What is Electromotive Force (EMF) in the context of Edexcel International A Levels Physics?

Electromotive Force (EMF) is a fundamental concept in the study of electricity and circuits, particularly in the Edexcel International A Levels Physics curriculum. It refers to the energy provided by a source, such as a battery or generator, per unit charge. EMF is measured in volts and is responsible for driving the current around a circuit. It is important to note that EMF is not a force but a potential difference that causes the flow of electrons.

How does internal resistance affect the performance of a battery in a circuit?

Internal resistance is an inherent property of all batteries and affects their performance in a circuit. It refers to the opposition to the flow of current within the battery itself. When a current flows, some energy is lost as heat due to this resistance, reducing the voltage available to the external circuit. This means that the terminal voltage of a battery is always less than its EMF when a current is flowing. Understanding internal resistance is crucial for accurately modeling and predicting the behavior of circuits in Edexcel International A Levels Physics.

How can you calculate the total resistance in a series circuit?

In a series circuit, the total resistance is the sum of the individual resistances of all components connected in series. This is because the current has only one path to follow, and it encounters each resistance sequentially. The formula to calculate the total resistance (R_total) is R_total = R1 + R2 + R3 + ..., where R1, R2, R3, etc., are the resistances of individual components. This concept is essential for solving circuit problems in the Edexcel International A Levels Physics syllabus.

What is the difference between EMF and terminal voltage?

EMF (Electromotive Force) and terminal voltage are related but distinct concepts in circuit analysis. EMF is the total energy supplied per unit charge by a source, such as a battery, when no current flows. Terminal voltage, on the other hand, is the voltage output of a battery when a current is flowing and is always less than the EMF due to the voltage drop across the internal resistance of the battery. Understanding this difference is crucial for analyzing circuits in the Edexcel International A Levels Physics curriculum.

Why is modelling resistance important in understanding electrical circuits?

Modelling resistance is vital for understanding and predicting the behavior of electrical circuits. Resistance determines how much current will flow for a given voltage, according to Ohm's Law (V = IR). By accurately modeling resistance, students can analyze how different components affect the overall circuit performance, including power consumption and heat generation. This understanding is crucial for solving complex circuit problems and is a key part of the Edexcel International A Levels Physics curriculum on Waves and Electricity.

Why is "Confusing e.m.f. with terminal potential difference (treating them as the same 'battery voltage')" a common mistake in EMF and Modelling Resistance?

Why it happens: On open circuit they are equal and a labelled cell quotes its e.m.f., so students assume the terminals always sit at the e.m.f. How to avoid it: **The e.m.f. is the energy supplied per coulomb; the terminal p.d. is what is delivered to the external circuit.** Whenever a current flows, V = - Ir < . They are equal only when I = 0. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "On a V–I graph, quoting the internal resistance as the gradient value (e.g. $r = -0.50\ \Omega$) instead of recognising gradient $= -r$" a common mistake in EMF and Modelling Resistance?

Why it happens: Students read the (negative) gradient straight off and forget the equation V = - Ir makes the gradient -r, not r. How to avoid it: Compare V = - Ir with y = mx + c: the **gradient is -r**, so the internal resistance is the **magnitude (positive value) of the gradient**, r = |gradient|. Resistance can never be negative. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Misunderstanding 'lost volts' — thinking it is energy that vanishes, or computing it from the wrong quantity" a common mistake in EMF and Modelling Resistance?

Why it happens: The phrase 'lost' suggests the energy disappears, and students sometimes use the terminal p.d. or e.m.f. rather than Ir. How to avoid it: The **'lost volts' is Ir, the p.d. across the internal resistance** — energy is not lost but dissipated as heat *inside* the cell. Compute it as - V or as Ir, and note it grows as the current increases. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Assuming a voltmeter across the cell reads the e.m.f. even while a current flows" a common mistake in EMF and Modelling Resistance?

Why it happens: Students forget that the voltmeter is connected across the terminals, so it reads the terminal p.d., not the e.m.f. How to avoid it: A voltmeter across the cell reads the **terminal p.d.** (V = - Ir). It reads (approximately) the **e.m.f. only on open circuit** — when the rest of the circuit is disconnected so that I 0. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Using only the external resistance $R$ (forgetting the internal resistance $r$) when calculating the current" a common mistake in EMF and Modelling Resistance?

Why it happens: In earlier circuits the cell was treated as ideal, so students apply I = /R instead of I = /(R + r). How to avoid it: The current depends on the **total** resistance: I = /R + r. Always include the internal resistance in the denominator unless the question states the cell is ideal (r = 0). Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Using the terminal p.d. or e.m.f. instead of $I^2 r$ to find the power dissipated *inside* the cell" a common mistake in EMF and Modelling Resistance?

Why it happens: Students reach for P = VI with whichever voltage is to hand, without checking which resistance the power is dissipated in. How to avoid it: Power **inside** the cell is P_r = I^2 r (heating the cell); power **to the load** is P_R = I^2 R = VI; total power supplied is I = P_R + P_r. Match the resistance to the location of the energy transfer. Source: Edexcel IAL Physics Unit 2 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

EMF and Modelling Resistance

Q: Why is "On a V–I graph, quoting the internal resistance as the gradient value (e.g. $r = -0.50\ \Omega$) instead of recognising gradient $= -r$" a common mistake in EMF and Modelling Resistance?

Why it happens: Students read the (negative) gradient straight off and forget the equation V = - Ir makes the gradient -r, not r. How to avoid it: Compare V = - Ir with y = mx + c: the **gradient is -r**, so the internal resistance is the **magnitude (positive value) of the gradient**, r = |gradient|. Resistance can never be negative. Source: Edexcel IAL Physics Unit 2 examiner reports

Q: Why is "Misunderstanding 'lost volts' — thinking it is energy that vanishes, or computing it from the wrong quantity" a common mistake in EMF and Modelling Resistance?

Why it happens: The phrase 'lost' suggests the energy disappears, and students sometimes use the terminal p.d. or e.m.f. rather than Ir. How to avoid it: The **'lost volts' is Ir, the p.d. across the internal resistance** — energy is not lost but dissipated as heat *inside* the cell. Compute it as - V or as Ir, and note it grows as the current increases. Source: Edexcel IAL Physics Unit 2 examiner reports

Q: Why is "Assuming a voltmeter across the cell reads the e.m.f. even while a current flows" a common mistake in EMF and Modelling Resistance?

Why it happens: Students forget that the voltmeter is connected across the terminals, so it reads the terminal p.d., not the e.m.f. How to avoid it: A voltmeter across the cell reads the **terminal p.d.** (V = - Ir). It reads (approximately) the **e.m.f. only on open circuit** — when the rest of the circuit is disconnected so that I 0. Source: Edexcel IAL Physics Unit 2 examiner reports

Q: Why is "Using only the external resistance $R$ (forgetting the internal resistance $r$) when calculating the current" a common mistake in EMF and Modelling Resistance?

Why it happens: In earlier circuits the cell was treated as ideal, so students apply I = /R instead of I = /(R + r). How to avoid it: The current depends on the **total** resistance: I = /R + r. Always include the internal resistance in the denominator unless the question states the cell is ideal (r = 0). Source: Edexcel IAL Physics Unit 2 examiner reports

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Short Notes - EMF and Modelling Resistance

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Detailed Study Notes

Detailed notes on Waves and Electricity for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

EMF and Modelling Resistance — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 2)

Electromotive force (e.m.f.) as energy supplied per unit charge; the internal resistance of a real cell; terminal p.d., the 'lost volts' and the equation $\varepsilon = I(R+r)$ ; combining cells; and Core Practical 8 — determining e.m.f. and internal resistance by plotting terminal p.d. against current (intercept = e.m.f., gradient = −r).

At a glance

e.m.f. $\varepsilon$ is the energy transferred to each coulomb of charge by the source, $\varepsilon = \dfrac{W}{Q}$ (unit: volt = J C⁻¹). It is not a force.
A real cell has internal resistance $r$ : some energy is dissipated inside the cell, so the terminal p.d. delivered to the circuit is less than the e.m.f.
Key equation: $\varepsilon = V + Ir = I(R + r)$ , so the terminal p.d. is $V = \varepsilon - Ir$ .
'Lost volts' $= Ir = \varepsilon - V$ — the p.d. dropped across the internal resistance (energy turned to heat inside the cell).
Terminal p.d. falls as current rises because the lost volts $Ir$ grows. They are equal to the e.m.f. only on open circuit ( $I = 0$ ).
Core Practical 8: plot terminal p.d. $V$ (y) against current $I$ (x). The y-intercept = e.m.f. and the gradient = $-r$ (so $r = |\text{gradient}|$ ).
Short-circuit current ( $R = 0$ ) is $I_{\text{max}} = \dfrac{\varepsilon}{r}$ — limited only by the internal resistance.
Cells in series: e.m.f.s and internal resistances add. Identical cells in parallel: e.m.f. unchanged, internal resistance falls to $r/n$ .

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Define electromotive force (e.m.f.) as the energy transferred per unit charge by a source, $\varepsilon = W/Q$ , and distinguish it from terminal potential difference.
Understand that a source has internal resistance $r$ , and use $\varepsilon = I(R + r)$ and $V = \varepsilon - Ir$ to model real circuits.
Interpret the 'lost volts' $Ir$ as the energy dissipated per coulomb inside the source.
Determine the e.m.f. and internal resistance of a cell from a graph of terminal p.d. against current (intercept = $\varepsilon$ , gradient = $-r$ ).
Calculate the effect of combining cells in series and in parallel on e.m.f. and internal resistance.
CP8 — Determine the e.m.f. and internal resistance of a cell using a variable load resistor, and evaluate the main sources of error.

What e.m.f. really means

e.m.f. is energy given to each coulomb by the source — not a force.

The electromotive force, or e.m.f. ( $\varepsilon$ ), of a source is the energy transferred to each unit of charge when charge passes through it. In a cell, chemical energy is converted into electrical energy; the e.m.f. tells you how many joules are given to every coulomb that flows:

$\varepsilon = \frac{W}{Q}$

where $W$ is the energy supplied and $Q$ is the charge. The unit is the volt — and since this is a joule per coulomb, an e.m.f. of $1.5\ \text{V}$ means the cell gives $1.5\ \text{J}$ of energy to every coulomb that passes through it.

Two warnings that earn (or lose) marks:

e.m.f. is NOT a force. The name is historical. It is an energy per charge, measured in volts, exactly like a potential difference.
e.m.f. is about energy given to the charge by the source, whereas a potential difference (p.d.) is about energy transferred from the charge to a component (e.g. a resistor turning electrical energy into heat). Both are measured in volts and both equal $W/Q$ , but the direction of energy transfer is opposite.

So a battery is an energy pump: the e.m.f. is the "push per coulomb" it provides to drive charge round the whole circuit, including back through itself.

$\varepsilon = W/Q$ — energy supplied to each coulomb by the source.
Unit: volt = joule per coulomb (J C⁻¹).
e.m.f. is NOT a force, despite the name.
e.m.f. = energy given to charge; p.d. = energy taken from charge by a component.

See the full worked example for emf and modelling resistance →

Internal resistance, terminal p.d. and 'lost volts'

Real cells dissipate energy inside themselves, so V = ε − Ir.

An ideal cell would deliver all its e.m.f. to the external circuit. A real cell, however, has internal resistance $r$ — the resistance of its own materials (for a chemical cell, mainly the electrolyte). As current flows, some energy is dissipated as heat inside the cell, so less is left for the external components.

Modelling the real cell as an ideal source of e.m.f. $\varepsilon$ in series with a small resistor $r$ , the e.m.f. is shared between the external load $R$ and the internal resistance $r$ :

$\varepsilon = IR + Ir = I(R + r)$

The terminal potential difference $V$ is the p.d. across the external circuit (what a voltmeter on the terminals reads):

$V = IR = \varepsilon - Ir$

The term $Ir$ is the "lost volts" — the p.d. dropped across the internal resistance, equal to the energy dissipated inside the cell per coulomb. Note that the energy is not really "lost"; it is converted to heat inside the cell (which is why a hard-working battery gets warm). So:

$\boxed{\;\varepsilon = V + Ir \quad\text{and}\quad V = \varepsilon - Ir\;}$

A real cell modelled as an ideal e.m.f. ε in series with internal resistance r. The e.m.f. is shared between the 'lost volts' Ir inside the cell and the terminal p.d. V across the external load R. A voltmeter on the terminals reads V = ε − Ir, not ε.

Because $Ir$ grows as the current increases, the terminal p.d. falls when more current is drawn. This is exactly why a car's headlights dim for a moment when the starter motor (which draws a huge current) is engaged: the large $Ir$ drops the terminal p.d. across everything connected to the battery.

Model a real cell as an ideal e.m.f. $\varepsilon$ in series with internal resistance $r$ .
$\varepsilon = I(R + r)$ ; terminal p.d. $V = \varepsilon - Ir$ .
Lost volts $= Ir = \varepsilon - V$ (heat dissipated inside the cell).
More current ⇒ larger $Ir$ ⇒ smaller terminal p.d.

See the full worked example for emf and modelling resistance →

The V–I graph: reading off ε and r

Plot V against I: intercept is the e.m.f., gradient is −r.

The single most examined skill in this subtopic is reading the e.m.f. and internal resistance straight off a graph. Start from the terminal-p.d. equation and rearrange it into straight-line form:

$V = \varepsilon - Ir \quad\Longrightarrow\quad V = (-r)\,I + \varepsilon$

Compare this with $y = mx + c$ , plotting terminal p.d. $V$ on the y-axis against current $I$ on the x-axis:

the y-intercept $c$ is the e.m.f. $\varepsilon$ (the terminal p.d. when $I = 0$ );
the gradient $m$ is $-r$ — i.e. the line slopes downward, and the internal resistance is the magnitude of the gradient, $r = |\text{gradient}|$ .

Terminal p.d. V plotted against current I gives a straight line of negative gradient. The y-intercept (at I = 0) is the e.m.f. ε; the gradient is −r, so the internal resistance is the magnitude of the gradient. Use a large triangle (at least half the line) to read the gradient accurately.

Exam technique for "Determine ε and r from the graph":

Read the y-intercept for $\varepsilon$ (extend the best-fit line back to $I = 0$ if needed).
Draw a large gradient triangle (at least half the line), read the coordinates carefully, and compute $\dfrac{\Delta V}{\Delta I}$ with units of $\Omega$ .
The internal resistance is the magnitude of that gradient: $r = |\text{gradient}|$ . State it as a positive value — resistance is never negative.

The classic mistake is quoting $r$ with a minus sign because the gradient is negative. The gradient is $-r$ , so a gradient of $-0.50\ \Omega$ means $r = +0.50\ \Omega$ .

Rearrange: $V = (-r)I + \varepsilon$ , so plot $V$ (y) vs $I$ (x).
y-intercept = e.m.f. $\varepsilon$ (terminal p.d. at $I = 0$ ).
gradient = $-r$ , so $r = |\text{gradient}|$ (positive).
Use a large triangle and give the gradient units of $\Omega$ .

See the full worked example for emf and modelling resistance →

Core Practical 8 — determining e.m.f. and internal resistance

Vary the load, record V and I, plot V against I, then read ε and r.

Core Practical 8 determines a cell's e.m.f. and internal resistance experimentally. The circuit is a cell in series with an ammeter and a variable resistor (rheostat), with a voltmeter connected across the cell's terminals (i.e. across the variable resistor) to read the terminal p.d.

CP8 circuit: a cell in series with an ammeter and a variable resistor, with a voltmeter across the terminals reading the terminal p.d. Changing the variable resistor gives many (I, V) pairs to plot.

Method

Set the variable resistor and record the current $I$ (ammeter) and terminal p.d. $V$ (voltmeter) together.
Change the variable resistor and repeat, taking a range of 6–8 readings over as wide a current range as the cell safely allows.
Plot $V$ (y-axis) against $I$ (x-axis) and draw the best-fit straight line.
Read the e.m.f. from the y-intercept and the internal resistance from the magnitude of the gradient ( $r = |\text{gradient}|$ ).

Reducing and judging error

Cell warming. Drawing current heats the cell, which changes its internal resistance during the experiment. Take readings quickly and switch off between them (and avoid very large currents) so $r$ stays constant.
Contact and lead resistance. The resistance of the connecting leads, switch and crocodile clips is in series with the cell and is measured together with $r$ , making the apparent internal resistance too large. Keep leads short and connections clean.
Voltmeter loading. Use a high-resistance (digital) voltmeter so it draws negligible current and the ammeter reads the true load current.
Repeat and best-fit. Several readings and a best-fit line average out random error and reveal anomalies.

Circuit: cell + ammeter + variable resistor in series; voltmeter across the terminals.
Vary the resistor; record matching $I$ and $V$ pairs (6–8 readings).
Plot $V$ vs $I$ : intercept = $\varepsilon$ , gradient = $-r$ .
Take readings quickly to stop the cell warming and changing $r$ .

See the full worked example for emf and modelling resistance →

Combining cells, short circuits and power

Series adds, parallel halves r; short circuit gives ε/r; max power when R = r.

Combining cells. When sources are joined together, treat them like resistors plus e.m.f.s:

In series (end to end, same direction): the e.m.f.s add and the internal resistances add. Three $1.5\ \text{V}$ , $0.30\ \Omega$ cells in series give $4.5\ \text{V}$ with $0.90\ \Omega$ . This is how you get a higher voltage.
Identical cells in parallel (side by side): the e.m.f. is unchanged (still one cell's worth), but the internal resistances combine like parallel resistors, so $n$ identical cells give $r_{\text{total}} = \dfrac{r}{n}$ . This lowers the internal resistance, letting the battery supply a larger current with smaller lost volts and last longer.

Short-circuit current. If the terminals are connected by something with negligible resistance, $R = 0$ , so all the e.m.f. is dropped internally:

$I_{\text{max}} = \frac{\varepsilon}{r}$

A battery with a very small $r$ (like a car battery, $r \sim 0.04\ \Omega$ ) can therefore push a huge short-circuit current — hundreds of amps — which dissipates power $I^2 r$ as intense heat. That is why dropping a spanner across a car battery's terminals is dangerous.

Power and maximum power transfer. The total power supplied by the source is $\varepsilon I$ , split between the load and the cell:

$\varepsilon I = \underbrace{I^2 R}_{\text{to load}} + \underbrace{I^2 r}_{\text{lost in cell}}$

As the load resistance is varied, the power delivered to the load is greatest when the load resistance equals the internal resistance, $R = r$ — the maximum power transfer condition. (At that point exactly half the supplied power reaches the load; with $R \gg r$ the cell is efficient but delivers little power, and with $R \ll r$ most power is wasted inside the cell.)

Series cells: e.m.f.s add, internal resistances add.
Identical parallel cells: e.m.f. unchanged, $r_{\text{total}} = r/n$ .
Short circuit ( $R = 0$ ): $I_{\text{max}} = \varepsilon/r$ .
Maximum power to the load when $R = r$ .

See the full worked example for emf and modelling resistance →

Quick recap

e.m.f. $\varepsilon = W/Q$ is the energy supplied to each coulomb by the source (volts, J C⁻¹); it is not a force.
A real cell has internal resistance $r$ ; terminal p.d. $V = \varepsilon - Ir$ , where $Ir$ is the 'lost volts' dissipated inside the cell.
$\varepsilon = I(R + r)$ , so the current is $I = \varepsilon/(R + r)$ — always include $r$ unless the cell is ideal.
Plot $V$ against $I$ : the y-intercept is the e.m.f. and the gradient is $-r$ , so $r = |\text{gradient}|$ .
Terminal p.d. falls as current rises because $Ir$ grows; they equal the e.m.f. only on open circuit ( $I = 0$ ).
Short-circuit current $I_{\text{max}} = \varepsilon/r$ ; maximum power is transferred to the load when $R = r$ .
Series cells add e.m.f. and internal resistance; identical parallel cells keep the e.m.f. but give $r/n$ .

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

e.m.f.: $\varepsilon = \dfrac{W}{Q}$ — energy transferred to each coulomb by the source (volts).
$\varepsilon = V + Ir = I(R + r)$ and terminal p.d. $V = \varepsilon - Ir$ .
Lost volts $= Ir = \varepsilon - V$ (energy dissipated inside the cell per coulomb).
V–I graph: y-intercept = e.m.f.; gradient = $-r$ ; so $r = |\text{gradient}|$ .
Terminal p.d. = e.m.f. only when $I = 0$ (open circuit).
Short-circuit current $I_{\text{max}} = \dfrac{\varepsilon}{r}$ (when $R = 0$ ).
Series cells: e.m.f.s add, internal resistances add. Identical parallel cells: e.m.f. unchanged, $r \to r/n$ .
Maximum power transfer to the load when $R = r$ ; total power $\varepsilon I = I^2R + I^2r$ .

How it’s examined

EMF and internal resistance is core Unit 2 content (WPH12, Electric Circuits) and appears on the Unit 2 paper as short calculations, a 'show that', a graph-determination task and an extended/practical question. Typical formats: define e.m.f. or distinguish it from terminal p.d. (1–3 marks, AO1) — always frame e.m.f. as 'energy transferred per unit charge by the source'; single-step uses of $V = \varepsilon - Ir$ or $\varepsilon = I(R + r)$ (2–3 marks, M and A marks); the central 'determine the e.m.f. and internal resistance from the graph' item, where you must link $V = \varepsilon - Ir$ to $y = mx + c$ , read the intercept for $\varepsilon$ and a large-triangle gradient for $-r$ (4–5 marks, AO3); a CP8 'describe the experiment' or 'suggest improvements/sources of error' question (4–6 marks); and synoptic power/efficiency parts using $I^2r$ versus $I^2R$ . ALWAYS show the equation, substitution with units and a 3 s.f. answer — 'no working = risk zero'. The recurring discriminators are: gradient is $-r$ (so $r$ is positive); the voltmeter reads terminal p.d., not e.m.f.; and each practical improvement must state its mechanism.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 2 examiner reports; Edexcel IAL Physics Unit 2 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

Step-by-step worked examples — EMF and Modelling Resistance

Step-by-step solutions to past-paper-style questions on emf and modelling resistance, written exactly the way a tutor would explain them at the board.

Question type:

1e.m.f. from energy and charge (2 marks)
Getting startedDirect calculation• emf, AO2, Unit 2
Question
A battery transfers 36 J of energy to the charge carriers when 4.0 C of charge passes through it. Calculate the electromotive force (e.m.f.) of the battery. (2 marks)
Step-by-step solution
1. Step 1
  e.m.f. is the energy transferred to each unit of charge by the source, $\varepsilon = W/Q$ .
  $\varepsilon = \frac{W}{Q}$
2. Step 2
  Substitute the energy and charge with units.
  $\varepsilon = \frac{36}{4.0} = 9.0\ \text{V}$
Answer
$\varepsilon = 9.0\ \text{V}$ .
Examiner tip
Mark scheme: (1) use of $\varepsilon=W/Q$ (M); (2) $9.0\ \text{V}$ (A). The volt is a joule per coulomb, so an e.m.f. of $9.0\ \text{V}$ means $9.0\ \text{J}$ of energy is given to every coulomb that passes through the source.
2Terminal p.d. and lost volts (3 marks)
Getting startedDirect calculation• terminal pd, internal resistance, Unit 2
Question
A cell of e.m.f. 1.5 V has an internal resistance of 0.50 Ω. It drives a current of 0.40 A through an external circuit. Calculate (a) the 'lost volts' inside the cell and (b) the terminal potential difference. (3 marks)
Step-by-step solution
1. Step 1
  The 'lost volts' is the p.d. dropped across the internal resistance, $Ir$ .
  $\text{lost volts} = Ir = 0.40 \times 0.50 = 0.20\ \text{V}$
2. Step 2
  The terminal p.d. is the e.m.f. minus the lost volts, $V = \varepsilon - Ir$ .
  $V = \varepsilon - Ir = 1.5 - 0.20$
3. Step 3
  Evaluate the terminal p.d.
  $V = 1.3\ \text{V}$
Answer
(a) Lost volts $= 0.20\ \text{V}$ . (b) Terminal p.d. $= 1.3\ \text{V}$ .
Examiner tip
Mark scheme: (1) lost volts $=Ir=0.20\ \text{V}$ ; (2) use of $V=\varepsilon-Ir$ (M); (3) $1.3\ \text{V}$ (A). The terminal p.d. is always less than the e.m.f. whenever a current flows, because energy is dissipated inside the cell.
3Combining identical cells in series (3 marks)
Getting startedDirect calculation• cells in series, emf, Unit 2
Question
Three identical cells, each of e.m.f. 1.5 V and internal resistance 0.30 Ω, are connected in series to form a battery. State (a) the total e.m.f. and (b) the total internal resistance of the battery. (3 marks)
Step-by-step solution
1. Step 1
  Cells in series add their e.m.f.s (they 'push' in the same direction).
  $\varepsilon_{\text{total}} = 3 \times 1.5 = 4.5\ \text{V}$
2. Step 2
  Their internal resistances are in series, so they also add.
  $r_{\text{total}} = 3 \times 0.30 = 0.90\ \Omega$
3. Step 3
  State both results clearly — the battery behaves as a single source of e.m.f. $4.5\ \text{V}$ with internal resistance $0.90\ \Omega$ .
Answer
(a) Total e.m.f. $= 4.5\ \text{V}$ . (b) Total internal resistance $= 0.90\ \Omega$ .
Examiner tip
Mark scheme: (1) e.m.f.s add to $4.5\ \text{V}$ ; (2) internal resistances add in series; (3) $0.90\ \Omega$ . Contrast with $n$ identical cells in parallel, where the e.m.f. stays $1.5\ \text{V}$ but the internal resistance falls to $r/n = 0.10\ \Omega$ .
4Finding internal resistance from one measurement (4 marks)
Building confidenceMulti-step problem• Adapted from Edexcel IAL Physics Unit 2• internal resistance, AO2, Unit 2
Question
A cell of e.m.f. 1.50 V is connected to a 4.5 Ω resistor. A voltmeter across the resistor reads 1.35 V. Calculate (a) the current in the circuit and (b) the internal resistance of the cell. (4 marks)
Step-by-step solution
1. Step 1
  The voltmeter reads the terminal p.d., which is the p.d. across the external resistor. Find the current from $V = IR$ .
  $I = \frac{V}{R} = \frac{1.35}{4.5} = 0.300\ \text{A}$
2. Step 2
  The lost volts equals the e.m.f. minus the terminal p.d.
  $\text{lost volts} = \varepsilon - V = 1.50 - 1.35 = 0.15\ \text{V}$
3. Step 3
  The lost volts is $Ir$ , so rearrange for the internal resistance $r$ .
  $r = \frac{\varepsilon - V}{I} = \frac{0.15}{0.300}$
4. Step 4
  Evaluate the internal resistance.
  $r = 0.500\ \Omega$
Answer
(a) Current $= 0.300\ \text{A}$ . (b) Internal resistance $= 0.500\ \Omega$ .
Examiner tip
Mark scheme: (1) current from $I=V/R=0.300\ \text{A}$ (M+A); (2) lost volts $=\varepsilon-V=0.15\ \text{V}$ ; (3) use of $r=(\varepsilon-V)/I$ (M); (4) $0.500\ \Omega$ (A). A frequent slip is using the e.m.f. $1.50\ \text{V}$ (not the terminal p.d.) to find the current — the voltmeter reads the terminal p.d. across the load.
5Using ε = I(R + r) to find the current (3 marks)
Building confidenceDirect calculation• emf, internal resistance, AO2, Unit 2
Question
A cell of e.m.f. 6.0 V and internal resistance 0.80 Ω is connected to an external resistor of 2.2 Ω. Calculate (a) the current in the circuit and (b) the terminal potential difference. (3 marks)
Step-by-step solution
1. Step 1
  The e.m.f. is shared between the external and internal resistance: $\varepsilon = I(R + r)$ . Rearrange for the current.
  $I = \frac{\varepsilon}{R + r} = \frac{6.0}{2.2 + 0.80}$
2. Step 2
  Evaluate the current.
  $I = \frac{6.0}{3.0} = 2.0\ \text{A}$
3. Step 3
  The terminal p.d. is the p.d. across the external resistor, $V = IR$ (equivalently $V = \varepsilon - Ir$ ).
  $V = IR = 2.0 \times 2.2 = 4.4\ \text{V}$
Answer
(a) Current $= 2.0\ \text{A}$ . (b) Terminal p.d. $= 4.4\ \text{V}$ .
Examiner tip
Mark scheme: (1) use of $\varepsilon=I(R+r)$ (M); (2) $I=2.0\ \text{A}$ (A); (3) $V=IR=4.4\ \text{V}$ (or $V=\varepsilon-Ir=6.0-1.6=4.4\ \text{V}$ ). Check: lost volts $=Ir=2.0\times0.80=1.6\ \text{V}$ , and $4.4+1.6=6.0\ \text{V}$ recovers the e.m.f.
6Determining ε and r from a V–I graph (5 marks)
Building confidenceGraph or diagram• Adapted from Edexcel IAL Physics Unit 2 (CP8)• CP8, graph, AO3, Unit 2
Question
In an experiment the terminal p.d. V of a cell is plotted against the current I. The straight-line graph cuts the V-axis at 1.50 V, and when the current is 0.600 A the terminal p.d. has fallen to 1.20 V. Determine (a) the e.m.f. and (b) the internal resistance of the cell. (5 marks)
Step-by-step solution
1. Step 1
  Rearranging $V = \varepsilon - Ir$ as $V = -rI + \varepsilon$ and comparing with $y = mx + c$ : the y-intercept is the e.m.f. and the gradient is $-r$ .
  $V = \varepsilon - Ir \;\Rightarrow\; \text{intercept} = \varepsilon,\ \text{gradient} = -r$
2. Step 2
  The intercept on the V-axis (where $I = 0$ ) is the e.m.f.
  $\varepsilon = 1.50\ \text{V}$
3. Step 3
  Find the gradient using the two points $(0,\ 1.50)$ and $(0.600,\ 1.20)$ .
  $\text{gradient} = \frac{1.20 - 1.50}{0.600 - 0} = \frac{-0.30}{0.600} = -0.50\ \Omega$
4. Step 4
  The internal resistance is the magnitude of the gradient (gradient $= -r$ ).
  $r = -\text{gradient} = 0.50\ \Omega$
5. Step 5
  State both results: the cell behaves as a source of e.m.f. $1.50\ \text{V}$ with internal resistance $0.50\ \Omega$ .
Answer
(a) e.m.f. $= 1.50\ \text{V}$ . (b) Internal resistance $= 0.50\ \Omega$ .
Examiner tip
Mark scheme: (1) link $V=\varepsilon-Ir$ to $y=mx+c$ — intercept $=\varepsilon$ , gradient $=-r$ ; (2) $\varepsilon=1.50\ \text{V}$ ; (3) gradient $=-0.50\ \Omega$ from a large triangle (M); (4) $r=0.50\ \Omega$ (A); (5) clear final statement. The dominant error is quoting $r$ as the gradient value $-0.50$ (with a sign) instead of recognising gradient $=-r$ , so $r=+0.50\ \Omega$ .
7Short-circuit current of a cell (4 marks)
StretchMulti-step problem• short circuit, internal resistance, AO2, Unit 2
Question
A car battery has an e.m.f. of 12.0 V and an internal resistance of 0.040 Ω. (a) Calculate the maximum current the battery can deliver (when its terminals are short-circuited). (b) Explain why a battery with a very low internal resistance can be dangerous if its terminals are accidentally connected by a spanner. (4 marks)
Step-by-step solution
1. Step 1
  A short circuit means the external resistance $R = 0$ , so all the e.m.f. is dropped across the internal resistance. From $\varepsilon = I(R + r)$ with $R = 0$ , $I = \varepsilon/r$ .
  $I_{\text{max}} = \frac{\varepsilon}{r} = \frac{12.0}{0.040}$
2. Step 2
  Evaluate the short-circuit current.
  $I_{\text{max}} = 300\ \text{A}$
3. Step 3
  Because the internal resistance is very small, the short-circuit current is very large.
4. Step 4
  This very large current dissipates power $P = I^2 r$ in the cell and in the spanner, generating intense heat very quickly — the spanner can become red hot, the battery can overheat and the electrolyte can boil, risking burns or an explosion.
Answer
(a) Short-circuit current $= 300\ \text{A}$ . (b) The very large current produces rapid resistive heating ( $P = I^2r$ ) in the spanner and battery, causing burns/fire/explosion.
Examiner tip
Mark scheme: (1) $R=0$ so $I=\varepsilon/r$ (M); (2) $300\ \text{A}$ (A); (3) low $r$ → large current; (4) heating effect $P=I^2r$ → danger. The internal resistance is what limits the current in a short circuit; without it the current would be theoretically infinite.
8Internal resistance from two load settings (6 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Physics Unit 2 (CP8)• CP8, internal resistance, AO3, Unit 2
Question
A student investigates a cell using a variable resistor. At a current of 0.20 A the terminal p.d. is 1.42 V; at a current of 0.50 A the terminal p.d. is 1.30 V. Using these two data points and the equation V = ε − Ir, calculate (a) the internal resistance and (b) the e.m.f. of the cell. (6 marks)
Step-by-step solution
1. Step 1
  Write $V = \varepsilon - Ir$ for each reading. Reading 1: $1.42 = \varepsilon - 0.20r$ . Reading 2: $1.30 = \varepsilon - 0.50r$ .
  $1.42 = \varepsilon - 0.20r \quad\text{and}\quad 1.30 = \varepsilon - 0.50r$
2. Step 2
  Subtract the second equation from the first to eliminate $\varepsilon$ . This is equivalent to finding the gradient $-r$ between the two points.
  $1.42 - 1.30 = (\varepsilon - 0.20r) - (\varepsilon - 0.50r)$
3. Step 3
  Simplify both sides.
  $0.12 = 0.30r$
4. Step 4
  Rearrange for the internal resistance $r$ .
  $r = \frac{0.12}{0.30} = 0.40\ \Omega$
5. Step 5
  Substitute $r = 0.40\ \Omega$ back into the first equation to find $\varepsilon$ .
  $\varepsilon = 1.42 + 0.20 \times 0.40 = 1.42 + 0.080$
6. Step 6
  Evaluate the e.m.f.
  $\varepsilon = 1.50\ \text{V}$
Answer
(a) Internal resistance $= 0.40\ \Omega$ . (b) e.m.f. $= 1.50\ \text{V}$ .
Examiner tip
Mark scheme: (1) two simultaneous equations from $V=\varepsilon-Ir$ ; (2)/(3) subtract to eliminate $\varepsilon$ (M); (4) $r=0.40\ \Omega$ (A); (5) back-substitute (M); (6) $\varepsilon=1.50\ \text{V}$ (A). The subtraction is exactly the graphical gradient method ( $r=-\Delta V/\Delta I$ ) done algebraically; check with the second point: $\varepsilon-0.50\times0.40=1.50-0.20=1.30\ \text{V}$ ✔.
9Power delivered to the load and lost in the cell (6 marks)
StretchMulti-step problem• power, internal resistance, AO2, Unit 2
Question
A battery of e.m.f. 9.0 V and internal resistance 1.5 Ω is connected to a 3.0 Ω resistor. Calculate (a) the current, (b) the power delivered to the external resistor, and (c) the power dissipated inside the battery. (d) Hence state the efficiency of energy transfer to the load. (6 marks)
Step-by-step solution
1. Step 1
  Find the current using $\varepsilon = I(R + r)$ .
  $I = \frac{\varepsilon}{R + r} = \frac{9.0}{3.0 + 1.5} = \frac{9.0}{4.5} = 2.0\ \text{A}$
2. Step 2
  Power delivered to the external resistor is $P_R = I^2 R$ .
  $P_R = I^2 R = (2.0)^2 \times 3.0 = 12\ \text{W}$
3. Step 3
  Power dissipated inside the battery is $P_r = I^2 r$ .
  $P_r = I^2 r = (2.0)^2 \times 1.5 = 6.0\ \text{W}$
4. Step 4
  The total power supplied by the e.m.f. is $\varepsilon I = 9.0 \times 2.0 = 18\ \text{W}$ , which equals $P_R + P_r = 12 + 6.0 = 18\ \text{W}$ ✔. The efficiency is the useful power to the load divided by the total power.
  $\eta = \frac{P_R}{\varepsilon I} = \frac{12}{18}$
5. Step 5
  Evaluate the efficiency as a percentage.
  $\eta = 0.667 = 66.7\%$
Answer
(a) Current $= 2.0\ \text{A}$ . (b) Power to load $= 12\ \text{W}$ . (c) Power lost in cell $= 6.0\ \text{W}$ . (d) Efficiency $= 66.7\%$ .
Examiner tip
Mark scheme: (1) $I=2.0\ \text{A}$ (M+A); (2) $P_R=I^2R=12\ \text{W}$ ; (3) $P_r=I^2r=6.0\ \text{W}$ ; (4) total power and efficiency ratio (M); (5) $66.7\%$ (A). Maximum power transfer to the load occurs when $R=r$ ; here $R>r$ so the cell is reasonably efficient but not at peak power.

Model Answers — EMF and Modelling Resistance

High-scoring sample answers for emf and modelling resistance on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define the electromotive force (e.m.f.) of a source. Include a word equation in your answer.
Model answer
The electromotive force (e.m.f.) of a source is the energy transferred to each unit of charge (per coulomb) by the source when charge passes through it — that is, the chemical (or other) energy converted into electrical energy per coulomb.

In symbols, $\varepsilon = \dfrac{W}{Q}$ , where $W$ is the energy transferred and $Q$ is the charge. It is measured in volts (joules per coulomb).
Why this scores
Why this scores 2/2. (1) energy transferred to the charge by the source; (2) per unit charge (or the equation $\varepsilon=W/Q$ ). Saying e.m.f. is 'the voltage of the battery' is not credited — the defining idea is energy supplied per coulomb, not the p.d. measured across the terminals.
Question 2
3 marks
Q (3 marks). Explain the difference between the e.m.f. of a cell and its terminal potential difference, and state the condition under which they are equal.
Model answer
The e.m.f. is the total energy supplied to each coulomb of charge by the cell (the energy converted from chemical to electrical form per unit charge). The terminal potential difference is the energy delivered to the external circuit per unit charge — i.e. the p.d. measured across the cell's terminals.

When a current flows, some energy is dissipated inside the cell across its internal resistance (the 'lost volts', $Ir$ ), so the terminal p.d. is less than the e.m.f.: $V = \varepsilon - Ir$ .

They are equal only when no current flows ( $I = 0$ ) — for example when the cell is on open circuit, or measured by a very high-resistance voltmeter, so $Ir = 0$ .
Why this scores
Why this scores 3/3. (1) e.m.f. = energy supplied per coulomb (total); (2) terminal p.d. = energy delivered to the external circuit per coulomb, smaller because of $Ir$ ; (3) equal when $I=0$ (open circuit / no current). The condition mark is the discriminator — many candidates describe the difference but omit when they are equal.
Question 3
3 marks
Q (3 marks). A cell of e.m.f. 1.50 V delivers a current of 0.25 A to a circuit and the terminal p.d. is measured as 1.35 V. Calculate the internal resistance of the cell, showing your working.
Model answer
The 'lost volts' is the difference between the e.m.f. and the terminal p.d.:

$V_{\text{lost}} = \varepsilon - V = 1.50 - 1.35 = 0.15\ \text{V}$

This lost p.d. is dropped across the internal resistance, so $V_{\text{lost}} = Ir$ . Rearranging for $r$ :

$r = \frac{\varepsilon - V}{I} = \frac{0.15}{0.25} = 0.60\ \Omega$

The internal resistance of the cell is therefore 0.60 Ω.
Why this scores
Why this scores 3/3. (1) lost volts $=\varepsilon-V=0.15\ \text{V}$ ; (2) use of $r=(\varepsilon-V)/I$ (M); (3) $0.60\ \Omega$ (A). Watch the arithmetic: $0.15/0.25=0.60$ , not $0.06$ . Full working secures both calculation marks even if the final division is slipped.
Question 4
4 marks
Q (4 marks). Using the equation $V = \varepsilon - Ir$ , explain why the terminal potential difference of a cell falls as the current drawn from it increases. Hence explain why a car's headlights dim momentarily when the starter motor is engaged.
Model answer
From $V = \varepsilon - Ir$ , the terminal p.d. equals the (fixed) e.m.f. minus the 'lost volts' $Ir$ across the internal resistance. The e.m.f. $\varepsilon$ and internal resistance $r$ are constant, so as the current $I$ increases the term $Ir$ increases, and therefore the terminal p.d. $V$ decreases. More energy per coulomb is dissipated inside the cell, leaving less for the external circuit.

When the starter motor is engaged it draws a very large current from the battery. This makes the lost volts $Ir$ large, so the terminal p.d. drops sharply. The headlights are connected across the same terminals, so the reduced terminal p.d. means less power is delivered to them and they dim momentarily. Once the engine starts and the starter disengages, the current falls, $Ir$ becomes small again, the terminal p.d. recovers and the lights brighten.
Why this scores
Why this scores 4/4. (1) $\varepsilon$ and $r$ constant; (2) as $I$ increases, $Ir$ increases so $V$ decreases; (3) starter draws a large current → large $Ir$ → terminal p.d. drops; (4) headlights across the terminals receive less p.d./power and dim. Linking the equation to the real context (not just reciting it) is what earns the application marks.
Question 5
4 marks
Q (4 marks). A battery of e.m.f. 12.0 V and internal resistance 0.50 Ω is connected to an external resistor of 5.5 Ω. (a) Calculate the current in the circuit. (b) Calculate the terminal potential difference. (c) Calculate the energy lost inside the battery in 60 s.
Model answer
(a) The e.m.f. drives the current through the total resistance $R + r$ :

$I = \frac{\varepsilon}{R + r} = \frac{12.0}{5.5 + 0.50} = \frac{12.0}{6.0} = 2.0\ \text{A}$

(b) The terminal p.d. is the e.m.f. minus the lost volts:

$V = \varepsilon - Ir = 12.0 - (2.0)(0.50) = 12.0 - 1.0 = 11.0\ \text{V}$

(c) The power dissipated inside the battery is $P_r = I^2 r = (2.0)^2 \times 0.50 = 2.0\ \text{W}$ . Over 60 s the energy lost is:

$E = P_r t = 2.0 \times 60 = 120\ \text{J}$
Why this scores
Why this scores 4/4. (a) $I=\varepsilon/(R+r)=2.0\ \text{A}$ (M+A); (b) $V=\varepsilon-Ir=11.0\ \text{V}$ ; (c) $P_r=I^2r=2.0\ \text{W}$ then $E=Pt=120\ \text{J}$ . Equivalently the energy lost $=I^2rt$ . A common error in (c) is using the terminal p.d. or e.m.f. instead of $I^2r$ for the internal losses.
Question 6
Adapted from Edexcel IAL Physics Unit 2 (CP8)5 marks
Q (5 marks). A student plots terminal p.d. $V$ (y-axis) against current $I$ (x-axis) for a cell and obtains a straight line. (a) Use $V = \varepsilon - Ir$ to state what the y-intercept and the gradient of this graph represent. (b) The line has a y-intercept of 1.55 V and passes through the point (0.80 A, 1.15 V). Determine the e.m.f. and the internal resistance of the cell.
Model answer
(a) Writing the equation as $V = -rI + \varepsilon$ and comparing with the straight-line form $y = mx + c$ (with $y = V$ , $x = I$ ):

the y-intercept $c$ is the e.m.f. $\varepsilon$ (the terminal p.d. when $I = 0$ );

the gradient $m$ is $-r$ (negative the internal resistance).

(b) The y-intercept gives the e.m.f. directly:

$\varepsilon = 1.55\ \text{V}$

The gradient is found from the intercept point $(0, 1.55)$ and the given point $(0.80, 1.15)$ :

$\text{gradient} = \frac{1.15 - 1.55}{0.80 - 0} = \frac{-0.40}{0.80} = -0.50\ \Omega$

Since the gradient equals $-r$ , the internal resistance is the magnitude:

$r = 0.50\ \Omega$
Why this scores
Why this scores 5/5. (a) 2 marks: intercept $=\varepsilon$ AND gradient $=-r$ (both needed). (b) 3 marks: $\varepsilon=1.55\ \text{V}$ ; gradient $=-0.50\ \Omega$ from a large triangle; $r=0.50\ \Omega$ . The sign is the key discriminator — the gradient is negative, but $r$ itself is a positive $0.50\ \Omega$ because gradient $=-r$ .
Question 7
Adapted from Edexcel IAL Physics Unit 2 (CP8)5 marks
Q (5 marks). Describe an experiment (Core Practical) to determine the e.m.f. and internal resistance of a cell. Include the circuit, the measurements taken, the graph plotted, and how you would obtain ε and r from it.
Model answer
Circuit. Connect the cell in series with an ammeter and a variable resistor (rheostat), and connect a voltmeter across the cell's terminals (i.e. across the variable resistor) to read the terminal p.d.

Measurements. Vary the variable resistor to change the current. For each setting, record the current $I$ (ammeter) and the corresponding terminal p.d. $V$ (voltmeter). Take a range of readings (e.g. 6–8 values) over as wide a current range as the cell safely allows.

Graph. Plot terminal p.d. $V$ on the y-axis against current $I$ on the x-axis. The points should lie on a straight line with a negative gradient.

Analysis. Comparing $V = \varepsilon - Ir$ with $y = mx + c$ : the y-intercept (at $I = 0$ ) gives the e.m.f. $\varepsilon$ , and the gradient gives $-r$ , so the internal resistance $r$ is the magnitude of the gradient (read from a large triangle).

To protect the cell, take readings quickly and switch off between them so it does not warm up (a warming cell changes $r$ ).
Why this scores
Why this scores 5/5. (1) circuit: cell, ammeter and variable resistor in series, voltmeter across the terminals; (2) vary $R$ , record $I$ and $V$ for several settings; (3) plot $V$ against $I$ ; (4) intercept $=\varepsilon$ ; (5) gradient $=-r$ so $r=|\text{gradient}|$ . Naming the variable resistor and stating the voltmeter is across the terminals (not just one component) are the precision marks.
Question 8
6 marks
Q (6 marks). A cell of e.m.f. 1.50 V and internal resistance 0.50 Ω is connected to a variable load resistor $R$ . (a) Show that when $R = 0.50\ \Omega$ the power delivered to the load is 1.125 W. (b) Calculate the power delivered to the load when $R = 2.0\ \Omega$ , and comment on which load receives more power. (c) State the general condition for maximum power transfer to the load.
Model answer
(a) When $R = 0.50\ \Omega$ , the current is

$I = \frac{\varepsilon}{R + r} = \frac{1.50}{0.50 + 0.50} = \frac{1.50}{1.00} = 1.50\ \text{A}$

The power delivered to the load is

$P_R = I^2 R = (1.50)^2 \times 0.50 = 2.25 \times 0.50 = 1.125\ \text{W}$

as required.

(b) When $R = 2.0\ \Omega$ :

$I = \frac{1.50}{2.0 + 0.50} = \frac{1.50}{2.5} = 0.60\ \text{A}, \qquad P_R = (0.60)^2 \times 2.0 = 0.36 \times 2.0 = 0.72\ \text{W}$

The $R = 0.50\ \Omega$ load receives more power (1.125 W) than the $R = 2.0\ \Omega$ load (0.72 W), even though it has the smaller resistance.

(c) The maximum power is transferred to the load when the load resistance equals the internal resistance, $R = r$ . That is exactly the case in part (a), which is why the smaller load delivers the greater power here.
Why this scores
Why this scores 6/6. (a) 3 marks: $I=1.50\ \text{A}$ , $P=I^2R$ , work through to $1.125\ \text{W}$ (a 'show that' must reach the value). (b) 2 marks: $I=0.60\ \text{A}$ , $P=0.72\ \text{W}$ , and identify that the matched load wins. (c) 1 mark: maximum power transfer when $R=r$ . Quoting the answer in (a) without the working would lose the 'show that' marks.
Question 9
6 marks
Q (6 marks). Two identical cells, each of e.m.f. 1.50 V and internal resistance 0.40 Ω, are connected to a 3.2 Ω resistor. (a) Calculate the current when the cells are connected in series. (b) Calculate the current when the cells are connected in parallel. (c) Explain, in terms of e.m.f. and internal resistance, the advantage of each arrangement.
Model answer
(a) Series. The e.m.f.s add and the internal resistances add:

$\varepsilon_{\text{series}} = 2 \times 1.50 = 3.00\ \text{V}, \qquad r_{\text{series}} = 2 \times 0.40 = 0.80\ \Omega$

$I = \frac{\varepsilon_{\text{series}}}{R + r_{\text{series}}} = \frac{3.00}{3.2 + 0.80} = \frac{3.00}{4.0} = 0.75\ \text{A}$

(b) Parallel. For identical cells the e.m.f. stays the same, but the two equal internal resistances combine in parallel:

$\varepsilon_{\text{parallel}} = 1.50\ \text{V}, \qquad r_{\text{parallel}} = \frac{0.40}{2} = 0.20\ \Omega$

$I = \frac{1.50}{3.2 + 0.20} = \frac{1.50}{3.4} = 0.44\ \text{A}$

(c) The series arrangement gives a larger e.m.f. (3.00 V), and so a larger current and p.d. across the load — useful when a higher voltage is needed. The parallel arrangement keeps the e.m.f. at 1.50 V but halves the internal resistance (0.20 Ω), which reduces the lost volts at a given current and lets the battery supply a larger current for longer without its terminal p.d. dropping as much — useful for high-current demands and longer life.
Why this scores
Why this scores 6/6. (a) 2 marks: series $\varepsilon=3.00\ \text{V}$ , $r=0.80\ \Omega$ , $I=0.75\ \text{A}$ . (b) 2 marks: parallel $\varepsilon=1.50\ \text{V}$ , $r=0.20\ \Omega$ , $I=0.44\ \text{A}$ . (c) 2 marks: series → higher e.m.f./voltage; parallel → lower internal resistance / higher current capacity. The discriminator is correctly stating that the parallel e.m.f. is unchanged while only $r$ falls.
Question 10
Adapted from Edexcel IAL Physics Unit 2 (CP8)6 marks
Q (6 marks). In a CP8 experiment to find the internal resistance of a cell, a student's $V$ – $I$ graph is a straight line but does not pass exactly through their plotted points, and the value of $r$ obtained is unexpectedly large. (a) Give two sources of error that could increase the apparent internal resistance, explaining each. (b) Describe two improvements the student could make to obtain a more reliable value of $r$ .
Model answer
(a) Sources of error.

Contact/connecting-lead resistance. The resistance of the connecting wires, switch and crocodile-clip contacts is in series with the cell, so it is measured together with the true internal resistance. This makes the gradient steeper and the apparent $r$ larger than the cell's real internal resistance.

The cell warming up as current flows. Drawing a large current heats the cell, which changes its internal resistance (and can shift its e.m.f.) during the experiment. The readings then do not all lie on one straight line, and the value of $r$ is unreliable.

(b) Improvements.

Use a voltmeter with very high resistance (or a digital meter) so that negligible current is diverted through it; this ensures the ammeter reads the true load current and the terminal p.d. is measured accurately.

Take readings quickly and switch off the circuit between readings (and keep the current modest) so the cell does not heat up, keeping $r$ constant. Repeating readings and using a best-fit line through many points further reduces the effect of random error.
Why this scores
Why this scores 6/6. (a) up to 3 marks: contact/lead resistance adds in series → larger apparent $r$ ; cell warming changes $r$ → scatter. (b) up to 3 marks: high-resistance voltmeter to avoid diverting current; switch off between readings/short bursts to prevent warming; best-fit line/repeats. Each point must include the mechanism — 'be more careful' is not credited.

Key Formulae — EMF and Modelling Resistance

The formulae you need to memorise for emf and modelling resistance on the Cambridge IGCSE paper, with every variable defined in plain English and a note on when to use it.

Electromotive force (definition)
$\varepsilon = \frac{W}{Q}$
$\varepsilon$
electromotive force / V
$W$
energy transferred to the charge by the source / J
$Q$
charge that passes / C
When to use
Defining e.m.f. as the energy given to each coulomb by the source. Must be recalled (a volt is a joule per coulomb).
Example
$36\ \text{J}$ to $4.0\ \text{C}$ : $\varepsilon=36/4.0=9.0\ \text{V}$ .
e.m.f., terminal p.d. and lost volts
$\varepsilon = V + Ir$
$\varepsilon$
e.m.f. of the source / V
$V$
terminal potential difference / V
$I$
current / A
$r$
internal resistance / $\Omega$
When to use
Relating the e.m.f. to the terminal p.d. delivered to the external circuit plus the 'lost volts' $Ir$ inside the source. In the Edexcel formulae booklet.
Example
$\varepsilon=1.5\ \text{V},\ I=0.40,\ r=0.50$ : $V=1.5-0.20=1.3\ \text{V}$ .
Terminal potential difference
$V = \varepsilon - Ir$
$V$
terminal p.d. (p.d. across the external load) / V
$\varepsilon$
e.m.f. / V
$Ir$
'lost volts' across the internal resistance / V
When to use
Finding the terminal p.d. when a current flows, or as a straight-line equation $V=-rI+\varepsilon$ for the CP8 graph (intercept $=\varepsilon$ , gradient $=-r$ ). Rearrangement of the booklet formula.
Example
Graph: V-intercept $=1.50\ \text{V}=\varepsilon$ ; gradient $=-0.50\ \Omega=-r$ .
e.m.f. as current × total resistance
$\varepsilon = I(R + r)$
$\varepsilon$
e.m.f. / V
$I$
current / A
$R$
external (load) resistance / $\Omega$
$r$
internal resistance / $\Omega$
When to use
Finding the current in a single-loop circuit with internal resistance, or the short-circuit current $I=\varepsilon/r$ when $R=0$ . Rearrangement of the booklet formula.
Example
$\varepsilon=6.0,\ R=2.2,\ r=0.80$ : $I=6.0/3.0=2.0\ \text{A}$ .
Lost volts
$V_{\text{lost}} = Ir = \varepsilon - V$
$V_{\text{lost}}$
p.d. dropped across the internal resistance / V
$I$
current / A
$r$
internal resistance / $\Omega$
When to use
Finding the energy lost per coulomb inside the cell — the difference between the e.m.f. and the terminal p.d. Must be recalled.
Example
$I=0.40,\ r=0.50$ : lost volts $=0.20\ \text{V}$ , so $V=\varepsilon-0.20$ .
Power dissipated in internal resistance
$P_r = I^2 r$
$P_r$
power dissipated inside the source / W
$I$
current / A
$r$
internal resistance / $\Omega$
When to use
Calculating the power wasted as heat inside a cell (causing it to warm up), or comparing it with the useful power $P_R=I^2R$ delivered to the load. Recall (from $P=I^2R$ ).
Example
$I=2.0,\ r=1.5$ : $P_r=4.0\times1.5=6.0\ \text{W}$ heating the cell.

Key Definitions and Keywords — EMF and Modelling Resistance

Definitions to memorise and the exact keywords mark schemes credit for emf and modelling resistance answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Electromotive force (e.m.f.)
Examiner keyword
The energy transferred to each unit of charge (per coulomb) by a source when charge passes through it: $\varepsilon = W/Q$ . Measured in volts. It is NOT a force.
Terminal potential difference
Examiner keyword
The energy delivered to the external circuit per unit charge — the p.d. measured across the terminals of the source: $V = \varepsilon - Ir$ . Equals the e.m.f. only when no current flows.
Potential difference (p.d.)
Examiner keyword
The energy transferred from electrical to other forms per unit charge between two points in a circuit: $V = W/Q$ . Measured in volts (J C⁻¹).
Internal resistance (r)
Examiner keyword
The resistance of the materials inside a source (e.g. the electrolyte of a cell), which causes energy to be dissipated within the source itself. Measured in ohms.
Lost volts
Examiner keyword
The potential difference dropped across the internal resistance, $Ir$ , equal to the difference between the e.m.f. and the terminal p.d.: $V_{\text{lost}} = \varepsilon - V = Ir$ . Represents energy dissipated inside the source per coulomb.
Volt
The SI unit of e.m.f. and potential difference. One volt is one joule of energy transferred per coulomb of charge ( $1\ \text{V} = 1\ \text{J C}^{-1}$ ).
EMF equation
Examiner keyword
$\varepsilon = I(R + r)$ (equivalently $\varepsilon = V + Ir$ ): the e.m.f. equals the current times the total circuit resistance, sharing the energy between the load ( $IR$ ) and the internal resistance ( $Ir$ ).
Source of e.m.f.
Any device that converts another form of energy into electrical energy and drives charge round a circuit — e.g. a cell (chemical), a generator (mechanical), a solar cell (light) or a thermocouple (thermal).
Open-circuit voltage
Examiner keyword
The terminal p.d. of a source when no current is drawn ( $I = 0$ ). Because $Ir = 0$ , the open-circuit voltage equals the e.m.f. of the source.
Short-circuit current
Examiner keyword
The maximum current a source can deliver, obtained when the external resistance is zero ( $R = 0$ ): $I_{\text{max}} = \varepsilon/r$ . It is limited only by the internal resistance.
Terminal p.d.–current graph
Examiner keyword
A plot of terminal p.d. $V$ (y-axis) against current $I$ (x-axis). From $V = \varepsilon - Ir$ , the y-intercept equals the e.m.f. and the gradient equals $-r$ (negative the internal resistance).
Cells in series
Cells connected end to end: their e.m.f.s add ( $\varepsilon_{\text{total}} = \varepsilon_1 + \varepsilon_2 + \dots$ ) and their internal resistances add. Gives a larger total e.m.f.
Identical cells in parallel
Identical cells connected side by side: the combined e.m.f. is unchanged (equal to that of one cell), but the total internal resistance falls to $r/n$ for $n$ cells, allowing a larger current.
Maximum power transfer
The condition under which the power delivered to the external load is greatest, which occurs when the load resistance equals the internal resistance ( $R = r$ ).
Core Practical 8
Examiner keyword
The Edexcel practical to determine the e.m.f. and internal resistance of a cell, using a variable load resistor while measuring the terminal p.d. and current and plotting a $V$ – $I$ graph.

Common Mistakes and Misconceptions — EMF and Modelling Resistance

The traps other students keep falling into on emf and modelling resistance questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Confusing e.m.f. with terminal potential difference (treating them as the same 'battery voltage')
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
On open circuit they are equal and a labelled cell quotes its e.m.f., so students assume the terminals always sit at the e.m.f.
How to avoid it
The e.m.f. is the energy supplied per coulomb; the terminal p.d. is what is delivered to the external circuit. Whenever a current flows, $V = \varepsilon - Ir < \varepsilon$ . They are equal only when $I = 0$ .
✕On a V–I graph, quoting the internal resistance as the gradient value (e.g. $r = -0.50\ \Omega$ ) instead of recognising gradient $= -r$
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students read the (negative) gradient straight off and forget the equation $V = \varepsilon - Ir$ makes the gradient $-r$ , not $r$ .
How to avoid it
Compare $V = \varepsilon - Ir$ with $y = mx + c$ : the gradient is $-r$ , so the internal resistance is the magnitude (positive value) of the gradient, $r = |\text{gradient}|$ . Resistance can never be negative.
✕Misunderstanding 'lost volts' — thinking it is energy that vanishes, or computing it from the wrong quantity
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
The phrase 'lost' suggests the energy disappears, and students sometimes use the terminal p.d. or e.m.f. rather than $Ir$ .
How to avoid it
The 'lost volts' is $Ir$ , the p.d. across the internal resistance — energy is not lost but dissipated as heat inside the cell. Compute it as $\varepsilon - V$ or as $Ir$ , and note it grows as the current increases.
✕Assuming a voltmeter across the cell reads the e.m.f. even while a current flows
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students forget that the voltmeter is connected across the terminals, so it reads the terminal p.d., not the e.m.f.
How to avoid it
A voltmeter across the cell reads the terminal p.d. ( $V = \varepsilon - Ir$ ). It reads (approximately) the e.m.f. only on open circuit — when the rest of the circuit is disconnected so that $I \approx 0$ .
✕Using only the external resistance $R$ (forgetting the internal resistance $r$ ) when calculating the current
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
In earlier circuits the cell was treated as ideal, so students apply $I = \varepsilon/R$ instead of $I = \varepsilon/(R + r)$ .
How to avoid it
The current depends on the total resistance: $I = \dfrac{\varepsilon}{R + r}$ . Always include the internal resistance in the denominator unless the question states the cell is ideal ( $r = 0$ ).
✕Using the terminal p.d. or e.m.f. instead of $I^2 r$ to find the power dissipated inside the cell
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students reach for $P = VI$ with whichever voltage is to hand, without checking which resistance the power is dissipated in.
How to avoid it
Power inside the cell is $P_r = I^2 r$ (heating the cell); power to the load is $P_R = I^2 R = VI$ ; total power supplied is $\varepsilon I = P_R + P_r$ . Match the resistance to the location of the energy transfer.

EMF and Modelling Resistance — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

EMF and Modelling Resistance

Short Notes - EMF and Modelling Resistance

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1e.m.f. from energy and charge (2 marks)

2Terminal p.d. and lost volts (3 marks)

3Combining identical cells in series (3 marks)

4Finding internal resistance from one measurement (4 marks)

5Using ε = I(R + r) to find the current (3 marks)

6Determining ε and r from a V–I graph (5 marks)

7Short-circuit current of a cell (4 marks)

8Internal resistance from two load settings (6 marks)

9Power delivered to the load and lost in the cell (6 marks)

Electromotive force (definition)

e.m.f., terminal p.d. and lost volts

Terminal potential difference

e.m.f. as current × total resistance

Lost volts

Power dissipated in internal resistance

Electromotive force (e.m.f.)

Terminal potential difference

Potential difference (p.d.)

Internal resistance (r)

Lost volts

Volt

EMF equation

Source of e.m.f.

Open-circuit voltage

Short-circuit current

Terminal p.d.–current graph

Cells in series

Identical cells in parallel

Maximum power transfer

Core Practical 8

✕Confusing e.m.f. with terminal potential difference (treating them as the same 'battery voltage')

✕On a V–I graph, quoting the internal resistance as the gradient value (e.g. r=−0.50 Ωr = -0.50\ \Omegar=−0.50 Ω) instead of recognising gradient =−r= -r=−r

✕Misunderstanding 'lost volts' — thinking it is energy that vanishes, or computing it from the wrong quantity

✕Assuming a voltmeter across the cell reads the e.m.f. even while a current flows

✕Using only the external resistance RRR (forgetting the internal resistance rrr) when calculating the current

✕Using the terminal p.d. or e.m.f. instead of I2rI^2 rI2r to find the power dissipated inside the cell

EMF and Modelling Resistance — frequently asked questions

✕On a V–I graph, quoting the internal resistance as the gradient value (e.g. $r = -0.50\ \Omega$ ) instead of recognising gradient $= -r$

✕Using only the external resistance $R$ (forgetting the internal resistance $r$ ) when calculating the current

✕Using the terminal p.d. or e.m.f. instead of $I^2 r$ to find the power dissipated inside the cell