 AC power

AC power describes energy transfer associated with time-varying electric potential difference and electric current, including the distinction between instantaneous transfer and its average. With the passive sign convention, current enters the terminal marked positive, and instantaneous power absorbed by the element is $p(t)=v(t)i(t)$. A positive value means energy entering the element; a negative value means energy returning through that terminal pair.

For sinusoidal steady operation, choosewhere $V$ and $I$ are root mean square magnitudes. The cosine-product identity givesAveraging over a complete period removes the second term, giving active power $P=VI\cos \phi$. This derivation separates the energy-transfer average from a twice-frequency oscillation.^[ref]

For an original example, $V=120\mathrm{V}$, $I=5\mathrm{A}$ and $\cos \phi =0.8$ give active power480W. The oscillating term has amplitude600W, so instantaneous power ranges from−120 to1080W. Over ten complete cycles at50Hz, the net energy absorbed is $480(10/50)=96\mathrm{J}$. Temporary negative power can accompany positive net transfer when an element stores and returns energy.

The product $VI=600$ in volt-amperes is the apparent power. In the single-frequency sinusoidal convention, reactive power is $Q=VI\sin \phi$, giving360 in reactive volt-amperes for a lagging current with $\sin \phi =0.6$. Complex electrical power packages $P$ and $Q$, while the power factor $P/(VI)=0.8$ distinguishes net transfer from the terminal magnitude product. The primary MIT treatment derives these quantities and makes the peak-versus-RMS convention explicit.^[ref]

A measurement must retain the relative phase of the two channels. Multiplying separately measured RMS magnitudes gives apparent power, not generally active power. For nonsinusoidal signals, the time average of $v(t)i(t)$ remains the definition of active power, but one phase angle need not describe every harmonic contribution.