°
S = +jV×I

Thus the inductance element absorbs reactive power. Most loads are net inductive and so they require reactive power to be supplied by the source. Similarly overhead lines are net absorbers of reactive power, but cables, with their high capacitance are net generators of reactive power and this has to be handled by the source.

The presence of reactive power in a load means that the power factor is reduced from unity and so it is best to operate at high power factor.

In principle the solution of the reactive power problem is obvious: it is to install additional inductance or capacitance as required to alleviate the supply of the need to handle the reactive power. This is the general principle of power factor correction.

For a general (lagging) power factor cosΦ , the current lags the voltage by Φ, while for a leading power factor the current leads the voltage by Φ. The various phase relationships and the corresponding phasor diagrams are shown below.

Lagging circuit:
Lagging circuit
Lagging circuit – phasor diagram
Leading circuit:
Leading circuit
Leading circuit – phasor diagram

Note that (the convention says) reactive power is positive (absorbed) in lagging power factor circuits and is negative (generated) in leading power factor circuits.

Note also that:

Total Current |I| = √((I cosφ)2 + (I sinφ)2)
Apparent power |S| = √(P2 + Q2) = √((VI cosφ)2 + (VI sinφ)2)

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2. The effects of low power factor

There are two main problems associated with low power factor (or the presence of reactive power) in a load:

  1. Voltage drop
  2. Efficiency (and cost of energy supply)

2.1 Voltage Drop

The reactive component of current, I × sinφ , causes unwanted voltage drop that affects the regulation at the load.

Effect of reduced power factor
Effect of reduced power factor

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2.2 Efficiency and cost

The reactive current (I sinφ ) also generates additional heat loss in the connecting lines and for this reason the supplier will impose limits on the load power factor by imposing an additional cost premium on low power factor loads.

In its most general form this will be by way of a charge on the peak kVA requirement of a load so that reduction of the kVA by reduction of power factor is an economic proposition for the customer.

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3. Power Factor Correction

The traditional means of correcting power factor for the typical lagging circuit is by installing capacitance to generate some of the reactive power required by the load so as to remove the necessity of the supply from providing it.

There are two configurations available for capacitance installation: series and shunt connection.

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3.1 Series Connection

Series connection of capacitance
Series connection of capacitance

In the diagram below, Vs is the source voltage without capacitance and Vs is the source voltage with capacitance correction.

Series connection diagram
Series connection diagram

The voltage regulation is:

ΔV = |VS| − |VR|

Thus the added VAR generation is QC = I2 XC and the resulting improvement (decrease) in the voltage regulation is: QC sinΦ / I volts

Series connection has some advantages and disadvantages:

  1. The reduction in total I is small so that the improvement in loss in the line is very small.
  2. Series capacitance is very good for voltage regulation and for improving (smoothing) voltage fluctuations.
  3. The series capacitance helps to improve stability if the line reactance is high.
  4. Series connection is much less flexible than shunt connection in that the capacitance value is not easily able to be changed.

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3.2 Shunt Connection of Capacitance

Shunt connection of capacitance
Shunt connection of capacitance

Note that:

I’ = IR + IC

Shunt capacitance connection diagram
Shunt capacitance connection diagram

Regulation in this case is:

ΔV = IRRcosΦ + IRXLsinΦ − ICXL

And the voltage boost is:


The advantages and disadvantages of shunt connection are:

  1. Improves power factor and reduces line current significantly and thus reduces line losses significantly.
  2. Not as good as series connection for voltage smoothing.
  3. Has much greater flexibility of use than series as change of the value has no effect on load current.

In general, series connection is rarely used, except in some very high voltage long transmission lines where better voltage regulation is needed.

The use of shunt capacitors is almost universal for distribution systems and loads, where the flexibility of shunt connection allows automatic compensation using switched capacitor banks (static VAR compensators) that can adapt to changing load conditions.

Important Note:

In the above, the discussion has been for lagging power factor loads, the most common type. For leading power factor loads it is necessary to add in some means of absorbing reactive power such as series inductance.

Alternatively, an option used on some transmission lines is the use of unloaded synchronous motors at the load end of the line:

  1. When over-excited the synchronous motor will operate at leading power factor and generate VARs
  2. When under-excited the motor will operate at lagging power factor and absorb VARs.

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4. Power Factor improvement at the load

The improvement of power factor at the consumers load will improve voltage regulation and the power transfer efficiency.

The basic aim is to increase cosΦ (typically 0.55-0.8) to some higher value cosΦ’ (typically about 0.9-0.95).

Power factor improvement diagram
Power factor improvement diagram

This will reduce the current magnitude from I to I’ and hence the line losses from I2R to I’2R. Similarly, the reactive current component IQ will reduce to IQ , where:

IC = IQ − IQ

This can be achieved in either of two ways, depending on load requirements:

  1. At constant total real power (P): This is done by reducing the total line current for the same total load P.
  2. At constant total apparent power (S): This is done by allowing the same total line current but increasing the total real power (P) at the load.

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4.1 Correction at constant real power

Lagging power factor load
Lagging power factor load

|S1| = VI1
|S2| = VI2
P = VI1 cosΦ1 = VI2 cosΦ2
Q1 = VI1 sinΦ1
Q2 = VI2 sinΦ2

We need installed capacitance C to generate reactive power of Q1 − Q2 VARs. Thus we need:

ΔQ = Q1 − Q2 = P (tanΦ1 − tanΦ2)

For shunt capacitance:

ΔQ = IC2XC = VC2 / XC = ωCVC2

The reduction in line current is:

I2 = I1 × cosΦ1 / cosΦ2

For example, for cosΦ1 = 0.65 and cosΦ1 = 0.90,

I2 = 0.72 I1 [a reduction of 28%]
I2 = 0.52 I12 [a reduction of 48%]

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4.2 Correction at constant S

This method increases availability of real power for possible increase of load capacity for a fixed source current.

Correction at constant S
Correction diagram at constant S

Initially, at cosΦ1, power P1 is available:

P1 = S cosΦ1
Q1 =S sinΦ1

After correction, at cosΦ2, P2 is available:

P2 =  S cosΦ2
Q2 = S sinΦ2

To achieve this we require reactive power generation: ΔQ = S(sinΦ1 − sinΦ2)

In this case there is of course no reduction of loss in the line. However the regulation is improved.

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5. Capacitor Installations

PFC installations (Static VAR Compensators or SVCs) are extensively used in transmission and distribution systems and in industry and commercial buildings where reactive power may be high. The capacitors are installed in banks made up in modular and separately switchable units which can range from 7.5kVAR to 100kVAR or higher.

Automatic power factor sensing is used and appropriate capacitance values are switched in and out as required, usually by thyristor switches.

Synchronous motor compensation is mostly used on transmission systems (when it is used).

One particular practical problem with such capacitor systems is the potential for resonance with inductance in the electrical system. The potential for resonant operation must be analysed before installing such capacitors and if resonance may occur at harmonics of the power frequency, harmonic filtering in the form of inductance must be added to the installation.

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6. Energy Saving Costs using PFC

Most items of large and extensive distributed equipment operate at relatively low power factors and thus benefit from power factor correction. To determine the benefit requires some study of the cost-benefit analysis of such capacitor installations.

Table 1 – Typical power factors

SourcePower factor
Incandescent loads1.0
Heating loads0.95-1.0
Fluorescent lighting0.6 (about 0.95-0.97 when individually corrected)
Large induction motors0.75-0.9 at full load
Small induction motors0.6-0.85 at full load (fractional horsepower size)

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6.1 Energy Metering and charging methods

Small consumers of electrical energy pay for their power by the individual kilowatthour (kWh) usage. There is no significant benefit from power factor improvement for small consumers.

For large users of energy, such as industry and commerce and similar establishments, the method of assessing charge may be one of a number of options. These are all based on a capacity charge basis where the peak demand is monitored and is used to assess the charge.

They may be:

  • A daily capacity rate per kW peak ($/kW/day)
  • A daily capacity rate per kVA peak ($/kVA/day)
The peak demand is determined from the maximum half hourly kW or kVA power reading over the last 12 months. Furthermore, the chargeable kW or kVA reading can only occur in peak times (2pm−8pm on working weekdays).

Note that whenever kVA or kVARs are used as the charge basis there is some potential for cost reduction by installing capacitors banks, or installations of VARs.

Typically, the cost of capacitors may be about $20-$40 per kVAR. This cost must be used with the amortized saving to determine the potential economic benefit. Normally a cost recovery time of a year or so would be aimed for. Any longer time would not be considered adequate benefit, although the added bonus of reduction of line losses should be included.

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6.2 Loss Reduction

The other savings benefit that comes with power factor correction occurs with the reduction of losses in the feeder connections. The per unit reduction in ohmic (I2R ) loss due to reduction of line current is:

P U loss reduction = 1 − (PFold / PFnew)2

Typically, line losses are about 3–7% of total load power usage. Any reduction of the 3–7% will give substantial benefit if the load power consumption is high.


A 1000 kW load is operated for 60 hours per week from a 415V supply with 7% loss in the supply lines. The load PF is 0.75 and the energy charge is $0.10 per kWh. What are the cost benefits of increasing PF to 0.95?

If the load power factor is increased to 0.95, the loss improvement is:

1 − (0.75 / 0.95)2 = 0.377 = 37.7%

Thus the new losses are 0.623 × 7% = 4.36 %
Old kWh loss per week was: 0.07 × 1000 × 60 = 4200 kWh
Energy cost was 4200 × $0.1 = $420 per week
New kWh loss per week: 0.0436 × 1000 × 60 = 2618 kWh
Energy cost is now (2618) × $0.1 = $261.8 per week

Saving is 420 – 261.8 = $158.2 per week or $8,226 per annum. [$685.5 /month]

For PF = 0.75, S = 1333 kVA, Q = 882 kVAR
For PF = 0.95, S = 1053 kVA, Q = 329 kVAR

Thus, need 882 − 329 = 553 kVAR of installed capacitance. If the cost of the capacitors is $20/kVAR, installation cost is $11,060. Time to achieve cost recovery is 11,060 / 686 = 16.1 months.

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6.3 Inrush current

In addition to the harmonics problem, use of capacitor banks will also cause transient HF inrush currents when they are energized. These may cause some problems.

For a capacitor installation with reactance XC, installed in shunt where there is inductive reactance XL between the capacitor and the source, the inrush current is:

Transient HF inrush current

and frequency is:

Frequency formulae

Shunt capacitors are protected by fuses. Fuse links are usually sized at 125% to 135% of the rated capacitor current. This is to allow for overvoltage conditions, transient currents, harmonic. The rated current for a 3-phase capacitor bank is:

Rated current for a 3-phase capacitor bank

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  1. Reactive Power and Power Factor Correction Lecture
  2. Power factor correction and harmonic filtering in electrical plants by ABB

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More Information

Edvard Csanyi

Electrical engineer, programmer and founder of EEP. Highly specialized for design of LV/MV switchgears and LV high power busbar trunking (<6300A) in power substations, commercial buildings and industry facilities. Professional in AutoCAD programming.


  1. Mallikarjuna
    Aug 16, 2022

    Am test engineer for Solar inverter and to test our inverter we use Inverter back to back,one inverter work as converter and other in Inverter mode,where the Losses supplied by a DC sources and both converter and Inverter are isolated with isolattion transformer
    Recently test rig was upgraded with 7.5MVA Transformer, but transformer kept away from 100 meters away due to oil type transformer,this is connected with bus duct 7000A rated,now we measure lot of reactive component added due to the bus duct,Please give your thought to make it PF-1, is the only way to add capacitors in the network

  2. Gyanendra Upadhyay
    May 28, 2022

    Is it necessary to correct power factor in hydropower plant. If yes; on what condition and at what locations?

    • Jim Slater
      Jul 16, 2022

      What device are you using, an induction generator? If so is it grid tied? IG’s produce power with a leading PF. This means the usual method of adding capacitance won’t work.

  3. vishal
    May 03, 2021

    how to detect harmonics and whats the remedy to overcome it .I am going to install capacitor bank at stone crusher plant which has induction motors with dol and star Delta starter no vfds there .will it cause harmonics
    capacitor banks are enough ??

  4. Oeer
    Jan 01, 2021

    Knowledge is power….

  5. Tererai
    Nov 14, 2020

    Very informative indeed. Thank you!

  6. Ayman Elkashef
    May 02, 2020

    Very valuable information

    Mar 09, 2020

    In above calculation you are using the value of 0.623 and 0.436. I want know how they are getting and why. please inform me.
    With regardas

  8. Golla subhakar
    Sep 03, 2019


  9. Golla subhakar
    Sep 03, 2019

    Valuable material.

  10. zuproc66
    Sep 03, 2019

    nice article specially for the students.. but why i cant Save to PDF.. i am a premium member. thank you

  11. Okhakume asekhakume
    Sep 02, 2019

    Kind of confuse with meter billing methods as regards instrumentation and calibration if the energy meters mostly in kwh..
    My question is how can a power factor correction be done on billings or meter readings in kwh…when no p.f or reactive componenet was not measured/bill…

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