A real battery is not a perfect source of voltage. The chemistry inside it resists the flow of charge a little, so a battery behaves like an ideal one, its EMF $\varepsilon$, with a small internal resistance $r$ in series. The loop rule shows what that does.
Why it works
Walk around a loop with the battery and a load $R$: $\varepsilon - Ir - IR = 0$, so $I = \dfrac{\varepsilon}{R + r}$. The voltage you can measure across the battery's terminals is what is left after the internal drop: $V = \varepsilon - Ir$. With no current, $V = \varepsilon$. The more current you draw, the more voltage is lost inside, and the battery warms up.
Key idea
$$I = \frac{\varepsilon}{R + r}, \qquad V_{\text{terminal}} = \varepsilon - Ir = IR$$ The terminal voltage equals the EMF only when no current flows.
Worked examples
Worked example: A loaded battery
A battery with $\varepsilon = 12.0\,\text{V}$ and $r = 0.50\,\Omega$ drives a $5.5\,\Omega$ load. Find the current and the terminal voltage.
- $I = \dfrac{12.0}{5.5 + 0.50} = 2.0\,\text{A}$.
- $V = \varepsilon - Ir = 12.0 - (2.0)(0.50) = 11.0\,\text{V}$. Check: $IR = (2.0)(5.5) = 11.0\,\text{V}$.
Answer: $2.0\,\text{A}$; $11.0\,\text{V}$
Worked example: Measuring r
A battery reads $9.0\,\text{V}$ with no load and $8.4\,\text{V}$ while supplying $1.5\,\text{A}$. Find $r$.
- No current, so $\varepsilon = 9.0\,\text{V}$.
- $r = \dfrac{\varepsilon - V}{I} = \dfrac{9.0 - 8.4}{1.5} = 0.40\,\Omega$.
Answer: $0.40\,\Omega$
Worked example: Where the power goes
For the first example, find the power delivered to the load and the power lost inside the battery.
- Load: $I^2R = (2.0)^2(5.5) = 22\,\text{W}$.
- Inside: $I^2r = (2.0)^2(0.50) = 2.0\,\text{W}$, which heats the battery. Total: $\varepsilon I = 24\,\text{W}$.
Answer: $22\,\text{W}$ and $2.0\,\text{W}$
Common mistake
Thinking a battery always delivers the voltage printed on it. The label gives the EMF, the voltage with no current. Under load, part of it is lost inside: when a car's starter motor draws a huge current, the headlights dim because the terminal voltage sags. Check: is current flowing? Then $V < \varepsilon$.
The most current a battery can supply is $\varepsilon/r$, when its terminals are joined by a wire with almost no resistance. Then the terminal voltage is zero, and all of the battery's power is turned into heat inside it.