What is resistivity and how is it different from resistance in the context of Edexcel International A Levels Physics?

Resistivity is a fundamental property of a material that quantifies how strongly it opposes the flow of electric current. It is denoted by the symbol ρ (rho) and is measured in ohm-meters (Ω·m). Unlike resistance, which depends on the dimensions of the conductor, resistivity is an intrinsic property of the material itself. In the Edexcel International A Levels Physics curriculum, understanding resistivity helps in analyzing how different materials affect the flow of electricity in circuits.

How do you calculate the resistance of a wire using resistivity?

The resistance (R) of a wire can be calculated using the formula R = ρ(L/A), where ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area. This formula is crucial in the Edexcel International A Levels Physics syllabus as it helps students understand how the physical dimensions and material properties of a wire influence its resistance.

What is a potential divider and how is it used in circuits?

A potential divider is a simple circuit that uses two or more resistors to produce a specific output voltage from a larger input voltage. It is commonly used in electronic circuits to adjust signal levels, bias active devices, or measure voltages. In the context of Edexcel International A Levels Physics, understanding potential dividers is essential for analyzing and designing circuits that require precise voltage control.

How does temperature affect resistivity and resistance in materials?

Temperature has a significant impact on resistivity and resistance. For most conductors, as temperature increases, resistivity and resistance also increase due to increased atomic vibrations that impede electron flow. Conversely, in semiconductors, resistivity decreases with temperature rise. This concept is important in Edexcel International A Levels Physics as it helps students predict and manage the behavior of materials in different thermal conditions.

Can you explain the role of a potential divider in sensor circuits?

In sensor circuits, a potential divider is often used to convert a change in resistance (from a sensor) into a change in voltage, which can then be easily measured or processed. For example, in a temperature sensor circuit, the sensor's resistance changes with temperature, and the potential divider converts this change into a corresponding voltage change. This application is a key part of the Edexcel International A Levels Physics curriculum, illustrating practical uses of potential dividers in real-world scenarios.

Why is "Putting the diameter straight into $\pi r^2$ (treating the diameter as the radius) when finding the cross-sectional area" a common mistake in Resistivity and Potential dividers?

Why it happens: The micrometer gives the diameter, but the familiar circle formula is r^2, so students forget to halve it (or to use d^2/4). How to avoid it: **Use A= d^2/4 directly with the diameter, or halve the diameter first (r=d/2) for r^2.** Forgetting this makes the area four times too big and every resistivity answer wrong. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Putting the wrong resistor on top of the potential-divider equation (not the one the output is taken across)" a common mistake in Resistivity and Potential dividers?

Why it happens: Students memorise V_out=V_inR_2/(R_1+R_2) without checking which resistor the output is actually measured across in the given circuit. How to avoid it: **The resistor the output is taken across goes ON TOP of the fraction.** Read the circuit: trace where the output terminals connect, and put that resistance in the numerator. Sanity-check that the larger resistor gets the larger share of the voltage. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Thinking an NTC thermistor's resistance INCREASES as it gets hotter" a common mistake in Resistivity and Potential dividers?

Why it happens: Students assume thermistors behave like metals (resistance up with temperature) and forget they are semiconductors. How to avoid it: **For an NTC thermistor, resistance FALLS as temperature rises** — heating releases more charge carriers (n increases), so from I=nqvA the resistance drops. 'NTC' = negative temperature coefficient. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Quoting the unit of resistivity as $\Omega\,\text{m}^{-1}$ (or $\Omega/\text{m}$) instead of $\Omega\,\text{m}$" a common mistake in Resistivity and Potential dividers?

Why it happens: Students guess the unit by analogy with 'per metre' quantities instead of deriving it from =RA/l. How to avoid it: **Derive it: =RA/l gives ×m^2÷m=\,m.** Always attach \,m to a resistivity answer — the accuracy mark depends on it. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "Forgetting to convert millimetre diameters/lengths to metres before substituting" a common mistake in Resistivity and Potential dividers?

Why it happens: Diameters are quoted in mm (e.g. 0.30 mm) and lengths sometimes in cm, but the resistivity equation needs SI metres. How to avoid it: **Convert everything to metres first** (1\ mm=10^-3\ m, so 0.30\ mm=3.0×10^-4\ m), then square for the area. A single unconverted mm typically throws the answer out by a factor of 10^6 in the area. Source: Edexcel IAL Physics Unit 2 examiner reports

Why is "In CP7, reading a single (R, l) point off the graph instead of using the gradient to find resistivity" a common mistake in Resistivity and Potential dividers?

Why it happens: Students do not connect R=Al to y=mx and so miss that the gradient is /A. How to avoid it: **Plot R against l and use the GRADIENT: gradient =/A, so =gradient× A.** The gradient averages out random error and cancels a constant contact-resistance offset, which a single point cannot. Source: Edexcel IAL Physics Unit 2 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Resistivity and Potential dividers

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Short Notes - Resistivity and Potential dividers

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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.

Resistivity and Potential Dividers — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 2)

Resistivity $R=\rho l/A$ and the diameter-to-area trap; the charge-carrier equation $I=nqvA$ and why metals conduct better than semiconductors; the temperature dependence of resistivity; the potential divider $V_{out}=V_{in}\,R_2/(R_1+R_2)$ ; thermistor and LDR sensor circuits and switches; and Core Practical 7 — determining the resistivity of a wire.

At a glance

Resistivity $\rho$ is a material property: $R=\dfrac{\rho l}{A}$ , so resistance rises with length and falls with cross-sectional area. Unit of $\rho$ is the ohm metre ( $\Omega\,\text{m}$ ).
Cross-sectional area of a wire: $A=\dfrac{\pi d^2}{4}=\pi r^2$ . The micrometer gives the diameter — halve it, or use $\pi d^2/4$ , or the area comes out four times too big.
Charge carriers: $I=nqvA$ . Metals have a huge carrier density $n\;(\sim10^{28}\,\text{m}^{-3})$ , so they conduct well; semiconductors have far fewer carriers and higher resistivity.
Drift velocity is tiny (a fraction of a mm s⁻¹), even though the current is established almost instantly.
Temperature: a metal's resistivity rises with temperature (more lattice vibration → more collisions); a semiconductor's falls (many more carriers released, $n$ rises sharply).
Potential divider: two series resistors split the supply. Output across $R_2$ : $V_{out}=V_{in}\,\dfrac{R_2}{R_1+R_2}$ . The resistor the output is taken across goes on top.
Sensor circuits: swap in a thermistor (NTC: resistance falls as it warms) or an LDR (resistance falls in bright light) to make a temperature- or light-controlled output.
Core Practical 7: plot $R$ against $l$ ; the gradient is $\rho/A$ , so $\rho=\text{gradient}\times A$ — and the gradient cancels constant contact-resistance errors.

What you’ll learn

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

2.1 — Use $R=\dfrac{\rho l}{A}$ to relate the resistance of a uniform conductor to its length, cross-sectional area and resistivity, and recall that the unit of resistivity is the ohm metre.
2.2 — Calculate the cross-sectional area of a wire from its diameter using $A=\dfrac{\pi d^2}{4}$ .
2.3 — Use $I=nqvA$ to relate current to the number density, charge and drift velocity of charge carriers, and explain differences in conductivity between metals and semiconductors.
2.4 — Describe and explain the temperature dependence of the resistivity of metals and of semiconductors (e.g. NTC thermistors).
2.5 — Use the potential-divider equation $V_{out}=V_{in}\,\dfrac{R_2}{R_1+R_2}$ to obtain a required output voltage.
2.6 — Explain the action of a thermistor or LDR in a potential divider and design temperature- or light-activated sensing circuits.
CP7 — Determine the resistivity of a wire using a micrometer to measure its diameter and obtaining resistance from voltage and current; evaluate sources of uncertainty and systematic error.

Resistivity — a property of the material

Resistance depends on shape; resistivity is the shape-free property of the substance itself.

Two wires of the same metal can have very different resistances — make one longer or thinner and its resistance goes up. To compare materials fairly, we need a quantity that does not depend on the size of the sample. That quantity is the resistivity, $\rho$ .

For a uniform conductor:

$R = \frac{\rho l}{A}$

$R$ = resistance (Ω), $\rho$ = resistivity (Ω m), $l$ = length (m), $A$ = cross-sectional area (m²).
Resistance is proportional to length ( $R\propto l$ ): a longer wire means carriers travel further and suffer more collisions.
Resistance is inversely proportional to area ( $R\propto 1/A$ ): a fatter wire gives more parallel paths for the carriers.

Rearranging gives $\rho=\dfrac{RA}{l}$ , from which the unit of resistivity drops out: $\dfrac{\Omega\times\text{m}^2}{\text{m}}=\boldsymbol{\Omega\,\text{m}}$ . (It is ohm metre, not $\Omega\,\text{m}^{-1}$ — a very common unit slip.)

The diameter-to-area trap. A wire has a circular cross-section, so its area is

$A = \frac{\pi d^2}{4} = \pi r^2,$

where $d$ is the diameter and $r=d/2$ the radius. A micrometer measures the diameter, so you must either use $\pi d^2/4$ directly or halve $d$ first. Putting the diameter into $\pi r^2$ makes the area four times too large and ruins every resistivity answer.

Resistance grows with the length l and shrinks as the cross-sectional area A grows. For a round wire the area comes from the diameter: A = πd²/4 (equivalently πr², with r = d/2). The micrometer measures d, so always halve it or use πd²/4.

$R=\dfrac{\rho l}{A}$ : resistance ∝ length, ∝ 1/area.
Resistivity $\rho$ is a material property; unit $\Omega\,\text{m}$ (derive from $\rho=RA/l$ ).
Wire area $A=\dfrac{\pi d^2}{4}=\pi r^2$ — the micrometer gives the diameter.
Putting the diameter into $\pi r^2$ makes A four times too big.

See the full worked example for resistivity and potential dividers →

Charge carriers and I = nqvA

Current is carriers in motion; metals win because they have far more of them.

A current is just charge carriers drifting through a conductor. Counting how much charge passes per second gives the charge-carrier equation:

$I = nqvA$

$n$ = number density of carriers (carriers per cubic metre, m⁻³),
$q$ = charge on each carrier (for an electron, $q=1.6\times10^{-19}\ \text{C}$ ),
$v$ = drift velocity (mean speed of the carriers along the wire, m s⁻¹),
$A$ = cross-sectional area (m²).

Why metals conduct so well. A metal has a vast number density of free electrons, around $n\approx10^{28}$ – $10^{29}\ \text{m}^{-3}$ . With so many carriers, a sizeable current flows even at a modest drift velocity — so metals have a low resistivity. A semiconductor has a far smaller $n$ (perhaps a million times fewer carriers), so for the same current the few carriers must drift faster, and the resistivity is much higher.

Drift velocity is astonishingly slow. Rearranging, $v=\dfrac{I}{nqA}$ . Because $n$ is so huge, a typical drift velocity is well under a millimetre per second — the electrons crawl. Yet a light turns on instantly when you flip the switch, because the electric field (and so the push on all the electrons at once) propagates near the speed of light. The carriers barely move; the signal moves fast.

Current is the flow of charge carriers: I = nqvA. Metals have an enormous carrier density n, so a large current flows even though each electron drifts very slowly. Semiconductors have far fewer carriers and therefore a higher resistivity.

$I=nqvA$ links current to carrier density, charge, drift velocity and area.
Metals: very high $n$ → low resistivity; semiconductors: low $n$ → high resistivity.
Drift velocity $v=\dfrac{I}{nqA}$ is tiny (well under 1 mm s⁻¹).
Current starts almost instantly because the field propagates near light speed.

See the full worked example for resistivity and potential dividers →

Temperature: why metals and semiconductors behave oppositely

Heat a metal and resistance rises; heat a semiconductor and it falls — because of n vs collisions.

The temperature dependence of resistivity is a favourite explain question, and it comes straight out of $I=nqvA$ .

Metal — resistance increases with temperature.

The number density $n$ of free electrons is essentially constant (all the conduction electrons are already free).
Heating makes the lattice ions vibrate more, so the drifting electrons collide with them more often.
More frequent collisions reduce the mean drift velocity for a given field, so the current falls and the resistance (resistivity) rises.

Semiconductor (e.g. an NTC thermistor) — resistance decreases with temperature.

Here the key effect is on $n$ : heating gives many more electrons enough energy to break free and become carriers, so the number density $n$ rises sharply.
From $I=nqvA$ , far more carriers means much more current can flow, so the resistance (resistivity) falls.
Increased lattice vibration would tend to raise the resistance (as in a metal), but the explosive growth in $n$ dominates, so the net effect is a fall.

This is why an NTC thermistor ("negative temperature coefficient") loses resistance as it warms, and why a metal wire gains resistance as a current heats it.

Metal: $n$ ≈ constant; more lattice vibration → more collisions → resistivity ↑.
Semiconductor: $n$ rises sharply with temperature → resistivity ↓.
NTC thermistor: resistance falls as temperature rises.
In a semiconductor the rise in $n$ outweighs the increased vibration.

See the full worked example for resistivity and potential dividers →

The potential divider

Two series resistors split the supply; put the output resistor on top.

A potential divider is two (or more) resistors in series across a supply, used to tap off a chosen fraction of the supply voltage. Because the same current flows through resistors in series, the supply voltage splits in the ratio of the resistances ( $V=IR$ ).

If the output is taken across $R_2$ :

$V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2}$

The resistor the output is taken across goes on top of the fraction.
The larger resistor gets the larger share of the voltage. Quick check: if $R_1=R_2$ the output is half the supply.
Replacing $R_1$ or $R_2$ with a variable resistor (potentiometer) lets you dial the output anywhere from 0 up to (almost) $V_{in}$ .

A potential divider: R1 and R2 are in series across V_in, and the output is tapped off across R2. The supply splits in the ratio of the resistances, so V_out = V_in × R2 / (R1 + R2). The resistor the output is measured across is always on top of the fraction.

$V_{out}=V_{in}\,\dfrac{R_2}{R_1+R_2}$ — output resistor on top.
Series voltages split in the ratio of the resistances ( $V\propto R$ ).
Larger resistor → larger share of the supply voltage.
A potentiometer gives a continuously variable output from 0 to ≈ $V_{in}$ .

See the full worked example for resistivity and potential dividers →

Sensor circuits — thermistors and LDRs

Swap a fixed resistor for a thermistor or LDR and the output responds to temperature or light.

If one resistor in a potential divider is a component whose resistance changes with conditions, the output voltage becomes a sensor signal.

Thermistor (NTC): resistance falls as temperature rises.
LDR (light-dependent resistor): resistance falls as light intensity rises.

The behaviour of the output depends entirely on which component the output is taken across — so always reason from $V_{out}=V_{in}\,\dfrac{R_{out}}{R_{total}}$ .

Worked logic — a temperature sensor. Put an NTC thermistor as the top resistor and take the output across the fixed lower resistor:

Room warms → thermistor resistance falls → it takes a smaller share of the supply → the fixed resistor's share, the output, rises.
So a rising temperature gives a rising $V_{out}$ — feed this to a switching circuit (transistor or comparator) set to trigger above a threshold, and you have a heat-activated switch (e.g. switching on a cooling fan).

Swap which resistor the output is across, or use the divider the other way up, and you can make the output fall with temperature instead — design choices follow from the divider equation. The same reasoning with an LDR gives light-activated switches (e.g. a street light that comes on at dusk).

A temperature-sensing divider: an NTC thermistor on top, output across the fixed resistor. As the temperature rises the thermistor's resistance falls, so it takes less of the supply and the output rises — ideal for triggering a heat-activated switch. An LDR in place of the thermistor gives a light-sensing circuit.

Thermistor (NTC): resistance falls as temperature rises.
LDR: resistance falls as light intensity rises.
The output's behaviour depends on which component it is taken across — reason from the divider equation.
Thermistor/LDR dividers drive temperature- and light-activated switches.

See the full worked example for resistivity and potential dividers →

Core Practical 7 — measuring the resistivity of a wire

Micrometer for d, metre rule for l, R from V/I; plot R vs l and ρ = gradient × A.

Core Practical 7 determines the resistivity of a metal wire. You need the wire's resistance, its length, and its cross-sectional area.

Measurements

Diameter $d$ — use a micrometer screw gauge at several points along the wire and in different orientations, then take the mean (the wire may not be perfectly round or uniform). Compute $A=\dfrac{\pi d^2}{4}$ .
Length $l$ — set with a metre rule (to the nearest mm). Vary it: take readings at several lengths.
Resistance $R$ — for each length, read the voltage $V$ across the wire and the current $I$ through it, and use $R=\dfrac{V}{I}$ .

Getting ρ from a graph (the reliable way). Since $R=\dfrac{\rho}{A}\,l$ , plotting $R$ (y-axis) against $l$ (x-axis) gives a straight line of gradient $\rho/A$ . Then:

$\rho = \text{gradient} \times A.$

Plotting R against l gives a straight line of gradient ρ/A. Reading the gradient from a large triangle and multiplying by the cross-sectional area gives the resistivity. The gradient averages out random error and cancels a constant contact-resistance offset.

Reducing and judging uncertainty

Random error: repeat each diameter and timing reading and take means; use several lengths and a best-fit line so one bad point cannot distort the result.
Systematic error — contact/lead resistance. The crocodile clips and connecting leads add a small constant resistance to every reading, which would raise the line by a fixed amount (a positive intercept). Using the gradient (not a single $R$ ) removes this, because a constant offset shifts the intercept, not the slope.
Systematic error — wire heating. A current warms the wire and raises its resistance (metal). Keep the current low and switch off between readings so the wire stays near room temperature.
Non-uniform wire: measuring the diameter at several places and averaging guards against a wire that is not perfectly uniform.

Compare your value with the data-book figure for the metal and quote a percentage difference when evaluating accuracy.

Micrometer (mean of several readings) for $d$ → $A=\pi d^2/4$ .
Metre rule for $l$ ; $R=V/I$ at several lengths.
Plot $R$ vs $l$ : gradient $=\rho/A$ , so $\rho=\text{gradient}\times A$ .
Gradient cancels constant contact resistance; keep current low to avoid heating.

See the full worked example for resistivity and potential dividers →

Quick recap

$R=\dfrac{\rho l}{A}$ : resistance ∝ length and ∝ 1/area; resistivity $\rho$ is a material property with unit $\Omega\,\text{m}$ .
Wire area $A=\dfrac{\pi d^2}{4}=\pi r^2$ — the micrometer gives the diameter, so halve it (or use $\pi d^2/4$ ).
$I=nqvA$ : metals have huge $n$ (low resistivity); semiconductors have small $n$ (high resistivity); drift velocity is tiny.
Metal resistivity rises with temperature (more collisions, $n$ constant); semiconductor resistivity falls ( $n$ rises sharply).
Potential divider: $V_{out}=V_{in}\,\dfrac{R_2}{R_1+R_2}$ — the resistor the output is across goes on top.
Thermistor (resistance ↓ as T ↑) and LDR (resistance ↓ as light ↑) make temperature- and light-sensing dividers; reason from the divider equation.
CP7: plot $R$ vs $l$ , gradient $=\rho/A$ , $\rho=\text{gradient}\times A$ ; the gradient removes constant contact resistance and keeping the current low avoids heating.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

$R = \dfrac{\rho l}{A}$ — resistance ∝ length, ∝ 1/area; resistivity unit is the ohm metre ( $\Omega\,\text{m}$ ).
$A = \dfrac{\pi d^2}{4} = \pi r^2$ — the micrometer measures the DIAMETER (halve it for the radius).
$I = nqvA$ — current = number density × charge × drift velocity × area.
$V_{out} = V_{in}\,\dfrac{R_2}{R_1+R_2}$ — the output resistor goes ON TOP.
Metal: resistivity rises with temperature (more lattice collisions, $n$ constant).
Semiconductor / NTC thermistor: resistivity falls with temperature ( $n$ rises sharply).
Thermistor resistance falls as temperature rises; LDR resistance falls as light intensity rises.
CP7: plot $R$ vs $l$ ; gradient $=\rho/A$ ; $\rho = \text{gradient} \times A$ .

How it’s examined

Resistivity and potential dividers are core Unit 2 content and appear on the Unit 2 paper (WPH12, Waves and Electricity), which mixes short calculations, 'show that' items, an explain/levels question and a practical/data question. Typical formats: define resistivity and give its unit (1–2 marks, AO1); calculate a resistance or resistivity, where the cross-sectional area must be found from the diameter via $A=\pi d^2/4$ (3–4 marks, M and A marks — the diameter-to-area trap and the $\Omega\,\text{m}$ unit are heavily penalised); a drift-velocity calculation from $I=nqvA$ (3–4 marks); a potential-divider calculation, often 'find the fixed resistor needed for a given output' (3–5 marks — put the output resistor on top); an explain question on a thermistor or LDR sensor circuit, requiring a reasoned link from the resistance change to the output via the divider equation (often the extended 6-marker); and a Core Practical 7 data question (plot $R$ vs $l$ , gradient $=\rho/A$ , contact-resistance and heating errors). ALWAYS show the equation, substitution with units and a 3 s.f. answer — 'no working = risk zero'. For 'show that' work to one extra s.f.; for graph determinations use a large triangle and state the gradient units.

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 — Resistivity and Potential dividers

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

Question type:

1Cross-sectional area of a wire from its diameter (2 marks)
Getting startedDirect calculation• resistivity, cross-sectional area, Unit 2
Question
A copper wire has a diameter of 0.40 mm. Calculate its cross-sectional area in m². (2 marks)
Step-by-step solution
1. Step 1
  Convert the diameter to metres and use the circle area in terms of diameter: $A=\dfrac{\pi d^2}{4}$ . (Using the radius, $r=d/2$ , gives the same thing.) Here $d=0.40\ \text{mm}=4.0\times10^{-4}\ \text{m}$ .
  $A = \frac{\pi d^2}{4} = \frac{\pi (4.0\times10^{-4})^2}{4}$
2. Step 2
  Square the diameter first ( $(4.0\times10^{-4})^2 = 1.6\times10^{-7}\ \text{m}^2$ ), then evaluate.
  $A = \frac{\pi \times 1.6\times10^{-7}}{4} = 1.257\times10^{-7}\ \text{m}^2$
Answer
$A = 1.26\times10^{-7}\ \text{m}^2$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) correct conversion + $A=\pi d^2/4$ (or $\pi r^2$ with $r=2.0\times10^{-4}$ m); (2) $1.26\times10^{-7}\ \text{m}^2$ . The dominant error is putting the diameter into $\pi r^2$ — that quadruples the area and ruins every later resistivity answer.
2Resistance of a wire from resistivity (3 marks)
Getting startedDirect calculation• resistivity, AO2, Unit 2
Question
A length of nichrome wire is 1.5 m long with a cross-sectional area of $8.0\times10^{-8}\ \text{m}^2$ . The resistivity of nichrome is $1.1\times10^{-6}\ \Omega\,\text{m}$ . Calculate the resistance of the wire. (3 marks)
Step-by-step solution
1. Step 1
  Select the resistivity equation, rearranged for resistance $R$ .
  $R = \frac{\rho l}{A}$
2. Step 2
  Substitute with units, keeping the powers of ten.
  $R = \frac{(1.1\times10^{-6})(1.5)}{8.0\times10^{-8}}$
3. Step 3
  Evaluate: numerator $=1.65\times10^{-6}$ , divide by $8.0\times10^{-8}$ .
  $R = \frac{1.65\times10^{-6}}{8.0\times10^{-8}} = 20.6\ \Omega$
Answer
$R = 20.6\ \Omega$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) correct equation $R=\rho l/A$ ; (2) correct substitution (M); (3) $20.6\ \Omega$ (A). Keep the powers of ten lined up — mishandling $10^{-6}\div10^{-8}=10^{2}$ is the usual slip.
3Output voltage of a potential divider (3 marks)
Getting startedDirect calculation• potential divider, AO2, Unit 2
Question
Two resistors $R_1=4.0\ \text{k}\Omega$ and $R_2=2.0\ \text{k}\Omega$ are connected in series across a 12 V supply. The output is taken across $R_2$ . Calculate the output voltage. (3 marks)
Step-by-step solution
1. Step 1
  The output is taken across $R_2$ , so use the potential-divider equation with $R_2$ on top.
  $V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2}$
2. Step 2
  Substitute (the kΩ cancel in the ratio, so no conversion is needed for the ratio itself).
  $V_{out} = 12 \times \frac{2.0}{4.0 + 2.0} = 12 \times \frac{2.0}{6.0}$
3. Step 3
  Evaluate the ratio and multiply.
  $V_{out} = 12 \times 0.333 = 4.0\ \text{V}$
Answer
$V_{out} = 4.0\ \text{V}$ .
Examiner tip
Mark scheme: (1) correct equation with $R_2$ on top (output is across $R_2$ ); (2) substitution; (3) $4.0\ \text{V}$ . The classic error is putting $R_1$ on top — always check which resistor the output is taken across. Sanity check: the volts split in the ratio of the resistances, so $R_2$ (smaller) gets the smaller share.
4Resistivity of a wire from a measured resistance (4 marks)
Building confidenceMulti-step problem• Adapted from Edexcel IAL Physics Unit 2 (Core Practical 7)• resistivity, Core Practical 7, AO2, Unit 2
Question
A metal wire of length 0.80 m and diameter 0.50 mm has a resistance of 1.7 Ω. Calculate the resistivity of the metal. (4 marks)
Step-by-step solution
1. Step 1
  First find the cross-sectional area from the diameter. $d=0.50\ \text{mm}=5.0\times10^{-4}\ \text{m}$ .
  $A = \frac{\pi d^2}{4} = \frac{\pi (5.0\times10^{-4})^2}{4} = 1.963\times10^{-7}\ \text{m}^2$
2. Step 2
  Rearrange $R=\rho l/A$ for the resistivity $\rho$ .
  $\rho = \frac{R A}{l}$
3. Step 3
  Substitute the measured resistance, the area and the length.
  $\rho = \frac{(1.7)(1.963\times10^{-7})}{0.80}$
4. Step 4
  Evaluate: numerator $=3.337\times10^{-7}$ , divide by $0.80$ .
  $\rho = \frac{3.337\times10^{-7}}{0.80} = 4.17\times10^{-7}\ \Omega\,\text{m}$
Answer
$\rho = 4.17\times10^{-7}\ \Omega\,\text{m}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) area from $\pi d^2/4$ with correct conversion (M); (2) rearrangement $\rho=RA/l$ ; (3) substitution (M); (4) $4.17\times10^{-7}\ \Omega\,\text{m}$ with the unit $\Omega\,\text{m}$ (A). The unit of resistivity is $\Omega\,\text{m}$ , NOT $\Omega\,\text{m}^{-1}$ — losing it costs the accuracy mark.
5Drift velocity from I = nqvA (4 marks)
Building confidenceMulti-step problem• charge carriers, drift velocity, AO2, Unit 2
Question
A copper wire of cross-sectional area $1.0\times10^{-6}\ \text{m}^2$ carries a current of 2.0 A. Copper has a free-electron number density of $8.5\times10^{28}\ \text{m}^{-3}$ and the electron charge is $1.6\times10^{-19}\ \text{C}$ . Calculate the drift velocity of the electrons. (4 marks)
Step-by-step solution
1. Step 1
  Select the charge-carrier equation and rearrange for the drift velocity $v$ .
  $I = nqvA \;\Rightarrow\; v = \frac{I}{nqA}$
2. Step 2
  Substitute the values with their units.
  $v = \frac{2.0}{(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6})}$
3. Step 3
  Evaluate the denominator: $(8.5\times10^{28})(1.6\times10^{-19})(1.0\times10^{-6}) = 1.36\times10^{4}$ .
  $nqA = 1.36\times10^{4}\ \text{(units combine to A s m}^{-1}\text{)}$
4. Step 4
  Divide to find $v$ .
  $v = \frac{2.0}{1.36\times10^{4}} = 1.47\times10^{-4}\ \text{m s}^{-1}$
Answer
$v = 1.47\times10^{-4}\ \text{m s}^{-1}$ (about 0.15 mm s⁻¹).
Examiner tip
Mark scheme: (1) rearrange to $v=I/(nqA)$ (M); (2) substitution; (3) denominator $=1.36\times10^{4}$ (M); (4) $1.47\times10^{-4}\ \text{m s}^{-1}$ (A). Students are often surprised the drift velocity is so tiny — the signal travels near light speed but the electrons themselves crawl.
6Output of a thermistor potential divider (4 marks)
Building confidenceMulti-step problem• potential divider, thermistor, AO2, Unit 2
Question
An NTC thermistor is in series with a fixed $2.0\ \text{k}\Omega$ resistor across a 6.0 V supply. The output is taken across the fixed resistor. At 20 °C the thermistor resistance is $4.0\ \text{k}\Omega$ . (a) Calculate the output voltage at 20 °C. (b) State and explain how the output changes when the thermistor is warmed. (4 marks)
Step-by-step solution
1. Step 1
  Output is across the fixed resistor $R_2=2.0\ \text{k}\Omega$ ; the thermistor is $R_1=4.0\ \text{k}\Omega$ .
  $V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2} = 6.0 \times \frac{2.0}{4.0 + 2.0}$
2. Step 2
  Evaluate the ratio $2.0/6.0=0.333$ and multiply.
  $V_{out} = 6.0 \times 0.333 = 2.0\ \text{V}$
3. Step 3
  Warming an NTC thermistor lowers its resistance ( $R_1$ falls). Since $R_2$ is fixed, the thermistor takes a smaller share of the supply voltage, so the share across $R_2$ — the output — rises.
Answer
(a) $V_{out}=2.0\ \text{V}$ . (b) As the thermistor warms its resistance falls, so $V_{out}$ (across the fixed resistor) increases.
Examiner tip
Mark scheme: (a) 2 marks: correct equation with $R_2$ on top and $2.0\ \text{V}$ . (b) 2 marks: thermistor resistance falls as temperature rises, AND a correct reasoned link to $V_{out}$ rising. A bare 'output increases' with no mechanism scores only one of the two marks.
7Determining resistivity from the gradient of an R–l graph (6 marks)
StretchGraph or diagram• Adapted from Edexcel IAL Physics Unit 2 (Core Practical 7)• Core Practical 7, resistivity, graph, AO3, Unit 2
Question
In Core Practical 7 a student measures the resistance R of a wire for several lengths l and plots R (y-axis) against l (x-axis), obtaining a straight line through the origin. A large gradient triangle gives ΔR = 6.40 Ω for Δl = 0.800 m. The wire diameter, measured with a micrometer, is 0.30 mm. Determine the resistivity of the wire. (6 marks)
Step-by-step solution
1. Step 1
  Write $R=\dfrac{\rho}{A}\,l$ and compare with $y=mx$ : the gradient equals $\rho/A$ .
  $R = \frac{\rho}{A}\,l \;\Rightarrow\; \text{gradient} = \frac{\rho}{A}$
2. Step 2
  Calculate the gradient from the triangle, with units.
  $\text{gradient} = \frac{\Delta R}{\Delta l} = \frac{6.40}{0.800} = 8.00\ \Omega\,\text{m}^{-1}$
3. Step 3
  Find the cross-sectional area from the diameter $d=0.30\ \text{mm}=3.0\times10^{-4}\ \text{m}$ .
  $A = \frac{\pi d^2}{4} = \frac{\pi (3.0\times10^{-4})^2}{4} = 7.069\times10^{-8}\ \text{m}^2$
4. Step 4
  Since gradient $=\rho/A$ , the resistivity is gradient $\times A$ .
  $\rho = \text{gradient} \times A = 8.00 \times 7.069\times10^{-8}$
5. Step 5
  Evaluate the resistivity.
  $\rho = 5.66\times10^{-7}\ \Omega\,\text{m}$
Answer
$\rho = 5.66\times10^{-7}\ \Omega\,\text{m}$ (to 3 s.f.).
Examiner tip
Mark scheme: (1) gradient $=\rho/A$ (linking to $y=mx$ ); (2) gradient $=8.00\ \Omega\,\text{m}^{-1}$ from a large triangle; (3) area $=\pi d^2/4$ with correct conversion; (4) $\rho=\text{gradient}\times A$ ; (5) $5.66\times10^{-7}\ \Omega\,\text{m}$ . Using a single ( $R$ , $l$ ) point instead of the gradient throws away the whole point of the graph (averaging out error); forgetting to convert the micrometer reading is the next-most-common slip.
8LDR light-activated switch design (6 marks)
StretchMulti-step problem• potential divider, LDR, sensor circuit, AO2, Unit 2
Question
An LDR and a fixed resistor $R$ form a potential divider across a 9.0 V supply. The output, taken across the LDR, must reach 6.0 V in bright light, when the LDR resistance is $1.0\ \text{k}\Omega$ . (a) Calculate the value of the fixed resistor $R$ required. (b) State and explain what happens to the output voltage when the light level falls. (6 marks)
Step-by-step solution
1. Step 1
  Output is across the LDR, so $V_{out}=V_{in}\,\dfrac{R_{LDR}}{R+R_{LDR}}$ . Substitute the bright-light values.
  $6.0 = 9.0 \times \frac{1.0\times10^3}{R + 1.0\times10^3}$
2. Step 2
  Divide both sides by 9.0 to isolate the ratio.
  $\frac{6.0}{9.0} = 0.667 = \frac{1.0\times10^3}{R + 1.0\times10^3}$
3. Step 3
  Rearrange: $R + 1.0\times10^3 = \dfrac{1.0\times10^3}{0.667}$ .
  $R + 1.0\times10^3 = 1.50\times10^3\ \Omega$
4. Step 4
  Subtract to find $R$ .
  $R = 1.50\times10^3 - 1.0\times10^3 = 5.0\times10^2\ \Omega$
5. Step 5
  When the light falls, the LDR resistance rises. With the output taken across the LDR, the LDR now takes a larger share of the 9.0 V, so the output voltage increases towards 9.0 V. (If a transistor/comparator switch is triggered above a threshold, the device switches on in the dark — a useful nightlight circuit.)
Answer
(a) $R = 500\ \Omega$ . (b) As the light dims, $R_{LDR}$ rises, so $V_{out}$ (across the LDR) increases.
Examiner tip
Mark scheme: (a) 4 marks: correct divider equation with the LDR on top, substitution, rearrangement, $R=500\ \Omega$ . (b) 2 marks: LDR resistance rises as light falls AND a reasoned link to the output rising. Mixing up which component the output is across reverses the whole behaviour — read the circuit carefully.
9Why a metal's resistance rises and a semiconductor's falls (6 marks)
StretchWord problem• resistivity, temperature, charge carriers, AO1, Unit 2
Question
Using ideas about charge carriers, explain (a) why the resistivity of a metal increases as its temperature rises, and (b) why the resistivity of a semiconductor (such as the material of an NTC thermistor) decreases as its temperature rises. Refer to $I=nqvA$ in your answer. (6 marks)
Step-by-step solution
1. Step 1
  Start from $I=nqvA$ : for a fixed current and geometry, the resistance depends on how easily carriers move ( $v$ for a given field) and on how many there are ( $n$ ).
  $I = nqvA$
2. Step 2
  Metal (a): the number density $n$ of free electrons is essentially constant with temperature. As temperature rises, the lattice ions vibrate more, so electrons collide with them more frequently. The mean drift velocity for a given field falls, current falls, so resistance (and resistivity) increases.
3. Step 3
  Semiconductor (b): here the dominant effect is that raising the temperature releases many more charge carriers (electrons gain enough energy to become free), so the number density $n$ rises sharply. From $I=nqvA$ , a larger $n$ means more carriers available to carry current, so the current rises and the resistance (and resistivity) decreases.
4. Step 4
  The two effects compete in a semiconductor — increased lattice vibration would tend to raise resistance — but the huge increase in $n$ dominates, so the net resistivity falls. This is why an NTC thermistor's resistance drops as it warms.
Answer
(a) Metal: $n$ roughly constant, but more lattice vibration → more collisions → lower drift velocity → higher resistivity. (b) Semiconductor: many more charge carriers released ( $n$ rises sharply) → resistivity falls; this increase in $n$ outweighs the increased vibration.
Examiner tip
Why this scores 6/6. (a) 3 marks: $n$ constant in a metal; lattice vibrates more; more collisions reduce drift velocity → resistivity up. (b) 3 marks: in a semiconductor $n$ increases greatly with temperature; from $I=nqvA$ more carriers → resistivity down; (credit) this dominates the opposing vibration effect. A levels-based 6-marker needs linked prose referencing $n$ , $v$ and $I=nqvA$ — not a bare 'metals go up, semiconductors go down'.

Model Answers — Resistivity and Potential dividers

High-scoring sample answers for resistivity and potential dividers 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). State what is meant by the resistivity of a material and give its SI unit.
Model answer
Resistivity $\rho$ is a property of a material defined by $R=\dfrac{\rho l}{A}$ — numerically it is the resistance of a sample of the material of unit length and unit cross-sectional area. Its SI unit is the ohm metre ( $\Omega\,\text{m}$ ).
Why this scores
Why this scores 2/2. (1) resistivity is a material property linked by $R=\rho l/A$ (or 'resistance of a 1 m length of 1 m² cross-section'); (2) unit $\Omega\,\text{m}$ . The unit is a frequent loss — it is $\Omega\,\text{m}$ , not $\Omega\,\text{m}^{-1}$ and not $\Omega/\text{m}$ .
Question 2
2 marks
Q (2 marks). State what a potential divider does, and write down the equation for the output voltage taken across the resistor $R_2$ of two series resistors $R_1$ and $R_2$ across a supply $V_{in}$ .
Model answer
A potential divider uses two (or more) resistors in series to produce a chosen fraction of the supply voltage as its output. With the output taken across $R_2$ :

$V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2}$
Why this scores
Why this scores 2/2. (1) divides the supply voltage / gives a fraction of $V_{in}$ ; (2) correct equation with $R_2$ (the output resistor) on top. Writing $R_1$ on top loses the second mark.
Question 3
3 marks
Q (3 marks). State how the resistance of an NTC thermistor changes as its temperature rises, and explain this change in terms of charge carriers.
Model answer
As temperature rises, the resistance of an NTC thermistor decreases.

The thermistor is made of a semiconductor. Raising the temperature gives more electrons enough energy to break free and become charge carriers, so the number density of charge carriers $n$ increases. From $I=nqvA$ , more carriers means more current can flow for the same voltage, so the resistance falls.
Why this scores
Why this scores 3/3. (1) resistance decreases (NTC); (2) more charge carriers are released → $n$ increases; (3) link via $I=nqvA$ (more carriers → lower resistance). 'NTC' literally means negative temperature coefficient — resistance down as temperature up.
Question 4
4 marks
Q (4 marks). A constantan wire of length 2.0 m and diameter 0.45 mm has a resistance of 6.3 Ω. Calculate the resistivity of constantan, showing your working.
Model answer
First find the cross-sectional area from the diameter, $d=0.45\ \text{mm}=4.5\times10^{-4}\ \text{m}$ :

$A = \frac{\pi d^2}{4} = \frac{\pi (4.5\times10^{-4})^2}{4} = 1.590\times10^{-7}\ \text{m}^2$

Rearranging $R=\dfrac{\rho l}{A}$ for the resistivity:

$\rho = \frac{RA}{l} = \frac{(6.3)(1.590\times10^{-7})}{2.0} = 5.01\times10^{-7}\ \Omega\,\text{m}$

The resistivity of constantan is therefore about $5.0\times10^{-7}\ \Omega\,\text{m}$ .
Why this scores
Why this scores 4/4. (1) area from $\pi d^2/4$ with correct mm→m conversion; (2) rearrangement $\rho=RA/l$ ; (3) substitution (M); (4) $5.01\times10^{-7}\ \Omega\,\text{m}$ with the correct unit (A). Squaring the diameter and converting units are where most marks are lost.
Question 5
4 marks
Q (4 marks). A 5.0 V supply is connected across two resistors in series. The output is taken across one of them and must be 1.5 V. The other (top) resistor is $R_1=7.0\ \text{k}\Omega$ . Calculate the resistance $R_2$ across which the output is taken.
Model answer
The output is across $R_2$ , so:

$V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2} \;\Rightarrow\; 1.5 = 5.0 \times \frac{R_2}{7.0 + R_2}$

(working in kΩ). Dividing both sides by 5.0:

$0.30 = \frac{R_2}{7.0 + R_2}$

Rearranging: $0.30(7.0 + R_2) = R_2$ , so $2.1 + 0.30R_2 = R_2$ , giving $2.1 = 0.70R_2$ :

$R_2 = \frac{2.1}{0.70} = 3.0\ \text{k}\Omega$
Why this scores
Why this scores 4/4. (1) correct equation with $R_2$ on top; (2) isolate the ratio $0.30$ ; (3) algebraic rearrangement; (4) $R_2=3.0\ \text{k}\Omega$ . Keeping consistent units (kΩ throughout) avoids slips; an alternative is $V_1=5.0-1.5=3.5$ V so $R_1/R_2=V_1/V_2=3.5/1.5$ giving $R_2=7.0\times1.5/3.5=3.0$ kΩ.
Question 6
5 marks
Q (5 marks). A copper wire and a tungsten wire of the same cross-sectional area carry the same current. Copper has a free-electron number density of $8.5\times10^{28}\ \text{m}^{-3}$ and tungsten $3.4\times10^{28}\ \text{m}^{-3}$ . (a) Using $I=nqvA$ , explain which wire has the greater electron drift velocity. (b) The copper wire has area $2.0\times10^{-6}\ \text{m}^2$ and carries 5.0 A. Calculate the drift velocity of its electrons.
Model answer
(a) From $I=nqvA$ , rearranged $v=\dfrac{I}{nqA}$ . With $I$ , $q$ and $A$ the same for both wires, the drift velocity $v$ is inversely proportional to the number density $n$ . Tungsten has the smaller $n$ , so the tungsten wire has the greater drift velocity.

(b) Rearranging for $v$ :

$v = \frac{I}{nqA} = \frac{5.0}{(8.5\times10^{28})(1.6\times10^{-19})(2.0\times10^{-6})}$

The denominator is $(8.5\times10^{28})(1.6\times10^{-19})(2.0\times10^{-6}) = 2.72\times10^{4}$ , so:

$v = \frac{5.0}{2.72\times10^{4}} = 1.84\times10^{-4}\ \text{m s}^{-1}$
Why this scores
Why this scores 5/5. (a) 2 marks: $v\propto1/n$ for fixed $I,q,A$ , so tungsten (smaller $n$ ) has the larger $v$ . (b) 3 marks: rearrange to $v=I/(nqA)$ , correct denominator $2.72\times10^{4}$ , $1.84\times10^{-4}\ \text{m s}^{-1}$ . The comparison hinges on holding $I$ , $q$ and $A$ constant — say so explicitly.
Question 7
Adapted from Edexcel IAL Physics Unit 2 (Core Practical 7)5 marks
Q (5 marks). A student measures the resistance R of a wire of diameter 0.28 mm for several lengths l and plots R against l. The best-fit line passes through the origin with a gradient of 12.0 Ω m⁻¹. (a) Explain why the graph is a straight line through the origin. (b) Determine the resistivity of the wire.
Model answer
(a) From $R=\dfrac{\rho l}{A}$ , the resistance is directly proportional to the length ( $R\propto l$ ) because $\rho$ and $A$ are constant for the wire. This has the form $y=mx$ with no constant term, so the graph is a straight line through the origin, and its gradient equals $\dfrac{\rho}{A}$ .

(b) First find the cross-sectional area from the diameter, $d=0.28\ \text{mm}=2.8\times10^{-4}\ \text{m}$ :

$A = \frac{\pi d^2}{4} = \frac{\pi (2.8\times10^{-4})^2}{4} = 6.158\times10^{-8}\ \text{m}^2$

Since the gradient is $\rho/A$ , the resistivity is the gradient multiplied by the area:

$\rho = \text{gradient} \times A = 12.0 \times 6.158\times10^{-8} = 7.39\times10^{-7}\ \Omega\,\text{m}$
Why this scores
Why this scores 5/5. (a) 2 marks: $R\propto l$ (with $\rho$ , $A$ constant) so it fits $y=mx$ → straight line through origin, gradient $=\rho/A$ . (b) 3 marks: area from $\pi d^2/4$ with correct mm→m conversion, $\rho=\text{gradient}\times A$ , $7.39\times10^{-7}\ \Omega\,\text{m}$ with the unit. A frequent error is quoting the gradient itself as the resistivity — the gradient is $\rho/A$ , so you must multiply by A.
Question 8
Adapted from Edexcel IAL Physics Unit 2 (Core Practical 7)6 marks
Q (6 marks). Describe how you would determine the resistivity of a metal wire in the laboratory (Core Practical 7). Your answer should state the measurements you would take and the instruments used, explain how you would obtain a reliable value of resistivity from the data, and identify one source of systematic error and how to reduce it.
Model answer
Measurements and instruments. Measure the diameter of the wire at several points and in different orientations with a micrometer screw gauge (or digital calliper), and take the mean — then compute the area $A=\pi d^2/4$ . Connect a measured length of the wire into a circuit with a battery, ammeter and voltmeter, and for each of several lengths $l$ (set with a metre rule) record the current $I$ and the voltage $V$ across that length. The resistance for each length is $R=V/I$ .

Obtaining a reliable value. Rather than using a single length, plot $R$ against $l$ . From $R=\dfrac{\rho}{A}\,l$ this is a straight line through the origin of gradient $\rho/A$ . Draw a best-fit line, find the gradient from a large triangle, and obtain the resistivity from:

$\rho = \text{gradient} \times A$

Using the gradient of the whole line averages out random error and exposes anomalies, giving a more reliable value than one reading.

Systematic error. Contact/lead resistance at the connecting terminals (and resistance of the crocodile clips) adds a constant amount to every $R$ , which would shift the line up. Using the gradient (not a single $R$ ) removes this, because the constant offset only changes the intercept, not the slope. Also keep the current low / switch off between readings so the wire does not heat up and change its resistance.
Why this scores
Why this scores 6/6. (≈2) micrometer for diameter (repeat + mean) and metre rule for length, with $R=V/I$ ; (≈2) plot $R$ vs $l$ , gradient $=\rho/A$ , $\rho=\text{gradient}\times A$ , best-fit line for reliability; (≈2) a named systematic error (contact resistance OR wire heating) with a valid mitigation (use the gradient / keep current low). Vague 'measure carefully' statements are not credited — the mechanism is required.
Question 9
6 marks
Q (6 marks). A potential divider is to be used as a temperature-activated switch that turns on a cooling fan when a room gets too hot. The divider uses an NTC thermistor and a fixed resistor across a 12 V supply, and the fan circuit is triggered when the output voltage exceeds a set threshold. (a) Explain, with reference to charge carriers, how the thermistor's resistance changes as the room heats up. (b) Explain whether the output should be taken across the thermistor or across the fixed resistor so that the output rises with temperature. (c) State one advantage of using a thermistor in a potential divider rather than measuring temperature by hand.
Model answer
(a) The thermistor is a semiconductor. As the room heats up, more electrons gain enough energy to break free and act as charge carriers, so the number density $n$ of carriers increases. From $I=nqvA$ , more carriers allow more current for the same voltage, so the thermistor's resistance falls as temperature rises.

(b) The output should be taken across the fixed resistor. Using $V_{out}=V_{in}\,\dfrac{R_{fixed}}{R_{therm}+R_{fixed}}$ : as the room warms, $R_{therm}$ falls while $R_{fixed}$ stays constant, so the fixed resistor takes a larger share of the 12 V and the output rises with temperature — exactly what is needed to cross the threshold and switch the fan on. (If the output were taken across the thermistor instead, $V_{out}$ would fall as it warmed.)

(c) It gives a continuous electrical output that can automatically trigger the fan without a person reading a thermometer — the response is fast, can be set to a precise threshold, and works unattended.
Why this scores
Why this scores 6/6. (a) 2 marks: semiconductor → $n$ increases with temperature → resistance falls (link to $I=nqvA$ ). (b) 3 marks: output across the FIXED resistor; reasoned divider argument that $R_{therm}$ falling raises the fixed resistor's share → output rises. (c) 1 mark: automatic/continuous/fast electrical control. The discriminating step is correctly choosing which component the output is taken across and justifying it from the divider equation.
Question 10
6 marks
Q (6 marks). A wire of resistance 4.0 Ω is stretched uniformly so that its length doubles, while the volume of metal (and the resistivity) stays the same. (a) Explain what happens to the cross-sectional area when the length doubles. (b) Determine the new resistance of the stretched wire, showing your reasoning. (c) Comment on why this result follows from $R=\rho l/A$ .
Model answer
(a) The volume $V=A l$ is constant. If the length $l$ doubles, then to keep $V=Al$ the same the cross-sectional area $A$ must halve (since $A=V/l$ and $l$ has doubled).

(b) Using $R=\dfrac{\rho l}{A}$ with $\rho$ unchanged: the length goes $\times 2$ and the area goes $\times \tfrac{1}{2}$ . Resistance is proportional to $l$ and inversely proportional to $A$ , so:

$R_{new} = R_{old} \times \frac{l_{new}}{l_{old}} \times \frac{A_{old}}{A_{new}} = 4.0 \times 2 \times 2 = 16\ \Omega$

The new resistance is 16 Ω — four times the original.

(c) From $R=\rho l/A$ , doubling $l$ alone doubles $R$ , and halving $A$ alone doubles $R$ again; the two effects multiply to give a factor of four. This is why stretching a wire raises its resistance much faster than the length increase alone suggests.
Why this scores
Why this scores 6/6. (a) 2 marks: constant volume $Al$ → area halves when length doubles. (b) 3 marks: $R\propto l/A$ , factor $2\times2=4$ , $16\ \Omega$ . (c) 1 mark: both effects act in the same direction, multiplying. The trap is changing only the length (factor 2) and forgetting the area also changes — a synoptic favourite.

Key Formulae — Resistivity and Potential dividers

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

Resistivity
$R = \frac{\rho l}{A}$
$R$
resistance / Ω
$\rho$
resistivity / Ω m (a property of the material)
$l$
length of the conductor / m
$A$
cross-sectional area / m²
When to use
Linking the resistance of a uniform conductor to its dimensions and material. Rearrange to $\rho=RA/l$ to find resistivity from a measured resistance. In the Edexcel formulae booklet (Appendix 7).
Example
$\rho=1.1\times10^{-6}$ , $l=1.5$ m, $A=8.0\times10^{-8}$ m²: $R=20.6\ \Omega$ .
Cross-sectional area of a wire
$A = \frac{\pi d^2}{4} = \pi r^2$
$A$
cross-sectional area / m²
$d$
diameter of the wire / m
$r$
radius of the wire ( $r=d/2$ ) / m
When to use
Finding the area of a circular wire before using $R=\rho l/A$ . The micrometer measures the DIAMETER, so use $\pi d^2/4$ (or halve it first for $\pi r^2$ ). Must be recalled.
Example
$d=0.40$ mm $=4.0\times10^{-4}$ m: $A=\pi(4.0\times10^{-4})^2/4=1.26\times10^{-7}$ m².
Current and charge carriers
$I = nqvA$
$I$
current / A
$n$
number density of charge carriers / m⁻³
$q$
charge on each carrier / C
$v$
mean drift velocity / m s⁻¹
$A$
cross-sectional area / m²
When to use
Relating current to the motion of charge carriers; finding drift velocity, or explaining why metals (high $n$ ) conduct better than semiconductors (lower $n$ ). In the formulae booklet.
Example
$I=2.0$ A, $n=8.5\times10^{28}$ , $q=1.6\times10^{-19}$ , $A=1.0\times10^{-6}$ : $v=1.47\times10^{-4}$ m s⁻¹.
Potential divider output
$V_{out} = V_{in}\,\frac{R_2}{R_1 + R_2}$
$V_{out}$
output voltage, taken across $R_2$ / V
$V_{in}$
supply (input) voltage / V
$R_2$
resistance the output is taken across / Ω
$R_1$
the other series resistor / Ω
When to use
Finding the output of two resistors in series across a supply. The resistor whose voltage you want goes ON TOP. With a thermistor/LDR as one resistor, this becomes a sensor circuit. Must be recalled (follows from $V=IR$ in series).
Example
$V_{in}=12$ V, $R_1=4.0$ kΩ, $R_2=2.0$ kΩ: $V_{out}=12\times2.0/6.0=4.0$ V.
Series voltage ratio
$\frac{V_1}{V_2} = \frac{R_1}{R_2}$
$V_1,\ V_2$
voltages across the two series resistors / V
$R_1,\ R_2$
the two series resistances / Ω
When to use
A quick consequence of the divider: in series the same current flows, so the voltage splits in the same ratio as the resistances. Handy for sanity-checking a divider answer. Follows from $V=IR$ .
Example
If $R_1=2R_2$ then $V_1=2V_2$ , so $R_1$ takes two-thirds of the supply.

Key Definitions and Keywords — Resistivity and Potential dividers

Definitions to memorise and the exact keywords mark schemes credit for resistivity and potential dividers answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Resistivity (ρ)
Examiner keyword
A property of a material, defined by $R=\dfrac{\rho l}{A}$ ; numerically the resistance of a sample of unit length and unit cross-sectional area. Unit: ohm metre ( $\Omega\,\text{m}$ ). It depends on the material and on temperature, not on the shape of the sample.
Resistance (R)
Examiner keyword
The opposition of a component to current, $R=\dfrac{V}{I}$ (unit ohm, Ω). For a uniform wire it depends on the material (ρ), the length and the cross-sectional area.
Cross-sectional area of a wire
Examiner keyword
The area of the circular face of the wire, $A=\dfrac{\pi d^2}{4}=\pi r^2$ , where $d$ is the diameter (as measured by a micrometer) and $r=d/2$ is the radius.
Number density of charge carriers (n)
Examiner keyword
The number of free charge carriers per unit volume (unit m⁻³). Metals have a very high $n$ (~ $10^{28}$ m⁻³); semiconductors have a much lower $n$ that rises with temperature.
Drift velocity (v)
Examiner keyword
The mean velocity of the charge carriers along a conductor due to the applied field, given by $v=\dfrac{I}{nqA}$ . It is very small (typically a fraction of a mm s⁻¹) even though the current itself is established almost instantly.
Charge carrier
A mobile charged particle that carries current — free electrons in a metal, or electrons and holes in a semiconductor. Each carries charge $q$ (for an electron, $q=1.6\times10^{-19}\ \text{C}$ ).
Current–carrier equation (I = nqvA)
Examiner keyword
Relates the current to the carriers: $I=nqvA$ , where $n$ is the number density, $q$ the charge per carrier, $v$ the drift velocity and $A$ the cross-sectional area.
Potential divider
Examiner keyword
Two (or more) resistors in series across a supply, used to produce a chosen fraction of the supply voltage as an output. Output across $R_2$ : $V_{out}=V_{in}\,\dfrac{R_2}{R_1+R_2}$ .
Output voltage (Vout)
Examiner keyword
The voltage tapped off across one of the resistors of a potential divider. It equals the supply voltage multiplied by the fraction $\dfrac{R_{out}}{R_{total}}$ , i.e. the share of the resistance the output is taken across.
Thermistor (NTC)
Examiner keyword
A semiconductor component whose resistance changes markedly with temperature. For the common NTC (negative temperature coefficient) type, resistance falls as temperature rises (more carriers released).
Light-dependent resistor (LDR)
Examiner keyword
A semiconductor component whose resistance falls as the light intensity on it rises (more carriers released by the incident light). Used in light-sensing potential dividers.
Sensor (sensing) circuit
A potential divider in which one resistor is a thermistor or LDR, so that the output voltage varies with temperature or light. Used to drive temperature- or light-activated switches.
Temperature dependence of resistivity
Examiner keyword
How resistivity changes with temperature: for a metal it increases (more lattice vibration → more collisions, $n$ roughly constant); for a semiconductor it decreases (many more carriers released, $n$ rises sharply).
Systematic error (in CP7)
An error that shifts every reading the same way — e.g. contact/lead resistance adding a constant to each measured R, or wire heating raising resistance. Using the gradient of an R–l graph removes a constant offset.
Micrometer screw gauge
A precision instrument (resolution typically 0.01 mm) used to measure the small diameter of a wire. Repeat at several points and orientations and take the mean before computing $A=\pi d^2/4$ .
Ohm metre (Ω m)
The SI unit of resistivity. Comes from rearranging $\rho=RA/l$ : $[\Omega][\text{m}^2]/[\text{m}]=\Omega\,\text{m}$ . Note it is NOT $\Omega\,\text{m}^{-1}$ .

Common Mistakes and Misconceptions — Resistivity and Potential dividers

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

✕Putting the diameter straight into $\pi r^2$ (treating the diameter as the radius) when finding the cross-sectional area
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
The micrometer gives the diameter, but the familiar circle formula is $\pi r^2$ , so students forget to halve it (or to use $\pi d^2/4$ ).
How to avoid it
Use $A=\dfrac{\pi d^2}{4}$ directly with the diameter, or halve the diameter first ( $r=d/2$ ) for $\pi r^2$ . Forgetting this makes the area four times too big and every resistivity answer wrong.
✕Putting the wrong resistor on top of the potential-divider equation (not the one the output is taken across)
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students memorise $V_{out}=V_{in}R_2/(R_1+R_2)$ without checking which resistor the output is actually measured across in the given circuit.
How to avoid it
The resistor the output is taken across goes ON TOP of the fraction. Read the circuit: trace where the output terminals connect, and put that resistance in the numerator. Sanity-check that the larger resistor gets the larger share of the voltage.
✕Thinking an NTC thermistor's resistance INCREASES as it gets hotter
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students assume thermistors behave like metals (resistance up with temperature) and forget they are semiconductors.
How to avoid it
For an NTC thermistor, resistance FALLS as temperature rises — heating releases more charge carriers ( $n$ increases), so from $I=nqvA$ the resistance drops. 'NTC' = negative temperature coefficient.
✕Quoting the unit of resistivity as $\Omega\,\text{m}^{-1}$ (or $\Omega/\text{m}$ ) instead of $\Omega\,\text{m}$
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students guess the unit by analogy with 'per metre' quantities instead of deriving it from $\rho=RA/l$ .
How to avoid it
Derive it: $\rho=\dfrac{RA}{l}$ gives $\Omega\times\text{m}^2\div\text{m}=\Omega\,\text{m}$ . Always attach $\Omega\,\text{m}$ to a resistivity answer — the accuracy mark depends on it.
✕Forgetting to convert millimetre diameters/lengths to metres before substituting
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Diameters are quoted in mm (e.g. 0.30 mm) and lengths sometimes in cm, but the resistivity equation needs SI metres.
How to avoid it
Convert everything to metres first ( $1\ \text{mm}=10^{-3}\ \text{m}$ , so $0.30\ \text{mm}=3.0\times10^{-4}\ \text{m}$ ), then square for the area. A single unconverted mm typically throws the answer out by a factor of $10^{6}$ in the area.
✕In CP7, reading a single (R, l) point off the graph instead of using the gradient to find resistivity
Edexcel IAL Physics Unit 2 examiner reports
Why it happens
Students do not connect $R=\tfrac{\rho}{A}l$ to $y=mx$ and so miss that the gradient is $\rho/A$ .
How to avoid it
Plot R against l and use the GRADIENT: gradient $=\rho/A$ , so $\rho=\text{gradient}\times A$ . The gradient averages out random error and cancels a constant contact-resistance offset, which a single point cannot.

Resistivity and Potential dividers — frequently asked questions

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

Resistivity and Potential dividers

Short Notes - Resistivity and Potential dividers

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Cross-sectional area of a wire from its diameter (2 marks)

2Resistance of a wire from resistivity (3 marks)

3Output voltage of a potential divider (3 marks)

4Resistivity of a wire from a measured resistance (4 marks)

5Drift velocity from I = nqvA (4 marks)

6Output of a thermistor potential divider (4 marks)

7Determining resistivity from the gradient of an R–l graph (6 marks)

8LDR light-activated switch design (6 marks)

9Why a metal's resistance rises and a semiconductor's falls (6 marks)

Resistivity

Cross-sectional area of a wire

Current and charge carriers

Potential divider output

Series voltage ratio

Resistivity (ρ)

Resistance (R)

Cross-sectional area of a wire

Number density of charge carriers (n)

Drift velocity (v)

Charge carrier

Current–carrier equation (I = nqvA)

Potential divider

Output voltage (Vout)

Thermistor (NTC)

Light-dependent resistor (LDR)

Sensor (sensing) circuit

Temperature dependence of resistivity

Systematic error (in CP7)

Micrometer screw gauge

Ohm metre (Ω m)

✕Putting the diameter straight into πr2\pi r^2πr2 (treating the diameter as the radius) when finding the cross-sectional area

✕Putting the wrong resistor on top of the potential-divider equation (not the one the output is taken across)

✕Thinking an NTC thermistor's resistance INCREASES as it gets hotter

✕Quoting the unit of resistivity as Ω m−1\Omega\,\text{m}^{-1}Ωm−1 (or Ω/m\Omega/\text{m}Ω/m) instead of Ω m\Omega\,\text{m}Ωm

✕Forgetting to convert millimetre diameters/lengths to metres before substituting

✕In CP7, reading a single (R, l) point off the graph instead of using the gradient to find resistivity

Resistivity and Potential dividers — frequently asked questions

✕Putting the diameter straight into $\pi r^2$ (treating the diameter as the radius) when finding the cross-sectional area

✕Quoting the unit of resistivity as $\Omega\,\text{m}^{-1}$ (or $\Omega/\text{m}$ ) instead of $\Omega\,\text{m}$