Three-phase power calculator
Enter your values. The answer and the full worked solution update as you type.
Result
Real power14.72 kW
Apparent power17.32 kVA
Reactive power9.124 kvar
Phase voltage (Y)230.9 V
Step-by-step solution
Apparent power
S = √3·VL·IL = √3·400·25 = 17.32 kVA
Real and reactive power
P = S·cos φ = 17.32·0.85 = 14.72 kW
Q = S·sin φ = 9.124 kvar
Key formulas
- Y connection
- VL = √3·Vph, IL = Iph
- Δ connection
- VL = Vph, IL = √3·Iph
- Total real power
- P = √3·VL·IL·cos φ = 3·Iph²·R
- Apparent and reactive power
- S = √3·VL·IL, Q = P·tan φ
- Correction capacitors
- QC = P(tan φ1 − tan φ2), C = QC/(3ωV²)
- Voltage across each capacitor
- Δ: VL Y: VL/√3
- Motor input
- Pin = Pout/η
Practise Three-phase power
A calculator gives the answer. In the exam you need the method: practise it with new problems, hints and worked solutions.
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