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]. Substituting $\theta = 0$ into Eq. (1), we obtain 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 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 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, 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: where P is the average value and t is the period of the applied voltage; that is, or, since $T_1 = 1/f_1$,