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Apparent Power (AC)

Complex power in AC circuits (volt-amperes).

Written by the Vorzep Team

ElectricalPower SystemsAC

Apparent Power Calculator AC · RMS Values

S = V · I
S = apparent power (VA)  ·  V = RMS voltage (V)  ·  I = RMS current (A)
⟹ Solve ⟷ S, V, I
VA
V
A
Please fix the errors above.
Solve for:
Presets:
Apparent Power —
—
S: — V: — I: —
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Apparent Power Gauge —
Low (< 500 VA) Moderate (500–5000 VA) High (> 5000 VA)
S = V · I (RMS values)  ·  Apparent power is the product of RMS voltage and RMS current, independent of phase angle.
S = V · I*
Apparent Power (AC)

Variables

SymbolQuantityUnit
SApparent powerVA
VVoltage phasorVolts
I*Conjugate current phasorAmperes

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
Scenario: An air conditioner draws 10 A at 240 V AC. Calculate the apparent power (VA) to size the circuit breaker.
ParameterValue
V240 V
I10 A
FormulaS = V · I
1Apply formula:S = 240 V × 10 A
2Calculate:S = 2400 VA
Final Design 2400 VA (2.4 kVA) ✓ Use 15 A breaker
Why: Apparent power indicates the total load on the supply, including reactive power. A 2400 VA load requires a 15 A circuit breaker for safety.

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