Active power P
Average power converted into useful work or heat.
Electricity · Industrial electricity
Single-phase power, P–Q–S and current
Calculate P, Q and S for a single-phase circuit from RMS voltage, RMS current and power factor, or solve current from kW or kVA.
Created v1.0
By Thibaut Grzelak, Mechanical Analysis Engineer
02
Results
Reactive power Q
Magnitude of reactive power associated with the entered power factor.
Apparent power S
Single-phase product U·I of RMS voltage and current.
RMS current I
RMS current in the single-phase circuit.
Relation and numerical substitution
S = U · I ; P = S · cosφ ; Q = S · √(1 − cos²φ)—
Single-phase circuit in sinusoidal steady state. U and I are RMS voltage and RMS current. Power factor is entered as 0 < cos φ ≤ 1. Q is reported as a magnitude; the calculator does not distinguish leading from lagging power factor. Harmonics, cable voltage drop and conductor losses are not included.
01Formulas and symbols
Formulas used
S
U · ISingle-phase apparent powerP
S · cosφActive powerQ
S · √(1 − cos²φ)Reactive power magnitudeI
P / (U · cosφ)Current from active powerI
S / UCurrent from apparent power02Assumptions and limits
Scope of validity
- Single-phase circuit in sinusoidal steady state.
- U and I are RMS voltage and RMS current.
- Power factor is entered as 0 < cos φ ≤ 1.
- Q is reported as a magnitude; the calculator does not distinguish leading from lagging power factor.
- Harmonics, cable voltage drop and conductor losses are not included.
03Validation example
Reference numerical case
- U = 230 V, I = 10 A and cos φ = 0.80.
- S = 230 × 10 = 2.30 kVA.
- P = 2.30 × 0.80 = 1.84 kW.
- Q = 2.30 × √(1 − 0.80²) = 1.38 kvar.
04References
Technical references
- IEC 60050-131 — International Electrotechnical Vocabulary: circuit theory.
- BIPM — The International System of Units (SI), 9th edition.