EngineeringFormulaElectrical Engineering
Single-Phase Power Calculator
Calculate real, reactive and apparent power for a single-phase AC circuit from voltage, current and power factor.
In an AC circuit with inductance or capacitance, current shifts out of phase with voltage, so simple P = VI overstates the power actually doing useful work. Single-phase AC power splits into three related quantities - real power (the watts that do useful work), reactive power (power that oscillates between source and load without being consumed), and apparent power (what a meter reading volts × amps actually shows). Enter voltage, current and power factor, or any three of voltage, current, power factor and real power, to solve for the fourth and see the full power triangle.
V · RMS voltage across the circuit.
I · RMS current drawn by the circuit.
PF · cos φ - the fraction of apparent power that is real power. A plain decimal from 0 to 1, not a percentage.
Solution
Enter the required values to calculate real power.
P = V × I × PF
Formula Sheet
- PReal Power
- VVoltage
- ICurrent
- PFPower Factor
- SApparent Power
- QReactive Power
Variables & Units
| Symbol | Variable | Description | Common Units |
|---|---|---|---|
| P | Real Power | Power actually converted into useful work - heat, light, motion. | mW, W, kW, MW, hp |
| V | Voltage | RMS voltage across the circuit. | V, mV, kV |
| I | Current | RMS current drawn by the circuit. | A, mA |
| PF | Power Factor | cos φ - the fraction of apparent power that is real power. A plain decimal from 0 to 1, not a percentage. | |
| S | Apparent Power | V × I, ignoring phase - what a volt-amp meter reads. | VA |
| Q | Reactive Power | Power that oscillates between source and load without being consumed, from any inductance or capacitance. | VAR |
How to Use This Calculator
- 01Select which quantity you want to solve for - Real Power, Voltage, Current, or Power Factor.
- 02Enter the other three values, choosing the correct unit for each.
- 03Power factor is entered as a plain decimal between 0 and 1 (for example, 0.85), not a percentage or an angle.
- 04Select Calculate to see the solved value, along with the apparent power (VA) and reactive power (VAR) that go with it.
How the Formula Works
Real power P = V × I × PF is the rate of energy actually converted into useful work - heat, light, motion. Apparent power S = V × I is what you'd get from just multiplying the RMS voltage and current readings, ignoring how in-phase they are. Power factor PF = cos φ (where φ is the phase angle between voltage and current) measures how much of the apparent power is real power: PF = P / S. A purely resistive load has PF = 1, so P = S; an inductive or capacitive load has PF < 1, so S is larger than P.
The remaining piece is reactive power, Q = S × sin φ - power that flows back and forth between the source and any inductance or capacitance in the load without being consumed. P, Q and S form a right triangle (the power triangle) with S as the hypotenuse: S² = P² + Q². This calculator uses that relationship to report all three quantities together, whichever one you solve for.
Worked Example 01
Real power from voltage, current and power factor
Known
- Voltage (V): 230 V
- Current (I): 10 A
- Power Factor (PF): 0.8
Formula
P = V × I × PF
Substitution
P = 230 × 10 × 0.8
Result
P = 1840 W (S = 2300 VA, Q = 1380 VAR)
A 230 V, 10 A load at 0.8 power factor delivers 1840 W of real power, drawing 2300 VA of apparent power - 1380 VAR of that is reactive power the source supplies but the load doesn't consume.
Worked Example 02
Current from real power, voltage and power factor
Known
- Real Power (P): 480 W
- Voltage (V): 100 V
- Power Factor (PF): 0.6
Formula
I = P / (V × PF)
Substitution
I = 480 / (100 × 0.6)
Result
I = 8 A (S = 800 VA, Q = 640 VAR)
A load rated for 480 W at 100 V with a 0.6 power factor draws 8 A - noticeably more current than a purely resistive 480 W load would, because the low power factor means a larger apparent power is needed to deliver the same real power.
Worked Example 03
Power factor from real power, voltage and current
Known
- Real Power (P): 800 W
- Voltage (V): 100 V
- Current (I): 10 A
Formula
PF = P / (V × I)
Substitution
PF = 800 / (100 × 10)
Result
PF = 0.8 (S = 1000 VA, Q = 600 VAR)
A meter reading 100 V, 10 A and 800 W implies a power factor of 0.8 - the load is drawing 1000 VA of apparent power to deliver 800 W of real power.
Worked Example 04
Voltage from real power, current and power factor
Known
- Real Power (P): 640 W
- Current (I): 8 A
- Power Factor (PF): 0.8
Formula
V = P / (I × PF)
Substitution
V = 640 / (8 × 0.8)
Result
V = 100 V (S = 800 VA, Q = 480 VAR)
A load drawing 8 A at 0.8 power factor to deliver 640 W of real power must be operating at 100 V.
Applications
- 01Sizing a generator, UPS, or transformer in kVA rather than just kW, since apparent power is what determines current-carrying and heating requirements
- 02Estimating the reactive power (kVAR) a capacitor bank needs to supply for power-factor correction
- 03Checking a motor or appliance's rated power factor against its real and apparent power nameplate values
Assumptions
- 01The circuit is single-phase AC in sinusoidal steady state - voltage and current are both at the same frequency.
- 02Power factor is treated as a plain magnitude between 0 and 1; this calculator doesn't track whether it's leading (capacitive) or lagging (inductive).
- 03Voltage and current are RMS values, which is the standard convention for AC power calculations.
Where This Model Stops
- 01Not valid for three-phase circuits - three-phase power uses a separate formula with a √3 factor, not this single-phase relationship.
- 02Does not distinguish leading from lagging power factor. For most single-load sizing questions the magnitude is what matters, but if you're correcting power factor with capacitors, the sign (direction) matters too.
- 03Assumes an ideal sinusoidal supply. Harmonic distortion in the voltage or current waveform is not modeled and would make a true power-quality meter reading differ from this result.
References
- [1]
AC power - real, reactive and apparent power
Standard electrical engineering fundamentals
P = VI cos φ, S = VI, Q = VI sin φ, with the power-triangle relationship S² = P² + Q².
Frequently Asked Questions
What's the difference between this and the Electrical Power Calculator?
The Electrical Power Calculator assumes a purely resistive load, where P = VI is the whole story. This calculator is for AC loads with a power factor below 1 - motors, transformers, and anything with inductance or capacitance - where voltage and current aren't perfectly in phase, so real, reactive and apparent power differ.
Why can't power factor be greater than 1?
Power factor is cos φ, the cosine of the phase angle between voltage and current, and cosine never exceeds 1. A power factor of 1 means voltage and current are perfectly in phase (a purely resistive load); anything less means some of the apparent power is reactive rather than real.
Why does apparent power (VA) matter if only real power (W) does useful work?
Because generators, transformers, wiring and circuit breakers are rated by the current they can carry, and current is set by apparent power, not real power. A low-power-factor load draws more current - and needs correspondingly larger equipment - to deliver the same real power as a load with a power factor near 1.