\subsection{Internal Resistance and Measurement Devices}

This subsection introduces the real-battery model (ideal emf in series with an internal resistance), derives the terminal-voltage relation under load, and discusses how practical measurement devices (ammeters and voltmeters) affect the circuits they measure.

\dfn{Real battery (emf and internal resistance)}{A \emph{real battery} is modelled as an ideal electromotive-force source $\mathcal{E}$ in series with an \emph{internal resistance} $r$. The emf $\mathcal{E}$ represents the work per unit charge the battery can deliver when no current flows. The internal resistance $r$ accounts for the finite conductivity of the electrolyte, electrode materials, and other dissipative processes inside the battery. The SI unit of emf is the volt (V), which is dimensionally equivalent to $\mathrm{J/C}$.}

\thm{Terminal voltage of a real battery}{Let a real battery with emf $\mathcal{E}$ and internal resistance $r$ deliver a current $I$ to an external load. The \emph{terminal voltage} $V$ across the battery's terminals is
\[
V = \mathcal{E} - Ir.
\]
When the battery is delivering current (discharging), the terminal voltage is \emph{less} than the emf by the voltage drop $Ir$ across the internal resistance. When $I = 0$ (open circuit), the terminal voltage equals the emf: $V = \mathcal{E}$. If the current is driven backwards through the battery (charging), then $I$ is negative and $V > \mathcal{E}$.}

\pf{Derivation of the terminal-voltage relation}{Consider a real battery connected to an external load resistor $R$. The equivalent circuit consists of the emf $\mathcal{E}$, the internal resistance $r$, and the load $R$ all in series. By Kirchhoff's loop rule, traversing the loop in the direction of current $I$:
\[
\mathcal{E} - Ir - IR = 0.
\]
Solving for the current gives
\[
I = \frac{\mathcal{E}}{r + R}.
\]
The terminal voltage $V$ is the potential drop across the load, which also equals the potential drop from the battery's positive to negative terminal:
\[
V = IR = \mathcal{E} - Ir.
\]
This proves the relation $V = \mathcal{E} - Ir$.}

\thm{Power delivered by a real battery}{The power $P$ delivered to an external load $R$ by a real battery of emf $\mathcal{E}$ and internal resistance $r$ is
\[
P = I^2 R = \mathcal{E}\,I - I^2 r,
\]
where $I = \mathcal{E}/(r + R)$. The first term $\mathcal{E}I$ is the total rate at which the battery converts chemical energy to electrical energy; the second term $I^2 r$ is the rate of internal dissipation as heat within the battery. The difference is the power delivered to the external circuit.}

\thm{Power transfer theorem (maximum power transfer)}{For a real battery with fixed $\mathcal{E}$ and $r$ connected to a variable load $R$, the power delivered to the load is
\[
P(R) = \frac{\mathcal{E}^2\,R}{(r + R)^2}.
\]
This power is maximized when $R = r$, giving $P_{\max} = \mathcal{E}^2 / (4r)$.}

\pf{Maximum power transfer}{From the preceding theorem, $P(R) = I^2 R$ with $I = \mathcal{E}/(r + R)$, so
\[
P(R) = \frac{\mathcal{E}^2 R}{(r + R)^2}.
\]
To find the maximum, differentiate with respect to $R$ and set the derivative to zero:
\[
\frac{dP}{dR} = \mathcal{E}^2 \,\frac{(r + R)^2 - R \cdot 2(r + R)}{(r + R)^4}
= \mathcal{E}^2 \,\frac{(r + R) - 2R}{(r + R)^3}
= \mathcal{E}^2 \,\frac{r - R}{(r + R)^3}.
\]
Setting $dP/dR = 0$ gives $R = r$. For $R < r$ the derivative is positive (power increases); for $R > r$ it is negative (power decreases). Hence $R = r$ is a maximum. Substituting $R = r$ into the power expression gives $P_{\max} = \mathcal{E}^2/(4r)$.}

\cor{Short circuit and open circuit limits}{When the load is zero ($R = 0$, \emph{short circuit}), the current is $I_{\text{sc}} = \mathcal{E}/r$ and the terminal voltage is $V = 0$. All power is dissipated internally: $P_{\text{load}} = 0$, and the battery heats up. When the load is infinite ($R \to \infty$, \emph{open circuit}), the current is $I = 0$ and $V = \mathcal{E}$. No power is delivered.}

\dfn{Ammeter}{An \emph{ammeter} measures the current through a branch of a circuit. It is inserted in \emph{series} with the branch. An ideal ammeter has zero resistance so it does not affect the circuit. A real ammeter has a small but finite resistance $R_A$ (the \emph{ammeter resistance}).}

\dfn{Voltmeter}{A \emph{voltmeter} measures the potential difference between two points in a circuit. It is connected in \emph{parallel} between those points. An ideal voltmeter has infinite resistance so no current flows through it. A real voltmeter has a large but finite resistance $R_V$ (the \emph{voltmeter resistance}).}

\thm{Loading effects of measurement devices}{When an ammeter of resistance $R_A$ is inserted in series with a circuit of total resistance $R_{\text{eq}}$ (not including $R_A$), the current measured is
\[
I_{\text{measured}} = \frac{I_{\text{true}}\,R_{\text{eq}}}{R_{\text{eq}} + R_A},
\]
where $I_{\text{true}}$ is the current that would flow without the ammeter. The reading is smaller than the true value by a factor of $R_{\text{eq}}/(R_{\text{eq}} + R_A)$.

When a voltmeter of resistance $R_V$ is connected across a component of resistance $R$ that has voltage $V$ across it (without the voltmeter), the measured voltage is
\[
V_{\text{measured}} = V \,\frac{R\,R_V/(R + R_V)}{R\,R_V/(R + R_V) + R_{\text{series}}}
= V \,\frac{R_V}{R_V + R_{\text{series}}\,(1 + R_V/R)}^{-1},
\]
where $R_{\text{series}}$ is the resistance in series with the component. In the common case where the component of interest is connected to a source with small internal resistance $r \ll R_V$, the correction is small: $V_{\text{measured}} \approx V\,(1 - r/R_V)$. The reading is smaller than the true voltage.}

\thm{Efficiency of power transfer}{The \emph{efficiency} $\eta$ of a real battery delivering power to a load $R$ is the ratio of power delivered to the load to the total power generated by the emf:
\[
\eta = \frac{P_{\text{load}}}{P_{\text{total}}}
= \frac{I^2 R}{\mathcal{E}I}
= \frac{IR}{\mathcal{E}}
= \frac{R}{R + r}.
\]
Efficiency increases as $R$ becomes large compared to $r$. When $R = r$, the efficiency is $50\%$: half the power is dissipated in the internal resistance.}

\nt{An ideal ammeter ($R_A = 0$) does not change the current in the branch it measures. A real ammeter always slightly \emph{reduces} the current. An ideal voltmeter ($R_V = \infty$) draws no current and does not disturb the circuit. A real voltmeter always slightly \emph{lowers} the voltage it is measuring because it provides an additional parallel current path. In well-designed circuits, $R_A \ll R_{\text{branch}}$ and $R_V \gg R_{\text{parallel}}$, so these perturbations are negligible.}

\ex{Illustrative example}{A $9\,\mathrm{V}$ battery with internal resistance $r = 1\,\Omega$ is connected to a load $R = 8\,\Omega$. The current is $I = 9\,\mathrm{V}/(1\,\Omega + 8\,\Omega) = 1.0\,\mathrm{A}$, and the terminal voltage is $V = 9\,\mathrm{V} - (1.0\,\mathrm{A})(1\,\Omega) = 8\,\mathrm{V}$.}

\qs{Worked example}{A battery has an emf
\[
\mathcal{E} = 12.0\,\mathrm{V}
\]
and an internal resistance
\[
r = 2.0\,\Omega.
\]
The battery is connected to an external load resistor $R = 10.0\,\Omega$, as shown in the circuit diagram below.

\begin{center}
\emph{(Diagram description: A single loop consisting of an ideal emf source $\mathcal{E}$, an internal resistor $r$ in series, and an external load resistor $R$ in series, forming a closed circuit.)}
\end{center}

Find:
\begin{enumerate}[label=(\alph*)]
\item the current $I$ in the circuit,
\item the terminal voltage $V$ across the battery,
\item the power $P_{\text{load}}$ delivered to the load resistor,
\item the power $P_{\text{int}}$ dissipated in the internal resistance,
\item the total power $P_{\text{total}}$ generated by the emf,
\item the efficiency $\eta$ of the battery, and
\item the value of $R$ that maximizes the power delivered to the load.
\end{enumerate}}

\sol \textbf{Part (a).} The circuit consists of $\mathcal{E}$, $r$, and $R$ in series. By Kirchhoff's loop rule,
\[
\mathcal{E} - Ir - IR = 0,
\]
so
\[
I = \frac{\mathcal{E}}{r + R}
= \frac{12.0\,\mathrm{V}}{2.0\,\Omega + 10.0\,\Omega}
= \frac{12.0\,\mathrm{V}}{12.0\,\Omega}
= 1.0\,\mathrm{A}.
\]

\textbf{Part (b).} The terminal voltage is
\[
V = \mathcal{E} - Ir
= 12.0\,\mathrm{V} - (1.0\,\mathrm{A})(2.0\,\Omega)
= 12.0\,\mathrm{V} - 2.0\,\mathrm{V}
= 10.0\,\mathrm{V}.
\]
Equivalently, $V = IR = (1.0\,\mathrm{A})(10.0\,\Omega) = 10.0\,\mathrm{V}$, which confirms the result.

\textbf{Part (c).} The power delivered to the load resistor is
\[
P_{\text{load}} = I^2 R
= (1.0\,\mathrm{A})^2(10.0\,\Omega)
= 10.0\,\mathrm{W}.
\]
Alternatively, $P_{\text{load}} = VI = (10.0\,\mathrm{V})(1.0\,\mathrm{A}) = 10.0\,\mathrm{W}$, giving the same result.

\textbf{Part (d).} The power dissipated in the internal resistance is
\[
P_{\text{int}} = I^2 r
= (1.0\,\mathrm{A})^2(2.0\,\Omega)
= 2.0\,\mathrm{W}.
\]

\textbf{Part (e).} The total power generated by the emf is
\[
P_{\text{total}} = \mathcal{E} I
= (12.0\,\mathrm{V})(1.0\,\mathrm{A})
= 12.0\,\mathrm{W}.
\]
Check: $P_{\text{total}} = P_{\text{load}} + P_{\text{int}} = 10.0\,\mathrm{W} + 2.0\,\mathrm{W} = 12.0\,\mathrm{W}$, so energy is conserved.

\textbf{Part (f).} The efficiency of the battery is
\[
\eta = \frac{P_{\text{load}}}{P_{\text{total}}}
= \frac{10.0\,\mathrm{W}}{12.0\,\mathrm{W}}
= 0.833
= 83.3\%.
\]
Equivalently, $\eta = R/(R + r) = 10.0\,\Omega/(10.0\,\Omega + 2.0\,\Omega) = 10/12 = 0.833 = 83.3\%$.

\textbf{Part (g).} By the maximum power transfer theorem, the power delivered to the load is maximized when the load resistance equals the internal resistance:
\[
R = r = 2.0\,\Omega.
\]
At this value, the maximum power delivered to the load is
\[
P_{\max} = \frac{\mathcal{E}^2}{4r}
= \frac{(12.0\,\mathrm{V})^2}{4(2.0\,\Omega)}
= \frac{144\,\mathrm{V}^2}{8.0\,\Omega}
= 18.0\,\mathrm{W}.
\]

Therefore,
\[
I = 1.0\,\mathrm{A},
\qquad
V = 10.0\,\mathrm{V},
\qquad
P_{\text{load}} = 10.0\,\mathrm{W},
\]
\[
P_{\text{int}} = 2.0\,\mathrm{W},
\qquad
P_{\text{total}} = 12.0\,\mathrm{W},
\qquad
\eta = 83.3\%,
\qquad
R_{\text{max power}} = 2.0\,\Omega.
\]
