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Apparent Power⚠ unverified

Electrical / Power · Compute apparent power

Parameters

InputSymbolUnitDefaultDescription
VVV1.0RMS voltage
IIA1.0RMS current
OutputSymbolUnitDescription
resultSVAApparent power, in volt-amperes (VA)

The science & history

Understanding the Parameters

Derivation (Approaching a Proof)

Represent the load with the complex power $\mathbf{S} = \mathbf{V}\,\mathbf{I}^{*}$ (voltage phasor times the conjugate of the current phasor). Writing the current at phase angle $\varphi$ behind the voltage, this separates into real and reactive parts:

$$\mathbf{S} = P + jQ, \qquad P = VI\cos\varphi, \quad Q = VI\sin\varphi.$$

The apparent power is its magnitude:

$$S = |\mathbf{S}| = \sqrt{P^2 + Q^2} = V I.$$

So $P$ (real, watts — see Watt's Law), $Q$ (reactive, VAR — see Reactive Power) and $S$ form a right triangle, the power triangle, with $S = VI$ as its hypotenuse and $\cos\varphi = P/S$ the Power Factor.

History

Apparent, real, and reactive power were unified by Charles Proteus Steinmetz in the 1890s through his complex‑power formalism, which let engineers of the young AC industry size transformers and lines by $S$ while billing and analysing useful work through $P$. The distinction became economically important as inductive motor loads proliferated.

Related Concepts: Watt's Law, Power Factor, Reactive Power, Three-Phase Power

Notes: Registry calculator apparent-power (unverified). Uses RMS quantities; for three‑phase use $S = \sqrt{3}\,V_L I_L$ (see Three-Phase Power).

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