Single-Phase Power Calculator
Compute active, reactive and apparent power for a single-phase AC load. Enter any three of voltage, current and power factor to get the fourth — plus the power triangle, phasor diagram, and companion metrics.
| Appliance | Typical P | Current @ 230 V | Typical PF | Notes |
|---|
| Load type | Typical PF | Phase angle θ | Nature |
|---|
| Quantity | Small unit | Mid unit | Large unit |
|---|
The power triangle
Active power P (W), reactive power Q (VAR), and apparent power S (VA) form a right triangle: S² = P² + Q². The angle between P and S is θ, and cos θ = PF. Only P does real work; Q just oscillates between source and load, wasting capacity.
Why utilities charge for poor PF
A load with PF = 0.7 draws 43% more current than an equivalent load at unity PF for the same useful power. That extra current means thicker cables, bigger transformers, and more losses — costs the utility passes back to industrial customers via PF penalties.
Capacitor correction
Most low-PF loads are inductive (motors, transformers). Adding a parallel capacitor supplies the reactive power locally, raising PF toward unity without changing the load's real power. The capacitor value follows Qc = P·(tan θ₁ − tan θ₂), with C = Qc/(2πf·V²).
Single-phase vs three-phase
This calculator handles single-phase loads. For balanced three-phase, active power is P = √3·V_LL·I_L·cos θ. Never apply the single-phase formula to a three-phase load — you will under-estimate by a factor of √3.
How to Use This Calculator
Pick which of the four quantities you want to solve for; the other three become editable inputs. Enter any two inputs to see the third — the fourth is always computed.
1. Choose what to solve for
Select P, V, I, or cos θ. That field becomes read-only with a SOLVED badge and updates live as you edit the other three.
2. Enter any two of the three remaining values
Type a value, drag the slider, or tap a preset. The third field is computed to keep P = V·I·cos θ.
3. Read the result card
The solved value appears large with unit; below it, all four values plus derived S, Q, and θ. The context banner classifies the load (resistive, inductive, etc.).
4. Check the charts
Power triangle draws P, Q and S to scale. P vs I plots active power against current for several PF values. PF zones shows your PF on a colour-coded quality bar. Phasor diagram shows V and I phasors with the phase angle θ.
5. Reference tables
Table 1 lists typical appliance power ratings. Table 2 lists power factors by load type. Table 3 (Advanced) converts your values across engineering units.
6. Advanced mode
See companion metrics: S, Q, θ, energy per hour/day, cost estimates at typical tariffs, unity-PF current comparison, and capacitor sizing for 0.95 PF correction.
What This Tool Offers
A complete workbench for single-phase AC power calculations — the same relations used in electrical design, energy audits, and load analysis.
Four-way solve
Solve for P, V, I, or power factor from the other three. No algebra required.
Four interactive views
Power triangle, P-vs-I curves, PF quality zones, and phasor diagram — all driven by your live values.
Per-field unit switching
Voltage in mV / V / kV; current in mA / A / kA; power in W / kW / MW / hp.
Reference tables
Typical appliance ratings, PF values by load type, and unit conversions for your exact values.
Full power triangle
Every solve returns S (VA), Q (VAR), and θ (°) alongside the requested quantity.
PF correction sizing
Recommended capacitor VAR and capacitance at 50 Hz / 60 Hz to raise the current PF to 0.95.
What Is Single-Phase Active Power?
In a single-phase AC circuit, the average real power delivered to a load — the power that actually does useful work, is dissipated as heat, or is converted to another form — is given by P = V·I·cos θ, where V and I are the RMS voltage and current, and cos θ is the power factor. The power factor is the cosine of the phase angle between the voltage and current phasors: unity for a purely resistive load, less than unity when the load is inductive (motors, transformers) or capacitive (long cables, power supplies with poor input filtering). A load with low power factor draws more current than a resistive load of the same real power, forcing the supply to provide more apparent power S = V·I and reactive power Q = V·I·sin θ. Utilities therefore measure both active power (which they bill for, in kWh) and reactive power (which they penalise, because it wastes distribution capacity). Power factor correction — usually by adding parallel capacitance to inductive loads — is one of the most cost-effective improvements available to industrial electrical systems.
Formula and Variables
S = V · I (VA) · Q = V · I · sin θ (VAR) · S² = P² + Q²
θ = arccos(PF) · Qc = P · (tan θ1 − tan θ2) · C = Qc / (2πf · V²)
| Symbol | Meaning | Notes |
|---|---|---|
| P | Active power | Watts (also kW, MW, hp) — does real work |
| V | RMS voltage | Volts — phase-to-neutral for single-phase |
| I | RMS current | Amperes — peak = √2 × RMS for sinusoidal |
| cos θ | Power factor | 0 to 1 (magnitude); leading or lagging by convention |
| S | Apparent power | Volt-amperes — what the wiring must carry |
| Q | Reactive power | VAR — oscillates between source and load |
| θ | Phase angle | Degrees — positive for inductive, negative for capacitive |
| Qc | Correction VAR | Capacitive VAR to raise PF to a target value |
Typical Power Factors by Load
| Load | Typical PF | Nature |
|---|---|---|
| Incandescent lamp | 1.00 | Pure resistive |
| Resistive heater / kettle | 1.00 | Pure resistive |
| Modern LED lamp (with PFC) | 0.90–0.98 | Near-unity |
| Cheap LED lamp (no PFC) | 0.50–0.70 | Capacitive input |
| Computer SMPS (with PFC) | 0.95–0.99 | Active PFC |
| Induction motor (loaded) | 0.80–0.90 | Inductive |
| Induction motor (idling) | 0.20–0.40 | Inductive, low |
| Fluorescent lamp (magnetic ballast) | 0.50–0.70 | Inductive |
| Fluorescent lamp (electronic) | 0.90–0.98 | Near-unity |
| Arc welder (transformer) | 0.35–0.60 | Inductive, poor |
| Transformer (loaded) | 0.80–0.95 | Inductive |
Frequently Asked Questions
Because the formula V·I·cos θ gives power in watts, and the metric prefix "kW" means 1000 watts. So 230 V × 10 A × 0.9 = 2070 W = 2.070 kW. The division by 1000 is just the conversion from W to kW. If you keep your units consistent, the physical formula stays P = V·I·cos θ.
kW (kilowatts) is real power — the power that does useful work and is billed by electricity meters. kVA (kilovolt-amperes) is apparent power — the total power the wiring and transformers must carry, including the reactive component that just oscillates. The ratio kW/kVA is the power factor. Utilities charge industrial customers for both, with a penalty when PF drops below a threshold (typically 0.90 or 0.95).
No. Power factor is defined as the cosine of a real phase angle, so its magnitude is at most 1. PF = 1 corresponds to a purely resistive load (θ = 0°). PF = 0 corresponds to a purely reactive load (θ = 90°). A PF greater than 1 in a measurement usually indicates a metering error or a distorted non-sinusoidal waveform where the true power factor is being mis-estimated.
Correcting PF from 0.7 to 0.95 reduces the current by about 26% for the same real power — cutting I²R losses in cables by ~45%, freeing capacity on transformers, and avoiding utility PF penalties. The correction is usually done by adding a parallel capacitor bank sized to cancel the inductive VAR. A typical industrial payback period is 12 to 24 months.
No — this calculator is for single-phase loads only. For a balanced three-phase load, active power is P = √3 · VLL · IL · cos θ, where VLL is the line-to-line voltage and IL is the line current. Applying the single-phase formula to a three-phase load under-estimates by a factor of √3 ≈ 1.732.
What it means
The real power in a single‑phase AC circuit is given by P(kW) = (V × I × cosθ) / 1000, where V is RMS voltage, I is RMS current, and cosθ is the power factor. This formula calculates the actual power consumed, which is the product of apparent power and power factor. It is used for sizing fuses, wires, and circuit breakers. Example: A 230V, 10A resistive heater (cosθ=1) draws P = 230 × 10 × 1 / 1000 = 2.3 kW. If the load is a motor with power factor 0.8, the real power is 230×10×0.8/1000 = 1.84 kW, though the current remains 10A. This is why power factor correction is important to reduce current and losses.
Worked example
Single‑Phase Power – Practical Example
Real‑World| Parameter | Value |
|---|---|
| V | 230 V |
| I | 10 A |
| cos θ | 0.85 |
| Formula | P(kW) = (V × I × cos θ) / 1000 |
Common mistakes
Watch for unit mismatches (W vs kW, single- vs three-phase) and remember to include power factor or efficiency where the formula requires it.Applications
Power in kW (single‑phase): P(kW) = (V × I × cosθ)/1000 calculates real power for single‑phase AC circuits. This is used to determine the active power consumption of single‑phase loads, such as lighting and small appliances. Engineers use it for load calculations, circuit design, and energy auditing.
- Residential and commercial load calculation
- Circuit breaker and cable sizing for single‑phase
- Energy consumption estimation
- Power factor analysis and correction
- Educational understanding of single‑phase power
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