 Complex power (electrical engineering)

Complex electrical power represents the active and reactive parts of sinusoidal electrical energy transfer in one complex number. Using phasors $V$ and $I$ whose magnitudes are root mean square values, with the passive sign convention, defineThe star denotes a complex conjugate. The real part is active power; the imaginary part is reactive power. The magnitude $|S|=|V||I|=\sqrt{P^2+Q^2}$ is apparent power. A source using peak-amplitude phasors instead has the factor $1/2$: $S=V_{\mathrm{peak}}{I}_{\mathrm{peak}}^{\ast }/2$.^[ref]

To see why the conjugate matters, write $V=|V|e^{j\alpha }$ and $I=|I|e^{j\beta }$. Then $S=|V||I|e^{j(\alpha -\beta )}$ depends on the relative phase. Changing the common time reference multiplies both phasors by $e^{j\theta }$, leaving $V{I}^{\ast }$ unchanged. The product $VI$ would instead acquire $e^{2j\theta }$ and would not represent a time-reference-independent average-power quantity.

For an original example, choose $V=100\mathrm{V}$ and $I=(4-j3)\mathrm{A}$. ThenThus active power is400W, reactive power is300 reactive volt-amperes and power factor is0.8. The corresponding electrical impedance is $V/I=16+j12\Omega$. With RMS current magnitude5A, the resistive term absorbs $16(5^2)=400\mathrm{W}$, agreeing with the real part.

Under the stated sign convention and $e^{j\omega t}$ time convention, an inductive reactance gives positive $Q$ and a capacitive reactance negative $Q$. Negative $Q$ does not mean a negative dissipated-energy average. For example, adding an ideal parallel capacitive branch contributing $-j300\mathrm{VA}$ cancels this example's reactive part, leaving terminal complex power400VA and RMS source current4A at100V, while the original resistive dissipation stays400W.

This one-frequency representation must not be applied indiscriminately to a distorted waveform. Different harmonics need their own phase relationships; the definition of active power remains the time average of the actual instantaneous power.