Single-Phase AC Power

Single-phase AC power

Instantaneous Power

Theory behind this

Let the instantaneous voltage be:

v(t)=Vmcos⁡(ωt+θv)(1)\begin{equation} v(t) = V_m \cos(\omega t + \theta_v) \tag{1} \end{equation}

The instantaneous current is:

i(t)=Imcos⁡(ωt+θi)(2)\begin{equation} i(t) = I_m \cos(\omega t + \theta_i) \tag{2} \end{equation}

The instantaneous power is:

p(t)=v(t)i(t)=VmImcos⁡(ωt+θv)cos⁡(ωt+θi)\begin{equation} p(t) = v(t)i(t) = V_m I_m \cos(\omega t + \theta_v)\cos(\omega t + \theta_i) \end{equation}

Using the identity:

cos⁡Acos⁡B=12cos⁡(A−B)+12cos⁡(A+B)\begin{equation} \cos A \cos B = \frac{1}{2}\cos(A – B) + \frac{1}{2}\cos(A + B) \end{equation}

we get:

p(t)=12VmIm[cos⁡(θv−θi)+cos⁡(2ωt+θv+θi)]\begin{equation} p(t) = \frac{1}{2} V_m I_m \left[\cos(\theta_v – \theta_i) + \cos(2\omega t + \theta_v + \theta_i)\right] \end{equation}
cos⁡(2ωt+θv+θi)=cos⁡[2(ωt+θv)−(θv−θi)]\begin{equation} \cos(2\omega t + \theta_v + \theta_i) = \cos\left[2(\omega t + \theta_v) – (\theta_v – \theta_i)\right] \end{equation}
cos⁡(A−B)=cos⁡Acos⁡B+sin⁡Asin⁡B\begin{equation} \cos(A – B) = \cos A \cos B + \sin A \sin B \end{equation}
cos⁡[2(ωt+θv)−(θv−θi)]=cos⁡2(ωt+θv)cos⁡(θv−θi)+sin⁡2(ωt+θv)sin⁡(θv−θi)\begin{equation} \cos\left[2(\omega t + \theta_v) – (\theta_v – \theta_i)\right] = \cos 2(\omega t + \theta_v)\cos(\theta_v – \theta_i) + \sin 2(\omega t + \theta_v)\sin(\theta_v – \theta_i) \end{equation}

Substituting Back

p(t)=12VmIm[cos⁡(θv−θi)+cos⁡2(ωt+θv)cos⁡(θv−θi)+sin⁡2(ωt+θv)sin⁡(θv−θi)]\begin{align} p(t) = \frac{1}{2} V_m I_m \Big[ &\cos(\theta_v – \theta_i) \\ &+ \cos 2(\omega t + \theta_v)\cos(\theta_v – \theta_i) \\ &+ \sin 2(\omega t + \theta_v)\sin(\theta_v – \theta_i) \Big] \end{align}

The voltage is and varies with time.

The voltage is V(t)=sin⁡(ωt)V(t) = \sin(\omega t) and varies with time.

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