Electromotive Force (EMF) and Terminal Potential Difference Revision Notes | A Level AQA Physics

1. Why Batteries Are Not Perfect

You have probably noticed that a torch dims as its batteries age, or that a car battery struggles on a cold morning even though it still has charge. The reason is not that the battery is running out — it is that every real source of electrical energy has resistance inside itself. That internal resistance is the hidden player in every circuit, and understanding it is the key to this topic.

At GCSE level, batteries were treated as ideal — a fixed voltage source with nothing else to worry about. At A Level, that simplification is removed. You need to account for the energy lost inside the source, and that changes every calculation you do with real circuits.

2. Electromotive Force — What EMF Actually Means

The term electromotive force is misleading — it is not a force at all. EMF is a measure of energy. Specifically, it is the energy transferred per unit charge by the source as charge passes through it. The symbol is $\varepsilon$ (the Greek letter epsilon).

Think of EMF as the source's promise: for every coulomb of charge that passes through it, the source will transfer this many joules of energy. A 9 V battery promises to transfer 9 J for every coulomb. The question is — where does that energy actually go?

This equation is clean and direct. If a battery transfers 45 J of energy when 5 C of charge flows through it, its EMF is $\varepsilon = 45/5 = 9$ V. The equation also tells you that EMF is measured in volts — the same unit as potential difference — because both are energy per unit charge.

3. Internal Resistance and the Full EMF Equation

Now here is where the real circuit behaviour emerges. Every source — a battery, a power supply, even a solar cell — has some resistance within its own structure. This is the internal resistance, $r$. When current flows, some energy is dissipated across $r$ before it even reaches the external circuit.

Picture the circuit as two resistors in series: the external resistance $R$ (the load — a lamp, a motor, whatever is connected) and the internal resistance $r$ inside the source. The same current $I$ flows through both. The EMF drives current through the total resistance $(R + r)$.

This equation can be expanded and rearranged in a way that is extremely useful. Expanding: $\varepsilon = IR + Ir$. The term $IR$ is the potential difference across the external resistance — this is the terminal potential difference, $V$. The term $Ir$ is the potential difference dropped across the internal resistance — these are the lost volts. So the full picture is:

This is the central relationship of the topic. The terminal pd is always less than the EMF when current flows, because some energy is always lost inside the source. The bigger the current, the greater the lost volts $Ir$, and the lower the terminal pd.

4. Terminal Potential Difference — The Voltage You Can Actually Use

The terminal potential difference is the pd measured directly across the terminals of the source while current is flowing. It is the voltage that the external circuit actually receives. A voltmeter placed across the battery terminals reads the terminal pd, not the EMF.

There is one special case worth knowing: when the circuit is open (no current flows, $I = 0$), the lost volts $Ir = 0$, so the terminal pd equals the EMF. This is how you measure the EMF of a source experimentally — use a high-resistance voltmeter across the terminals with no other load connected, so the current drawn is negligible.

5. Worked Calculations — Applying the Equations

The equations look straightforward, but exam questions mix up the given quantities deliberately. The reliable approach is always to write out what you know, identify which form of the equation fits, and substitute carefully with units at every step.

Worked example: Finding Internal Resistance from EMF and Terminal pd

A battery has an EMF of 12.0 V. When a current of 3.0 A flows through an external resistance of 3.5 $\Omega$, the terminal pd is measured as 10.5 V. Calculate the internal resistance of the battery. [3 marks]

1. $\varepsilon = 12.0$ V, $V = 10.5$ V, $I = 3.0$ A, $R = 3.5\ \Omega$ — Listing knowns prevents substitution errors and earns the method mark [M1].
2. Lost volts $= \varepsilon - V = 12.0 - 10.5 = 1.5$ V — The lost volts equal $Ir$, so this intermediate step isolates the $Ir$ term.
3. $r = \dfrac{\text{lost volts}}{I} = \dfrac{1.5}{3.0}$ — Rearranging $Ir = 1.5$ V gives $r = 1.5 / 3.0$ [M1 for correct rearrangement].
4. $r = 0.50\ \Omega$ — Check: $\varepsilon = I(R + r) = 3.0 \times (3.5 + 0.5) = 3.0 \times 4.0 = 12.0$ V. Confirmed. [A1]

Answer: Internal resistance $r = 0.50\ \Omega$

Worked example: Finding Current and Terminal pd from EMF and Resistances

A cell of EMF 1.5 V and internal resistance 0.20 $\Omega$ is connected to an external resistor of 2.80 $\Omega$. Calculate (a) the current in the circuit and (b) the terminal potential difference. [4 marks]

1. $\varepsilon = I(R + r)$ — This is the starting equation for any circuit with internal resistance. [M1]
2. $I = \dfrac{\varepsilon}{R + r} = \dfrac{1.5}{2.80 + 0.20}$ — Substitute all known values with units. The denominator is $2.80 + 0.20 = 3.00\ \Omega$.
3. $I = \dfrac{1.5}{3.00} = 0.50$ A — $1.5 \div 3.00 = 0.50$. Current is 0.50 A. [A1]
4. $V = \varepsilon - Ir = 1.5 - (0.50 \times 0.20) = 1.5 - 0.10 = 1.40$ V — Lost volts $= 0.50 \times 0.20 = 0.10$ V. Terminal pd $= 1.5 - 0.10 = 1.40$ V. [A1] Alternatively: $V = IR = 0.50 \times 2.80 = 1.40$ V — both routes give the same answer.

Answer: (a) $I = 0.50$ A; (b) Terminal pd $V = 1.40$ V

6. The V-I Graph for a Real Source

One of the most important ways this topic appears in exams is through a graph of terminal pd ($V$) against current ($I$) for a source. Rearranging $V = \varepsilon - Ir$ into the form $y = mx + c$ gives $V = -rI + \varepsilon$. This is a straight-line graph with a negative gradient.

From this graph, the y-intercept (where $I = 0$) gives the EMF directly — it is the terminal pd when no current flows. The gradient of the line is $-r$, so the magnitude of the gradient gives the internal resistance. This is the standard experimental method for determining both $\varepsilon$ and $r$ from a single graph.

To find the gradient, choose two well-separated points on the line, calculate $\Delta V / \Delta I$, and take the magnitude. For the data above: gradient $= (4.5 - 6.0)/(3.0 - 0.0) = -1.5/3.0 = -0.50$, so $r = 0.50\ \Omega$ and $\varepsilon = 6.0$ V.

7. Required Practical — Determining EMF and Internal Resistance

The standard practical for this topic involves measuring the terminal pd of a cell at different currents, then using a $V$-$I$ graph to extract $\varepsilon$ and $r$. The method is clean and directly tests the equation $V = \varepsilon - Ir$.

8. Exam Technique and Common Mistakes

This topic rewards students who are precise with language and careful with signs. The most common errors in exam scripts are not conceptual — they are presentational. Here is how to avoid losing marks that you have genuinely earned.

Pulling the whole topic together: EMF is the energy per unit charge supplied by the source ($\varepsilon = E/Q$); internal resistance causes a voltage drop $Ir$ inside the source; and the terminal pd — the voltage available to the external circuit — is $V = \varepsilon - Ir$. Every calculation in this topic is a variation on these three linked ideas.