Resistivity and Resistance

AP Physics 2· AP Physics 2 CED — Electric Circuits· 14 min read

1. Core Definitions: Resistance vs Resistivity★★☆☆☆⏱ 3 min

Resistance (symbol $R$ , unit ohm $Ω$ ) is a measure of how much a material component opposes the flow of electric current through it, conventionally defined as the ratio of the potential difference across the component to the current passing through it. Resistivity (symbol $ρ$ , unit ohm-meter $Ω \cdot m$ ) is an intensive intrinsic property of a material, meaning it does not depend on the size or shape of the sample, only on the type of material and its temperature.

This topic accounts for approximately 12% of the AP Physics 2 Unit 4 exam weight, and appears regularly in both multiple-choice and free-response sections, often combined with other circuit concepts like power dissipation or equivalent resistance. AP exam questions frequently test the ability to distinguish between the two properties and relate changes in wire dimensions or temperature to changes in resistance.

📘 Definition

Intensive vs Extensive Properties

Intensive properties (like resistivity) do not depend on the amount of material present. Extensive properties (like resistance) depend on the amount and geometry of material.

Example:

Cutting a wire in half halves the resistance (extensive) but leaves resistivity (intensive) unchanged.

2. The Resistance-Resistivity Relationship★★☆☆☆⏱ 4 min

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The fundamental relationship between the resistance of a uniform sample and its resistivity depends directly on the sample's geometry. For a wire of uniform cross-sectional area $A$ and length $L$ , resistance is given by:

R = \rho \frac{L}{A}

Increasing the length of the wire means current passes through more resistive material, so resistance increases linearly with $L$ . Increasing the cross-sectional area gives more space for charge carriers to flow, so resistance decreases inversely with $A$ . This formula only applies to uniform, isotropic materials, which is the only case tested on AP Physics 2.

📐 Worked Example

A copper wire of length 2.0 m, cross-sectional diameter 1.0 mm, has resistivity $ρ = 1.7 \times 1 0^{- 8} Ω \cdot m$ . What is the resistance of the wire?

1
Convert diameter to meters and calculate radius:
$d = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m}, \quad r = d/2 = 0.5 \times 10^{-3} \text{ m}$
2
Calculate cross-sectional area:
$A = \pi r^2 = \pi (0.5 \times 10^{-3})^2 \approx 7.85 \times 10^{-7} \text{ m}^2$
3
Substitute into the resistance formula:
$R = \rho \frac{L}{A} = \frac{(1.7 \times 10^{-8} \Omega \cdot \text{m})(2.0 \text{ m})}{7.85 \times 10^{-7} \text{ m}^2}$
4
Calculate the final value:
$R \approx 0.043 \Omega$

3. Temperature Dependence of Resistivity★★★☆☆⏱ 4 min

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Resistivity of a material depends on temperature, because temperature changes the motion of charge carriers and the atoms they collide with. For most metallic conductors, increasing temperature increases the kinetic energy of lattice atoms, leading to more frequent collisions between free electrons and the lattice, so resistivity increases. For semiconductors, increasing temperature releases more charge carriers from the lattice, which more than offsets the increased collision rate, so resistivity decreases as temperature increases.

The approximate linear relationship for small temperature changes around a reference temperature $T_{0}$ is:

\rho(T) = \rho_0 \left[ 1 + \alpha (T - T_0) \right]

Where $ρ_{0}$ is the resistivity at reference temperature $T_{0}$ (usually 20°C), and $α$ is the temperature coefficient of resistivity. For metals, $α$ is positive; for semiconductors, $α$ is negative. Since resistance is proportional to resistivity, the same relationship applies directly to resistance:

R(T) = R_0 [1 + \alpha (T - T_0)]

📐 Worked Example

A tungsten filament in a light bulb has a resistance of 0.50 Ω at 20°C. The temperature coefficient of resistivity for tungsten is $α = 4.5 \times 1 0^{- 3} °C^{- 1}$ . What is the resistance of the filament when it heats up to 2500°C?

1
Identify known values: $R_{0} = 0.50Ω$ , $T_{0} = 20° C$ , $T = 2500° C$ , $α = 4.5 \times 1 0^{- 3} °C^{- 1}$
2
Calculate the temperature change:
$\Delta T = T - T_0 = 2500 - 20 = 2480 °C$
3
Substitute into the resistance temperature formula:
$R = 0.50 \left[ 1 + (4.5 \times 10^{-3})(2480) \right]$
4
Simplify the term inside the brackets:
$(4.5 \times 10^{-3})(2480) = 11.16, \quad 1 + 11.16 = 12.16$
5
Calculate final resistance:
$R = 0.50 \times 12.16 \approx 6.1 \Omega$

4. Ohmic vs Non-Ohmic Materials★★★☆☆⏱ 3 min

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A material or component is classified as Ohmic if it obeys Ohm's law, meaning that the resistance of the component is constant regardless of the potential difference applied across it or the current passing through it, at constant temperature. For an Ohmic material, a graph of potential difference $V$ vs current $I$ is a straight line passing through the origin, with slope equal to the constant resistance $R$ .

Non-Ohmic materials do not have a constant resistance; resistance changes with applied voltage or current. The most common examples are filament light bulbs (resistance increases as current heats the filament) and diodes (high resistance in one direction, low in the other). A critical point to remember: the definition of resistance as $R = V / I$ still applies to non-Ohmic materials at any given point—resistance is just not constant for different operating points.

📐 Worked Example

A student measures voltage and current for an unknown component at constant temperature, collecting the following data: $V = 2.0 V, I = 0.50 A$ ; $V = 4.0 V, I = 1.0 A$ ; $V = 6.0 V, I = 2.0 A$ . Is the component Ohmic? What is the resistance of the component at 6.0 V?

1
For a component to be Ohmic, $R = V / I$ must be constant for all data points at constant temperature.
2
Calculate $R$ for each point:
$R_{2V} = 2.0 / 0.50 = 4.0 \Omega; \quad R_{4V} = 4.0 / 1.0 = 4.0 \Omega; \quad R_{6V} = 6.0 / 2.0 = 3.0 \Omega$
3
Resistance is not constant across operating points, so the component is non-Ohmic.
4
Even for non-Ohmic components, resistance at a specific point is still $V / I$ , so resistance at 6.0 V is 3.0 Ω.

5. Additional AP-Style Worked Problems★★★★☆⏱ 7 min

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📐 Worked Example

A uniform wire of resistance $R$ is stretched uniformly to three times its original length, with no change in density. What is the resistance of the stretched wire?

Options: A) $\frac{R}{3}$ , B) $3 R$ , C) $9 R$ , D) $27 R$

1
When stretched uniformly, volume remains constant because mass and density do not change. Original volume $V = L A$ .
2
New length $L^{'} = 3 L$ , so solve for new cross-sectional area:
$A' = V / L' = \frac{LA}{3L} = \frac{A}{3}$
3
Substitute into the resistance formula:
$R' = \rho \frac{L'}{A'} = \rho \frac{3L}{A/3} = 9 \left(\frac{\rho L}{A}\right) = 9R$
4
The correct answer is C.

📐 Worked Example

A student is testing the resistivity of an unknown cylindrical material sample. The sample has length 10.0 cm and diameter 2.0 cm. The student measures current through the sample for different potential differences at constant room temperature, with results: $V = 0.5 V, I = 0.20 A$ ; $V = 1.0 V, I = 0.41 A$ ; $V = 1.5 V, I = 0.59 A$ ; $V = 2.0 V, I = 0.81 A$ .

(a) Determine if the sample is Ohmic. (b) Calculate average resistivity. (c) Heating by 40°C increases resistance by 12%, find $α$ and identify material type.

1
Part (a): Calculate $R = V / I$ for each point:
$0.5/0.20 = 2.5 \Omega, \quad 1.0/0.41 \approx 2.4 \Omega, \quad 1.5/0.59 \approx 2.5 \Omega, \quad 2.0/0.81 \approx 2.5 \Omega$
2
Resistance is constant within experimental uncertainty, so the sample is Ohmic.
3
Part (b): Convert dimensions to SI units and calculate area:
$L = 0.100 \text{ m}, \quad r = 0.010 \text{ m}, \quad A = \pi r^2 \approx 3.14 \times 10^{-4} \text{ m}^2$
4
Average $R = 2.5Ω$ , rearrange to solve for $ρ$ :
$\rho = \frac{RA}{L} = \frac{(2.5 \Omega)(3.14 \times 10^{-4} \text{ m}^2)}{0.100 \text{ m}} \approx 7.9 \times 10^{-3} \Omega \cdot \text{m}$
5
Part (c): Use $R = R_{0} (1 + α Δ T)$ , with $R = 1.12 R_{0}$ :
$1.12 = 1 + 40\alpha \implies \alpha = 0.12/40 = 3.0 \times 10^{-3} ^\circ\text{C}^{-1}$
6
Positive $α$ means resistance increases with temperature, so the material is most likely a metallic conductor.

6. Common Pitfalls

Wrong move:

Assuming stretching a wire uniformly by a factor of $k$ only changes length, so resistance increases by a factor of $k$ .

Why:

Students forget that volume is conserved when stretching, so cross-sectional area also decreases by a factor of $k$ to keep volume constant.

Correct move:

Always apply volume conservation $L_{1} A_{1} = L_{2} A_{2}$ when a wire is stretched or compressed before calculating new resistance.

Wrong move:

Claiming that cutting a wire in half halves its resistivity.

Why:

Students mix up the definitions of intensive (resistivity) and extensive (resistance) properties.

Correct move:

Always remember: only changes to the material or temperature change resistivity; changes to size/shape only change resistance.

Wrong move:

Generalizing that all materials have increasing resistance as temperature increases.

Why:

Students mostly work with metallic conductors in examples, so they forget the opposite behavior of semiconductors.

Correct move:

Always check the sign of $α$ : positive $α$ (metals, resistance increases with T), negative $α$ (semiconductors, resistance decreases with T).

Wrong move:

Stating that $R = V / I$ does not apply to non-Ohmic materials.

Why:

Students confuse the definition of resistance with Ohm's law (which requires constant resistance).

Correct move:

Use $R = V / I$ to find resistance at any point for any component, whether it is Ohmic or non-Ohmic.

Wrong move:

Using millimeters for diameter to calculate area without converting to meters, leading to resistance that is six orders of magnitude too large.

Why:

Units of resistivity are given in $Ω \cdot m$ , so all input lengths must match this unit.

Correct move:

Always convert all length units to meters before substituting into $R = ρ L / A$ .

7. Quick Reference Cheatsheet

Category	Formula / Property	Key Notes
Resistance from resistivity	$R = ρ \frac{L}{A}$	Applies to uniform isotropic materials; $L$ = length, $A$ = cross-sectional area, units of $ρ$ : $Ω \cdot m$
Resistivity property	Intensive material property	Does not change when sample size/shape changes; only changes with material or temperature
Definition of resistance	$R = \frac{V}{I}$	Always valid for any component, Ohmic or non-Ohmic
Temperature dependence of resistance	$R (T) = R_{0} [1 + α (T - T_{0})]$	Linear approximation for small $Δ T$ ; $α > 0$ for conductors, $α < 0$ for semiconductors
Temperature dependence of resistivity	$ρ (T) = ρ_{0} [1 + α (T - T_{0})]$	Identical form to resistance relationship, since $R \propto ρ$ at fixed dimensions
Ohmic material	$R = constant (fixed T)$	V-I graph is a straight line through the origin
Volume conservation (stretched wire)	$L_{1} A_{1} = L_{2} A_{2}$	Use for uniform stretching/compression; volume is constant for fixed mass and density

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · AP Physics 2
MCQ: Stretched wire resistance change
2023 · AP Physics 2
FRQ: Resistivity calculation from data

What's Next

This topic is the foundation for all further work in electric circuits, as all circuit components have resistance that contributes to overall circuit behavior. Next you will apply the concepts of resistance and resistivity to analyze power dissipation in resistors, calculate equivalent resistance for series and parallel resistor combinations, and solve for currents and voltages in complex DC circuits using Kirchhoff's laws. Without a solid understanding of how resistance depends on material, dimensions, and temperature, you will not be able to correctly interpret circuit behavior or solve common FRQ problems requiring reasoning about changing circuit conditions. This topic also connects to broader concepts across AP Physics 2, including the behavior of semiconductors in modern electronic devices.