AS Level · Topic 10.1
A-Level 9702 / Topic 10 / AS

The cost of being a battery.

A real cell is not a perfect source. Some of the energy it gives each coulomb is spent inside the cell itself, on its own internal resistance, so the voltage you measure across the terminals drops as soon as current flows.

The key idea

The electromotive force (e.m.f.) is the energy supplied per unit charge by a source, the total energy each coulomb gains. A real source has internal resistance r, so when a current I flows, some energy is lost inside it: the terminal potential difference is V = E − Ir. The e.m.f. equals the terminal p.d. only when no current flows.

real cell E r R V I
Fig. 1 — A real cell: e.m.f. E and internal resistance r. Terminal p.d. V = E − Ir
Section 01

Load the cell.

Draw more current from the cell and watch the terminal voltage fall below the e.m.f. by the lost volts Ir. Change the internal resistance and see how a high-r cell sags under load.

Section 02

E.M.F. versus terminal p.d.

Energy is conserved around the whole loop, inside the cell included.

RelationMeaningUnit
E = energy / chargeelectromotive forcevolt (V)
V = E − Irterminal potential differencevolt (V)
lost volts = Irenergy spent inside the sourcevolt (V)
Stage 1 · Learn

Check what the sim just showed you

Four quick checks tied to this lesson. Each correct answer earns XP and lights this skill on your star map.

Quick check+10 XP

The electromotive force of a source is the:

Quick check+10 XP

The terminal potential difference of a cell is given by:

Quick check+10 XP

A cell of e.m.f. 1.5 V has internal resistance 0.50 Ω and supplies 2.0 A. The lost volts are:

Quick check+10 XP

The e.m.f. of a cell equals its terminal potential difference only when:

Section 03

Why batteries get warm.

Internal resistance is where a cell loses energy as heat.

Examiner trap

The e.m.f. is not simply the voltage you read across a battery in use; that is the terminal p.d., which is lower by Ir. Use a minus sign: V = E − Ir. The internal resistance is in series with the external circuit, so the same current flows through it, and the cell is warmest when delivering a large current into a small external resistance.

Stage 2 · Exam

Exam-style questions

Unlocks once the checks above are done. Worth more XP, written to AS Paper 1 and 2 standard.

Finish the checks above to unlock the exam questions
Exam style+20 XP

A cell of e.m.f. 6.0 V and internal resistance 0.50 Ω drives a current of 2.0 A. The terminal potential difference is:

Exam style+20 XP

A battery of e.m.f. 12 V and internal resistance 1.0 Ω is connected to a 5.0 Ω resistor. The current is:

Exam style+20 XP

When a graph of terminal p.d. against current is plotted for a cell, the magnitude of the gradient gives the:

Skill unlocked

E.M.F. and internal resistance, mastered.

This skill is now lit gold on your star map. Keep the chain going.

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Go deeper · practice
Six original Cambridge-style questions
Using V = E − Ir, finding current with internal resistance, lost volts, and reading a terminal-p.d. against current graph. Attempt each, then reveal the worked solution.
Stage 3 · Paper 1 readiness
D.C. circuits · Paper 1 Practice
A bank of original multiple-choice questions across the whole topic, in the style of Paper 1. Start this once you are confident across e.m.f. and internal resistance, Kirchhoff laws, and potential dividers.
Start Paper 1 Practice →