# Capacitor Calculation for Power Factor Improvement

Written On Monday, November 23rd, 2009 At 11:18:27 am By Sunil Saharan
2280 Times
This article explain detailed procedure to calculate numerical value of a capacitor's capacitance to be connected with an Inductive load like induction motor to improve power factor from one value to other given value.

Capacitor calculation or capacitance calculation for a capacitor bank to improve power factor is very important topic for an electrical engineer. Capacitor bank help to reduce reactive power. Here is one numerical problem on capacitor or capacitance calculation for an induction motor.
Here is a Numerical Problem : -
A single phase 220 volts, 50 Hz, motor takes a supply current of 50 Amperes at a power factor of 0.5 lagging. The motor power factor has been improved to 0.9 lagging by connecting a condenser in parallel. Calculate the capacitance of the capacitor required ?
Solution : -
Please note that current power factor is 0.5 it indicates that motor is induction motor or an inductive load is connected to the system.
Data extracted from above problem : -
Motor Type : - 1 Phase, Induction Motor,
Voltage Applied : - 220 volts,
Current from Supply : - 50 Amp,
Supply Frequency : - 50 Hz,
Current Power Factor =cos Φ=0.5 ( lagging ),
Φ = cos-10.5=60˚.
Active Component of current, say Ia = I * cos Φ = 50 * 0.5 = 25 Amp,
One thing you must remember regarding power factor improvement
` Improving power factor does not change active power consumed in the appliance. `
Since the voltage also same so only current will be changed after power factor improvement. Hence only total current will be changed but active component will be same i.e. 25 Amp.
Tan Φ = tan 60 = 1.732,
Reactive component of current, say Ir1 = I tan Φ = 25 * 1.732 = 43.3 Amp
Let the capacitor of C capacitance is connected in parallel with motor then,
New Power Factor = 0.9,
cos Φ1 = 0.9,
tan Φ1= 0.4843
New reactive component = I tanΦ1 = 25 * 0.4843 = 12.1 Amp,
Thus, current through capacitor = 43.3 - 12.1 = 31.2 Amp,
As we know,
`Ic = V/Xc = 2*pi*f*C*V`
C = 451 * 10-6 F

Tags :Power Factor Improvement, Numerical Calculation of Capacitance for Power Factor Improvement
Article Was Last Updated on Saturday, March 10th, 2018 At 12:57:15 pm Sunil SaharanAn Engineer I'm Sunil Saharan, An Electrical Engineer by chance and a Web developer by choice. I love HTML, CSS, JS, Bootstrap and PHP off course. This site is my effort to share my little knowlege with the world.

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