= Volt-ampere
{title2=VA}
The <volt-ampere>, symbolVA, is the unit commonly used to express <apparent power>. At a terminal pair, RMS potential difference in <volts> multiplied by RMS current in <amperes> gives the numerical value in <volt-amperes>. The unit product has the same coherent SI dimensions as the <watt>:
$$
1\,\mathrm{V\,A}=1\,\mathrm{kg\,m^2\,s^{-3}}=1\,\mathrm W.
$$
This unit identity does not make <apparent power> and <active power> the same quantity. The SI text explicitly warns that different physical quantities can share a unit and that the quantity must be specified.<The International System of Units (SI), ninth edition, SI Brochure>[NIST SP330 parallel text, §2.3.4, coherent derived units and quantity identification]{ref}
For an original example,250V RMS at8A RMS gives2000VA, or2kVA. If the absorbing load has <power factor>0.75, its <active power> is1500W. A1500W value cannot be inferred from the2kVA alone without the phase/waveform information summarized by that factor.
Hypothetical equipment can have separate apparent- and active-power limits. A1000VA/600W limit would permit a500W, factor0.5 load only at the apparent-power boundary, since it requires1000VA. A700W, factor-one load would require only700VA but exceed the separate600W limit. These examples interpret stated numerical constraints rather than specify an actual product's behavior.
For single-frequency sinusoids, the magnitude of <complex electrical power> is expressed in <volt-amperes>, while its real part is commonly expressed in <watts> and its imaginary part in <reactive volt-amperes>. Those names help keep the quantities distinct even though their base-unit dimensions agree.<AC Power Flow in Linear Networks by James L. Kirtley Jr.>[Energy and Power, equations61–63; §7.1 RMS Amplitude]{ref} With distorted signals, retain the actual RMS-product definition of <apparent power> rather than assuming a single fundamental phasor accounts for all current.
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