Apparent Power Calculator
Apparent power is the product of the RMS voltage and RMS current in an AC circuit, measured in volt-amperes. Solve for S, V or I, for single-phase and three-phase supplies, and add the power factor to see real and reactive power.
S = V · I
Inputs
Pick what to solve for and the supply type, then drag a slider or type an exact value, and the results update as you drag. The sliders are logarithmic, so one drag covers both small and large values. The power factor is optional: include it to get real power P and reactive power Q.
Results
Enter values and press Calculate.
Power triangle
Real power P does useful work, reactive power Q is exchanged back and forth with the supply, and apparent power S is what the source, cables and breakers must actually carry. S² = P² + Q².
Reference tables
Common examples (illustrative values, computed by the same engine)
| Case | Supply | V | I | S |
|---|
Current drawn for a given apparent power (I = S / V single-phase, I = S / (√3·V) three-phase)
| S | I at 230 V single-phase | I at 120 V single-phase | I at 400 V three-phase | I at 480 V three-phase |
|---|
Typical power factors (rough ranges, depend on the specific equipment)
| Load | Typical cos θ |
|---|---|
| Resistive heater, incandescent lamp | about 1.0 |
| Induction motor at full load | 0.8 – 0.9 |
| Induction motor lightly loaded | 0.2 – 0.5 |
| Fluorescent lamp with a magnetic ballast | 0.5 – 0.6 |
| Power supply without power-factor correction | 0.5 – 0.7 |
| Power supply with active power-factor correction | above 0.95 |
Common transformer and generator ratings (kVA)
| 25 · 50 · 100 · 160 · 250 · 400 · 630 · 1000 · 1600 · 2500 |
Key relations
| Quantity | Relation |
|---|---|
| Apparent power, single-phase | S = V · I |
| Apparent power, three-phase | S = √3 · VL · IL = 3 · Vph · Iph |
| Voltage | V = S / I · V = S / (√3 · I) |
| Current | I = S / V · I = S / (√3 · V) |
| Real power | P = S · cos θ (W) |
| Reactive power | Q = S · sin θ (VAr) |
| Triangle | S² = P² + Q² |
| Units | 1 kVA = 1000 VA · 1 MVA = 10⁶ VA · 1 kV = 1000 V · 1 kA = 1000 A |
Formula and notes
S = √3 · VL · IL (three-phase)
- S: apparent power, in volt-amperes (VA)
- V: RMS voltage; I: RMS current
- cos θ: power factor, used to split S into real power P and reactive power Q
Example. A 230 V single-phase load draws 10 A. S = 230 × 10 = 2300 VA = 2.3 kVA. At a power factor of 0.9 the real power is P = 2300 × 0.9 = 2070 W and the reactive power is Q = 2300 × sin(25.84°) = 1003 VAr.
Three-phase example. A 400 V three-phase supply with a 50 A line current gives S = √3 × 400 × 50 = 34 641 VA, about 34.6 kVA.
Apparent power sets the size of cables, breakers, transformers, generators and UPS units, because the current they carry depends on S and not only on the useful power P. That is why these devices are rated in kVA rather than kW. A load with a low power factor needs more current for the same real power. S = V·I uses RMS values, and for loads that distort the current waveform it still holds, with the extra distortion adding to the reactive part of the triangle.
Disclaimer. For education and estimation. Sizing or wiring real equipment must follow the ratings and local electrical rules and be done by a qualified person.
References
- Wikipedia, AC power: real, reactive and apparent power
- Wikipedia, Volt-ampere
- Wikipedia, Power factor
Variables
| Symbol | Quantity | Unit |
|---|---|---|
| S | Apparent power | VA |
| V | Voltage phasor | Volts |
| I* | Conjugate current phasor | Amperes |
What it means
In AC circuits, apparent power S is the product of RMS voltage and RMS current, without considering the phase angle between them. It is expressed as S = V_rms * I_rms and measured in volt‑amperes (VA). Apparent power is the magnitude of the complex power, which includes both real power (P, in watts) and reactive power (Q, in volt‑amperes reactive, VAR). The relationship is S = √(P² + Q²). Apparent power represents the total power that must be supplied by the source; it is larger than the real power unless the power factor is unity. Utility companies often charge for apparent power (kVA) because it determines the sizing of transformers and transmission lines. Understanding apparent power is crucial for sizing electrical equipment and for power factor correction. In a purely resistive load, S = P; in inductive or capacitive loads, S > P. The formula S = V·I* (where I* is the conjugate of current) is used in complex power calculations. Example: With RMS voltage 230V and RMS current 10A, the apparent power is 230 * 10 = 2300 VA, regardless of the phase angle.
Worked example
Apparent Power (AC) – Practical Example
Real‑World| Parameter | Value |
|---|---|
| V | 240 V |
| I | 10 A |
| Formula | S = V · I |
Common mistakes
- RMS values: Use RMS voltage and current – not peak values.
- Complex conjugate: S = V·I* (where I* is the complex conjugate of I) – this gives apparent power with phase information.
- Units: S in volt‑amperes (VA), not watts.
- Power factor: The real power is P = S·cosθ, and S is the magnitude of complex power.
- Sign: The angle of S indicates whether the load is inductive (positive angle) or capacitive (negative).
Applications
Apparent power (S) in AC circuits is the product of RMS voltage and RMS current, expressed in volt‑amperes (VA), without considering the phase angle. It is the magnitude of complex power and combines real power (P) and reactive power (Q). Electrical engineers use apparent power to size transformers, generators, and transmission lines, as these must be rated to handle both the active and reactive components. By calculating S, they can determine the total current drawn from the grid and the required capacity of equipment. Apparent power is also used in power factor correction, where the goal is to reduce the current for a given real power, thereby improving system efficiency. Understanding S is essential for utility and industrial power system design.
- Sizing of transformers, generators, and switchgear
- Power factor correction and efficiency improvement
- Load flow analysis in power systems
- Design of AC power distribution networks
- Understanding the difference between real and apparent power
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