** How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement (Easiest way ever)**

Hi there! With a very important tutorial.. I hope you will find it very useful because I have already spent two days to prepare this article. I think all of those who have sent messages and mails about the topic will never ask again if they follow these simple methods to calculate the proper Size of Capacitor bank in kVAR and micro-farads for power factor correction and improvement in both single phase and three phase circuits. I think it’s too much..

Now let’s begin…

Consider the following Examples.

**Example: 1**

**A 3 Phase, 5 kW Induction Motor has a P.F (Power factor) of 0.75 lagging. What size of Capacitor in kVAR is required to improve the P.F (Power Factor) to 0.90?**

**Solution #1 (By Simple Table Method)**

Motor Input = 5kW

From Table, Multiplier to improve PF from 0.75 to 0.90 is .398

Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = kW x Table 1 Multiplier of 0.75 and 0.90

= 5kW x .398

=

**1.99 kVAR**And Rating of Capacitors connected in each Phase

1.99/3 =

**0.663 kVAR****Solution # 2 (Classical Calculation Method)**

Motor input = P = 5 kW

Original P.F = Cosθ

_{1}= 0.75Final P.F = Cosθ

_{2}= 0.90θ

_{1}= Cos^{-1}= (0.75) = 41°.41; Tan θ_{1 }= Tan (41°.41) = 0.8819θ

_{2 }= Cos^{-1}= (0.90) = 25°.84; Tan θ_{2 }= Tan (25°.50) = 0.4843Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = P (Tan θ

_{1 }– Tan θ_{2})= 5kW (0.8819 – 0.4843)

=

**1.99 kVAR**And Rating of Capacitors connected in each Phase

1.99/3 =

**0.663 kVAR****Tables (Capacitor sizing in kVAr and Farads for PF correction)**

The following tables have been prepared to simplify kVAR calculation for power factor improvement. The size of capacitor in kVAR is the kW multiplied by factor in table to improve from existing power factor to proposed power factor. Check the others Examples below.

*Table – from 0.01 to 0.09 (Click image to enlarge)*

*Table – from 0.10 to 0.30 (Click image to enlarge)*

*Table – from 0.31 to 0.49 (Click image to enlarge)*

*Table – from 0.50 to 0.74 (Click image to enlarge)*

*Table – from 0.75 to 1.00 (Click image to enlarge)*

*The whole Table – from 0.10 to 1.0 (Click image to enlarge)*

**Example 2:**

**An Alternator is supplying a load of 650 kW at a P.F (Power factor) of 0.65. What size of Capacitor in kVAR is required to raise the P.F (Power Factor) to unity (1)? And how many more kW can the alternator supply for the same kVA loading when P.F improved.**

**Solution #1 (By Simple Table Method)**

Supplying kW = 650 kW

From Table 1, Multiplier to improve PF from 0.65 to unity (1) is 1.169

Required Capacitor kVAR to improve P.F from 0.65 to unity (1)

Required Capacitor kVAR = kW x Table 1 Multiplier of 65 and 100

= 650kW x 1.169

=

**759.85 kVAR**We know that P.F = Cosθ = kW/kVA . . .or

kVA = kW / Cosθ

= 650/0.65 = 1000 kVA

When Power Factor is raised to unity (1)

No of kW = kVA x Cosθ

= 1000 x 1 = 1000kW

Hence increased Power supplied by Alternator

**1000kW – 650kW = 350kW**

**Solution # 2 (Classical Calculation Method)**

Supplying kW = 650 kW

Original P.F = Cosθ

_{1}= 0.65Final P.F = Cosθ

_{2}= 1θ

_{1}= Cos^{-1}= (0.65) = 49°.45; Tan θ_{1 }= Tan (41°.24) = 1.169θ

_{2 }= Cos^{-1}= (1) = 0°; Tan θ_{2 }= Tan (0°) = 0Required Capacitor kVAR to improve P.F from 0.75 to 0.90

Required Capacitor kVAR = P (Tan θ

_{1 }– Tan θ_{2})= 650kW (1.169– 0)

=

**759.85 kVAR***How to Calculate the Required Capacitor bank value in both kVAR and Farads?*

*How to Calculate the Required Capacitor bank value in both kVAR and Farads?*

*(How to Convert Farads into kVAR and Vice Versa)*

*(How to Convert Farads into kVAR and Vice Versa)*

**Example: 3**

**A Single phase 400V, 50Hz, motor takes a supply current of 50A at a P.F (Power factor) of 0.6. The motor power factor has to be improved to 0.9 by connecting a capacitor in parallel with it. Calculate the required capacity of Capacitor in both kVAR and Farads.**

**Solution.:**

**(1) To find the required capacity of Capacitance in kVAR to improve P.F from 0.6 to 0.9 (Two Methods)**

**Solution #1 (By Simple Table Method)**

Motor Input = P = V x I x Cosθ

= 400V x 50A x 0.6

= 12kW

From Table, Multiplier to improve PF from 0.60 to 0.90 is 0.849

Required Capacitor kVAR to improve P.F from 0.60 to 0.90

Required Capacitor kVAR = kW x Table Multiplier of 0.60 and 0.90

= 12kW x 0.849

=

**10.188 kVAR****Solution # 2 (Classical Calculation Method)**

Motor Input = P = V x I x Cosθ

= 400V x 50A x 0.6

= 12kW

Actual

**P.F = Cosθ**_{1}= 0..6Required P.F = Cosθ

_{2}= 0.90θ

_{1}= Cos^{-1}= (0.60) = 53°.13; Tan θ_{1 }= Tan (53°.13) = 1.3333θ

_{2 }= Cos^{-1}= (0.90) = 25°.84; Tan θ_{2 }= Tan (25°.50) = 0.4843Required Capacitor kVAR to improve P.F from 0.60 to 0.90

Required Capacitor kVAR = P (Tan θ

_{1 }– Tan θ_{2})= 5kW (1.3333– 0.4843)

=

**10.188 kVAR****(2) To find the required capacity of Capacitance in Faradsto improve P.F from 0.6 to 0.9 (Two Methods)**

**Solution #1 (Using a Simple Formula)**

We have already calculated the required Capacity of Capacitor in kVAR, so we can easily convert it into Farads by using this simple formula

Required Capacity of Capacitor in Farads/Microfarads

**C = kVAR / (2 π f V**

^{2}) in microfaradPutting the Values in the above formula

= (10.188kVAR) / (2 x π x 50 x 400

^{2})= 2.0268 x 10

^{-4}= 202.7 x 10

^{-6}**= 202.7μF**

**Solution # 2 (Simple Calculation Method)**

kVAR = 10.188 … (i)

We know that;

I

_{C}= V/ X_{C}Whereas X

_{C}= 1 / 2 π F CI

_{C}= V / (1 / 2 π F C)I

_{C}= V 2 F C= (400) x 2π x (50) x C

I

_{C}= 125663.7 x CAnd,

kVAR = (V x I

_{C}) / 1000 … [kVAR =( V x I)/ 1000 ]= 400 x 125663.7 x C

I

_{C}= 50265.48 x C … (ii)Equating Equation (i) & (ii), we get,

50265.48 x C = 10.188C

C = 10.188 / 50265.48

C = 2.0268 x 10

^{-4}C = 202.7 x 10

^{-6}C

**= 202.7μF****Example 4**

**What value of Capacitance must be connected in parallel with a load drawing 1kW at 70% lagging power factor from a 208V, 60Hz Source in order to raise the overall power factor to 91%.**

**Solution:**

You can use either Table method or Simple Calculation method to find the required value of Capacitance in Farads or kVAR to improve Power factor from 0.71 to 0.97. So I used table method in this case.

P = 1000W

Actual Power factor = Cosθ

_{1 }= 0.71Desired Power factor = Cosθ

_{2 }= 0.97From Table, Multiplier to improve PF from 0.71 to 0.97 is 0.783

Required Capacitor kVAR to improve P.F from 0.71 to 0.97

Required Capacitor kVAR = kW x Table Multiplier of 0.71 and 0.97

= 1kW x 0.783

**=783 VAR**(required Capacitance Value in kVAR)

Current in the Capacitor =

I

_{C }= Q_{C }/ V= 783 / 208

= 3.76A

And

X

_{C }= V / I_{C}= 208 / 3.76 = 55.25Ω

C = 1/ (2 π f X

_{C})C = 1 (2 π x 60 x 55.25)

**C = 48 μF**(required Capacitance Value in Farads)

**Good to Know:**

**Important formulas which is used for Power factor improvement calculation as well as used in the above calculation**

**Power in Watts**

kW = kVA x Cosθ

kW = HP x 0.746 or (HP x 0.746) / Efficiency … (HP = Motor Power)

kW = √ ( kVA

^{2}– kVAR^{2})kW = P = VI Cosθ … (Single Phase)

kW = P =√3x V x I Cosθ … (Three Phase)

**Apparent Power in VA**

kVA= √(kW

^{2}+ kVAR^{2})kVA = kW/ Cosθ

**Reactive Power in VA**

kVAR= √(kVA

^{2}– kW^{2})kVAR = C x (2 π f V

^{2})**Power factor (from 0.1 to 1)**

Power Factor = Cosθ = P / V I … (Single Phase)

Power Factor = Cosθ = P / (√3x V x I) … (Three Phase)

Power Factor = Cosθ = kW / kVA … (Both Single Phase & Three Phase)

Power Factor = Cosθ = R/Z … (Resistance / Impedance)

Power Factor = Cosθ = kW / kVA … (Both Single Phase & Three Phase)

Power Factor = Cosθ = R/Z … (Resistance / Impedance)

X

_{C}= 1/ (2 π f C) … (X_{C }= Capacitive reactance)I

_{C}= V/ X_{C }… (I = V / R)Required Capacity of Capacitor in Farads/Microfarads

**C = kVAR / (2 π f V**

^{2}) in microfaradRequired Capacity of Capacitor in kVAR

**kVAR = C x (2 π f V**

^{2})You May Read More:

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- How to Convert Capacitor Farads into kVAR & Vice Versa (For Power factor improvement)

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sir u have biven best examples<br /> and i have one dought how we r going to get 1.169 value<br /> From Table 1, Multiplier to improve PF from 0.65 to unity (1) is 1.169

Check in the table… the existing PF is 0.65 and required one is unity (i.e, 1)<br />so the connecting point number between these two numbers (i.e, 0.65 and 1) is 1.169. this is multilier…<br />or<br />you can find it easily by scientific calculator as;<br />θ1 = Cos-1 = (0.65) = 49°.45; Tan θ1 = Tan (41°.24) = 1.169<br />θ2 = Cos-1 = (1) = 0°; Tan θ2 = Tan (0°) = 0<br />(Check solution number

Required Capacitor kVAR = P (Tan θ1 – Tan θ2)

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Nice Article.<br /><br />Can you also teach us on how to size up the capacitors in capacitor start – capacitor run motor?<br /><br />Thanks..

I am Mechanical Engineer But i love to do Electrical Job to, Love this Page

sir …could u plzz help me on this topic – transformer protection with wireless alert (gsm based)??

sir, i need your help in calculating the kvar, the power factor and also the capacitor size together with how many step power factor board should i make. iam very new to this so i need your help as soon as possible so that i can quote for the job.

details are as follows:

1) volts – 3300v

amps – 107amps x2

kw – 525kw

2) volts – 3300

amps – 101amps x2

kw – 485kw

3) volts – 415v

hp – 20hp x3

4) volts – 415v

hp – 20hp x3

your help will be highly appreciated.

thanks

Abdul

How to compute required capacitor voltage, Farads for an 8 KW alternator

Please help, I have an 8 kW generator that don’t produce output voltage, when tested, a capacitor I believe acts as a voltage regulator don’t reads any farads value. “INFINITY”. Any specs/ratings that supposed to be the in the capacitor body was erased due to advance rusting.

Generator components is still good as it was maintain properly, it’s only this capacitor part need to be replaced. Your help is highly appreciated.

Sincerely Yours,

i thank you for the valuable information about PF calculations

A friend has an AirFoxx 4000a carpet dryer that burnt up it’s Cap and most of wires. I assume it was a start Capacitor as it’s plastic with epoxy. Was hoping you could help me find out what size is needed to replace it.

This thing cost around $400 bucks but the company that made them is defunct.

It is a 120volt 60hz three speed motor. 4 pole 1HP amps are 8,9 and 11 amps.

There is no info left on the Cap as it was fried real good.

Any Help appreciated.

I want to know how many kvar capacitor bank would be required for 200 kw 3 phase 4 wire load to achieve a pf of 0.9 from.08 lagging for my office

hi Bro,

According to my Calculations you need 53kVAr.

Connected Load 200 kW

Voltage 400 v

Current PF 0.8 %

Desired PF 0.9 %

Orginal kVA 250.00 kVA

Desired kVA 222.22 kVA

Acutal kVAr 150.00 kVAr

Desired kVAr 96.86 kVAr

Improverd kVAr 53.136 kVAr

Capacitor Value 1062.71 uF

so you can use it in (6 x 10)kVAr for your load

Sir

Please specify a suitable capacitor specification to be installed for a load of 20 hp as the earlier capacitor is malfunctioning with a power factor of 0.8

sir, I want to know how to calculate a suitable size of a cable for a motor. thanks

want to know step by step of calculating the capacitance required to bring pf to a certain angle.thank you

An audit for one firm showed that the power factor is almost 70% and that the demand is 1000kw.

What capacitor size is need to correct power factor to 90%?

you need 535.882 kVAr for your load

Connected Load 1000 kW

Voltage 400 v

Current PF 0.7 %

Desired PF 0.9 %

Orginal kVA 1428.57 kVA

Desired kVA 1111.11 kVA

Acutal kVAr 1020.20 kVAr

Desired kVAr 484.32 kVAr

Improverd kVAr 535.882 kVAr

Capacitor Value 10717.64 uF

So you can use (2 x 200 + 2 x 100 + 2 x 25) = 650 kVAr

with 6 stage PFI Controller

converte uf microfare para kvar

Hey Tyrone,Great info for home owners and DIY’s, but I’m wdeonring if you could do a demo on the neutral strap (when it’s needed or not) for us apprentices. I would like to benefit from your experience.Thanks again for your efforts.Regards,Sean

Sir

How much rating kvar is to be adopted for getting 0.9 pf of 33 hp load of 0.8 pf ( industry)

is the formula is same for calculating kvar of a capacitor for two phase and three phase voltage

great great , it was very helpfull and clear, big thanks to you bro.

i made pf corrector but i have some dought becous i made pf corrector for 230 volt

what voltage rating cap i chose and its electrolatic

its very faterstic,thx.

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Dear Sir,

I would like to ask for your help to solve the question below. Thank you very much. I have been following your articles and I must confess that you are genius. Keep it up. I wish you all the best.

************************************************************************************

Question: A 11/0.433kV, 100kVA distribution transformer of 4.25% impedance is supplying a load of 0.6 power factor. It is required that the power factor be improved by a four – step capacitor bank. Reactive Power of each step of the capacitor bank is 50kVars.

i. Calculate resonant frequency (in harmonic order) of the circuit at every step of the capacitor bank.

ii. A measured harmonic content of the above circuit is given in Table-1 below. At what step of the capacitor bank is resonance condition possible and why.

Harmonic order Fundamental 3rd 5th 7th 9th 11th

Current (A) 35 16 13 9 7 3

.

Calculate the current total harmonic distortion of the circuit.

Sir,

We need your help to calculate the correct capacitor to complete a three phase power.

We have asingle phase power source at 220 v, single phase, 60 cycles. We want to connect our three phase motor to the single phase. What capacity of capacitor to be installed in the third line to have a three phase line & connect our three phase motor whichh is a three phase, 220v, 60 cycles.

We need your assistance.

Thanks.

Ed