AP Physics 2 · Topic 11.3

Topic 11.3: Resistance, Resistivity, and Ohm's Law

Unit 11: Electric Circuits15-18% of the multiple-choice section

Resistance measures how strongly an object opposes the movement of electric charge, in ohms. For a resistor of uniform geometry it rises with the material's resistivity and with length, and falls with cross-sectional area. The AP Physics 2 sheet prints Ohm's law as I = ΔV/R, keeping the delta.

AP Physics: Unit 11 (topics 11.3 Resistance, Resistivity, and Ohm's Law). AP Physics 2 Unit 11, Topic 11.3. Two learning objectives: 11.3.A, describe the resistance of an object using physical properties of that object, and 11.3.B, describe the electrical characteristics of elements of a circuit. Under 11.3.A: 11.3.A.1 defines resistance as a measure of the degree to which an object opposes the movement of electric charge; 11.3.A.2 makes the resistance of a resistor with uniform geometry proportional to resistivity and length and inversely proportional to cross-sectional area, with R = rho L / A as the relevant equation; 11.3.A.2.i covers resistivity as a fundamental property of a material and 11.3.A.2.ii the resistivity of a conductor typically increasing with temperature. Under 11.3.B: 11.3.B.1 gives Ohm's law as I = delta V / R, with 11.3.B.1.i defining ohmic materials as those with constant resistance for all currents, 11.3.B.1.ii stating that the resistivity of an ohmic material is constant regardless of temperature, 11.3.B.1.iii noting that resistors can convert electrical energy to thermal energy, and 11.3.B.1.iv allowing resistance to be found from the slope of a graph of current as a function of potential difference. The topic carries no boundary statement. Suggested skills are 1.B, 2.B, 2.D, 3.A and 3.B. Unit 11 is weighted at 15 to 18 percent of the exam with a suggested 12 to 20 class periods.

What Topic 11.3 requires

Topic 11.3 carries two learning objectives, three essential knowledge statements, and six sub-statements. Everything else here unpacks them.

11.3.A: describe the resistance of an object using physical properties of that object.

  • 11.3.A.1 Resistance is a measure of the degree to which an object opposes the movement of electric charge.
  • 11.3.A.2 The resistance of a resistor with uniform geometry is proportional to its resistivity and length and is inversely proportional to its cross-sectional area. The CED gives R=ρAR = \frac{\rho \ell}{A} as the relevant equation.
  • 11.3.A.2.i Resistivity is a fundamental property of a material that depends on its atomic and molecular structure and quantifies how strongly the material opposes the motion of electric charge.
  • 11.3.A.2.ii The resistivity of a conductor typically increases with temperature.

11.3.B: describe the electrical characteristics of elements of a circuit.

  • 11.3.B.1 Ohm's law relates current, resistance, and potential difference across a conductive element of a circuit. The CED gives I=ΔVRI = \frac{\Delta V}{R} as the relevant equation.
  • 11.3.B.1.i Materials that obey Ohm's law have constant resistance for all currents and are called ohmic materials.
  • 11.3.B.1.ii The resistivity of an ohmic material is constant regardless of temperature.
  • 11.3.B.1.iii Resistors can also convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment.
  • 11.3.B.1.iv The resistance of an ohmic circuit element can be determined from the slope of a graph of the current in the element as a function of the potential difference across the element.

Topic 11.3 prints no boundary statement of its own. The nearest one in the unit belongs to Topic 11.2 and says that unless otherwise specified, all circuit schematic diagrams will be drawn using conventional current.

The CED lists five suggested skills here: 1.B, create quantitative graphs with appropriate scales and units, including plotting data; 2.B, calculate or estimate an unknown quantity with units from known quantities, by selecting and following a logical computational pathway; 2.D, predict new values or factors of change of physical quantities using functional dependence between variables; 3.A, create experimental procedures that are appropriate for a given scientific question; and 3.B, apply an appropriate law, definition, theoretical relationship, or model to make a claim. Two of the five, 1.B and 3.A, are about producing and reading data rather than about algebra. Unit 11 carries 15 to 18 percent of the exam and a suggested 12 to 20 class periods.

Resistance belongs to the object, resistivity to the material

The Unit 11 opener makes an unusual request. Under Preparing for the AP Exam, the CED says students should know the differences in meaning between current, potential difference, resistance, resistivity, and capacitance. Topic 11.3 is where two of those five get mixed up, so start with the split.

Resistivity (ρ\rho) is a property of a substance. Graphite has one, copper has another, and a block of copper the size of a house has the same resistivity as a copper shaving. Essential knowledge 11.3.A.2.i says it depends on the material's atomic and molecular structure and quantifies how strongly the material opposes the motion of electric charge.

Resistance (RR) is a property of one particular object: this wire, of this length and this thickness, made of that substance. Essential knowledge 11.3.A.1 defines it as a measure of the degree to which an object opposes the movement of electric charge.

So two objects made of the same material can have very different resistances, and two objects with the same resistance can be made of different materials. Resistivity is what the material contributes, and the geometry supplies the rest.

ResistivityResistance
Symbolρ\rhoRR
UnitΩm\Omega \cdot \mathrm{m}Ω\Omega
Belongs toa materialan object
Cut the object shorterunchangedfalls
Draw it out thinnerunchangedrises
Heat it uprises, for a real conductor (11.3.A.2.ii)rises, through ρ\rho

Neither unit needs memorizing. Ohm's law gives the ohm: 1Ω=1V/A1 \, \Omega = 1 \, \mathrm{V/A}. Rearranging R=ρ/AR = \rho \ell / A gives ρ=RA/\rho = RA/\ell, so resistivity comes out in Ωm2/m\Omega \cdot \mathrm{m}^2 / \mathrm{m}, which is Ωm\Omega \cdot \mathrm{m}. Deriving a unit is faster than recalling it, and it checks that you wrote the equation the right way up.

Geometry sets the resistance (11.3.A.2)

R=ρAR = \frac{\rho \ell}{A}

This is one of the 20 equations in the Electricity group of the AP Physics 2 equation sheet, so you do not have to recall it. What you do have to recall is what each symbol points at.

  • ρ\rho is the resistivity of the material, in Ωm\Omega \cdot \mathrm{m}.
  • \ell is the length along the direction the charge travels.
  • AA is the cross-sectional area perpendicular to that direction.

Swap those two and the answer inverts. For a cylindrical wire, \ell is the length of the wire and A=πr2A = \pi r^2 is the area of the circular end face. A problem that hands you a diameter is asking you to halve it first, and that is where factor-of-four errors are born, because the area goes as the square of the radius.

Essential knowledge 11.3.A.2 states the dependence as a proportionality rather than as a calculation, which makes the equation a scaling rule and is what suggested skill 2.D asks you to use.

Change to a uniform resistorEffect on RR
double the length, area unchanged×2\times 2
double the diameter, length unchanged×14\times \frac{1}{4}
halve the cross-sectional area, length unchanged×2\times 2
stretch it to twice its length at constant volume×4\times 4
swap to a material of triple the resistivity×3\times 3

The fourth row is the one that catches people. Stretching conserves the volume AA\ell, so doubling \ell also halves AA, and the two effects multiply instead of cancelling.

The phrase "uniform geometry" in 11.3.A.2 matters: the equation describes an object whose cross-section does not change along its length, which is why AP problems are written around cylinders and rectangular blocks.

The same geometry turns up in Topic 9.5, where the sheet prints thermal conduction as QΔt=kAΔTL\frac{Q}{\Delta t} = \frac{k A \Delta T}{L}. There AA is on top because that equation describes a conductance: a short, fat slab passes more heat. Turn it over and the thermal resistance goes as L/(kA)L/(kA), the same skeleton as ρ/A\rho \ell / A.

Temperature, and the two CED statements that look opposed

Put 11.3.A.2.ii and 11.3.B.1.ii next to each other and they read like a contradiction.

  • 11.3.A.2.ii: the resistivity of a conductor typically increases with temperature.
  • 11.3.B.1.ii: the resistivity of an ohmic material is constant regardless of temperature.

They sit on facing pages of the CED and both are true, because they describe different things. The first is about real conductors. The second is part of the definition of the ohmic model: 11.3.B.1.i defines ohmic materials as those with constant resistance for all currents, and a resistance that does not move with current does not move as the element warms. Being ohmic is what rules the temperature dependence out.

Essential knowledge 11.3.B.1.iii is the bridge between the two: resistors can also convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment. Push current through a real resistor and it warms. Warm a real conductor and its resistivity climbs. So a real resistor's resistance drifts upward as it works, and calling something ohmic is the decision to ignore that drift.

An incandescent filament is the standard case where you cannot ignore the drift. It runs hot enough that a graph of its current against potential difference bends over instead of staying straight. That is not a faulty bulb, it is a conductor doing what 11.3.A.2.ii says conductors typically do. Which statement applies on an exam question is not left to you to guess, and the next section is where that gets settled.

Ohm's law, exactly as the sheet prints it

I=ΔVRI = \frac{\Delta V}{R}

The current in a conductive element equals the potential difference across it divided by its resistance. That is essential knowledge 11.3.B.1, and the same arrangement is printed in the Electricity group of the equation sheet.

Keep the delta. Most textbooks write V=IRV = IR. The AP Physics 2 sheet does not, and the reason is physical rather than cosmetic. Electric potential VV is a property of a single point in space, which is Topic 10.5. A resistor does not have a potential. It has two ends, each at its own potential, and what drives current through it is the difference between them. Writing ΔV\Delta V keeps that two-point nature visible, and on a free-response item that asks you to justify a claim the notation is doing part of the justifying.

The algebra is unchanged: ΔV=IR\Delta V = IR and R=ΔV/IR = \Delta V / I are the same statement rearranged. The Ohm's law guide works all three arrangements with a unit check for each, and the Ohm's law calculator will solve for whichever one you are missing.

What this topic asks for sits one level above that arithmetic: the discipline of pairing a potential difference with the element it was measured across. The ΔV\Delta V and the RR have to describe the same piece of circuit. Use a battery's potential difference with one resistor's resistance and you get a current that flows nowhere in the actual circuit. Topic 11.5 is where multi-element circuits get reduced so that the pairing becomes legal, and the series and parallel guide walks that reduction.

One piece of vocabulary is worth care. Ohm's law is a relationship that some materials obey and others do not, which makes it unlike conservation of charge or conservation of energy. The CED's wording is careful in the same way: 11.3.B.1 says Ohm's law relates current, resistance, and potential difference across a conductive element, and 11.3.B.1.i then names the materials that obey it. Nothing in physics obliges a device to.

What the ohmic convention buys you

The AP Physics 2 Table of Information prints a short list of assumptions, introduced as the conventions used in this exam unless otherwise stated. One line reads: "Resistors and lightbulbs are ohmic."

The "unless otherwise stated" clause is not decoration. The convention is a default, and questions are free to override it. While it holds, it buys you a great deal:

  • One number RR describes the element at every operating point, so a resistance measured at 2 V is still the resistance at 8 V.
  • The equivalent-resistance rules of Topic 11.5 are legitimate, because they assume each element has one fixed resistance, and the power forms in Topic 11.4 that carry an RR mean something, because there is a single RR to carry.
  • A lightbulb, genuinely non-ohmic on a laboratory bench, can be treated as a plain resistor. That is the convention doing its heaviest lifting, and it is why a bulb in an AP problem behaves nothing like the filament in the previous section.

The default comes off in two situations. The first is when a question says so: a line telling you the filament's resistance rises with temperature cancels the convention for that item. The second is when a question asks you to determine whether an element is ohmic, which the CED's own sample free-response Question 3 on experimental design does, giving a student ohmic resistors of known resistance and ideal meters and asking what to measure and graph. Its scoring guidelines list learning objectives 11.2.A, 11.3.B and 11.5.B. You cannot assume the answer to the question you were asked.

The same list also states that strings, springs, batteries, wires, and meters are ideal, and that current is conventional current. The circuit topics fill in what ideal means: ideal batteries and wires have negligible resistance (11.5.B.1), ideal ammeters have zero resistance (11.5.C.1.ii), and ideal voltmeters have infinite resistance (11.5.C.2.ii). Those defaults are what let a Topic 11.3 problem put all of a circuit's resistance into the one element you are studying.

Reading a current-versus-voltage graph (11.3.B.1.iv)

Essential knowledge 11.3.B.1.iv is specific about the axes, and the specificity is the point: the resistance of an ohmic circuit element can be determined from the slope of a graph of the current in the element as a function of the potential difference across the element.

Current as a function of potential difference means current on the vertical axis and potential difference on the horizontal axis. On those axes the slope is a change in current over a change in potential difference, which by Ohm's law is 1/R1/R. The resistance is the reciprocal of the slope, not the slope.

Vertical axisHorizontal axisSlope equalsSo the resistance is
IIΔV\Delta V1/R1/Rone divided by the slope
ΔV\Delta VIIRRthe slope itself

Both graphs are legitimate and both show up on exams; the CED's wording picks the first. Read the axis labels before you touch the slope, and carry a unit: a slope in A/V\mathrm{A/V} inverts to V/A\mathrm{V/A}, which is an ohm.

The shape of the plot carries the physics.

  • A straight line through the origin means the element is ohmic over the range you tested. It certifies nothing outside that range.
  • A curve that flattens as ΔV\Delta V grows means the current is rising more slowly than the potential difference, so the resistance is increasing. Self-heating, through 11.3.B.1.iii and then 11.3.A.2.ii, is the usual cause.
  • A straight line that misses the origin is a warning about the measurement rather than about the material. An ohmic element carries no current at zero potential difference.

Suggested skills 1.B and 3.A both live here. 1.B asks for quantitative graphs with appropriate scales and units, so label each axis with a symbol and a unit, pick a scale that spreads the points across the grid, and draw a best-fit line rather than joining the dots. 3.A asks for a procedure that answers a stated question, and the CED's Unit 11 sample activities include exactly that shape of task: measuring the potential difference across and the current in a length of mechanical pencil lead to find the resistivity of graphite.

How Topic 11.3 is tested, and where it leads

The five suggested skills map onto five recognizable question shapes.

  1. Compute a resistance or a resistivity from dimensions (2.B). Halve the diameter before you square it.
  2. Predict a factor of change (2.D). Often with no numbers at all: a wire is stretched, lengthened, or swapped for a thicker one, and you report the multiplier on RR.
  3. Apply Ohm's law to one element (3.B), and justify the pairing of the potential difference with that element.
  4. Plot and interpret data (1.B) to decide whether a device is ohmic and to pull RR out of a slope.
  5. Design the experiment that produces those data (3.A), naming the meters, where each one goes, and what you would repeat to reduce uncertainty.

The unit's opening pages are worth reading with that list in hand. The CED says Unit 11 requires more than calculating currents, resistances and potential differences in a simple circuit, and it names testing whether a light bulb is ohmic as an example of the experimental design work students should practice.

From here the unit runs two ways. Topic 11.4 takes the same II, ΔV\Delta V and RR and asks how fast energy moves, which is where the thermal conversion in 11.3.B.1.iii gets an equation. Topic 11.5 puts several resistors in one circuit and supplies the equivalent-resistance rules. Underneath both sit Topic 11.1, which defines the current in I=ΔV/RI = \Delta V / R, and Topic 11.2, which defines the circuit it flows around. The Unit 11 overview lays out all eight topics in order.

Measuring the resistivity of a pencil lead

A student connects a 60.0 mm length of 0.50 mm diameter mechanical pencil lead in series with a 1000Ω1000 \, \Omega resistor and a 1.5 V cell. An ammeter in the loop reads 1.49 mA and a voltmeter across the lead reads 7.5 mV. Take the battery, wires and meters as ideal. Find the resistance of the lead and the resistivity of the graphite.

  1. Both meters report on the same element, so Ohm's law applies to the lead on its own: R=ΔV/IR = \Delta V / I. Convert the readings to base units before dividing: ΔV=7.5mV=7.5×103V\Delta V = 7.5 \, \mathrm{mV} = 7.5 \times 10^{-3} \, \mathrm{V} and I=1.49mA=1.49×103AI = 1.49 \, \mathrm{mA} = 1.49 \times 10^{-3} \, \mathrm{A}.

  2. R=7.5×103V1.49×103A=5.03ΩR = \dfrac{7.5 \times 10^{-3} \, \mathrm{V}}{1.49 \times 10^{-3} \, \mathrm{A}} = 5.03 \, \Omega, which is 5.0Ω5.0 \, \Omega to two significant figures.

  3. Check that reading against the rest of the loop. The ammeter implies a total resistance of 1.5V/(1.49×103A)=1.01×103Ω1.5 \, \mathrm{V} / (1.49 \times 10^{-3} \, \mathrm{A}) = 1.01 \times 10^{3} \, \Omega, and the loop holds 1000Ω1000 \, \Omega plus about 5Ω5 \, \Omega. Those agree to the precision of the readings, so the measurement is self-consistent.

  4. Now bring in the geometry. The lead is a cylinder, so its cross-section is a circle of radius r=0.25mm=2.5×104mr = 0.25 \, \mathrm{mm} = 2.5 \times 10^{-4} \, \mathrm{m}, giving A=πr2=π(2.5×104m)2=1.96×107m2A = \pi r^2 = \pi (2.5 \times 10^{-4} \, \mathrm{m})^2 = 1.96 \times 10^{-7} \, \mathrm{m}^2. Halving the diameter first is the step to guard: treating 0.50 mm as the radius would make AA four times too large.

  5. Rearrange R=ρ/AR = \rho \ell / A for the resistivity, ρ=RA\rho = \dfrac{RA}{\ell}, then substitute, keeping the unrounded resistance and converting the length to meters: ρ=(5.03Ω)(1.96×107m2)0.0600m=1.6×105Ωm\rho = \dfrac{(5.03 \, \Omega)(1.96 \times 10^{-7} \, \mathrm{m}^2)}{0.0600 \, \mathrm{m}} = 1.6 \times 10^{-5} \, \Omega \cdot \mathrm{m}.

  6. Check the unit: Ωm2/m=Ωm\Omega \cdot \mathrm{m}^2 / \mathrm{m} = \Omega \cdot \mathrm{m}, which is what a resistivity is measured in. An answer in ohms means you divided by the area instead of multiplying.

R=5.0ΩR = 5.0 \, \Omega and ρ=1.6×105Ωm\rho = 1.6 \times 10^{-5} \, \Omega \cdot \mathrm{m}, both to two significant figures. The 1000Ω1000 \, \Omega resistor holds the current down, which means nearly all of the 1.5 V falls across it and the voltmeter has to resolve millivolts. Naming that limitation is the kind of uncertainty comment suggested skill 3.A asks for.

Which pencil lead has the greater resistance?

Two students argue about which mechanical pencil lead to use as a heating element in water. Student A wants the 0.70 mm lead so that charges flow more easily. Student B says the 0.50 mm lead has the greater resistance. Both leads are graphite and both are 60.0 mm long, and the 0.50 mm lead has the 5.0Ω5.0 \, \Omega resistance found above. Decide who is right about the resistance, and by what factor.

  1. The leads share a material and a length, so ρ\rho and \ell cancel in a ratio and only the areas matter: R0.50R0.70=A0.70A0.50\dfrac{R_{0.50}}{R_{0.70}} = \dfrac{A_{0.70}}{A_{0.50}}. The area is on top because RR is inversely proportional to it.

  2. Circular areas go as the square of the diameter, since A=πd2/4A = \pi d^2 / 4, so the ratio is (0.70mm0.50mm)2=(1.4)2=1.96\left(\dfrac{0.70 \, \mathrm{mm}}{0.50 \, \mathrm{mm}}\right)^2 = (1.4)^2 = 1.96. The thinner lead has 1.96 times the resistance of the thicker one, so Student B is right.

  3. Put a number on the thicker lead: R0.70=5.03Ω1.96=2.57ΩR_{0.70} = \dfrac{5.03 \, \Omega}{1.96} = 2.57 \, \Omega, which is 2.6Ω2.6 \, \Omega to two significant figures.

  4. Sanity check the direction. More cross-sectional area means more room for charge to move, so less resistance, and the thicker lead did come out smaller.

  5. Student A is not wrong either. At a fixed potential difference, the lead with the smaller resistance carries the larger current, which is what flowing more easily means. Both students have stated something true, and neither statement settles the argument they are having.

The 0.50 mm lead has 1.96 times the resistance of the 0.70 mm lead: 5.0Ω5.0 \, \Omega against 2.6Ω2.6 \, \Omega. Student B is right about the resistance and Student A is right about the current. Which lead heats the water faster is a question about power rather than resistance, and Topic 11.4 finishes the argument with the same two numbers.

Is it ohmic? Reading two sets of data

A student measures the current in two circuit elements at five potential differences. Element X gives 0.020 A at 1.00 V, 0.040 A at 2.00 V, 0.059 A at 3.00 V, 0.081 A at 4.00 V, and 0.100 A at 5.00 V. Element Y gives 0.050 A at 1.00 V, 0.080 A at 2.00 V, 0.100 A at 3.00 V, 0.115 A at 4.00 V, and 0.125 A at 5.00 V. Decide which element is ohmic over this range, and find its resistance.

  1. Test each element against the definition in 11.3.B.1.i, constant resistance for all currents, by computing R=ΔV/IR = \Delta V / I point by point.

  2. Element X: 1.00/0.020=50.0Ω1.00/0.020 = 50.0 \, \Omega, 2.00/0.040=50.0Ω2.00/0.040 = 50.0 \, \Omega, 3.00/0.059=50.8Ω3.00/0.059 = 50.8 \, \Omega, 4.00/0.081=49.4Ω4.00/0.081 = 49.4 \, \Omega, 5.00/0.100=50.0Ω5.00/0.100 = 50.0 \, \Omega.

  3. Element Y: 1.00/0.050=20.0Ω1.00/0.050 = 20.0 \, \Omega, 2.00/0.080=25.0Ω2.00/0.080 = 25.0 \, \Omega, 3.00/0.100=30.0Ω3.00/0.100 = 30.0 \, \Omega, 4.00/0.115=34.8Ω4.00/0.115 = 34.8 \, \Omega, 5.00/0.125=40.0Ω5.00/0.125 = 40.0 \, \Omega.

  4. X returns the same resistance at every point to within the scatter you would expect from meter readings, so it is ohmic across 1.00 V to 5.00 V. Y's resistance doubles over the same range, so it is not.

  5. Get X's resistance from the graph, the way 11.3.B.1.iv directs. Plot II vertically against ΔV\Delta V horizontally and draw a best-fit line through the origin. Using the first and last points, the slope is 0.100A0.020A5.00V1.00V=0.080A4.00V=0.0200A/V\dfrac{0.100 \, \mathrm{A} - 0.020 \, \mathrm{A}}{5.00 \, \mathrm{V} - 1.00 \, \mathrm{V}} = \dfrac{0.080 \, \mathrm{A}}{4.00 \, \mathrm{V}} = 0.0200 \, \mathrm{A/V}.

  6. Invert the slope rather than reporting it: R=1/(0.0200A/V)=50.0ΩR = 1 / (0.0200 \, \mathrm{A/V}) = 50.0 \, \Omega. A least-squares fit forced through the origin on all five points gives 0.02002A/V0.02002 \, \mathrm{A/V}, so R=50.0ΩR = 50.0 \, \Omega again.

  7. Element Y's climbing resistance is the signature of a filament: its resistivity is rising as it heats, which is 11.3.A.2.ii, and the heating is 11.3.B.1.iii. One resistance quoted for Y would be wrong at four of its five points.

Element X is ohmic with R=50.0ΩR = 50.0 \, \Omega, read as the reciprocal of the 0.0200A/V0.0200 \, \mathrm{A/V} slope. Element Y is not: its resistance climbs from 20.0Ω20.0 \, \Omega at 1.00 V to 40.0Ω40.0 \, \Omega at 5.00 V. Note what this depends on. An exam item would treat both elements as ohmic unless it said otherwise or, as here, explicitly asked you to check.

Frequently asked questions

What is the difference between resistance and resistivity?

Resistivity is a property of a material and resistance is a property of an object made from it. Essential knowledge 11.3.A.2.i in the AP Physics 2 CED describes resistivity as a fundamental property of a material that depends on its atomic and molecular structure, while 11.3.A.1 defines resistance as a measure of the degree to which an object opposes the movement of electric charge. Resistivity is in ohm meters, resistance in ohms, and R = rho times length divided by cross-sectional area links them. Cut a wire in half and its resistance halves while its resistivity does not change.

What is the formula for resistance in AP Physics 2?

Two are printed in the Electricity group of the AP Physics 2 equation sheet. R = rho L / A gives the resistance of a resistor with uniform geometry from its resistivity, its length along the current path, and its cross-sectional area. I = ΔV/R, Ohm's law, gives the current from the potential difference across an element and its resistance, and rearranges to R = ΔV/I. The first tells you what the object is; the second tells you what it does in a circuit.

Why does the AP sheet write Ohm's law as I = ΔV/R instead of V = IR?

Because voltage is a difference between two points, not a value at one point. Electric potential is defined at a location, so a resistor does not have a potential: it has two ends at two different potentials, and the difference between them drives the current. The algebra is identical, so ΔV = IR and R = ΔV/I are both fine to use, but the printed form on the AP Physics 2 sheet is I = ΔV/R, and free-response answers read more clearly when they keep the delta.

What does it mean for a material to be ohmic?

Essential knowledge 11.3.B.1.i says materials that obey Ohm's law have constant resistance for all currents and are called ohmic materials. In practice that means one value of R describes the element no matter what potential difference you put across it, so a graph of current against potential difference is a straight line through the origin. The AP Physics 2 conventions state that resistors and lightbulbs are ohmic unless otherwise stated, so you may assume it by default, but a question that asks you to test whether an element is ohmic has removed that default for you.

Does resistance depend on temperature in AP Physics 2?

It depends on which model the question is using, and the CED states both sides. Essential knowledge 11.3.A.2.ii says the resistivity of a conductor typically increases with temperature, and 11.3.B.1.iii says resistors can convert electrical energy to thermal energy and change the temperature of the resistor and its environment. But 11.3.B.1.ii says the resistivity of an ohmic material is constant regardless of temperature, because constant resistance is what ohmic means. Since the exam conventions assume resistors and lightbulbs are ohmic unless otherwise stated, treat resistance as fixed unless the question raises temperature itself.

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

Check the axes first. Essential knowledge 11.3.B.1.iv describes a graph of current as a function of potential difference, meaning current on the vertical axis. On those axes the slope equals 1/R, so the resistance is the reciprocal of the slope, not the slope itself: a slope of 0.0200 A/V means 50.0 ohms. Swap the axes so potential difference is vertical and the slope is R directly. Either way the line must pass through the origin for the element to count as ohmic over the range measured.

What happens to the resistance of a wire if you double its length?

Resistance is proportional to length, so doubling the length at unchanged cross-sectional area doubles the resistance. The common trap is a wire that is stretched to twice its length rather than simply being longer. Stretching keeps the volume fixed, so doubling the length halves the cross-sectional area, and since resistance is inversely proportional to area the two effects multiply: the resistance goes up by a factor of four, not two.