ExampleOfElectricalPower 2026-02-13 23:54:41 2026-02-13 23:54:41 Example power The line \= \frac{1}{T} \int\_0^T ... contains the command \= in math mode. % this is the default PhysicsLibrary preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them \usepackage{geometry} \usepackage{hyperref} % define commands here \section{Definition of Electrical Power} Electrical power is the rate at which electrical energy is delivered or consumed in a circuit. If a device has an electrical potential difference denoted by $V$ and current denoted by $I$, then the instantaneous electrical power $P(t)$ is defined as \begin{equation} P(t) = V(t) I(t). \end{equation} The SI unit of $P$ is the watt, denoted $\mathrm{W}$, where one watt equals one joule per second. \section{Ohms Law and Resistive Load} Consider a simple series circuit consisting of a voltage source $V(t)$ and a resistor $R$. For a purely resistive load, the current and voltage are in phase and Ohms law applies at every instant: \begin{equation} I(t) = \frac{V(t)}{R}. \end{equation} We will assume that the source voltage is sinusoidal of angular frequency $\omega$ and amplitude $V_0$, so that \begin{equation} V(t) = V_0 \sin(\omega t). \end{equation} Substituting into Ohms law gives the instantaneous current: \begin{equation} I(t) = \frac{V_0}{R} \sin(\omega t). \end{equation} \section{Instantaneous Power in a Resistor} Using the definitions above, the instantaneous power delivered to the resistor is \begin{align} P(t) &= V(t) I(t) \\ &= V_0 \sin(\omega t) \left(\frac{V_0}{R} \sin(\omega t)\right) \\ &= \frac{V_0^2}{R} \sin^2(\omega t). \end{align} This equation shows that the instantaneous power oscillates at twice the fundamental frequency. The power is always nonnegative for a resistive load, consistent with the fact that a resistor only consumes energy, it does not return energy to the source. \section{Average Power} The average power $\langle P \rangle$ over one full cycle is obtained by integrating $P(t)$ with respect to time over the period $T = 2\pi/\omega$ and dividing by the period: \begin{equation} \langle P \rangle = \frac{1}{T} \int_0^T P(t) \, dt. \end{equation} Substituting $P(t)$ yields \begin{equation} \langle P \rangle = \frac{1}{T} \int_0^T \frac{V_0^2}{R} \sin^2(\omega t) \, dt. \end{equation} Using the trigonometric identity $\sin^2(\theta) = \frac{1}{2} (1 - \cos(2\theta))$, we find \begin{align} \langle P \rangle &= \frac{V_0^2}{R} \frac{1}{T} \int_0^T \frac{1}{2} \left( 1 - \cos(2\omega t)\right) dt \\ &= \frac{V_0^2}{2R}. \end{align} The cosine term averages to zero over a full cycle, leaving one half of the peak squared divided by the resistance. \section{Root Mean Square Values} It is common to express average power in terms of root mean square (rms) values. Define the rms voltage $V_\text{rms}$ and rms current $I_\text{rms}$ as \begin{align} V_{rms} &= \frac{V_0}{\sqrt{2}}, \\ I_{rms} &= \frac{V_{rms}}{R} = \frac{V_0}{R\sqrt{2}}. \end{align} Using these definitions, the average power becomes \begin{equation} \langle P \rangle = V_{rms} I_{rms}, \end{equation} which is the standard form used in circuit analysis. \section{Interpretation and Units} In our example, the voltage and current are in phase, so the average power represents real energy delivered to the resistor each second. The average power in watts equals the rate of conversion of electrical energy into heat in the resistor. \section{Conclusion} We computed the instantaneous and average electrical power delivered to a resistive element in response to a sinusoidal source. The average power is one half the peak voltage squared divided by the resistance and may be expressed as the product of rms voltage and rms current. These results are standard in electrical engineering and physics and provide a basis for more complex analysis of circuits with time varying sources. \begin{thebibliography}{9} \bibitem{alexander} C. K. Alexander and M. N. O. Sadiku, \textit{Fundamentals of Electric Circuits}, 6th edition, McGraw Hill, 2016. \bibitem{nilsson} J. W. Nilsson and S. A. Riedel, \textit{Electric Circuits}, 10th edition, Pearson, 2019. \bibitem{boylestad} R. L. Boylestad, \textit{Introductory Circuit Analysis}, 14th edition, Pearson, 2018. \end{thebibliography}