IAL Physics (Edexcel) · Unit 2 · Topic 2.8AS≈ 25 min read

2.8 Resistivity and the potential divider

⤓ Presentation ⤓ PowerPointPresentation: opens offline in any browser, simulations live. PowerPoint: editable.

AS content in Unit 2 (Waves and Electricity, WPH12), specification points 71–72 and 74–76, including the core practical on the resistivity of a wire. Sources have negligible internal resistance unless stated.

What this page covers

6 points

Resistivity and the equation \(R = \rho l/A\); how resistance scales with length and diameter
Core practical: determining the resistivity of the metal of a wire from a graph of R against l
How the potential varies along a uniform wire that carries a current
Potential dividers: sharing a supply p.d. between resistors, and choosing a resistor for a given output
Potential dividers with a variable resistor, a thermistor or an LDR as sensor circuits
Why a voltmeter or other load of finite resistance lowers the output p.d.

In one minute

6 points

Resistivity \(\rho\) is a property of the material, \(R = \rho l/A\), with unit Ω m: a longer wire has more resistance and a thicker wire has less.
For a round wire \(A = \pi d^2/4\), so halving the diameter makes the resistance four times larger.
To measure \(\rho\): plot R against l; the gradient is \(\rho/A\), and A comes from the micrometer diameter.
Along a uniform wire carrying a steady current, the potential changes linearly with distance.
In a potential divider the same current passes through each resistor, so the supply p.d. is shared in the ratio of the resistances: \(V_{\text{out}} = \dfrac{R_2}{R_1 + R_2}V_{\text{in}}\).
Replace one resistor with a thermistor or LDR and the output responds to temperature or light; a variable resistor sets the switching point.

12 sections

What is resistivity, and how does resistance depend on size?
How do you determine the resistivity of the metal of a wire?
How does the potential vary along a uniform wire?
How does a potential divider share out a p.d.?
How do sensor circuits use thermistors, LDRs and variable resistors?
What happens when you connect a voltmeter or load to the output?
Exam tips
Common misconceptions
How this topic connects
Key equations
Check your understanding
Specification checklist

What is resistivity, and how does resistance depend on size?

Two wires of the same metal can have very different resistances, because resistance depends on the shape of the piece as well as the material. A longer wire has more resistance: the charge carriers pass more ions on the way, so \(R \propto l\). A thicker wire has less resistance: there are more parallel paths for the charge, so \(R \propto 1/A\). Resistivity is the constant of proportionality that is left. It depends only on the material (and its temperature), not on the shape of the sample.

The three quantities in R = ρl/A: the length l along which the current flows, and the cross-sectional area A, worked out from the measured diameter d.

Resistance from resistivity: \(R = \dfrac{\rho l}{A}\), where R is the resistance (Ω), \(\rho\) the resistivity of the material (Ω m), l the length (m) and A the cross-sectional area (m²). In the formulae list supplied in the exam. It applies to a uniform conductor at constant temperature. Rearranged, \(\rho = RA/l\), which gives the unit: Ω × m² ÷ m = Ω m.

Typical values (room temperature): copper \(1.7 \times 10^{-8}\) Ω m; aluminium \(2.8 \times 10^{-8}\) Ω m; constantan \(4.9 \times 10^{-7}\) Ω m; nichrome \(1.1 \times 10^{-6}\) Ω m; pure silicon about \(10^{3}\) Ω m; glass above \(10^{10}\) Ω m. Good conductors have tiny resistivities, insulators enormous ones, and semiconductors sit in between.
Round wires: \(A = \pi r^2 = \pi d^2/4\). Micrometers measure the diameter, so halve it before squaring.
Units: 1 mm = \(10^{-3}\) m but 1 mm² = \(10^{-6}\) m².

R against l for two constantan wires, plotted from R = ρl/A. Both lines pass through the origin. The thinner wire's gradient is 2.04 times larger, because (0.40 ÷ 0.28)² = 2.04.

Worked example 1 Easy · Calculate

An extension lead is 25 m long. The current flows out along one copper core and back along the other. Each core has a cross-sectional area of 1.5 mm². The resistivity of copper is \(1.7 \times 10^{-8}\) Ω m. (a) Calculate the total resistance of the two cores. (b) Calculate the p.d. across the lead when it carries a current of 10 A.

(a) Convert the area: \(A = 1.5 \times 10^{-6}\) m². The current passes through \(2 \times 25 = 50\) m of copper, so \(R = \dfrac{1.7 \times 10^{-8} \times 50}{1.5 \times 10^{-6}}\), giving \(R = 0.57\ \Omega\) (0.567 Ω).

(b) \(V = IR = 10 \times 0.567 = 5.7\) V. This p.d. is lost before the supply reaches the appliance, which is why long leads for high-current appliances use thicker cores.

Worked example 2 Medium · Calculate

A 2.0 m length of nichrome wire has a resistance of 14 Ω. (a) Calculate the resistance of a 0.50 m length of nichrome wire with half the diameter. (b) The original 2.0 m wire is drawn out uniformly until it is 3.0 m long. Its volume does not change. Calculate its new resistance.

(a) Scale each factor separately: length × 0.25; diameter × ½, so area × ¼ and resistance × 4. \(R = 14 \times 0.25 \times 4 = 14\ \Omega\).

(b) Volume \(Al\) is constant, so if l rises by a factor of 1.5 the area falls by a factor of 1.5. \(R = 14 \times 1.5 \times 1.5 = 31.5\ \Omega \approx 32\ \Omega\). Stretching a wire increases its resistance twice over: it gets longer and thinner.

Common error

Using the diameter as if it were the radius in \(\pi r^2\). The area comes out four times too big and the resistance (or resistivity) four times too small. Write \(A = \pi (d/2)^2\) as a separate line of working.

How do you determine the resistivity of the metal of a wire?

This is a core practical. Measure the resistance of several different lengths of the same wire, plot R against l, and use the gradient together with the cross-sectional area.

The ammeter measures the current through the wire; the voltmeter measures the p.d. across the length l between the two clips. R = V/I for each length.

Diameter: use a micrometer screw gauge (resolution 0.01 mm), checked for zero error when closed. Measure at several points along the wire, turning the micrometer through 90° at each point, and take the mean: the wire may be uneven or not perfectly round.
Set-up: tape the wire straight along a metre rule. Connect it in series with a battery (or low-voltage supply), a switch and an ammeter, with one crocodile clip fixed at the zero mark and a second clip that can be moved along the wire.
Readings: connect a voltmeter across the length l between the clips. For each length (for example every 0.10 m from 0.10 m to 1.00 m) record V and I and calculate \(R = V/I\).
Constant temperature: use a small current and switch off between readings. Resistivity rises with temperature, so a heated wire would give a wrong value.
Analysis: plot R (y-axis) against l (x-axis) and draw the line of best fit.
Safety: a short length of thin wire can become hot; avoid touching it while current flows.

Why the gradient gives the resistivity

Write the equation as \(R = \left(\dfrac{\rho}{A}\right) l\) and compare with \(y = mx\): the gradient is \(\rho/A\) (not in the formulae list: derive it by rearranging \(R = \rho l/A\), which is). So \(\rho = \text{gradient} \times A\). A small intercept, from the resistance of the clips and contacts or a zero error, does not change the gradient.

Worked example 3 Hard · Determine

A student measures the diameter of a wire at five places: 0.37 mm, 0.38 mm, 0.38 mm, 0.37 mm, 0.38 mm. Her values of R for lengths 0.200 m to 1.000 m (in steps of 0.200 m) are 1.06 Ω, 1.90 Ω, 2.82 Ω, 3.65 Ω and 4.57 Ω. Her best-fit line passes through 0.61 Ω at 0.10 m and 4.55 Ω at 1.00 m. (a) Calculate the mean diameter and its percentage uncertainty. (b) Determine the resistivity of the metal.

The best-fit line does not quite pass through the origin: its intercept of 0.17 Ω is the extra resistance at the clips. The gradient triangle is drawn as large as the line allows.

(a) \(d = 1.88 / 5 = 0.376\) mm. Uncertainty = half the range \(= (0.38 - 0.37)/2 = \pm 0.005\) mm, so the percentage uncertainty = \(0.005/0.376 \times 100 = 1.3\%\).

(b) Area: \(A = \pi (0.376 \times 10^{-3}/2)^2 = 1.110 \times 10^{-7}\) m². Gradient: \((4.55 - 0.61)/(1.00 - 0.10) = 3.94/0.90 = 4.38\ \Omega\ \text{m}^{-1}\). Then \(\rho = 4.38 \times 1.110 \times 10^{-7}\), so \(\rho = 4.9 \times 10^{-7}\) Ω m (\(4.86 \times 10^{-7}\) Ω m).

The diameter is the smallest quantity measured, so it usually has the largest percentage uncertainty, and it is squared in the area, so it is worth measuring carefully.

Exam tip

Explain why the graph method is better than a single reading: the gradient uses all the results, it shows up any anomalous point, and it is not affected by a constant extra resistance at the contacts, which only shifts the intercept.

How does the potential vary along a uniform wire?

When a steady current I flows through a uniform wire (same material and cross-section all the way along), every centimetre has the same resistance. The p.d. across any section is IR, and R ∝ length, so the p.d. across a section is proportional to its length. Taking one end as 0 V, the potential rises linearly with distance towards the other end. The potential gradient (V m⁻¹) is the p.d. across the whole wire divided by its length. A sliding contact can therefore pick off any p.d. between zero and the full p.d. across the wire: a continuously variable potential divider.

The circuit of worked example 4. A 0.50 A current through the 4.0 Ω wire gives 2.0 V across it, and the voltmeter picks off part of that.

With B at 0 V the potential rises in a straight line to 2.0 V at A; with the sliding contact 0.65 m from B the voltmeter reads 1.3 V.

Worked example 4 Medium · Calculate

A uniform wire AB of length 1.00 m and resistance 4.0 Ω is connected in series with an 8.0 Ω resistor and a 6.0 V supply of negligible internal resistance. End B is connected to the negative terminal. (a) Calculate the p.d. across the wire. (b) Calculate the p.d. between two points on the wire 0.30 m and 0.85 m from B. (c) How far apart must two contacts be to pick off 0.50 V?

(a) \(I = 6.0/(8.0 + 4.0) = 0.50\) A, so \(V = 0.50 \times 4.0 = 2.0\) V.

(b) The potential gradient is \(2.0/1.00 = 2.0\) V m⁻¹ and the points are 0.55 m apart: \(V = 2.0 \times 0.55 = 1.1\) V.

(c) Distance = p.d. ÷ potential gradient: \(0.50/2.0 = 0.25\) m. Only the separation of the two points matters, not where they are on the wire.

How does a potential divider share out a p.d.?

A potential divider is two (or more) resistors in series across a supply, giving an output p.d. that is a chosen fraction of the supply p.d. The same current flows through each resistor, so each p.d. is proportional to its resistance and the larger resistor takes the larger share. The p.d.s add up to the supply p.d. (Kirchhoff's second law).

Potential divider: \(I = \dfrac{V_{\text{in}}}{R_1 + R_2}\), so \(V_{\text{out}} = IR_2 = \dfrac{R_2}{R_1 + R_2} \times V_{\text{in}}\), and \(\dfrac{V_1}{V_2} = \dfrac{R_1}{R_2}\). \(V_{\text{in}}\) is the supply p.d. across both resistors, \(V_{\text{out}}\) the p.d. across \(R_2\). Not in the formulae list: derive it from \(V = IR\) and the series rule \(R = R_1 + R_2\), both of which are. Valid when no current is drawn from the output, or the load's resistance is much larger than \(R_2\).

The divider of worked example 5. The 9.0 V supply is shared in the ratio 4.7 : 2.2, giving 6.1 V and 2.9 V, which add up to 9.0 V.

Worked example 5 Medium · Calculate

A 4.7 kΩ resistor and a 2.2 kΩ resistor are connected in series across a 9.0 V supply of negligible internal resistance. The output is taken across the 2.2 kΩ resistor. (a) Calculate the output p.d. (b) The 4.7 kΩ resistor is replaced so that the output becomes 6.0 V. Calculate the resistance of the new resistor.

(a) \(V_{\text{out}} = \dfrac{2.2}{4.7 + 2.2} \times 9.0\), so \(V_{\text{out}} = 2.9\) V (2.87 V).

(b) The other resistor must take \(9.0 - 6.0 = 3.0\) V, and p.d.s are in the ratio of resistances: \(R_1/2.2 = 3.0/6.0\), so \(R_1 = 1.1\) kΩ. Sense check: for the output to be the larger share, \(R_2\) must be the larger resistance.

Using a variable resistor

Replacing \(R_2\) with a variable resistor makes the output adjustable. With a 1.0 kΩ fixed resistor and a 0–5.0 kΩ variable resistor on 6.0 V, the output across the variable resistor can be set anywhere from 0 to \(6.0 \times 5.0/6.0 = 5.0\) V. A potentiometer (a resistance track with a sliding contact, like the wire above) gives the full range from 0 to \(V_{\text{in}}\).

Output across a 0–5.0 kΩ variable resistor in series with 1.0 kΩ on 6.0 V. The output rises quickly at first, then levels off below the supply p.d.: it never reaches 6.0 V.

Common error

Writing \(V_{\text{out}} = \dfrac{R_2}{R_1} \times V_{\text{in}}\). The denominator is the total resistance \(R_1 + R_2\). A quick check: the output can never be bigger than the supply.

How do sensor circuits use thermistors, LDRs and variable resistors?

An NTC thermistor's resistance falls as its temperature rises. A light-dependent resistor's (LDR's) resistance falls as the light intensity on it rises (why, is on page 2.10). Put one of them in a potential divider and the output p.d. changes with temperature or light: a sensor circuit. The output can switch a lamp, heater, fan or alarm through a switching circuit when it passes a set value.

To decide which way the output goes, reason in three steps: (1) the sensor's resistance changes; (2) it therefore takes a larger or smaller share of the supply p.d.; (3) the other resistor takes the rest.

Sensor in the top position, output across the fixed resistor: sensor resistance falls → its share falls → \(V_{\text{out}}\) rises.
Sensor in the bottom position, output across the sensor: sensor resistance falls → \(V_{\text{out}}\) falls.
Swapping the two components reverses the response.

A thermistor sensor on a 5.0 V supply, with the output taken across the fixed 10 kΩ resistor.

The thermistor's resistance falls from about 34 kΩ at 0 °C to 10 kΩ at 25 °C and 2.5 kΩ at 60 °C, so the output rises from 1.1 V to 2.5 V to 4.0 V. The curve is plotted from data for a typical thermistor.

Setting the switching point: make the fixed resistor a variable resistor. Adjusting it changes the temperature (or light level) at which the output reaches the switching value.
Sensitivity: the output changes most when the sensor's resistance is comparable to the other resistor's, so choose the fixed resistance close to the sensor resistance in the range you care about.

Worked example 6 Medium · Explain, Calculate

A garden lamp should switch on at dusk. An LDR and a fixed resistor R are in series across a 6.0 V supply of negligible internal resistance, and the output is taken across the LDR. The switching circuit turns the lamp on when the output reaches 2.0 V. At dusk the LDR's resistance is 8.0 kΩ; in bright daylight it is 0.50 kΩ. (a) Explain why the output p.d. rises as it gets darker. (b) Calculate the value of R needed. (c) Calculate the output p.d. in bright daylight.

In darkness the LDR's resistance rises, it takes a larger share of the 6.0 V, and the output across it rises to the 2.0 V that switches the lamp on.

(a) Less light → LDR resistance increases → the LDR takes a larger share of the 6.0 V (same current, p.d. ∝ resistance) → the output across the LDR rises.

(b) At switching, the LDR has 2.0 V and R has \(6.0 - 2.0 = 4.0\) V: \(R/8.0 = 4.0/2.0\), so \(R = 16\) kΩ.

(c) \(V_{\text{out}} = \dfrac{0.50}{16 + 0.50} \times 6.0\), so \(V_{\text{out}} = 0.18\) V, well below 2.0 V, so the lamp stays off. Using a variable resistor for R lets the user choose how dark it must be before the lamp comes on.

Exam tip

"Explain" questions on sensor circuits want the full chain: what changes in the surroundings, what happens to the sensor's resistance, what happens to its share of the p.d., and what happens to the output. Leaving out the middle step, the share of the p.d., leaves the explanation incomplete.

What happens when you connect a voltmeter or load to the output?

Anything connected to the output (a voltmeter, a lamp, a switching circuit) is in parallel with \(R_2\). The parallel combination has a lower resistance than \(R_2\) alone, so it takes a smaller share of the supply and the output p.d. falls. The effect is small only if the load resistance is much larger than \(R_2\). That is why an ideal voltmeter has infinite resistance, and why digital voltmeters (about 10 MΩ) are used to measure across large resistors.

A 50 kΩ voltmeter across one of two 20 kΩ resistors. The meter and resistor together make 14.3 kΩ, so the reading is 5.0 V instead of the 6.0 V that was there before the meter was connected.

Worked example 7 Hard · Calculate

Two 20 kΩ resistors are in series across a 12.0 V supply of negligible internal resistance. A voltmeter of resistance 50 kΩ is connected across one of them. (a) State the p.d. across each resistor before the voltmeter is connected. (b) Calculate the voltmeter reading. (c) Calculate the reading with a 10 MΩ digital voltmeter instead.

(a) Equal resistors share the supply equally: 6.0 V across each.

(b) \(1/R = 1/20 + 1/50\), so \(R = 14.3\) kΩ. \(V = \dfrac{14.3}{20 + 14.3} \times 12.0\), giving 5.0 V, 17% lower than the true 6.0 V.

(c) With 10 MΩ = 10 000 kΩ in parallel, \(R = 19.96\) kΩ and \(V = 5.994\) V, so \(V \approx 6.0\) V. A meter whose resistance is 500 times \(R_2\) barely disturbs the divider; one only 2.5 times \(R_2\) disturbs it badly.

Exam tips for 2.8

Exam tip 1 — convert areas properly

Convert the diameter to metres before squaring (0.38 mm = \(0.38 \times 10^{-3}\) m), or convert mm² with \(10^{-6}\). Using \(10^{-3}\) for an area is a factor of 1000 out.

Exam tip 2 — scaling: one factor at a time

For "twice the length and half the diameter" write the factors separately: × 2 for length, × 4 for area (½ squared, then inverted), so × 8 overall. Keep the material the same or the resistivity changes too.

Exam tip 3 — describe the practical in measurable detail

Name the instrument for each quantity (micrometer for diameter, metre rule for length, ammeter and voltmeter for R), say how you reduce uncertainty (repeat the diameter at several places and orientations, many lengths, graph), and how you keep the temperature constant (small current, switch off).

Exam tip 4 — check that p.d.s add up

After any potential divider calculation, add the p.d.s: they must equal the supply. It catches the common slip of calculating the p.d. across the wrong resistor.

Exam tip 5 — the load is in parallel

If a question mentions the resistance of a voltmeter or of a device connected to the output, combine it in parallel with the resistor it is across before using the divider equation.

Common misconceptions

Misconception	Why it is wrong	Correct idea
Resistivity and resistance are the same thing.	Two copper wires of different sizes have different resistances but the same resistivity.	Resistivity is a property of the material; resistance also depends on length and area.
Doubling the diameter halves the resistance.	Area depends on the diameter squared.	Doubling the diameter quarters the resistance.
The unit of resistivity is Ω m⁻¹.	From \(\rho = RA/l\): Ω × m² ÷ m.	The unit is Ω m (ohm metre). Ω m⁻¹ is the unit of the gradient of an R–l graph.
The R–l line must go through the origin, so the experiment failed if it does not.	Contact resistance or a zero error adds a constant to every reading.	A constant offset moves the intercept but not the gradient, which is what gives \(\rho\).
Current is "used up" along a wire, so the potential falls.	Current is the same at every point in a series circuit.	Energy per unit charge is transferred along the wire; the potential falls by IR for each section.
Connecting a voltmeter across a resistor does not change the circuit.	A real voltmeter has finite resistance in parallel with the resistor.	The combined resistance is lower, so the measured p.d. is lower; the effect is small only if the meter's resistance is much larger.

How this topic connects

2.6 Current, p.d. and resistance

\(R = V/I\) defines resistance; this page explains how that resistance depends on the material and the dimensions, and uses \(V = IR\) in every divider calculation.

2.7 Series and parallel circuits

A potential divider is a series circuit; the loading effect of a voltmeter is a parallel combination.

2.9 E.m.f. and internal resistance, 2.10 Resistance, temperature and light, and Unit 3

A real cell is a potential divider between its internal and external resistance. Thermistor and LDR behaviour is explained on page 2.10 using \(I = nqvA\). The planning, graph and uncertainty skills of the resistivity practical reappear in Unit 3.

Key equations

Equation	Use	In the formulae list?
\(R = \rho l/A\)	resistance from resistivity and dimensions (\(\rho\) in Ω m, l in m, A in m²)	Yes
\(R = V/I\)	resistance from p.d. and current	Yes
\(R = R_1 + R_2 + \dots\)	resistors in series (the divider chain)	Yes
\(1/R = 1/R_1 + 1/R_2 + \dots\)	resistors in parallel (a load across the output)	Yes
\(A = \pi d^2/4\)	cross-sectional area of a round wire	No — learn it
gradient of R–l graph \(= \rho/A\)	resistivity practical	No — derive it from \(R = \rho l/A\)
\(V_{\text{out}} = R_2V_{\text{in}}/(R_1 + R_2)\)	output of a potential divider	No — derive it from \(V = IR\)
\(V_1/V_2 = R_1/R_2\)	ratio of p.d.s in series	No — derive it (same current)

Check your understanding

1. Show that the unit of resistivity is Ω m.

\(\rho = RA/l\), so the unit is Ω × m² ÷ m = Ω m.

2. Wire Q is made of the same metal as wire P, with twice the length and twice the diameter. Compare their resistances.

Length × 2, area × 4, so \(R_Q = (2/4)R_P = 0.5R_P\): wire Q has half the resistance.

3. In the resistivity practical, why is the diameter measured at several places and in different directions?

The wire may not be uniform or perfectly round. The mean of several readings gives a better value for the area, and the spread gives the uncertainty in the diameter.

4. A 1.0 kΩ and a 2.0 kΩ resistor are in series across 6.0 V. Calculate the p.d. across each.

The 6.0 V is shared in the ratio 1 : 2, so 2.0 V across the 1.0 kΩ resistor and 4.0 V across the 2.0 kΩ resistor.

5. A thermistor is in the top position of a divider and the output is across the fixed resistor below it. What happens to the output as the temperature falls?

The thermistor's resistance increases, so it takes a larger share of the supply p.d. The fixed resistor's share, the output, falls.

6. A uniform wire of resistance 6.0 Ω and length 1.20 m has 3.0 V across it. What length of it has a p.d. of 1.0 V across it?

Potential gradient \(= 3.0/1.20 = 2.5\) V m⁻¹, so the length is \(1.0/2.5 = 0.40\) m (one third of the wire, as 1.0 V is one third of 3.0 V).

Specification checklist

(71) I can use \(R = \rho l/A\), including converting diameters and areas to SI units and scaling resistance with length and diameter.
(72) I can describe how to determine the resistivity of the metal of a wire, naming the instruments and how each source of uncertainty is reduced.
(72) I can find the resistivity from the gradient of a graph of R against l, and explain why a non-zero intercept does not affect it.
(74) I can explain why the potential varies linearly along a uniform wire carrying a steady current, and calculate the p.d. between two points on it.
(75) I can explain how a potential divider shares a supply p.d. and calculate p.d.s and resistances in one.
(76) I can explain how a potential divider containing a thermistor, an LDR or a variable resistor responds to changes in temperature or light, and how the switching point is set.
(76) I can explain and calculate how a voltmeter or load of finite resistance lowers the output of a potential divider.