Internal Resistance and EMF

Ideal batteries print their label on the side and deliver it no matter what you connect. Real batteries sag — connect a heavy load and the terminal voltage drops below the number on the packaging. The gap is explained by one extra element no data sheet hides but every textbook at this level conveniently ignores: a small resistor inside the battery called the internal resistance, $r$.

The two-parameter battery

A real battery has two parameters: its electromotive force $\mathcal{E}$ (open-circuit voltage — what a voltmeter reads when nothing is connected) and its internal resistance $r$. Model it as an ideal EMF source in series with a resistor. All the current that flows out of the battery passes through $r$ on the way, dropping a little voltage inside.

Terminal voltage under load

Apply KVL to the loop formed by the battery driving a single external load $R$. Walk from the − terminal through the EMF, through the internal resistance, through the load, and back:

The voltage you would measure at the terminals — the terminal voltage — is the external drop, $IR$:

Three cases walk through how this behaves:

• Open circuit: $R \to \infty$, $I = 0$, $V_{term} = \mathcal{E}$. The voltmeter reads the nameplate value.
• Light load: $R \gg r$, so $Ir$ is a small fraction of $\mathcal{E}$ and $V_{term} \approx \mathcal{E}$.
• Heavy load: $R \sim r$, $Ir$ eats a noticeable chunk of the EMF, and $V_{term}$ sags. This is why a car battery's voltage dips when you crank the starter — the starter motor is a heavy load and $r$ is suddenly visible.

Short-circuit current

Set $R = 0$ (a dead short across the battery) and the only resistance left in the loop is $r$:

A 9 V battery with $r = 1\,\Omega$ can in theory push 9 A into a short; a high-current battery with $r = 0.01\,\Omega$ would push 900 A. The short-circuit current is one of the quickest ways to estimate the internal resistance of a cell — but don't do it casually, since the power dissipated inside the battery is $I_{short}^2 \cdot r$, and the cell can get hot enough to vent or rupture.

Maximum power transfer

A surprising consequence: the load that extracts the most power from a given battery isn't the smallest possible resistance (which shorts it) or the largest (which draws no current). It's the one that matches the internal resistance: $R = r$. At that match, the battery delivers half its stored power to the load and dissipates the other half in $r$ — so this is the least efficient operating point, but the absolute-power champion. Real power systems run far from this match (load much larger than source resistance) for efficiency; signal systems sometimes match deliberately for maximum signal transfer.

Measuring r in practice

The classic lab procedure — two data points, one equation. Measure the terminal voltage at two different loads, $(V_1, I_1)$ and $(V_2, I_2)$. Since $V = \mathcal{E} - Ir$, the slope of the terminal-voltage vs. current line is $-r$, and the intercept is $\mathcal{E}$:

For a coin cell the result is usually a handful of ohms; for a good lithium-ion cell, milliohms; for a lead-acid car battery, tens of milliohms. Batteries age and $r$ climbs — a tired 9 V cell that still reads 8.5 V at open circuit can still be useless under load because $r$ has crept up into the tens of ohms.