Resistive (AC) Circuit Power Calculation

For a purely resistive circuit (such as that in [Fig. 1]), v and i are in phase, and $\theta = 0$, as appearing in [Fig. 2].
Fig. 1: Determining the power delivered to a purely resistive load.
Fig. 2: Power versus time for a purely resistive load.
Substituting $\theta = 0$ into Eq. (1),
$$p = VI(1 - \cos 2wt) \cos \theta + VI \sin \theta (\sin 2wt) \tag{1}$$
we obtain
$$ \begin{split} p_R &= VI(1 - \cos 2wt) \cos (0^\circ)+ VI \sin (0^\circ) (\sin 2wt)\\ &=VI(1 - \cos 2wt) + 0\\ \end{split} $$
$$\bbox[10px,border:1px solid grey]{p_R = VI - VI \cos 2wt)} \tag{2}$$
where $VI$ is the average or dc term and $VI \cos 2wt$ is a negative cosine wave with twice the frequency of either input quantity (v or i) and a peak value of $VI$.
Plotting the waveform for $p_R$ [Fig. 2], we see that
$$T_1 = \text{period of input quantities} $$
$$T_2 = \text{period off power curve p_R} $$
Note that in [Fig. 2] the power curve passes through two cycles about its average value of $VI$ for each cycle of either v or i ($T_1 = 2T_2$ or $f_2 = 2f_1$). Consider also that since the peak and average values of the power curve are the same, the curve is always above the horizontal axis. This indicates that
the total power delivered to a resistor will be dissipated in the form of heat.
The power returned to the source is represented by the portion of the curve below the axis, which is zero in this case. The power dissipated by the resistor at any instant of time $t_1$ can be found by simply substituting the time $t_1$ into Eq. (2) to find $p_1$, as indicated in [Fig. 2]. The average (real) power from Eq. (2), or [Fig. 2], is VI; or, as a summary,
$$ \bbox[10px,border:1px solid grey]{P = VI = {V_m I_m \over 2} = I^2R = {V^2 \over R}}$$
The energy dissipated by the resistor ($W_R$) over one full cycle of the applied voltage [Fig. 2] can be found using the following equation:
$$W = Pt$$
where P is the average value and t is the period of the applied voltage; that is,
$$\bbox[10px,border:1px solid grey]{W_R = VI T_1} \text{(joules, J)} $$
or, since $T_1 = 1/f_1$,
$$\bbox[10px,border:1px solid grey]{W_R = {VI \over f_1}} \text{(joules, J)} $$

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