Fiveable

🧲AP Physics 2 (2025) Unit 11 Review

QR code for AP Physics 2 (2025) practice questions

11.3 Resistance, Resistivity, and Ohm's Law

🧲AP Physics 2 (2025)
Unit 11 Review

11.3 Resistance, Resistivity, and Ohm's Law

Written by the Fiveable Content Team • Last updated September 2025
Verified for the 2026 exam
Verified for the 2026 examWritten by the Fiveable Content Team • Last updated September 2025
🧲AP Physics 2 (2025)
Unit & Topic Study Guides
Pep mascot

Resistance and Resistivity

Resistance is a fundamental property that describes how strongly a material opposes the flow of electric current. When electrons move through a material, they encounter obstacles in the form of atoms and impurities that impede their flow.

  • Resistance (R) is measured in ohms (Ω)
  • Higher resistance means less current flows for a given voltage
  • Resistance depends on both the material and its physical dimensions

For uniform materials like wires, resistance follows this relationship:

R=ρLAR = \rho \frac{L}{A}

Where:

  • ρ\rho (rho) is the resistivity (an intrinsic material property)
  • LL is the length of the material
  • AA is the cross-sectional area

This relationship shows that:

  • Doubling the length doubles the resistance
  • Doubling the cross-sectional area halves the resistance
  • Different materials have different resistivities (copper has lower resistivity than iron)

Temperature significantly affects resistivity in most materials. For most conductors:

  • Resistivity increases as temperature rises
  • This happens because increased thermal vibrations of atoms create more obstacles for electron flow
  • Some materials like semiconductors show the opposite behavior
Pep mascot
more resources to help you study

Ohm's Law and Electrical Characteristics

Ohm's Law describes the relationship between voltage, current, and resistance in an electrical circuit. This fundamental principle states that the current flowing through a conductor is directly proportional to the voltage and inversely proportional to the resistance.

V=IRV = IR

Where:

  • VV is the potential difference (voltage) measured in volts (V)
  • II is the current measured in amperes (A)
  • RR is the resistance measured in ohms (Ω)

This relationship can be rearranged to solve for any of the three variables:

  • I=VRI = \frac{V}{R} (current equals voltage divided by resistance)
  • R=VIR = \frac{V}{I} (resistance equals voltage divided by current)

Materials that follow Ohm's Law are called "ohmic" materials:

  • Their resistance remains constant regardless of the applied voltage
  • A current vs. voltage graph for an ohmic material is a straight line
  • The slope of this line equals the resistance

Many materials, however, are "non-ohmic":

  • Their resistance changes with voltage or current
  • Examples include diodes, transistors, and some semiconductors
  • Their current vs. voltage graphs are curved rather than linear

When current flows through a resistor, electrical energy is converted to thermal energy:

  • This is known as Joule heating or resistive heating
  • The power (rate of energy conversion) is given by P=IV=I2R=V2RP = IV = I^2R = \frac{V^2}{R}
  • This heating effect is useful in appliances like toasters and heaters
  • In electronic circuits, this heating is often undesirable and requires management

Practice Problem 1: Calculating Resistance

A copper wire has a length of 10 meters and a cross-sectional area of 2.0 × 10^-6 m². If the resistivity of copper is 1.7 × 10^-8 Ω·m, what is the resistance of the wire?

Solution

To find the resistance of the wire, we'll use the resistance equation that relates resistivity, length, and cross-sectional area:

R=ρLAR = \rho \frac{L}{A}

Substituting the given values:

  • Resistivity (ρ\rho) = 1.7 × 10^-8 Ω·m
  • Length (L) = 10 m
  • Cross-sectional area (A) = 2.0 × 10^-6 m²

R=(1.7×108Ωm)×10m2.0×106m2R = (1.7 \times 10^{-8} \Omega \cdot m) \times \frac{10 m}{2.0 \times 10^{-6} m^2}

R=(1.7×108Ωm)×(5.0×106m1)R = (1.7 \times 10^{-8} \Omega \cdot m) \times (5.0 \times 10^6 m^{-1})

R=0.085ΩR = 0.085 \Omega

Therefore, the resistance of the copper wire is 0.085 Ω.

Practice Problem 2: Applying Ohm's Law

A circuit contains a 12 V battery connected to a resistor. If the current flowing through the circuit is 0.5 A, what is the resistance of the resistor? If the resistor's temperature increases significantly, would you expect the current to increase, decrease, or stay the same?

Solution

First, let's find the resistance using Ohm's Law: V=IRV = IR

Rearranging to solve for resistance: R=VIR = \frac{V}{I}

Given:

  • Voltage (V) = 12 V
  • Current (I) = 0.5 A

R=12V0.5A=24ΩR = \frac{12 V}{0.5 A} = 24 \Omega

For the second part of the question, we need to consider how temperature affects resistance. For most conductors (like typical resistors), resistivity increases with temperature, which means resistance increases.

If the resistance increases while the voltage remains constant at 12 V, then according to Ohm's Law (I=VRI = \frac{V}{R}), the current must decrease. This is because current is inversely proportional to resistance when voltage is constant.

Therefore, as the resistor heats up, we would expect the current in the circuit to decrease.

Frequently Asked Questions

What is resistance and how does it work in circuits?

Resistance is how much an object opposes motion of electric charge—think of it as “electrical friction.” For a uniform resistor, R = ρℓ/A: it increases with the material’s resistivity ρ and length ℓ and decreases with cross-sectional area A. Resistivity is a material property (usually rises with temperature for conductors); if ρ varies along the length use R = ∫ρ(ℓ) dℓ / A. Ohm’s law links current, resistance, and potential difference: I = ΔV / R. Materials with constant R (for all ΔV and I) are ohmic; nonohmic devices (diodes, filament lamps) don’t follow that linear relation. On the AP exam you may be asked to find R from the slope of an I vs ΔV graph (slope = 1/R, conductance). Resistors convert electric energy to thermal energy (Joule heating), so power P = I²R = ΔV·I = ΔV²/R. For a focused review see the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa); more unit content and practice problems are at the Unit 11 overview (https://library.fiveable.me/ap-physics-2-revised/unit-11) and the practice bank (https://library.fiveable.me/practice/ap-physics-2-revised).

Why does a longer wire have more resistance than a shorter one?

A longer wire has more resistance because charge carriers have a longer path and therefore more chances to scatter off atoms—so the wire opposes current more. For a uniform wire the CED gives the exact relation: R = ρℓ/A. That means R is directly proportional to length ℓ (so doubling ℓ doubles R) and inversely proportional to cross-sectional area A (thicker wire → less R). ρ (resistivity) is a material property that can change with temperature (for most conductors ρ rises as T rises). Ohm’s law (I = ΔV/R) then tells you a longer wire draws less current for the same voltage. If you want practice applying this or seeing IV-graphs for ohmic materials, check the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and Unit 11 review (https://library.fiveable.me/ap-physics-2-revised/unit-11).

What's the difference between resistance and resistivity?

Resistance and resistivity are related but different. Resistivity (ρ) is a material property that quantifies how strongly a material opposes charge flow—it depends on the material’s atomic structure and usually changes with temperature (CED 11.3.A.2.i–ii). Resistance (R) is a property of a particular object or resistor and depends on both the material and its shape: R = ρℓ/A for a uniform resistor (CED 11.3.A.2). So for two wires made of the same metal (same ρ), the longer or thinner one has larger R. If ρ varies along the length, use R = ∫ρ(ℓ) dℓ / A (CED 11.3.A.2.iii). Ohm’s law (I = ΔV/R) connects R to circuit behavior; materials with constant ρ (over the tested range) are “ohmic” (CED 11.3.B.1.i–ii). For a quick review, see the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and more practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

How do I use Ohm's law to solve circuit problems?

Ohm’s law says I = ΔV / R for a conductive element. To solve circuit problems, follow these steps: 1. Identify the element(s) and whether they’re ohmic (constant R). Use R = ρℓ/A if you need resistance from geometry (CED 11.3.A.2). 2. Redraw the circuit, label ΔV across each resistor, and mark currents with directions. For series, Rtot = R1+R2+…; for parallel, 1/Rtot = Σ(1/Ri) (CED 11.5). 3. Use Ohm’s law: for a known ΔV find I = ΔV/R; for a known I find ΔV = I·R. For circuits with a battery+internal resistance use ε = I(R + r). 4. Check power: P = IΔV = I^2R = (ΔV)^2 / R (CED 11.4). 5. If asked for R from IV data, plot I vs ΔV; slope = 1/R (conductance) or plot ΔV vs I; slope = R (CED 11.3.B.1.iv). That graph/derivation style appears on the AP FRQ rubric. Practice problems and the Topic 11.3 study guide can help you apply these steps (Fiveable study guide: https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa; unit overview: https://library.fiveable.me/ap-physics-2-revised/unit-11).

I'm confused about the formula R = ρℓ/A - can someone break this down?

R = ρℓ/A just tells you what determines how hard a material makes it for charges to flow. - ρ (resistivity) is a material property (Ω·m). Metals have low ρ, insulators high ρ; ρ usually increases with temperature. - ℓ (length)—resistance is proportional to length. Double the length → double R (more collisions/opposition). - A (cross-sectional area)—resistance is inversely proportional to area. Double A → half R (more paths for charge to flow). So R = ρℓ/A means long, thin wires have high R; short, fat wires have low R. For nonuniform ρ(ℓ) use R = ∫ ρ(ℓ) dℓ / A. Combine with Ohm’s law I = ΔV/R to find currents. On the AP CED this is 11.3.A.2 (use functional dependence to predict changes on the exam). For a quick review and practice problems, see the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and Fiveable practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

What makes a material ohmic vs non-ohmic?

A material is called ohmic if the current through it is directly proportional to the potential difference across it—that means a constant resistance R = ΔV/I for all applied voltages. On an I vs. V graph an ohmic element makes a straight line; the slope is the conductance (1/R) and R is constant. The CED also notes an ohmic material has resistivity that doesn’t change with temperature. Non-ohmic materials don’t follow that linear relationship: their resistance changes with voltage, current, or temperature. Examples: diodes, incandescent lamp filaments (resistance rises as they heat), and thermistors. For AP style questions, you’ll often identify ohmic behavior by plotting I vs. V and checking linearity—slope gives R (CED 11.3.B.1.iv). For a quick review, see the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and practice lots of IV-graph problems (https://library.fiveable.me/practice/ap-physics-2-revised).

Why does the resistance of a conductor increase when it gets hot?

As temperature rises, the atoms in a conductor vibrate more. Those stronger lattice vibrations scatter the free electrons that carry current, so electrons have a shorter average time (and distance) between collisions. That makes it harder for charge to flow—i.e., the material’s resistivity ρ increases—and since R = ρℓ/A, the resistance R goes up (for fixed length ℓ and area A). This is why most metallic conductors show increasing resistance with temperature (CED: 11.3.A.2.ii). Note: on the AP exam you may need to cite R = ρℓ/A or explain scattering qualitatively; some problems also treat “ohmic” behavior as having constant resistance over the measured range (CED: 11.3.B.1.i–ii). For a quick review of these ideas, see the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and try more practice problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

How do you find resistance from a current vs voltage graph?

If your graph plots current (I) on the vertical axis and potential difference (ΔV) on the horizontal axis, use the slope. For an ohmic device (CED 11.3.B.1.i–iv) the I vs V plot is a straight line through the origin: - slope = ΔI / ΔV = 1/R (that slope is the conductance, in siemens S) - so R = 1 / (slope) = ΔV / ΔI (units: ohms, Ω) If the I–V curve is nonlinear (nonohmic), the resistance isn't constant. You can still find: - average resistance over a range: Ravg = ΔV / ΔI for that interval - instantaneous (dynamic) resistance: Rinst = 1 / (dI/dV) = dV/dI (reciprocal of the derivative at the point) Practice reading slopes and distinguishing ohmic vs nonohmic for the AP exam (see CED 11.3.B.1.iv). For extra practice and examples, check the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and more problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

What happens to resistance if I make a wire thicker but keep the same length?

Use R = ρℓ/A (CED 11.3.A.2). If you make the wire thicker but keep the same length and material, its cross-sectional area A increases, so R decreases in inverse proportion to A. Example: if you double the wire’s diameter, A goes up by 4 (because A ∝ diameter²), so the resistance drops to one quarter of the original value. Resistivity ρ is a property of the material (and can change with temperature), so if you change material or heat the wire the result can differ. For AP work, apply R = ρℓ/A to predict numerical factors of change (skill 2.D) and show reasoning on exams. For a quick review, see the Topic 11.3 study guide on Fiveable (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and more unit resources at (https://library.fiveable.me/ap-physics-2-revised/unit-11).

Can someone explain why resistors convert electrical energy to heat?

Resistors turn electrical energy into heat because moving charges lose kinetic energy to the material. Microscopically, free electrons drifting under a potential difference collide with ions/atoms in the resistor’s lattice; those collisions transfer energy to atomic vibrations (thermal energy). Macroscopically you can use Ohm’s law and power formulas: I = ΔV/R and P = IV = I^2R = V^2/R. That P is the rate electrical energy is converted to thermal (Joule) heating. Resistivity ρ (R = ρℓ/A) sets how strongly a material opposes charge motion, so materials with higher ρ produce more heating for the same current. Resistivity usually rises with temperature, so heating can change R (non-ohmic behavior if large). This concept appears in CED 11.3 (Resistance, resistivity, Ohm’s law) and in problems on electric power—practice these ideas on the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and unit review (https://library.fiveable.me/ap-physics-2-revised/unit-11). For extra practice, try the AP practice set (https://library.fiveable.me/practice/ap-physics-2-revised).

What's the deal with that integral formula for resistance when resistivity changes?

Think of the resistor as lots of tiny pieces in series. For a small slice of length dl the resistance is dR = ρ(l)·dl / A (from R = ρℓ/A applied to a differential piece). If ρ varies with position, you add (integrate) all those tiny dR’s along the whole length to get the total: R = ∫[0 to L] ρ(l) dl / A. That’s it—the integral just replaces the simple ρ·L/A when ρ isn’t constant. If the cross-sectional area also changes, use A(l) in the denominator: R = ∫ ρ(l) dl / A(l). Ohm’s law still uses the total R (I = ΔV/R)—the integral gives the effective resistance you plug into I = ΔV/R (CED: 11.3.A.2.iii and 11.3.B.1). Quick example: if ρ(l)=ρ0(1+αl) and A is constant, R = (ρ0/A)∫0^L(1+αl)dl = (ρ0L/A)(1 + αL/2). For more practice, check the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and AP practice problems (https://library.fiveable.me/practice/ap-physics-2-revised).

How do I know if a material obeys Ohm's law just by looking at data?

Look at V and I data and check whether V and I are proportional. Practically: - Compute R = V/I for several data points. If R is (approximately) the same for all V, the material is ohmic. - Or better, plot I vs. V (or V vs. I). If the points lie on a straight line through the origin, the element is ohmic. The slope of I vs. V is the conductance (1/R); the slope of V vs. I is the resistance R (this is exactly what the CED says: “Resistance … can be determined from the slope of a graph of current as a function of potential difference.”). - Watch out for temperature effects: heating can change resistivity, producing curved I–V behavior even for a metal. Also diode-like or filament devices show nonlinear (nonohmic) I–V curves. If you want practice identifying ohmic vs. nonohmic behavior, check the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and more problems at Fiveable practice (https://library.fiveable.me/practice/ap-physics-2-revised).

Why do we need to know about resistivity when we already have resistance?

Resistance (R) tells you how a particular object opposes current—it depends on both the material and the shape. Resistivity (ρ) is the material’s intrinsic property: R = ρℓ/A. That matters because two wires with the same R could be different materials (one long/thin, one short/thick). ρ lets you compare materials (metals vs. insulators), predict how R changes with length or area, and include temperature effects (ρ usually increases with T for conductors). For nonuniform materials you need the integral R = ∫ρ(ℓ) dℓ / A to get the total resistance. On the AP exam you might extract R from an I vs ΔV slope (Ohm’s law) but you’ll use ρ when the question asks about material properties, geometry changes, or temperature dependence (CED 11.3.A.2, 11.3.B.1). For a quick review check the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and try practice problems at (https://library.fiveable.me/practice/ap-physics-2-revised).

What's the relationship between current, voltage, and resistance in simple terms?

Think of an electric circuit like water in pipes: voltage (ΔV) is the “push” that makes charge flow, current (I) is how much charge flows per second, and resistance (R) is how much the material or wire resists that flow. Ohm’s law puts it simply: I = ΔV / R—more push or less resistance gives more current. For a physical resistor, R depends on the material and shape: R = ρℓ/A (ρ = resistivity, ℓ = length, A = cross-sectional area). Metals usually get more resistive as they heat up (ρ rises with temperature). Materials with constant R over different voltages are called ohmic; you find R from the slope of an I vs. ΔV graph. This is exactly what AP Topic 11.3 covers—review the CED points and the Topic 11.3 study guide on Fiveable (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and practice problems (https://library.fiveable.me/practice/ap-physics-2-revised) to prep for exam-style questions.

I missed the lab on measuring resistance - how do you actually do it experimentally?

You can measure a resistor’s resistance two easy ways in the lab. 1) Direct V/I method (good for AP labs): Put the ammeter in series with the resistor and the battery; connect the voltmeter across the resistor (voltmeter parallel). Vary the battery voltage (or use different known series resistors), measure V and I for several settings. For each run compute R = V/I (Ohm’s law). To test ohmic behavior, plot I vs V—a straight line through the origin means the element is ohmic; slope = 1/R (or plot V vs I and slope = R). Repeat measurements and average to reduce random error. Tips to reduce uncertainty: use multiple voltage points, avoid large currents (heating changes resistivity), use high-resistance voltmeter and low-resistance ammeter, record uncertainties, and correct for instrument internal resistance if needed. For more detail and AP-style lab guidance, see the Topic 11.3 study guide (https://library.fiveable.me/ap-physics-2-revised/unit-3/3-resistance-resistivity-and-ohms-law/study-guide/y0ZmKqhOPqeLWZFa) and the unit overview (https://library.fiveable.me/ap-physics-2-revised/unit-11). For extra practice, try problems at (https://library.fiveable.me/practice/ap-physics-2-revised).