Resistivity and Resistance — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Definition of resistance and resistivity, the dependence of resistance on material and dimensions, temperature dependence of resistivity, classification of Ohmic and non-Ohmic materials, and problem-solving for circuit component resistance calculations.

You should already know: Definition of electric current, definition of potential difference, basic properties of electric circuits.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Resistivity and Resistance?

Resistance (symbol $R$, unit ohm $\Omega$) 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 $\rho$, unit ohm-meter $\Omega \cdot \text{m}$) 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 (Electric Circuits) exam weight, and appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections, often combined with other circuit concepts like power dissipation or equivalent resistance. Students are expected to clearly distinguish between resistance, an extensive property that depends on the amount and geometry of the material, and resistivity, the intensive property that describes the material’s inherent opposition to current flow. AP exam questions frequently test the ability to relate changes in wire dimension (e.g., stretching a wire uniformly) to resulting changes in resistance, or predict how resistance changes with temperature for common conductors and semiconductors.

2. The Resistance-Resistivity Relationship

The fundamental relationship between the resistance of a uniform sample and its resistivity is derived directly from the geometry of the sample. For a wire of uniform cross-sectional area $A$ and length $L$, resistance is given by: $$R=\rho \frac{L}{A}$$

Intuition for this formula: Increasing the length of the wire means current has to pass 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 matches real-world behavior: a thick short wire has much lower resistance than a thin long wire of the same material at the same temperature. Because resistivity is intensive, cutting a wire in half does not change the resistivity of either piece—only the resistance is cut in half. This is a common point of confusion, so always remember: any change to the size or shape of a sample changes the resistance, but only a change in temperature or material changes the resistivity. This formula only applies to uniform, isotropic materials (same properties in all directions), 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 $\rho =1.7\times 10^{-8}\Omega \cdot \text{m}$. What is the resistance of the wire?

1. Convert diameter to meters and calculate radius: $d=1.0\text{ mm}=1.0\times 10^{-3}\text{ m}$, so radius $r=d/2=0.5\times 10^{-3}\text{ m}$.
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: $R\approx 0.043\Omega$.

Exam tip: Always convert all dimensions to meters and square meters before substituting into the formula; units of resistivity are in $\Omega \cdot \text{m}$, so mismatched units (like mm for diameter) will give an answer off by multiple orders of magnitude, which is a common MCQ distractor.

3. Temperature Dependence of Resistivity

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 the lattice atoms, leading to more frequent collisions between free electrons (charge carriers) and the lattice. This increases resistivity as temperature increases. For semiconductors (like silicon or germanium), 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 ${\rho }_0$ is the resistivity at reference temperature $T_0$ (usually 20°C), and $\alpha$ is the temperature coefficient of resistivity, with units of ${\text{°C}}^{-1}$. For metals, $\alpha$ is positive; for semiconductors, $\alpha$ is negative. Since resistance is proportional to resistivity, the same relationship applies directly to resistance: $R(T)=R_0[1+\alpha (T-T_0)]$, which is often easier to use for problems.

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 $\alpha =4.5\times 10^{-3}{\text{°C}}^{-1}$. What is the resistance of the filament when it heats up to 2500°C?

1. Identify known values: $R_0=0.50\Omega$, $T_0=20°C$, $T=2500°C$, $\alpha =4.5\times 10^{-3}{\text{°C}}^{-1}$.
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: $(4.5\times 10^{-3})(2480)=11.16$, so $1+11.16=12.16$.
5. Calculate final resistance: $R=0.50\times 12.16\approx 6.1\Omega$.

Exam tip: Always check the sign of $\alpha$ when working with semiconductors; a negative $\alpha$ means resistance decreases as temperature increases, which is the opposite of conductors, and is a common point tested in MCQ reasoning questions.

4. Ohmic vs Non-Ohmic Materials

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 $V$ (y-axis) vs current $I$ (x-axis) is a straight line passing through the origin, with slope equal to the constant resistance $R$. Examples of Ohmic components include metal wires at constant temperature and manufactured fixed resistors.

Non-Ohmic materials do not have a constant resistance; resistance changes with applied voltage or current. The most common example of a non-Ohmic component is a filament light bulb: as current increases, the filament heats up, resistance increases, so the V-I curve bends away from the straight Ohmic line. Another common example is a diode, which has very high resistance in one direction and low resistance in the other, so its V-I curve is highly non-linear. 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: $V=2.0\text{ V},I=0.50\text{ A}$; $V=4.0\text{ V},I=1.0\text{ A}$; $V=6.0\text{ V},I=2.0\text{ A}$. 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: $R_{2V}=2.0/0.50=4.0\Omega$; $R_{4V}=4.0/1.0=4.0\Omega$; $R_{6V}=6.0/2.0=3.0\Omega$.
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.

Exam tip: For a non-linear V-I graph, calculate resistance at a specific point as $V/I$ for that point, do not use the tangent slope unless explicitly asked for differential resistance, which is not tested on AP Physics 2.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Assuming that stretching a wire uniformly by a factor of $k$ only changes the 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 $V=L_1{A}_1=L_2{A}_2$ 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 only work with metallic conductors in most examples, so they forget the opposite behavior of semiconductors. Correct move: Always check the sign of $\alpha$: positive $\alpha$ (metals, resistance increases with T), negative $\alpha$ (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 $\Omega \cdot \text{m}$, so all input lengths must match this unit. Correct move: Always convert all length units to meters before substituting into $R=\rho L/A$.

6. Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

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? A) $\frac{R}{3}$ B) $3R$ C) $9R$ D) $27R$

Worked Solution: When a wire is stretched uniformly, its volume remains constant because mass and density do not change. Original volume $V=LA$, new length $L^{\prime }=3L$, so new cross-sectional area $A^{\prime }=V/L^{\prime }=LA/(3L)=A/3$. Original resistance is $R=\rho L/A$. New resistance is $R^{\prime }=\rho L^{\prime }/A^{\prime }=\rho (3L)/(A/3)=9(\rho L/A)=9R$. The correct answer is C.

Question 2 (Free Response)

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 the current through the sample for different potential differences across it, at constant room temperature, with the following results: $V=0.5\text{ V},I=0.20\text{ A}$; $V=1.0\text{ V},I=0.41\text{ A}$; $V=1.5\text{ V},I=0.59\text{ A}$; $V=2.0\text{ V},I=0.81\text{ A}$. (a) Determine whether the sample is Ohmic. Justify your answer. (b) Calculate the average resistivity of the sample from the data. (c) If the sample is heated by 40°C above room temperature, its resistance increases by 12%. What is the temperature coefficient of resistivity $\alpha$ for this material? What does this tell you about whether the material is more likely a conductor or a semiconductor?

Worked Solution: (a) Calculate $R=V/I$ for each point: $0.5/0.20=2.5\Omega$, $1.0/0.41\approx 2.4\Omega$, $1.5/0.59\approx 2.5\Omega$, $2.0/0.81\approx 2.5\Omega$. Resistance is constant within experimental uncertainty, so the sample is Ohmic. (b) Convert dimensions to SI units: $L=10.0\text{ cm}=0.100\text{ m}$, radius $r=1.0\text{ cm}=0.010\text{ m}$. Cross-sectional area $A=\pi r^2=\pi (0.010{)}^2\approx 3.14\times 10^{-4}{\text{ m}}^2$. Average $R=2.5\Omega$. Rearrange $R=\rho L/A$ to get $\rho =RA/L=(2.5\Omega )(3.14\times 10^{-4}{\text{ m}}^2)/0.100\text{ m}\approx 7.9\times 10^{-3}\Omega \cdot \text{m}$. (c) Use $R=R_0(1+\alpha \Delta T)$. We know $R=1.12R_0$, $\Delta T=40^{\circ }\text{C}$. Substitute: $1.12=1+40\alpha \to \alpha =0.12/40=3.0\times 10^{-3}\circ {\text{C}}^{-1}$. Positive $\alpha$ means the material is most likely a metallic conductor.

Question 3 (Application / Real-World Style)

A heating element for an electric portable water heater is made of nichrome wire, with resistivity $\rho =1.1\times 10^{-6}\Omega \cdot m$ at operating temperature. The heating element requires a total resistance of 12 Ω to produce the required 2000 W of power at 240 V. If the wire used has a diameter of 1.0 mm, what total length of nichrome wire is needed for the heating element?

Worked Solution: Rearrange $R=\rho L/A$ to solve for length: $L=RA/\rho$. Radius of the wire is $r=0.5\text{ mm}=0.5\times 10^{-3}\text{ m}$. Cross-sectional area $A=\pi r^2=\pi (0.5\times 10^{-3}{)}^2\approx 7.85\times 10^{-7}{\text{ m}}^2$. Substitute values: $L=(12\Omega )(7.85\times 10^{-7}{\text{ m}}^2)/(1.1\times 10^{-6}\Omega \cdot \text{m})\approx 8.6\text{ m}$. Interpretation: An 8.6 meter long piece of 1.0 mm diameter nichrome wire provides the 12 Ω resistance required for the water heater's heating element at operating temperature.

7. Quick Reference Cheatsheet

8. 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, such as how heating changes resistance and alters current and power output. This topic also feeds into broader concepts across AP Physics 2, including the behavior of semiconductors in modern electronic devices.

• Power in Electric Circuits
• Series and Parallel Resistors
• Kirchhoff's Circuit Laws
• Semiconductor Modern Physics