Wait, What? A 1.5 V cell can be working normally while your voltmeter reads less than 1.5 V.

The missing potential difference has not vanished. When current flows, some energy per unit charge is transferred inside the cell itself because real sources have internal resistance. The practical challenge is to separate the source’s EMF from the terminal p.d. available to the external circuit.

The scientific model

For a source with EMF ε and internal resistance r supplying current I,

V = ε − Ir

where V is terminal potential difference. The term Ir is sometimes called “lost volts”: it represents potential difference associated with energy transfer inside the source.

The Institute of Physics recommends measuring terminal V and current I and plotting V against I. The y-intercept gives ε and the gradient is −r. IOPSpark: EMF and internal resistance.

Why open-circuit voltage is only an approximation to EMF

With a high-resistance voltmeter across an unloaded cell, current is very small, so Ir is very small and terminal V approaches ε. But a real voltmeter still draws some current. “No current” is an idealisation; “negligible current compared with the experiment” is the practical statement.

How to generate a useful range of currents

Connect a variable external resistance and change it between readings. Measure I in series and V across the cell terminals. Choose enough settings to establish a trend without drawing excessive current that heats or rapidly discharges the cell.

The graph is the experiment

Rearrange V = ε − Ir into straight-line form y = c + mx:

V = ε + (−r)I.

Therefore the intercept at I = 0 estimates ε, and the magnitude of the negative gradient estimates r. A graph uses multiple data points and reveals whether the simple constant-r model remains adequate.

Quantitative window

Suppose the best-fit line is V = 1.54 − 0.82I, with V in volts and I in amperes.

Then ε ≈ 1.54 V and r ≈ 0.82 Ω.

At I = 0.50 A, the predicted terminal p.d. is 1.54 − 0.82(0.50) = 1.13 V. The internal drop is about 0.41 V.

Why heating can bend the graph

Internal resistance is not necessarily constant. Large currents can heat the cell; chemistry and polarisation can also change during use. If r changes while measurements are being taken, V against I may curve or later points may drift.

IOPSpark specifically warns that cells can run down or polarise under low load resistance. Switch the circuit off between readings where appropriate, work efficiently and avoid unnecessarily high currents.

Meter loading and contact resistance

A voltmeter should have high resistance so it draws little current. An ammeter should have low resistance so it adds little series resistance. Leads and contacts also contribute resistance. In most school setups these effects are small but they become important when r is small or precision is high.

Energy reasoning

EMF is energy supplied per unit charge by the source. Terminal p.d. is energy per unit charge transferred to the external circuit. The difference corresponds to internal energy transfer. This is why “lost volts” should not be interpreted as destroyed energy.

Observation versus inference

Observation: when I increased from 0.10 A to 0.60 A, terminal V fell.

Transformation: V plotted against I gave an approximately straight line.

Inference: the source behaves approximately as an EMF in series with a constant internal resistance over the measured range.

Model parameters: intercept estimates ε; negative gradient estimates r.

Failure modes

• Leaving high current flowing between readings, changing cell state.
• Using too narrow a current range, making gradient uncertain.
• Reading V as EMF at every load.
• Taking gradient from one pair of raw points instead of the best-fit line.
• Ignoring a visibly curved graph and forcing the constant-r model.
• Short-circuiting a cell to “find maximum current”—unsafe and unnecessary.

Checkpoint 1

A V-I graph has intercept 1.60 V and gradient −2.4 V A⁻¹. What are ε and r?

Checkpoint 2

Later readings lie systematically below the initial straight-line trend after the cell becomes warm. What assumption may have failed?

Answer key and WHY reasoning

1: ε = 1.60 V; r = 2.4 Ω because the gradient of V against I is −r.

2: internal resistance and/or effective EMF may have changed as the cell heated, discharged or polarised. The source is no longer behaving as a time-invariant ε-plus-r model.

Unfamiliar transfer: potato cell versus alkaline cell

A potato cell can have much larger internal resistance than a commercial cell. The same V-I method still works, but useful current ranges and meter-loading significance change. The transferable skill is to choose a load range appropriate to the source.

Secondary → JC → deeper Science

Secondary: terminal p.d., current, resistance, circuit meters and energy transfer.

JC: EMF, internal resistance, V-I linearisation, gradient/intercept, uncertainty and changing source behaviour.

Deeper Science: batteries require electrochemical kinetics, state-of-charge models, equivalent circuits, impedance and temperature-dependent internal resistance.

How to study this practical

Memorise V = ε − Ir as an energy model, not merely a graph formula. Practise translating it into graph features, then diagnose what would happen if r increased, the cell warmed, the voltmeter resistance fell or the current range became too small.

Evidence boundaries

The experiment estimates ε and an effective r over the tested current range and cell state. Real batteries can have nonlinear, frequency-dependent and state-dependent behaviour, so r is not always a universal fixed number.

Authoritative next steps

• Institute of Physics: EMF and internal resistance
• Institute of Physics: EMF and internal resistance collection

Teaching Guide

For teachers and parents: ask why terminal p.d. falls when current rises and require an energy explanation before algebra. Then ask the student to recover ε and r from an unfamiliar straight-line equation. This connects circuit behaviour, physical model and mathematical evidence.