Practical: Investigating EMF and Internal Resistance Revision Notes | A Level AQA Physics

1. Why a Battery Never Delivers Its Full Voltage

Connect a voltmeter directly across a fresh AA battery and you read 1.5 V. Now connect a bulb and measure again — the reading drops. The battery has not suddenly become weaker. Something inside it is consuming voltage, and that something is internal resistance.

This practical sits at the heart of AQA Required Practical 6. You will set up a circuit, vary the current drawn from a cell, and measure how the terminal pd changes. From those measurements you can extract two fundamental properties of any real cell: its emf and its internal resistance.

2. The Key Quantities: emf, Terminal pd, and Lost Volts

Picture the cell as a small pump pushing charge around a circuit. The emf is the total 'push' the pump can provide per unit of charge. However, the pump itself has friction — that friction is the internal resistance $r$. Every time current flows, some of the push is used up overcoming that internal friction before the charge even reaches the external circuit.

The quantity $Ir$ — current multiplied by internal resistance — is called the lost volts. It represents the energy per unit charge wasted inside the cell as heat. The larger the current drawn, the greater the lost volts, and the lower the terminal pd.

Rearranging gives $V = \varepsilon - Ir$, which is the form you will use most often in this practical. Notice it has the form $y = mx + c$ if you treat $V$ as $y$ and $I$ as $x$: the gradient is $-r$ and the $y$-intercept is $\varepsilon$. That straight-line relationship is exactly what you will plot from your experimental data.

3. Required Practical 6 — The Experimental Method

The aim is to measure how the terminal pd of a cell varies with the current drawn from it. By collecting multiple pairs of $V$ and $I$ values, you can plot a straight-line graph and use it to determine both the emf and the internal resistance of the cell.

One detail that students often overlook: the voltmeter must be connected directly across the cell terminals, not across the external resistor. The primary reason is definitional — the terminal pd is, by definition, the pd across the cell's own terminals, so the voltmeter must be placed there to measure it directly. Connecting it across the external resistor measures the pd across that component instead, which is not the same quantity. In addition, any voltage drop in the connecting leads between the cell and the resistor would be missed, though in a well-constructed circuit this is a minor secondary consideration.

4. Graphical Analysis — Reading emf and Internal Resistance

Once you have your data table, the graph does the heavy lifting. Plot terminal pd on the y-axis and current on the x-axis. A well-conducted experiment gives a straight line with a negative gradient — terminal pd falls as current rises, exactly as the equation $V = \varepsilon - Ir$ predicts.

From the illustrative data above, the y-intercept is 1.50 V — that is the emf. The gradient is $\frac{1.10 - 1.50}{1.0 - 0.0} = \frac{-0.40}{1.0} = -0.40$ V A$^{-1}$, so the internal resistance is 0.40 Ω. The worked example below takes you through this calculation step by step.

Worked example: Determining emf and Internal Resistance from a Graph

A student plots terminal pd against current for a cell and obtains a straight line passing through the points (0.20 A, 1.42 V) and (0.80 A, 1.18 V). Determine the emf and internal resistance of the cell.

1. $\text{gradient} = \frac{\Delta V}{\Delta I} = \frac{1.18 - 1.42}{0.80 - 0.20} = \frac{-0.24}{0.60}$ — Choose two points that are well separated on the best-fit line — not raw data points — to reduce uncertainty. [M1]
2. $\text{gradient} = \frac{-0.24}{0.60} = -0.40 \text{ V A}^{-1}$ — The negative sign confirms terminal pd decreases as current increases, consistent with $V = \varepsilon - Ir$. [A1]
3. $r = -\text{gradient} = -(-0.40) = 0.40 \ \Omega$ — Since gradient $= -r$, the internal resistance is the magnitude of the gradient. [A1]
4. Using $V = \varepsilon - Ir$: $1.42 = \varepsilon - (0.20 \times 0.40)$ $1.42 = \varepsilon - 0.08$ $\varepsilon = 1.42 + 0.08 = 1.50 \text{ V}$ — Substitute one known point and the value of $r$ to find $\varepsilon$. Alternatively, extend the best-fit line to $I = 0$ and read off the intercept directly. [M1, A1]
5. Check with second point: $V = 1.50 - (0.80 \times 0.40) = 1.50 - 0.32 = 1.18$ V ✓ — The predicted value matches the data point, confirming the calculation is internally consistent.

Answer: emf $\varepsilon = 1.50$ V; internal resistance $r = 0.40$ Ω

5. Evaluating the Experiment — Reliability, Validity, and Improvements

A-level examiners expect you to go beyond describing the method. You need to identify specific sources of error, explain how they affect your results, and suggest concrete improvements.

The most significant systematic error in this experiment is usually cell discharge. As the cell delivers charge, its emf gradually falls. This means your later data points are taken from a slightly different cell than your earlier ones, causing the graph to deviate from a perfect straight line. Opening the switch between readings is the most effective mitigation.

6. Exam Technique and Common Mistakes

Questions on this practical appear in two forms: data-analysis questions where you are given a table or graph, and written-method questions where you describe or evaluate the procedure. Both reward precise scientific language and explicit links to the underlying physics.

When a question asks you to describe the graph, state the shape (straight line), the direction of the gradient (negative), and identify both intercepts explicitly. When asked to explain the shape, link back to $V = \varepsilon - Ir$: the linear decrease in $V$ with $I$ follows directly from this equation because $\varepsilon$ and $r$ are both constant for a given cell at constant temperature.

For a 6-mark extended response on this practical, structure your answer as: state the aim and key equation, describe the circuit setup with correct meter placement, explain how you vary the independent variable, state what you plot and how you extract $\varepsilon$ and $r$ from the graph, then evaluate one significant source of error and its improvement. That structure reliably covers the marking points.

Every element of Required Practical 6 — the circuit setup, the graphical analysis using $V = \varepsilon - Ir$, and the critical evaluation of errors — maps directly onto the AQA specification requirement to investigate the variation of terminal pd with current. Mastering the graph interpretation and the error analysis gives you the tools to answer both data-handling and extended-writing questions on this topic with confidence.