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EMF and internal resistance

Every real cell resists the charge passing through its own chemicals, and that internal resistance is why the pd a circuit receives sags below the cell's emf as the current grows. Plotting terminal potential difference against current gives a near-straight line whose intercept is the emf and whose gradient is the internal resistance.

Year 12AQA 3.5.1.6 CIE 10.1.3, 10.1.4, 10.1.5 OCR A 4.2.2(b), 4.2.2(c), 4.2.2(d), 4.3.2(a), 4.3.2(b), 4.3.2(c)(i), 4.3.2(c)(ii)EDEXCEL 45, 46

Builds on Circuits and Kirchhoff's laws.

IN THIS TOPIC

Define emf as the energy given to each unit of charge, ε = E/Q, and keep it distinct from terminal pd.
Work circuits where r cannot be neglected, using ε = I(R + r) and V = ε − Ir.
Pull ε and r off a graph of terminal pd against current.

COMMON MISCONCEPTION

A 1.5 V battery gives you 1.5 volts.

What emf actually is

The electromotive force of a source is the energy it gives to each coulomb of charge passing through it.

\epsilon = \frac{E}{Q}

ON THE AQA DATA SHEET

That energy is measured in volts, joules per coulomb, exactly like pd. The name is a historical accident. An energy per unit charge is what it is, and no part of it is a force. It describes what the chemistry supplies; the terminal pd, coming next, describes what the outside world receives.

The resistance inside

Real cells are built of chemicals and electrodes that resist the very current they drive. Model this as a perfect source of emf in series with a small internal resistance r, both sealed inside the case.

FIG. 1The cell model: a perfect 1.5 V emf in series with an amber 0.50 Ω internal resistance, hidden in the case. Driving 0.50 A through a 2.5 Ω resistor leaves a terminal pd of 1.25 V.

The emf has to pay for both resistances in the loop.

\epsilon = I(R + r)

ON THE AQA DATA SHEET

What appears across the cell's terminals, the terminal pd, is the emf minus the part spent inside, V = ε − Ir. That same quantity is also IR, the pd across the external circuit.

Lost volts

That difference Ir goes by the nickname lost volts, the energy per coulomb spent crossing the cell's own innards. It warms the battery and does nothing for the circuit. Lost volts grow with the current, so the harder a cell works the further its terminal pd sags below the emf. Headlights dim while the starter motor turns for exactly this reason. The starter's huge current inflates Ir, and every other component feels the drop.

Only when no current flows does the sag vanish. A high-resistance voltmeter across an unused cell reads very nearly the full emf, because it draws almost no current and so leaves almost no lost volts to subtract.

The graph that measures the cell

Vary the external resistance, record the terminal pd and the current, and plot V against I. The model equation V = ε − Ir is a straight line, with intercept ε and gradient −r.

FIG. 2Terminal pd against current: a straight line whose intercept on the V axis is the emf and whose downward gradient is the internal resistance.

The graph earns its place because it measures the emf properly. Even a good voltmeter draws a whisper of current, so a direct reading sits a hair below the emf. Extrapolating the line to I = 0 gives ε with no lost volts in it at all, and the slope gives r at the same time. Required practical 6 is this experiment, with a variable resistor stepping the current and a dozen or so V and I pairs plotted by hand.

WORKED EXAMPLE

Reading the line

A cell's V–I graph is a straight line cutting the V axis at 1.58 V and falling to 1.10 V at a current of 2.0 A. Find the emf and the internal resistance.

Match the physics to the line before reaching for numbers. V = ε − Ir has the shape of y = c + mx, so the intercept gives ε and the gradient gives −r.

Read the intercept straight off, ε = 1.58 V. The gradient is (1.10 − 1.58)/2.0 = −0.24 V A⁻¹, so r = 0.24 Ω, quoted positive with the gradient described as negative.

Is that reasonable? A single dry cell reads a touch over 1.5 V with a fraction of an ohm inside, which is exactly what the line says.

GUIDED PRACTICE

The algebra-only route

A cell delivers a terminal pd of 1.35 V at 0.50 A, and 1.20 V at 1.25 A. Write V = ε − Ir once for each reading, subtract one equation from the other, and find r and then ε.

Show the working

Subtracting eliminates ε. From 1.35 − 1.20 = r × (1.25 − 0.50) comes 0.15 = 0.75r, so r = 0.20 Ω.

Substitute back into either reading. ε = 1.35 + 0.50 × 0.20 = 1.45 V, and the other reading agrees, 1.20 + 1.25 × 0.20 = 1.45 V. That agreement is the built-in check.

INDEPENDENT PRACTICE

The short circuit

A cell of emf 4.5 V and internal resistance 1.5 Ω is accidentally short-circuited by a wire of negligible resistance. Find the current, the terminal pd, and state where the energy is going.

Show the working

With R = 0 the loop equation ε = I(R + r) collapses to ε = Ir, giving I = 4.5/1.5 = 3.0 A.

Terminal pd is then V = ε − Ir = 4.5 − 3.0 × 1.5 = 0 V. Every volt is being lost inside.

All the power, P = εI = 13.5 W, is dissipated in the cell's own internal resistance. That is why a shorted cell gets dangerously hot. The working treats r as constant, which real cells only approximate at currents this size, and every exam question grants.

ASSESSMENT FOCUS

From a V–I graph the intercept gives ε and the gradient gives minus r. Quote r as a positive resistance and describe the gradient as negative.
Terminal pd equals the emf only when I = 0. A high-resistance voltmeter across an isolated cell reads ε for exactly that reason.
The cell's r sits in series with everything outside it, so the loop equation is ε = I(R + r) and never ε = IR.
Lost volts means Ir, which is no fixed number for a given cell. Double the current and you double the loss.
Despite its name, emf is an energy per unit charge, measured in volts. Call it a force and the definition mark goes.

CHECK YOURSELF

A battery of emf 9.0 V and internal resistance 0.60 Ω drives a 4.4 Ω resistor. Find the current, the terminal pd, and the lost volts.

Show a hint

The emf drives current through both resistances; the terminals only show the external share.

Show the answer

Current first. $I = \frac{\epsilon}{R + r}$ = 9.0 / (4.4 + 0.60) = 1.8 A.

Terminal pd is V = IR = 1.8 × 4.4 = 7.9 V, and ε − Ir gives the same.

Lost volts, Ir = 1.8 × 0.60 = 1.1 V. The 9.0 V splits as 7.9 outside plus 1.1 inside, and the books balance.

The emf is the energy per coulomb the cell supplies.

The terminal pd is what is left after the internal resistance.

The difference, Ir, grows with the current.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

19 questions on this topicAnswer them one at a time and mark yourself against the mark scheme.Practise this topic

Or read them with their mark schemes on the emf and internal resistance questions page.

6 flashcards on this topicDefinitions, off-sheet equations and a spot-the-error card, scheduled by spaced repetition in your browser.Revise with flashcards

WHERE TO GO NEXT

Required practical 6: emf and internal resistance of a cell puts this topic in the lab, and the written papers ask about it.
Gradients and areas under graphs is the maths this lesson leans on, worked through from GCSE.
Rearranging equations is the maths this lesson leans on, worked through from GCSE.

CHECK YOUR PROGRESS

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Define emf as the energy given to each unit of charge, ε = E/Q, and keep it distinct from terminal pd.
Work circuits where r cannot be neglected, using ε = I(R + r) and V = ε − Ir.
Pull ε and r off a graph of terminal pd against current.

Open the full revision checklist to track your progress across the whole unit.