Electromotive force, internal resistance, terminal potential difference and lost volts, and finding emf and internal resistance from experimental data.

What this key area is asking

The SQA wants you to define electromotive force and internal resistance, relate the terminal potential difference to the emf through the lost volts, and find the emf and internal resistance of a source from experimental data such as a graph of terminal voltage against current.

Emf, internal resistance and lost volts

Because the source has internal resistance, not all of the emf appears across the external load. The energy per coulomb splits between the external resistance $R$ and the internal resistance $r$.

The terminal potential difference $V$ is what a voltmeter across the source actually reads while current flows. The lost volts $Ir$ is the part of the emf used up inside the source. Rearranging, $V = E - Ir$: the more current the source supplies, the larger the lost volts and the lower the terminal voltage.

Finding emf and internal resistance from data

The standard experiment varies the external resistance, records the current $I$ and the terminal voltage $V$, and plots $V$ against $I$. Since $V = E - rI$, this is a straight line:

So the $V$-axis intercept gives the emf, and the magnitude of the gradient gives the internal resistance. This graphical method is favoured because it uses many data points and averages out random error.

Examples in context

When you start a car, the starter motor draws a current of hundreds of amps; the large $Ir$ term momentarily drops the terminal voltage, which is why the headlights dim. A fresh battery has low internal resistance, so it holds its terminal voltage well under load, while an old or cold battery has high internal resistance and sags badly when asked for current. A bench power supply quoted as "12 V" delivers slightly less under heavy load for the same reason. Designers minimise internal resistance where high current is needed (for example in electric vehicle battery packs) so the lost volts and the internal heating stay small.

Try this

Q1. A cell has emf $1.5\ \text{V}$ and internal resistance $0.40\ \Omega$. It supplies a current of $0.60\ \text{A}$. Calculate the terminal potential difference. [2 marks]

• Cue. $V = E - Ir = 1.5 - 0.60 \times 0.40 = 1.26$ V.

Q2. State what the terminal voltage equals when no current is drawn from a source. [1 mark]

• Cue. It equals the emf of the source.

Q3. A graph of terminal voltage against current has a $V$-intercept of $9.0\ \text{V}$ and a gradient of $-1.5\ \text{V A}^{-1}$. State the emf and the internal resistance. [2 marks]

• Cue. $E = 9.0$ V (intercept); $r = 1.5\ \Omega$ (magnitude of gradient).