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

**How to Convert Capacitor Farads into kVAR & Vice Versa for Power factor improvement?**

*How to Calculate the Required Capacitor bank value in both kVAR and Farads?***How to Convert Capacitor Farads into kVAR & Vice Versa (For Power factor improvement)**

In this simple Calculation tutorial, we will find the way “How to Convert Capacitor

**Farads**into**kVAR**and Vice Versa, usually used in Power Factor improvement Calculation and problems. We will use two simple methods for finding and Conversion for both quantities. For explanation, consider the following simple Example.**Example 1:**

**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 (i.e, Convert Capacitor Farads into 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

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

ActualP.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})= 12kW (1.3333– 0.4843)

=

**10.188 kVAR****(2) To find the required capacity of Capacitance in Farads**

**(i.e, Convert Capacitor Farads into kVAR)**to 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****Good to Know:**

These are the main Formulas to Convert Capacitor

**kVAR**into**Farads**and Vice VersaRequired Capacity of Capacitor in Farads/Microfarads

**(Convert Capacitor Farads into kVAR)****C = kVAR / (2 π f V**

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

**(Convert Capacitor kVAR into Farads)****kVAR = C x (2 π f V**

^{2})**You may also read about;**

- Power Factor
- Active, Reactive, Apparent and Complex Power. Simple explanation with formulas.
- Causes of low Power Factor
- Disadvantages of Low Power Factor
- Power Factor improvement Methods with Their advantages & Disadvantages
- Advantages of Power factor improvement and Correction
- How to Calculate the Suitable Capacitor Size in Farads & kVAR for Power factor Improvement (Easiest way ever)

thanks buddy ,this information ll use in my project !!!! …. 😀 :)<br /><br />

I have a question on the following lines of your calculations:

Solution # 2 (Classical Calculation Method)

Motor Input = P = V x I x Cosθ

= 400V x 50A x 0.6

= 12kW

.

.

.

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

= 5kW (1.3333– 0.4843)

Where did the P=5kW come from? I thought it was calculated to be 12kW. Is this an error?

Thanks for correction 🙂

yes bro it is given 12.. and you can see this problem in Vk Mehta book on Power distribution and utilization as well

I got some confusrion for refixing one Philips model HL1642 -600 watts mixie grinder motor wire connection with its Rotary switch connection having overload,INCHERswitch&line indicator .wiring diagrame may be sent to my E mail address for clearing my doubts&refixing the connections correctly…Tnx in anticipation…acn 1-2-15

Hello

How kvar came to 10.188 kvar.

If kw =12 and kva = 20, kvar will be 16 for sure.

Kvar = kva sin theta

= 400x50x 0.8

= 16

hence C will come 318.31 uf

Please clarify or correct.

Dear Jashwant,

We are not going to calculate the kVAR from kW or kVA here…

But want to improve the Power factor from 0.6 to 0.9.

That’s Why we didn’t use that (Which you have mentioned) formula… 🙂

let me know if you want to know more… Thanks

how to calculate the required capacitor to improve power factor by using

apparent power(kw)=3.66

voltage(v)=408V

current(i)=1640A

VAR=1403

pf=0.5.

it is a three phase

i did not understand you have a given exmple 400v and single phase what iis that is it 3phase .still iam new for this subject

This is a right or wrong formula to convert KVAR to microfarad.

C=KVAR/(2*1.732*F*V^2).

or any other solution is available?

how calculate

10.188/2*3.14*50*400*400=202/1000000

Thank you very much

Good reminder of what I learned in my first year of engineering.

the formula is not right, it should be:

KVAR = Q = Vrated^2/Zc

I need a power capacitor of UN 380V QN 150 micro F 50 Hz

In 17.9 /10.3 A but I cannot find similar rating in Local market.

Which KVAR and In rating may suitable

Let me know pls