Capacitor Calculation for Power Factor Improvement

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 I_a = 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 I_r1 = 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