EMF and internal resistance

Real cells are not ideal

• A real battery or cell is made of materials that themselves resist the flow of charge. This is its internal resistance, $r$.
• The consequence is that a real cell behaves like an ideal source of emf in series with a small resistor.
• Internal resistance is why:
  • a battery gets warm when supplying a large current,
  • a car's headlights dim while the starter motor is turning,
  • an old battery struggles to deliver current even though its voltage reads normally when nothing is connected.

EMF and terminal voltage

EMF ($\varepsilon$) — the electromotive force, the energy given to each coulomb of charge by the source.

• Measured in volts, despite the name — it is not a force.
• It is the voltage the cell would supply if it had no internal resistance.

Terminal voltage ($V$) — the voltage actually available across the cell's terminals, and therefore across the external circuit.

$$\varepsilon = V + Ir \qquad \text{or} \qquad V = \varepsilon - Ir$$

• $\varepsilon$ — emf of the source (V)

• $V$ — terminal voltage (V)

• $I$ — current drawn from the cell (A)

• $r$ — internal resistance (Ω)

• The term $Ir$ is the "lost volts" — the energy per coulomb dissipated inside the cell rather than delivered to the circuit.

What this predicts

• No current drawn ($I = 0$): $V = \varepsilon$. A high-resistance voltmeter across an unconnected cell reads the emf directly.

• Larger current drawn: more lost volts, so the terminal voltage falls. This is why the terminal voltage is not a fixed property of a cell.

• Short circuit ($R \to 0$): the current is limited only by the internal resistance, $I = \varepsilon / r$ — often dangerously large, and all the energy is dissipated inside the cell.

• With an external resistance $R$, the whole circuit obeys:

$$\varepsilon = I(R + r) \qquad \Rightarrow \qquad I = \frac{\varepsilon}{R + r}$$

Finding $\varepsilon$ and $r$ from a graph

This is a standard experiment and a standard exam question:

• Vary the external resistance and record the terminal voltage $V$ and the current $I$ at each setting.
• Plot $V$ (vertical) against $I$ (horizontal).
• Rearranged, $V = -rI + \varepsilon$ has the form $y = mx + c$, so:
  • the $y$-intercept is the emf $\varepsilon$,
  • the gradient is $-r$, so the internal resistance is the magnitude of the gradient.
• The graph is a straight line sloping downward — that downward slope is the visible signature of internal resistance.