EMF equation and internal resistance for A-Level Physics

A-LevelPhysicsScience9 min readBy Tom Mercer

Electromotive force (EMF) is the energy transferred per unit charge by a source, such as a cell or battery, as charge passes through it. Internal resistance is the resistance inside the source itself, which causes some energy to be lost as heat before any current reaches the external circuit.

This guide covers the EMF equation in the AQA A-Level Physics specification (3.5.1.6), the definitions the mark scheme expects, the standard required practical and the calculation patterns that come up in Paper 1.


EMF equation: ε = I(R + r)

Charge flows through both the external resistance R and the internal resistance r. Both share the total EMF of the source.

Required practical

Plot V (terminal pd) against I and read off ε as the y-intercept and r as the gradient (with a minus sign).

Definition marks

EMF and terminal pd both ask for energy per unit charge, but for different parts of the circuit. Examiners are strict on the wording.


Defining EMF and internal resistance

The mark-scheme definition of EMF is: The energy transferred from chemical (or other) energy to electrical energy per unit charge passing through the source. The unit is the volt (V), because 1 V = 1 J/C, the same as potential difference.

Internal resistance is the resistance of the source itself. In a chemical cell it comes from the electrolyte and electrode materials. Some of the energy supplied by the EMF is dissipated as heat across this internal resistance, which is why a battery gets warm under load and why the terminal pd is always less than the EMF when current flows.

Good to know

EMF is not a force Despite the name, electromotive force is not a force in the mechanical sense. It is an energy per unit charge, measured in volts. The name is historical and you should not mention forces in newtons in your answer.

The EMF equation

The standard EMF equation for a source with internal resistance r connected to a single external resistance R is:

ε = I(R + r)

This can be rearranged as ε = V + Ir, where V = IR is the terminal pd across the external circuit and Ir is the pd "lost" across the internal resistance.

SymbolMeaningUnit
ε (epsilon)EMF of the sourceVolts (V)
VTerminal pd across the external circuitVolts (V)
ICurrent in the circuitAmps (A)
RExternal resistanceOhms (Ω)
rInternal resistance of the sourceOhms (Ω)
All five quantities appear on the AQA equation sheet. You should still memorise the equation in both forms.
Tip

When V = ε exactly The only time the terminal pd equals the EMF is when no current flows, because then Ir = 0. This is why a high-resistance voltmeter connected directly across an isolated cell reads the EMF, not the terminal pd under load.

The required practical: Measuring ε and r

The standard experiment uses a cell, a variable resistor, an ammeter in series and a voltmeter across the cell's terminals. You vary the external resistance, record V and I pairs, then plot V on the y-axis against I on the x-axis.

Rearranging ε = V + Ir gives V = ε – Ir. This is a straight line of the form y = mx + c, with intercept ε on the V axis and gradient –r. So: The y-intercept gives the EMF, and the magnitude of the gradient gives the internal resistance.

StepWhat to doWhy
1Set up cell, variable resistor, ammeter (series), voltmeter (across cell)Voltmeter reads terminal pd, ammeter reads circuit current
2Vary the resistance and record paired V and I values (at least 6 points)Spread of readings gives a reliable line of best fit
3Plot V (y-axis) against I (x-axis)Line obeys V = ε – Ir, so intercept and gradient give the unknowns
4Read y-intercept = ε and gradient magnitude = rApply the linear equation directly
5Repeat and average to reduce random errorImproves precision
AQA required practical 6: Investigation of EMF and internal resistance using a graphical method.

Worked example: A cell under load

A cell has an EMF of 1.5 V and an internal resistance of 0.5 Ω. It is connected to an external resistor of 2.5 Ω. Calculate the current and the terminal pd.

Step 1: Use ε = I(R + r). So 1.5 = I × (2.5 + 0.5) = 3.0 I.

Step 2: I = 1.5 / 3.0 = 0.5 A.

Step 3: Terminal pd V = IR = 0.5 × 2.5 = 1.25 V.

Step 4: Check by adding the "lost" pd: Ir = 0.5 × 0.5 = 0.25 V, and 1.25 + 0.25 = 1.5 V, which equals the EMF.

Why batteries go flat

As a cell discharges, the internal resistance increases. The EMF stays roughly constant until very near the end of the cell's life, but the rising r means that when any meaningful current flows, more of the EMF is dropped internally and less reaches the external circuit.

This is why an "almost flat" battery can still read its full nominal voltage on a high-resistance voltmeter (because almost no current flows during the measurement) but fails to power a torch (because under load, the terminal pd collapses).

Good to know

Common mistakes that cost easy marks Forgetting to include r in the total resistance when calculating current. Confusing terminal pd with EMF in the wording of a definition. Reading the gradient as +r instead of –r in the graph. Using V = IR to find V when there is internal resistance, instead of V = ε – Ir.

EMF in multiple-cell circuits

Cells in series add their EMFs and their internal resistances. Two 1.5 V cells in series each with r = 0.5 Ω give a total ε of 3.0 V and a total r of 1.0 Ω.

Cells in parallel keep the same EMF (assuming identical cells), but the effective internal resistance falls because the current splits between the cells. Two identical cells in parallel give the same ε and half the r. This is why high-current devices sometimes use parallel cell packs.

Key facts to memorise for the exam

  • EMF is the energy transferred from chemical (or other) energy to electrical energy per unit charge
  • Internal resistance r is the resistance inside the source itself, measured in ohms
  • Main equation: ε = I(R + r), equivalent to ε = V + Ir
  • Terminal pd V is always less than the EMF whenever current flows
  • Graph V against I: Y-intercept = ε, gradient = –r
  • Cells in series add EMFs and add internal resistances
  • Cells in parallel keep the same EMF and reduce the effective internal resistance
  • EMF is not a force despite the name; it is measured in volts, not newtons

Frequently asked questions


Related articles

See all
Science5 min

Diffusion in biology for A-Level

Science5 min

Transcription and translation in DNA for A-Level Biology

Science5 min

The nitrogen cycle for A-Level Biology