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AP Course

AP Physics 2

Study AP Physics 2 Units 9–15, with the Fall 2026 radioactive-decay clarification applied and official exam updates linked separately.

Study Units 9–15 through original models, derivations, experiments, quizzes, and practice sets.

Lessons
Particle Motion Behind Temperature and PressureConnecting Gas Variables with the Ideal-Gas ModelHeat Flow and the Approach to Thermal BalanceEnergy Accounting for Thermodynamic ProcessesMaterial Response to Heating and Heat ConductionEntropy, Probability, and the Direction of Thermal ChangeCharge Interactions and Coulomb-Force ModelsTracking Charge During Contact and InductionMapping Electric Fields from Source ChargesEnergy of Configurations of ChargesElectric Potential and Equipotential ReasoningCapacitance and Energy Stored in Electric FieldsConserving Energy in Charged-Particle MotionCharge Flow and Conventional CurrentModeling Sources, Wires, and Loads in Simple NetworksHow Geometry and Material Set ResistanceElectrical Energy Transfer and PowerReducing Series-Parallel DC NetworksLoop Equations from Energy ConservationJunction Equations from Charge ConservationTransient Charging and Discharging in RC NetworksSources, Direction, and Strength of Magnetic FieldsMagnetic Forces and Curved Paths of ChargesForces Between Fields and Current-Carrying ConductorsChanging Magnetic Flux and Induced EMFReflection and Absorption at Material BoundariesLocating Images with Mirror Ray ModelsRefraction, Index, and Total Internal ReflectionLocating Images with Thin-Lens ModelsPulses, Wave Types, and Propagation SpeedPeriodic-Wave Measures and GraphsBoundary Changes and PolarizationElectromagnetic-Wave BehaviorFrequency Shifts from Relative MotionSuperposition, Interference, and Standing WavesDiffraction from Openings and EdgesTwo-Slit and Grating Pattern GeometryPhase Change and Thin-Film ColorQuantum Models and Dual Wave-Particle EvidenceEnergy Levels in a Hydrogen-Like AtomConnecting Spectral Lines to Energy TransitionsThermal Spectra and Blackbody CurvesPhoton Thresholds in the Photoelectric EffectPhoton-Electron Scattering and MomentumMass-Energy Accounting in Fission and FusionRandom Nuclear Decay, Activity, and Half-Life
Quizzes
Practice Problems Formula notes, diagrams, and practice sets will be added here.

AP Physics 2 · Unit 11 · Topic 11.3

How Geometry and Material Set Resistance

Connect the macroscopic resistance of an object to its dimensions, material response, and temperature. Derive \(R=\rho L/A\), use radius and volume constraints correctly, and distinguish constant-resistance ohmic behavior from devices whose operating resistance changes.

Learning Goals

  • Distinguish resistance \(R\) from resistivity \(\rho\) and conductivity \(\sigma\).
  • Derive and apply \(R=\rho L/A\) for a uniform conductor.
  • Predict resistance changes from length, cross-sectional area, radius, or fixed-volume reshaping.
  • Interpret voltage–current graphs for ohmic and nonohmic elements.
  • Estimate resistance changes with \(R=R_0[1+\alpha(T-T_0)]\) within its valid range.
  • Design measurements that separate material, geometry, and temperature effects.
\(R\)object resistanceohm, \(\Omega=\mathrm{V/A}\)
\(\rho\)material resistivity\(\Omega\cdot\mathrm m\)
\(L/A\)geometry factorlength per area
\(\alpha\)temperature coefficient\(\mathrm{K^{-1}}\) or \(\mathrm{^\circ C^{-1}}\)

1. Resistance Describes an Object

For an element at a specified operating state, resistance compares the potential difference across it with the current through it:

\(R=\frac{|\Delta V|}{|I|}\)

One ohm is one volt per ampere. Resistance can change if the object’s shape, material state, or temperature changes.

2. Resistivity Describes a Material

Resistivity \(\rho\) measures how strongly a material opposes current at a specified temperature and composition. It does not depend on the sample’s length or cross-sectional area.

\(\rho=\frac1{\sigma}\)

Conductivity \(\sigma\) is the reciprocal: good conductors have low \(\rho\) and high \(\sigma\).

Worked Example 1 · Operational Resistance

Use a Measured Voltage and Current

A device has \(12\,\mathrm V\) across it while carrying \(0.80\,\mathrm A\).

\(R=\frac{12}{0.80}=15\,\Omega\)

This ratio describes that operating point. More measurements are needed before claiming the device is ohmic with a constant \(15\,\Omega\) resistance.

3. Deriving \(R=\rho L/A\)

Consider a uniform conductor of length \(L\) and cross-sectional area \(A\), with a nearly uniform electric field and current density.

  1. The material relation is \(E=\rho J\).
  2. For a uniform field, \(|\Delta V|=EL\).
  3. For uniform current density, \(I=JA\), so \(J=I/A\).
  4. Substitute into the voltage relation:
\(|\Delta V|=(\rho J)L=\rho\frac{I}{A}L\)

Comparing with \(|\Delta V|=IR\) gives

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

This model assumes a uniform material, constant cross-section, and a temperature at which the chosen \(\rho\) applies.

Length and area affect resistance in opposite waysFor one material at one temperature, a longer path raises \(R\), while a larger cross-sectional area provides more parallel conducting space and lowers \(R\).

4. Why Length Raises Resistance

At fixed material and area, doubling \(L\) doubles \(R\). Consecutive equal wire segments behave like equal resistors in series because every carrier traverses each segment.

\(R\propto L\)

5. Why Area Lowers Resistance

At fixed material and length, doubling \(A\) halves \(R\). Side-by-side conducting regions behave like parallel paths for charge flow.

\(R\propto\frac1A\)

6. Radius Enters as a Square

For a cylindrical wire,

\(A=\pi r^2=\frac{\pi d^2}{4}\)

Therefore \(R\propto1/r^2\) and \(R\propto1/d^2\). Doubling the radius makes the area four times larger and the resistance one fourth as large—not one half.

Worked Example 2 · Copper Wire

Calculate Resistance from Material and Shape

A \(10.0\,\mathrm m\) copper wire has cross-sectional area \(1.00\,\mathrm{mm^2}=1.00\times10^{-6}\,\mathrm{m^2}\). Use \(\rho=1.68\times10^{-8}\,\Omega\cdot\mathrm m\) near \(20^\circ\mathrm C\).

\(R=\rho\frac{L}{A}=(1.68\times10^{-8})\frac{10.0}{1.00\times10^{-6}}=0.168\,\Omega\)

A small resistance can still produce significant voltage drop and heating when current is large.

Worked Example 3 · Radius and Length Scaling

Use Ratios Before Substituting Numbers

A wire is replaced by the same material at the same temperature with triple the length and double the radius.

\(\frac{R_f}{R_i}=\frac{\rho_f}{\rho_i}\frac{L_f}{L_i}\frac{A_i}{A_f}=1\cdot3\cdot\frac1{2^2}=\frac34\)

The longer replacement nevertheless has \(25\%\) less resistance because its fourfold area increase outweighs its threefold length increase.

7. Fixed-Volume Reshaping Couples \(L\) and \(A\)

If a wire is stretched without changing material volume, \(AL=\text{constant}\). Increasing length forces the area to decrease:

\(A_f=A_i\frac{L_i}{L_f}\)

For length factor \(k\), \(A_f=A_i/k\), so

\(\frac{R_f}{R_i}=k^2\)

This differs from a problem that independently changes length while holding area fixed.

Worked Example 4 · Stretched Wire

Double Length at Constant Volume

A uniform wire is stretched to twice its original length without a significant change in resistivity. Constant volume makes \(A_f=A_i/2\).

\(\frac{R_f}{R_i}=\frac{L_f}{L_i}\frac{A_i}{A_f}=2(2)=4\)

The new resistance is four times the original resistance.

8. Material Choice Matters

At room temperature, good metallic conductors such as copper and aluminum have resistivities on the order of \(10^{-8}\,\Omega\cdot\mathrm m\). Heating alloys such as nichrome are roughly two orders of magnitude more resistive, while insulators can be many orders of magnitude higher.

Quoted values are representative, not universal constants: alloy composition, impurities, mechanical condition, and temperature all affect \(\rho\).

Worked Example 5 · Nichrome Element

Combine Radius, Resistivity, and Voltage

A nichrome wire has \(\rho=1.10\times10^{-6}\,\Omega\cdot\mathrm m\), length \(2.0\,\mathrm m\), and radius \(0.25\,\mathrm{mm}\).

\(A=\pi(2.5\times10^{-4})^2=1.96\times10^{-7}\,\mathrm{m^2}\)
\(R=\frac{(1.10\times10^{-6})(2.0)}{1.96\times10^{-7}}\approx11.2\,\Omega\)
\(I=\frac{6.0\,\mathrm V}{11.2\,\Omega}\approx0.536\,\mathrm A\)

The calculation treats resistivity as constant; substantial self-heating would require updating the resistance.

9. Conductance

Conductance describes how readily an object carries current:

\(G=\frac1R\)

Its SI unit is the siemens, \(\mathrm S=\Omega^{-1}\). Resistance belongs to an object; conductivity \(\sigma=1/\rho\) belongs to a material at a specified state.

10. Series Length and Parallel Area

Equal wire pieces connected end-to-end add length and resistance. Equal pieces placed side-by-side add effective area and conductance. This gives a geometric explanation for resistor series and parallel rules.

Worked Example 6 · Design a Wire Diameter

Meet a Maximum Resistance Requirement

A \(20\,\mathrm m\) copper path must have \(R\le0.50\,\Omega\). Use \(\rho=1.68\times10^{-8}\,\Omega\cdot\mathrm m\).

\(A_{\min}=\frac{\rho L}{R_{\max}}=\frac{(1.68\times10^{-8})(20)}{0.50}=6.72\times10^{-7}\,\mathrm{m^2}\)
\(d_{\min}=2\sqrt{\frac{A_{\min}}{\pi}}\approx9.25\times10^{-4}\,\mathrm m=0.925\,\mathrm{mm}\)

A practical design would choose a standard diameter at least this large and also check current capacity and heating.

11. Ohmic Behavior Requires a Constant Operating State

An ohmic element has a proportional voltage–current relationship over the tested range:

\(\Delta V=IR\)

On a graph with \(V\) vertically and \(I\) horizontally, the slope is \(R\). A straight line through the origin indicates constant resistance. If the axes are reversed, the slope is \(I/V=1/R=G\).

Slope reveals operating resistanceA constant \(V/I\) produces a straight line. A curved response means the effective resistance changes with current, temperature, field, or device state.

12. Nonohmic Does Not Mean “No Resistance”

A filament lamp, diode, and thermistor can have nonlinear \(V\)-\(I\) behavior. The ratio \(V/I\) still describes a static resistance at one operating point, while the local slope \(dV/dI\) describes differential resistance. Neither must remain constant.

For a metal filament, increasing current can raise temperature and resistivity, causing the graph to steepen.

13. Approximate Temperature Dependence

Over a moderate temperature interval where \(\alpha\) can be treated as constant,

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

If thermal expansion changes \(L/A\) negligibly compared with the change in resistivity, then

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

Many metals have \(\alpha>0\), so resistance rises with temperature. Many semiconductors and thermistors can have a negative, strongly nonlinear temperature response. The linear equation should not be extrapolated far outside its stated range.

Worked Example 7 · Metal Temperature Change

Estimate Copper Resistance at a Higher Temperature

A copper coil has \(R_0=10.0\,\Omega\) at \(20^\circ\mathrm C\). Take \(\alpha=3.9\times10^{-3}\,\mathrm{^\circ C^{-1}}\) and estimate \(R\) at \(80^\circ\mathrm C\).

\(\Delta T=60^\circ\mathrm C\)
\(R=(10.0)[1+(3.9\times10^{-3})(60)]=12.34\,\Omega\)

The estimate assumes constant \(\alpha\) and negligible geometric expansion.

Worked Example 8 · Parallel Strands

Increase Effective Conducting Area

Four identical wire strands each have resistance \(2.0\,\Omega\) and connect side-by-side between the same two nodes.

\(R_{\mathrm{eq}}=\frac{R_{\mathrm{strand}}}{4}=0.50\,\Omega\)

Geometrically, four strands provide four times the effective cross-sectional area, matching \(R=\rho L/A\).

14. Self-Heating Creates Feedback

A current transfers thermal energy in a resistive element. If a metal’s temperature rises, its resistance often rises, which changes the current for a fixed-voltage source. A calculation using one room-temperature resistance may therefore fail at a hot operating state.

Always distinguish an externally controlled constant temperature from a device that is allowed to heat.

15. Reliable Resistance Workflow

  1. Identify whether the question asks for \(R\), \(\rho\), \(G\), or \(\sigma\).
  2. Convert length to meters and area to square meters.
  3. For a cylinder, calculate \(A=\pi r^2\) before using \(R=\rho L/A\).
  4. Use ratios when material and temperature stay unchanged.
  5. Check whether volume, area, or radius is actually held fixed.
  6. For graphs, read the axes before interpreting slope.
  7. Apply a temperature coefficient only within a reasonable range.

16. Common Traps

  • Using resistance and resistivity as interchangeable quantities.
  • Reporting resistivity in ohms instead of \(\Omega\cdot\mathrm m\).
  • Using diameter as radius or forgetting the square in \(A=\pi r^2\).
  • Converting \(\mathrm{mm^2}\) as though it were millimeters.
  • Assuming a longer wire always has larger resistance when its area also changes.
  • Using \(V/I\) from one point to prove a device is ohmic.
  • Calling the slope of an \(I\)-versus-\(V\) graph resistance.
  • Assuming every material’s resistance rises with temperature.
  • Ignoring self-heating or extrapolating the linear temperature model too far.
Mastery Check

1. A wire’s length doubles while area and material stay fixed. What happens to \(R\)?

Show reasoning and answer

\(R=\rho L/A\), so doubling \(L\) doubles \(R\).

2. A wire’s diameter triples while length and material stay fixed. What happens to \(R\)?

Show reasoning and answer

Area is proportional to diameter squared, so it becomes nine times larger. Resistance becomes \(R_i/9\).

3. Two samples have the same material and temperature but different dimensions. Must they have the same resistivity and resistance?

Show reasoning and answer

They have the same resistivity because \(\rho\) is a material property at that state. Their resistances can differ because \(R=\rho L/A\).

4. A \(V\)-versus-\(I\) graph is a straight line with slope \(25\,\mathrm{V/A}\). Find \(R\).

Show reasoning and answer

The vertical-over-horizontal slope is \(V/I=R=25\,\Omega\).

5. A constant-volume wire is stretched to three times its length. What is \(R_f/R_i\)?

Show reasoning and answer

Constant volume makes the area one third as large. Thus \(R_f/R_i=3\times3=9\).

6. Why can a filament lamp produce a curved \(V\)-\(I\) graph?

Show reasoning and answer

Increasing current heats the filament. Its resistivity and resistance change with temperature, so \(V/I\) is not constant across operating points.

Investigation: Separate Material and Geometry Effects

Use the open PhET Resistance in a Wire simulation. Hold two variables fixed while changing the third. Record \(R\) versus \(L\), \(R\) versus \(A\), and \(R\) versus \(\rho\). Test which plots are linear and whether \(R\) versus \(1/A\) passes through the origin.

Then use an instructor-approved low-voltage dataset or apparatus to measure \(V\) and \(I\) for a fixed resistor at several small currents. Plot \(V\) vertically against \(I\), determine resistance from slope, and discuss uncertainty, contact resistance, and whether self-heating is visible. Do not connect an unknown wire directly across a high-current source.

Official curriculum reference: College Board AP Physics 2 course page. The explanation and worked example are independently written for this study site.