Math Scrap 2

Consider an arbitrary load (potentially both resistive and reactive) where:

\begin{gather*} v(t) = V_m \cos{(\omega t + \theta_v)} \\ i(t) = I_m \cos{(\omega t + \theta_i)} \end{gather*}

Instantaneous power is given by:

p(t) = v(t)\ i(t) = V_m I_m \cos{(\omega t + \theta_v)} \cos{(\omega t + \theta_i)}

Find the average power delivered to this load.

We can apply the following product-to-sum identity:

\begin{gather*} \boxed{ 2 \cos{A} \cos{B} = \cos{(A+B)} + \cos{(A-B)} } \\ p(t) = \frac{V_m I_m}{2} \brackets{\cos{(2 \omega t + \theta_v + \theta_i)} + \cos{(\theta_v - \theta_i)}} \end{gather*}

To make things simpler, let’s define the power angle $\theta$ :

\theta := \theta_v - \theta_i \qquad \Longrightarrow \qquad \theta_i = \theta_v - \theta

Substituting into $p(t)$ then applying the following compound angle identity:

\begin{gather*} \boxed{ \cos{(\alpha - \beta)} = \cos{\alpha} \cos{\beta} + \sin{\alpha} \sin{\beta} } \\ \begin{aligned} p(t) &= \frac{V_m I_m}{2} \brackets{\cos{\parens{[2 \omega t + 2 \theta_v] - \theta}} + \cos{(\theta)}} \\ &= \frac{V_m I_m}{2} \brackets{\cos{(2 \omega t + 2 \theta_v)} \cos{(\theta)} + \sin{(2 \omega t + 2 \theta_v)} \sin{(\theta)} + \cos{(\theta)}} \\ &= \frac{V_m I_m}{2} \cos{(\theta)} \brackets{\underbrace{\cos{(2 \omega t + 2 \theta_v)}}_{\text{average value} = 0} + 1} + \frac{V_m I_m}{2} \sin{(\theta)} \underbrace{\sin{(2 \omega t + 2 \theta_v)}}_{\text{average value} = 0} \end{aligned} \end{gather*}

To find average power, we notice that two time-dependent sinusoids have average values of zero, therefore our average power $P$ is:

P = \frac{V_m I_m}{2} \cos{(\theta)}