Premium Membership ♕

CYBER WEEK OFFER 💥 Save 20% on PRO Membership Plan and Video Courses with the coupon CYM23 and learn from experienced engineers.

Home / Technical Articles / How capacitors improve the power factor and how to calculate them?

How to improve the power factor?

It’s quite simple. By installing capacitors or capacitor banks. Improving the power factor of an electrical installation consists of giving it the means to “produce” a certain proportion of the reactive energy it consumes itself.

How capacitors improve the power factor and how to calculate them?
How capacitors improve the power factor and how to calculate them? (photo credit:
There are various different systems for producing reactive energy, including in particular asynchronous compensators and shunt capacitors (or serial capacitors for large transmission systems).

The capacitor is most frequently used, given:

  1. Its non-consumption of active energy
  2. Its purchase cost
  3. Its ease of use
  4. Its service life (approximately 10 years)
  5. Its low maintenance (static device)

Power diagram

Power factor is the ratio of working power to apparent power. It measures how effectively electrical power is being used.

Low power factor means you’re not fully utilizing the electrical power you’re paying for. A high power factor signals efficient utilization of electrical power, while a low power factor indicates poor utilization of electrical power.

To determine power factor (PF), divide working power (kW) by apparent power (kVA). In a linear or sinusoidal system, the result is also referred to as the cosine θ.

PF = kW / kVA = cosine θ kVA

For example, if you had a boring mill that was operating at 100 kW and the apparent power consumed was 125 kVA, you would divide 100 by 125 and come up with a power factor of 0.80.

(kW) 100 / (kVA) 125 = (PF ) 0.80

Power diagram
Power diagram


  • P – Active power
  • S1 and S2 – apparent powers
    (before and after compensation)
  • Qc – capacitor reactive power
  • Q1 – reactive power without capacitor Q2: reactive power with capacitor


  • Q2 = Q1 – Qc
  • Qc = Q1 – Q2
  • Qc = P×tg φ1 – P×tgφ2
  • Qc = P×(tg φ1 – tg φ2)

Where φ1 is phase shift without capacitor and φ2 is phase shift with capacitor

The capacitor is a receiver composed of two conductive parts (electrodes) separated by an insulator. When this receiver is subjected to a sinusoidal voltage, the current and therefore its power (capacitive reactive) is leading the voltage by 90°.

Conversely, for all other receivers (motors, transformers, etc.) the current and therefore its power (reactive inductive) is lagging the voltage by 90°.

The vectorial composition of these currents or reactive powers (inductive and capacitive) gives a resulting current or power below the value which existed before the capacitors were installed.

In simple terms, it is said that inductive receivers (motors, transformers, etc.) consume reactive energy whereas capacitors (capacitive receivers) produce reactive energy.

How to calculate the power of capacitors

Based on electricity bills to calculate the capacitor banks to be installed, use the following method:

  • Select the month in which the bill is highest (kVArh to be billed)
  • Assess the number of hours the installation operates each month
  • Calculate the capacitor power Qc to be installed
Qc = kVArh to be billed (monthly) / No. of hours’ operation (monthly)

Example for the subscriber //

  • Highest reactive energy bill: December Number of kVArh to be billed: 70,000
  • Monthly operating times: High-load + peak times = 350 hours

Qc (bank to be installed) = 70,000 / 350 = 200 kVAr

Based on measurements taken on the HV/LV transformer secondary: PkW-cosFI

Example //

An establishment supplied from an 800 KVA HV/LV subscriber station wanting to change the power factor of its installation to:

  • Cosφ = 0.928 (tgφ = 0.4) at the primary
  • I.e. Cosφ = 0.955 (tgφ = 0.31) at the secondary, with the following readings:
    • Voltage: 400 V 3-phase 50 HZ
    • PkW = 475
    • Cos (secondary) = 0.75 (i.e. tg ø = 0.88)

Qc (bank to be installed) = PkW x (tgφ measured – tgφ to be obtained)
Qc = 475 x (0.88 – 0.31) = 270 kVAr

Calculation for future installations:


1000 kva transformer, Q capacitor = 250 kVAr

Note: This type of ratio corresponds to the following operating conditions:

  • 1000 kVA transformer
  • Actual transformer load = 75%
  • Cosφ of the load = 0.80 } k = 0.421
  • Cosφ to be obtained = 0.95 } – see table below

Qc = 1000 x 75% x 0.80 x 0.421 = 250 kVAr

Capacitor power calculation table

Conversion table

Based on the power of a receiver in kW, this table can be used to calculate the power of the capacitors to change from an initial power factor to a required power factor. It also gives the equivalence between cos ø and tg ø.

Power factor table
Power factor table

Example: 200 kW motor – cosφ = 0.75 – required cosφ = 0.93 – Qc = 200 x 0.487 = 98 kVAr

Reference // Reactive energy compensation and power quality monitoring by Legrand

Premium Membership

Get access to premium HV/MV/LV technical articles, electrical engineering guides, research studies and much more! It helps you to shape up your technical skills in your everyday life as an electrical engineer.
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. Peter Hughes
    Jun 22, 2023

    Many years ago I did an Mech and Elec apprenticeship, where I learnt about Power Factor improvements and Voltage management on industrial systems.
    Alas, that was a long time ago,
    Now, wanting to save pension money on my electric bills at home. I would appreciate some help on where to get advice on using today’s technology to do the similar efficiencies.
    Be very grateful for your input, Thanks.

    Best wishes,

  2. akash dandge
    May 26, 2023


  3. Sirelkhatim Nugud
    May 18, 2022

    A very useful article
    Eng sirelkhatim

  4. velmurugan ramakrishnan
    Oct 12, 2021

    very good

  5. Deepak Agrawal
    Aug 30, 2020


  6. Richmond
    Jul 19, 2020

    Very easy to understand. I really enjoy this article. Thanks a lot

  7. Davis chesoli
    Jul 18, 2020

    Good effort.

  8. Davis chesoli
    Jul 18, 2020

    Well explained.

  9. Duke Damptey Danquah
    Mar 30, 2019

    Very good and informative keep it up

    Mar 29, 2019

    dentro de su explicación es muy importante para la formación de los futuros ingenieros electromecánicos y los electricistas , hay otra forma practica para resolver la compensación reactiva capacitiva.

    por medio de inyección de reactivos con un banco de cualquier capacidad por ejemplo si tenemos 40 kvar los multiplicamos x 24 horas de funcionamiento esto nos da 960 kvarh diarios y los multiplicamos x 30 días de funcionamiento obtenemos 28800 kvarh inyectados al sistema eléctrico de la factoría estos se los restamos a los consumidos en un mes kvarh consumidos- los kavrh inyectados con el banco instalados y la división kvarh totales/kwh nos da el arc tg kvarh totales/ kwh totales y a este resultado del Angulo y se saca el fp mejorado

  11. Wryter
    Mar 29, 2019

    Pretty good

  12. AMIS Electricals
    May 23, 2018

    Nice Information. Well we can make excel sheet for this to calculate require capacitors for power factor improvement.

    Thanks for sharing

  13. Keith Simpson
    May 20, 2018

    Interesting, informative and well presented.

  14. hussein hakki
    Feb 11, 2018

    useful information

  15. Andrés Elías Pérez Laguna
    Feb 09, 2018

    Estoy copiando los temas que ha dejado el Ingeniero Edvard Scesanyi y lo que puedo decir es que se trata de una muy valiosa información, muchas gracias

  16. Praveen singh
    Feb 09, 2018

    it is very good information

  17. Steven
    Nov 24, 2016

    Well presented! tank you so much…

  18. Ameer
    Nov 19, 2016

    thank you so much

  19. Chidi
    Nov 16, 2016

    This information I readout in this your article is of great importance to my electrical power practices. Thank you and please do not hesitate to assist me when I need your help. God bless..

  20. Ivan
    Nov 16, 2016

    Clear and very illustrative this post.

Leave a Comment

Tell us what you're thinking. We care about your opinion! Please keep in mind that comments are moderated and rel="nofollow" is in use. So, please do not use a spammy keyword or a domain as your name, or it will be deleted. Let's have a professional and meaningful conversation instead. Thanks for dropping by!

23  +    =  twenty eight

Learn How to Design Power Systems

Learn to design LV/MV/HV power systems through professional video courses. Lifetime access. Enjoy learning!

Subscribe to Weekly Newsletter

Subscribe to our Weekly Digest newsletter and receive free updates on new technical articles, video courses and guides (PDF).
EEP Academy Courses - A hand crafted cutting-edge electrical engineering knowledge