A cell of EMF 1.5 V has a terminal pd of 1.2 V when supplying 0.60 A. Calculate the internal resistance. [2 marks]

AQA AQA 2018-style practice: A cell of EMF 1.5 V and internal resistance 0.50 is connected to a 2.5 resistor. Calculate the current in the circuit and the terminal potential difference across the cell.

Use the EMF equation = I(R + r) to find the current, including the internal resistance in the total. I = R + r = 1.52.5 + 0.50 = 1.53.0 = 0.50 A. Terminal pd is the voltage across the external resistor, V = IR = 0.50 × 2.5 = 1.25 V. Equivalently V = - Ir = 1.5 - (0.50)(0.50) = 1.25 V. Markers reward including the internal resistance in the total resistance and recognising that the terminal pd is less than the EMF because of the lost volts.

EnglandPhysicsSyllabus dot point

Why is the voltage across a battery's terminals less than its rated value when it supplies current?

Electromotive force as energy per unit charge, internal resistance, the equations linking EMF, terminal potential difference and lost volts, and measuring EMF and internal resistance experimentally.

A focused answer to AQA A-Level Physics 3.5.1.6, covering electromotive force as energy transferred per unit charge, internal resistance, the equations linking EMF, terminal potential difference and lost volts, and the experiment to measure EMF and internal resistance.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Electromotive force
Internal resistance and lost volts
Measuring EMF and internal resistance
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What this dot point is asking

AQA specification point 3.5.1.6 wants you to define EMF, explain internal resistance and lost volts, use the equations linking EMF, terminal potential difference and internal resistance, and describe how to measure EMF and internal resistance from a graph.

Electromotive force

The EMF is the maximum potential difference a source can deliver, which occurs only when no current is drawn. The energy comes from chemical reactions in a cell, from electromagnetic induction in a dynamo, or from incident light in a photovoltaic cell.

Internal resistance and lost volts

Every real source has some resistance to current within itself, the internal resistance $r$ , due to the resistance of its electrolyte, electrodes or windings. When a current flows, energy is dissipated inside the source as heat.

A source with a low internal resistance (such as a car battery) can deliver large currents with little drop in terminal voltage, whereas a high internal resistance limits the maximum useful current.

Measuring EMF and internal resistance

Finding EMF and internal resistance from a graph

A cell is connected to a variable resistor, an ammeter in series and a voltmeter across the terminals. The resistance is changed to give pairs of terminal pd $V$ and current $I$ .

step 1: Set up the rearranged equation

Rearrange the circuit equation to $V = \varepsilon - rI$ , which has the straight-line form $y = c + mx$ with $V$ on the vertical axis and $I$ on the horizontal axis.

step 2: Take readings and plot

Record several $(I, V)$ pairs by adjusting the variable resistor, then plot $V$ against $I$ and draw the best-fit straight line.

step 3: Read off the EMF

The y-intercept, where $I = 0$ , gives the EMF $\varepsilon$ , because with no current there are no lost volts and the terminal pd equals the EMF.

step 4: Read off the internal resistance

The gradient is $-r$ , so the internal resistance is the magnitude of the gradient, $r = |\text{gradient}|$ .

The EMF alone can be estimated directly as the terminal pd measured by a high-resistance voltmeter when negligible current flows.

Try this

Q1. Define the EMF of a cell. [1 mark]

Cue. The energy transferred to each coulomb of charge passing through the source.

Q2. A cell of EMF $1.5 \text{ V}$ has a terminal pd of $1.2 \text{ V}$ when supplying $0.60 \text{ A}$ . Calculate the internal resistance. [2 marks]

Cue. Lost volts $= 1.5 - 1.2 = 0.3 \text{ V}$ ; $r = \dfrac{0.3}{0.60} = 0.50 \text{ }\Omega$ .

Q3. State the terminal pd of a cell when no current is drawn from it. [1 mark]

Cue. It equals the EMF.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20184 marksA cell of EMF

1.5 \text{ V}

and internal resistance

0.50 \text{ }\Omega

is connected to a

2.5 \text{ }\Omega

resistor. Calculate the current in the circuit and the terminal potential difference across the cell.

Show worked answer →

Use the EMF equation $\varepsilon = I(R + r)$ to find the current, including the internal resistance in the total.

$I = \dfrac{\varepsilon}{R + r} = \dfrac{1.5}{2.5 + 0.50} = \dfrac{1.5}{3.0} = 0.50 \text{ A}$ .

Terminal pd is the voltage across the external resistor, $V = IR = 0.50 \times 2.5 = 1.25 \text{ V}$ . Equivalently $V = \varepsilon - Ir = 1.5 - (0.50)(0.50) = 1.25 \text{ V}$ .

Markers reward including the internal resistance in the total resistance and recognising that the terminal pd is less than the EMF because of the lost volts.

AQA 20225 marksA student measures the terminal potential difference of a cell for several values of current and plots terminal pd against current, obtaining a straight line. Explain how the EMF and internal resistance can be found from this graph, and state what the line shows about the cell.

Show worked answer →

The cell obeys $V = \varepsilon - Ir$ , which has the form $y = c + mx$ with $V$ on the y-axis and $I$ on the x-axis. The y-intercept (where $I = 0$ ) gives the EMF $\varepsilon$ , because with no current there are no lost volts.

The gradient is $-r$ , so the internal resistance is the magnitude of the gradient. A straight line confirms that $r$ is constant over the range tested.

The line slopes downwards: as the current rises, the lost volts $Ir$ grow, so the terminal pd falls below the EMF.

Markers reward identifying the y-intercept as the EMF, the gradient magnitude as the internal resistance, and explaining the negative slope through lost volts.

Related dot points

Sources & how we know this

AQA A-level Physics (7408) specification — AQA (2017)