\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.}
