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concepts/em/u11/e11-3-power.tex

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+\subsection{Electric Power and Dissipation}
+
+Electric power quantifies the rate at which electrical energy is transferred or dissipated in a circuit element. When a charge $q$ moves through a potential difference $\Delta V$, its electric potential energy changes by $\Delta U = q\,\Delta V$. The power delivered to (or dissipated by) a circuit element is the rate of this energy transfer.
+
+\dfn{Electric power}{Let a circuit element have a potential difference $\Delta V$ across it and carry a current $I$ through it. The \emph{electric power} delivered to the element is
+\[
+P = I\,\Delta V.
+\]
+The SI unit of power is the watt ($1\,\mathrm{W}=1\,\mathrm{J/s}=1\,\mathrm{A\!\cdot\!V}$). When the element is a resistor, electrical energy is dissipated as thermal energy (Joule heating).}
+
+\nt{Power is the rate of energy transfer. A battery \emph{supplies} power to the circuit, while resistors \emph{dissipate} it. In a steady circuit, the total power supplied equals the total power dissipated.}
+
+\ex{Illustrative example}{A $60\,\mathrm{W}$ incandescent lightbulb is connected to a $120\,\mathrm{V}$ household outlet. Find (a) the current through the bulb and (b) the resistance of its filament.
+
+\sol (a) From $P = I\,\Delta V$,
+\[
+I = \frac{P}{\Delta V} = \frac{60\,\mathrm{W}}{120\,\mathrm{V}} = 0.50\,\mathrm{A}.
+\]
+
+(b) From Ohm's law,
+\[
+R = \frac{\Delta V}{I} = \frac{120\,\mathrm{V}}{0.50\,\mathrm{A}} = 240\,\Omega. \qedhere
+\]}
+
+\nt{The $I^{2}R$ form shows why transmission lines use very high voltages: for a fixed power $P=IV$, raising the voltage lowers the current, and since resistive losses scale as $I^{2}R$, the dissipation drops dramatically.}
+
+\thm{Microscopic power density}{In a continuous medium, the local rate of energy dissipation per unit volume is the dot product of the current density $\vec{J}$ and the electric field $\vec{E}$ at that point:
+\[
+p = \vec{J}\cdot\vec{E}.
+\]
+The total power delivered to a volume $\mathcal{V}$ is the integral
+\[
+P = \iiint_{\mathcal{V}} \vec{J}\cdot\vec{E}\; d\tau.
+\]}
+
+\pf{Microscopic to macroscopic}{Consider a straight wire of length $L$ and cross-sectional area $A$, with a uniform electric field $\vec{E}$ along its axis and a uniform current $I$. The field is related to the potential difference by $\Delta V = E\,L$. The current density is $J = I/A$, and by Ohm's law $E = \rho\,J$ (where $\rho$ is the resistivity). The total power dissipated is
+\[
+P = \iiint_{\mathcal{V}} \vec{J}\cdot\vec{E}\; d\tau = J\,E\,(A\,L)
+\]
+since $\vec{J}$ and $\vec{E}$ are parallel and uniform. Substituting $J = I/A$ and $E = \Delta V/L$,
+\[
+P = \left(\frac{I}{A}\right)\!\left(\frac{\Delta V}{L}\right)(A\,L) = I\,\Delta V,
+\]
+recovering the macroscopic expression. Alternatively, using $E = \rho\,J = \rho\,(I/A)$ and $R = \rho\,L/A$,
+\[
+P = J\,E\,(A\,L) = \frac{I^{2}}{A^{2}}\cdot\rho\frac{I}{A}\cdot A\,L
+= I^{2}\,\frac{\rho\,L}{A} = I^{2}R. \qedhere
+\]
+
+\qs{Worked example}{A resistor is connected across a $12.0\,\mathrm{V}$ battery. The current through the resistor is measured to be $2.00\,\mathrm{A}$.
+
+Find:
+\begin{enumerate}[label=(\alph*)]
+\item the power dissipated by the resistor,
+\item the resistance of the resistor, and
+\item the total energy dissipated as heat over a time interval of $5.00\,\mathrm{min}$.
+\end{enumerate}}
+
+\sol \textbf{Part (a).} The power dissipated by the resistor is given by
+\[
+P = I\,\Delta V.
+\]
+Substitute the given values:
+\[
+P = (2.00\,\mathrm{A})(12.0\,\mathrm{V}) = 24.0\,\mathrm{W}.
+\]
+
+\textbf{Part (b).} By Ohm's law,
+\[
+R = \frac{\Delta V}{I}.
+\]
+Substitute the values:
+\[
+R = \frac{12.0\,\mathrm{V}}{2.00\,\mathrm{A}} = 6.00\,\Omega.
+\]
+As a check, verify using $P = I^{2}R$:
+\[
+P = (2.00\,\mathrm{A})^{2}(6.00\,\Omega) = 4.00 \times 6.00 = 24.0\,\mathrm{W},
+\]
+which agrees with part (a).}
+
+\textbf{Part (c).} The total energy dissipated is power multiplied by time:
+\[
+E = P\,t.
+\]
+Convert the time to seconds:
+\[
+t = 5.00\,\mathrm{min}\times 60.0\,\mathrm{s/min} = 300\,\mathrm{s}.
+\]
+Then
+\[
+E = (24.0\,\mathrm{W})(300\,\mathrm{s}) = 7200\,\mathrm{J}.
+\]
+Alternatively, using $E = I\,\Delta V\,t$ directly:
+\[
+E = (2.00\,\mathrm{A})(12.0\,\mathrm{V})(300\,\mathrm{s}) = 7200\,\mathrm{J}.
+\]
+
+\nt{Energy can also be written as $E = I^{2}Rt$ or $E = (\Delta V)^{2}t/R$. All three are equivalent and follow from $E = P\,t$ together with Ohm's law.}