Resistivity

Spec mapping: OCR H556 Module 4.2 — Energy, power and resistance (resistivity $\rho$ρ of a material; the equation $R = \rho L / A$R=ρL/A; temperature dependence; experimental determination as a PAG 4 anchor practical). Refer to the official OCR H556 specification document for exact wording.

Resistance is a property of a component: it depends on the material a wire is made from, its length, its cross-sectional area and its temperature. Resistivity is the property of the material itself: it is what is left after the geometry has been stripped away. "Copper is more conductive than nichrome" is a statement about resistivity, not about any particular resistor.

This lesson derives the resistivity equation $R = \rho L / A$R=ρL/A, places the resistivities of common materials on a thirty-orders-of-magnitude ladder, describes the OCR PAG 4 anchor practical (resistivity of a metal wire), and works through the standard mark-scheme moves — graphical analysis, area-from-diameter, uncertainty budget. By the end you should be able to compute $R$R from $(\rho, L, A)$(ρ,L,A) and $\rho$ρ from a measured gradient and a micrometer-derived area without confusion between intrinsic and geometric factors.

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From Resistance to Resistivity — Motivating the Equation

Take a uniform rectangular block of material of length $L$L along the current direction, with cross-sectional area $A$A perpendicular to the current. Apply a potential difference $V$V across its ends and drive a current $I$I. Intuitively:

• Doubling $L$L doubles the distance every conduction electron has to travel against scattering; the same accelerating field gives the same drift velocity but acts over twice the length, so the resistance doubles. Hence $R \propto L$R∝L.
• Doubling $A$A doubles the number of parallel conduction "lanes"; for a fixed pd the current doubles, so the resistance halves. Hence $R \propto 1/A$R∝1/A.

Combining: $R \propto L/A$R∝L/A. To convert proportionality to equality we introduce a constant of proportionality that depends only on the material — the resistivity, symbol $\rho$ρ (Greek "rho"):

$R = \frac{\rho L}{A}$R=AρL​

Symbol	Meaning	SI unit
$R$R	resistance of the sample	$\Omega$Ω
$L$L	length along current direction	m
$A$A	cross-sectional area perpendicular to current	m$^2$2
$\rho$ρ	resistivity of the material	$\Omega$Ω m

Rearranging:

$\rho = \frac{RA}{L}$ρ=LRA​

Resistivity $\rho$ρ is a material property: it is the same for a thin copper wire as for a copper bus-bar, provided the temperature is the same. Resistance $R$R is a component property: it depends on $L$L, $A$A and $\rho$ρ.

Definition. The resistivity of a material is the resistance of a sample of unit length and unit cross-sectional area at a stated temperature, with SI unit $\Omega$Ω m.

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Units — Why $\Omega$Ω m and Not $\Omega/$Ω/m

A frequent slip is to write the unit of resistivity as $\Omega /$Ω/m. Dimensional analysis of $\rho = RA/L$ρ=RA/L:

$[\rho] = \frac{[\Omega] \times [\text{m}^2]}{[\text{m}]} = \Omega \, \text{m}$[ρ]=[m][Ω]×[m2]​=Ωm

The unit is ohm-metre, not ohms-per-metre. A 1 $\Omega$Ω m resistivity means a 1 m cube of the material has a resistance of 1 $\Omega$Ω between opposite faces. By that benchmark, $\rho_{\text{copper}} = 1.7 \times 10^{-8}$ρcopper​=1.7×10−8 $\Omega$Ω m says that a 1 m cube of copper would have a between-faces resistance of $17$17 n$\Omega$Ω — vanishingly small.

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The Resistivity Ladder — Thirty Orders of Magnitude

Resistivity spans the largest dynamic range of any commonly encountered physical property. Memorise the order of magnitude, not the precise digits.

Material	Type	$\rho$ρ at 20 $^{\circ}$∘C ($\Omega$Ω m)
Silver	Metal	$1.6 \times 10^{-8}$1.6×10−8
Copper	Metal	$1.7 \times 10^{-8}$1.7×10−8
Gold	Metal	$2.4 \times 10^{-8}$2.4×10−8
Aluminium	Metal	$2.7 \times 10^{-8}$2.7×10−8
Tungsten	Metal	$5.5 \times 10^{-8}$5.5×10−8
Iron	Metal	$\sim 1.0 \times 10^{-7}$∼1.0×10−7
Constantan	Precision alloy	$4.9 \times 10^{-7}$4.9×10−7
Nichrome	Heating alloy	$1.1 \times 10^{-6}$1.1×10−6
Sea water	Electrolyte	$\sim 0.2$∼0.2
Pure silicon	Intrinsic semiconductor	$\sim 2 \times 10^{3}$∼2×103
Glass	Insulator	$10^{10}$1010 – $10^{14}$1014
PTFE (teflon)	Insulator	$\sim 10^{23}$∼1023

That is a range of about 31 orders of magnitude from silver to PTFE. Three consequences worth committing to memory:

• Silver has the lowest resistivity of any common element, but copper is only 6% worse and much cheaper — hence copper dominates electrical wiring.
• Nichrome has a resistivity nearly $65\times$65× that of copper, which is why heating elements use it: a useful amount of heat dissipates in a practical length of wire (lesson 8 worked example revisits this for a toaster).
• Constantan is chosen for precision resistors because its resistivity has an unusually low temperature coefficient — a precision resistor must not drift as the room warms up.

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Resistivity Depends on Temperature

A common A-Level error is to treat $\rho$ρ as a universal constant for a given material. It is not.

Metals

For a pure metal, resistivity increases approximately linearly with temperature over the range $-100$−100 $^{\circ}$∘C to $+500$+500 $^{\circ}$∘C:

$\rho(T) \approx \rho_0 \, [1 + \alpha (T - T_0)]$ρ(T)≈ρ0​[1+α(T−T0​)]

where $\alpha$α is the temperature coefficient of resistivity (about $+4 \times 10^{-3}$+4×10−3 K$^{-1}$−1 for copper). Physical reason: hotter lattice $\Rightarrow$⇒ larger amplitude thermal vibrations of the ion cores $\Rightarrow$⇒ more frequent scattering of conduction electrons $\Rightarrow$⇒ shorter mean free path $\Rightarrow$⇒ lower drift velocity $v_d$vd​ for a given field $\Rightarrow$⇒ smaller current $I = nAve$I=nAve for the same applied pd $\Rightarrow$⇒ higher $R$R. The current carrier density $n$n is essentially constant for a metal across the A-Level temperature range, so the entire change is through the mean free path.

Semiconductors

For a pure (intrinsic) semiconductor — silicon, germanium — resistivity decreases dramatically with temperature, roughly exponentially. Physical reason: thermal energy promotes more electrons from the valence band to the conduction band, so the carrier density $n$n rises sharply (Boltzmann factor $e^{-E_g/k_BT}$e−Eg​/kB​T with $E_g \sim 1$Eg​∼1 eV). The increase in scattering is overwhelmed by the increase in $n$n. This is the principle that drives the NTC thermistor (lesson 12 sensor circuits).

Superconductors

Below a critical temperature $T_c$Tc​ (which depends on the material — about $7$7 K for lead, up to about $138$138 K for some ceramic cuprates), $\rho$ρ drops to exactly zero — currents can circulate indefinitely with no dissipation. Beyond A-Level, but the existence of superconductors is examinable as a contrast with the linear-in-$T$T behaviour of normal metals.

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Worked Example 1 — Length of a Nichrome Resistor

You want to make a $5.0$5.0 $\Omega$Ω heating resistor from nichrome wire of cross-sectional area $0.20$0.20 mm$^2$2. Take $\rho = 1.1 \times 10^{-6}$ρ=1.1×10−6 $\Omega$Ω m. What length of wire is needed?

• Convert the area to SI: $A = 0.20$A=0.20 mm$^2 = 0.20 \times 10^{-6}$2=0.20×10−6 m$^2 = 2.0 \times 10^{-7}$2=2.0×10−7 m$^2$2.
• Rearrange $R = \rho L / A$R=ρL/A for $L$L: $L = RA/\rho$L=RA/ρ.
• $L = (5.0 \times 2.0 \times 10^{-7}) / (1.1 \times 10^{-6}) = (1.0 \times 10^{-6}) / (1.1 \times 10^{-6}) \approx \mathbf{0.91 \text{ m}}$L=(5.0×2.0×10−7)/(1.1×10−6)=(1.0×10−6)/(1.1×10−6)≈0.91 m.

Sanity check: a kilogram of nichrome wire at this gauge contains hundreds of metres, so 0.91 m is a sensible "kettle-coil" length.

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Worked Example 2 — Minimum Copper Cross-Section for a Cable

A $10$10 m run of copper mains flex carries the current to a kettle. The cable resistance must be no more than $0.10$0.10 $\Omega$Ω to keep transmission losses below 1% of the kettle's $\sim 2$∼2 kW rating. What minimum cross-sectional area is required? Take $\rho_{\text{Cu}} = 1.7 \times 10^{-8}$ρCu​=1.7×10−8 $\Omega$Ω m.

$A_{\min} = \frac{\rho L}{R} = \frac{(1.7 \times 10^{-8})(10)}{0.10} = 1.7 \times 10^{-6} \text{ m}^2 = \mathbf{1.7 \text{ mm}^2}$Amin​=RρL​=0.10(1.7×10−8)(10)​=1.7×10−6 m2=1.7 mm2

A standard UK 13 A mains flex has a conductor area of $1.25$1.25 or $1.5$1.5 mm$^2$2 — close to but below this calculated minimum. In practice, the $0.10$0.10 $\Omega$Ω allowance is generous; designers allow a few percent more drop because the alternative — heavier and more expensive copper — is wasteful. But the principle is exactly this calculation.

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Worked Example 3 — Diameter vs Area (the $\times 4$×4 Trap)

Two wires P and Q are made of the same material and have the same length. Wire P has twice the diameter of wire Q. What is the ratio of their resistances?

• Diameter doubles $\Rightarrow$⇒ area quadruples (since $A = \pi d^2 / 4$A=πd2/4, and $A \propto d^2$A∝d2).
• $R = \rho L / A$R=ρL/A, so $R \propto 1/A$R∝1/A, hence $R_P = R_Q / 4$RP​=RQ​/4.
• $R_P : R_Q = 1 : 4$RP​:RQ​=1:4.

The single most common A-Level resistivity slip is to write $A \propto d$A∝d instead of $A \propto d^2$A∝d2, giving a factor of $2$2 instead of $4$4. Whenever a question quotes a diameter, square it: $A = \pi d^2 / 4$A=πd2/4 (with diameter, not radius).

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Worked Example 4 — Two Materials, Same Resistor

A composite resistor is made by joining a 0.20 m length of copper wire ($\rho = 1.7 \times 10^{-8}$ρ=1.7×10−8 $\Omega$Ω m) to a 0.50 m length of nichrome wire ($\rho = 1.1 \times 10^{-6}$ρ=1.1×10−6 $\Omega$Ω m), end-to-end. Both wires have area $A = 1.0 \times 10^{-7}$A=1.0×10−7 m$^2$2. What is the total resistance?

The two segments are in series (current flows through one, then the other). Compute each separately and add.

• $R_{\text{Cu}} = (1.7 \times 10^{-8} \times 0.20) / (1.0 \times 10^{-7}) = 0.034$RCu​=(1.7×10−8×0.20)/(1.0×10−7)=0.034 $\Omega$Ω.
• $R_{\text{NC}} = (1.1 \times 10^{-6} \times 0.50) / (1.0 \times 10^{-7}) = 5.5$RNC​=(1.1×10−6×0.50)/(1.0×10−7)=5.5 $\Omega$Ω.
• $R_{\text{total}} = R_{\text{Cu}} + R_{\text{NC}} = 0.034 + 5.5 = \mathbf{5.534}$Rtotal​=RCu​+RNC​=0.034+5.5=5.534 $\Omega \approx \mathbf{5.5}$Ω≈5.5 $\Omega$Ω.

The copper segment contributes less than 1% to the total resistance — the heating happens almost entirely in the nichrome. This is exactly the design intent: copper leads connect to a nichrome element, so the copper wastes no energy and the nichrome glows.

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Required Practical — Resistivity of a Metal Wire (PAG 4 Anchor)

The OCR H556 specification requires students to perform the measurement of the resistivity of a metallic wire. This is one of the PAG 4 anchor practicals (electrical components and circuits) and a frequent indirectly assessed exam topic.

Apparatus

• Reel of uniform constantan (or nichrome) wire, $\sim 1$∼1 m.
• Metre ruler.
• Two crocodile clips, one mounted on a slider that moves along the ruler.
• Digital ammeter ($0$0 – $1$1 A) and digital voltmeter ($0$0 – $2$2 V).
• Low-voltage variable power supply or 1.5 V cell with rheostat.
• Switch.
• Micrometer screw gauge, resolution $0.01$0.01 mm.

Circuit

flowchart LR
  PS["Power supply / cell"] --> SW["Switch"]
  SW --> AM["Ammeter (A)"]
  AM --> WIRE["Wire under test (length L)"]
  WIRE --> PS
  WIRE -.-> VM["Voltmeter (V) across tested length"]

The voltmeter is connected directly across the tested length of wire (the section between the two clips), not across the whole circuit, so ammeter resistance and connection-wire resistance do not contaminate the reading.

Method

1. Take three micrometer readings of the wire diameter at different positions along the wire. Compute the mean diameter $\bar d$dˉ and the cross-sectional area $A = \pi \bar d^{\,2} / 4$A=πdˉ2/4.
2. Tape the wire taut along the metre ruler. Connect the fixed clip at one end and the slider clip so that an initial tested length $L \approx 1.00$L≈1.00 m sits between them.
3. Set the power supply to a low value so that the current heats the wire as little as possible (a few hundred milliamperes is typical). Close the switch only while taking readings.
4. Read $V$V across the tested length and $I$I through the circuit. Compute $R = V/I$R=V/I for that length.
5. Move the slider to shorten the tested length in $\sim 0.10$∼0.10 m steps. Record $V$V, $I$I, and compute $R$R at each length.
6. Plot $R$R on the $y$y-axis against $L$L on the $x$x-axis.
7. The graph is a straight line through the origin with gradient $\text{grad} = \rho / A$grad=ρ/A. Compute resistivity as $\rho = \text{grad} \times A$ρ=grad×A.

Why the Graph, Not a Single Reading?

You could compute $\rho$ρ from a single $V$V, $I$I, $L$L reading. Three reasons to prefer the gradient method:

• A single reading is dominated by contact resistance at the crocodile clips. The graph technique cancels this: the intercept (if non-zero) is the contact resistance, but the gradient depends only on the wire's intrinsic resistance per unit length.
• A gradient calculated from many points averages out random scatter; the resulting $\rho$ρ has a smaller percentage uncertainty than any single calculation.
• Plotting visually exposes any non-linearity (e.g. heating of the wire as the experiment progresses), prompting a re-design rather than a contaminated answer.

Worked Example 5 — Analysing Real Practical Data

A student measures a constantan wire of diameter $0.36$0.36 mm and obtains:

$L$L (m)	$R$R ($\Omega$Ω)
0.20	1.02
0.40	2.04
0.60	3.05
0.80	4.08
1.00	5.10

Calculate the resistivity.

• Gradient of the $R$R-$L$L graph: $\Delta R / \Delta L = (5.10 - 1.02) / (1.00 - 0.20) = 4.08 / 0.80 = \mathbf{5.10 \text{ }\Omega\text{ m}^{-1}}$ΔR/ΔL=(5.10−1.02)/(1.00−0.20)=4.08/0.80=5.10 Ω m−1.
• Cross-sectional area: $A = \pi (0.18 \times 10^{-3})^2 = \pi \times 3.24 \times 10^{-8} = \mathbf{1.02 \times 10^{-7} \text{ m}^2}$A=π(0.18×10−3)2=π×3.24×10−8=1.02×10−7 m2.
• Resistivity: $\rho = \text{grad} \times A = 5.10 \times 1.02 \times 10^{-7} = \mathbf{5.20 \times 10^{-7} \text{ }\Omega\text{ m}}$ρ=grad×A=5.10×1.02×10−7=5.20×10−7 Ω m.

Compared to the accepted value $\rho_{\text{constantan}} \approx 4.9 \times 10^{-7}$ρconstantan​≈4.9×10−7 $\Omega$Ω m, this is about 6% high — well within the combined micrometer and voltmeter uncertainty for a sixth-form practical.

Uncertainty Budget

For this experiment the dominant uncertainty contributions are:

Source	Typical $\Delta$Δ	Percentage of measured value
Micrometer on diameter	$\pm 0.01$±0.01 mm on $\sim 0.36$∼0.36 mm	2.8% in $d$d $\Rightarrow$⇒ 5.6% in $A$A (since $A \propto d^2$A∝d2)
Voltmeter (digital, 3 d.p.)	$\pm 0.005$±0.005 V on $\sim 1$∼1 V	0.5%
Ammeter (digital, 3 d.p.)	$\pm 0.005$±0.005 A on $\sim 0.20$∼0.20 A	2.5%
Metre rule on $L$L	$\pm 1$±1 mm on $\sim 500$∼500 mm	0.2%

Combined fractional uncertainty in $\rho$ρ:

$\frac{\Delta\rho}{\rho} = \sqrt{\left(\frac{\Delta A}{A}\right)^2 + \left(\frac{\Delta R}{R}\right)^2 + \left(\frac{\Delta L}{L}\right)^2} \approx \sqrt{(5.6\%)^2 + (3\%)^2 + (0.2\%)^2} \approx 6.3\%$ρΔρ​=(AΔA​)2+(RΔR​)2+(LΔL​)2​≈(5.6%)2+(3%)2+(0.2%)2​≈6.3%

So $\rho = (5.2 \pm 0.3) \times 10^{-7}$ρ=(5.2±0.3)×10−7 $\Omega$Ω m. The micrometer-diameter contribution dominates — diameter accuracy is the lever in this experiment.

Common Procedural Errors