AC Power

Suppose in an AC circuits, voltage and current appear in the following sinusoidal wave forms with a certain phase difference \phi between the two.

\begin{aligned} v(t) &= V_{peak}cos( \omega t) \\ i(t) &= I_{peak}cos( \omega t – \phi )\\ \end{aligned}

Plugging these values into the electric power equation, the instantaneous power can be calculated:

\begin{aligned} p(t) &= v(t) \times i(t) \\ p(t) &= V_{peak} cos(\omega t) \times I_{peak} cos(\omega t – \phi ) \\ p(t) &= V_{peak} I_{peak} cos(\omega t) cos(\omega t – \phi) \\ \end{aligned}

Applying trig identity:

cosAcosB = \frac{1}{2}(cos(A-B)+cos(A+B))

Where:

\begin{aligned} A &= \omega t \\ B &= \omega t – \phi \end{aligned}

thus the instantaneous power can be expanded:

\begin{aligned} p(t) &= \frac{V_{peak} I_{peak}}{2} (cos\phi + cos(2\omega t – \phi)) \end{aligned}

Applying trig identity:

cos(A-B) = cosAcosB + sinAsinB

Where:

\begin{aligned} A &= 2 \omega t \\ B &= \phi \end{aligned}

thus the instantaneous power can be expanded:

\begin{aligned} p(t) &= \frac{V_{peak} I_{peak}}{2} (cos\phi + cos(2\omega t)cos\phi + sin(2\omega t)sin\phi) \\ p(t) &= \frac{V_{peak} I_{peak}}{2}cos\phi + \frac{V_{peak} I_{peak}}{2} cos(2\omega t)cos\phi + \frac{V_{peak} I_{peak}}{2} sin(2\omega t)sin\phi \\ p(t) &= \frac{V_{peak} I_{peak} cos\phi}{2} (1+cos(2\omega t)) + \frac{V_{peak} I_{peak} sin\phi}{2} (sin(2\omega t)) \\ \end{aligned}

Where:

\frac{V_{peak} I_{peak} cos\phi}{2} = Average active power

\frac{V_{peak} I_{peak} sin\phi}{2} = Average reactive power

Interactive Demo

Edit the demo below to see the affects of phase difference.