- 2067 Asoj · 4 marks
Capacitor for 6 kW load pf improvement
Short answer
Active power is 6 kW. Required reactive compensation Qc = 6000×(tan(arccos 0.6) – tan(arccos 0.95)) = 6028 VAR. Capacitance C = Qc/(2πf V²) = 6028/(314.16×220²) = 396 μF.
An inductive load of 6 kW at power factor of 0.6 is connected across 220V, 50Hz supply. Calculate the value of capacitance to be connected in parallel with the load to bring the resultant power factor of 0.95 lagging.
Given P = 6 kW, V = 220 V, f = 50 Hz, cos φ₁ = 0.6 lagging, cos φ₂ = 0.95 lagging.
- tan φ₁ = tan(cos⁻¹ 0.6) = 1.3333
- tan φ₂ = tan(cos⁻¹ 0.95) = 0.3287
Reactive power
- Before: Q₁ = P tan φ₁ = 6000 × 1.3333 = 8000 VAR
- After: Q₂ = P tan φ₂ = 6000 × 0.3287 = 1972 VAR
- Capacitor: Q_c = 8000 − 1972 = 6028 VAR
Capacitance
C = Q_c / (ωV²) = 6028 / (314.16 × 220²) = 6028 / (1.5205 × 10⁷)
C = 396 μF
Supply current falls from 6000/(220 × 0.6) = 45.5 A to 6000/(220 × 0.95) = 28.7 A.
✓ Answer checked by Claude Opus 5.5 (high effort)
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