Motor input = P = 5 kW

Original P.F = Cosθ1 = 0.75

Final P.F = Cosθ2 = 0.90

θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819

θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843

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

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

= 5kW (0.8819 – 0.4843)

= 1.99 kVAR

And Rating of Capacitors connected in each Phase

1.99/3 = 0.663 kVAR.

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**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?**

Motor input = P = 5 kW

Original P.F = Cosθ1 = 0.75

Final P.F = Cosθ2 = 0.90

θ1 = Cos-1 = (0.75) = 41°.41; Tan θ1 = Tan (41°.41) = 0.8819

θ2 = Cos-1 = (0.90) = 25°.84; Tan θ2 = Tan (25°.50) = 0.4843

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

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

= 5kW (0.8819 – 0.4843)

= **1.99 kVAR**

And Rating of Capacitors connected in each Phase

1.99/3 = **0.663 kVAR**

**Step 1** **– Calculate Actual Load (kW)**

(Load) Power **kW** = Volts **V** x **√3** x Current **I** x Power factor **Pf**

**Step 2 – Calculate Required Power Factor Correction (kVAr)**

Power Factor Correction **kVAr** = Power **kW** (**TanΦi – TanΦd**)

**Φi** = Cos-1 Initial Power Factor Pf

**Φd** = Cos-1 Required Power Factor Pf

**Step 3** **– Calculate Actual Power Factor Correction [kAVrl**

Actual Power Factor Correction Pf = Cos (**Tan-1** (**TanΦi** – **Correction kVAr**/**Power kW** ))

Typical Power Factor Correction available in multiples of 25kVAR’s, for further details please Contact Blakley Electrics Technical Department.

You are welcome to read more Important Interview Questions and MCQs.