AP Physics C: E&M · Topic 11.3

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

Unit 11: Electric Circuits15-25% of the multiple-choice section

Resistance measures how strongly an object opposes charge movement; resistivity is a property of the material. For uniform geometry R = rho L / A. AP Physics C adds one calculus statement: if the resistivity varies along the length, integrate it. Ohm's law is printed as I = ΔV/R.

AP Physics: Unit 11 (topics 11.3 Resistance, Resistivity, and Ohm's Law). Topic 11.3 of the current AP Physics C: Electricity and Magnetism course and exam description, in Unit 11 Electric Circuits (15 to 25% of the multiple-choice section, about 12 to 24 class periods). Two learning objectives: 11.3.A, describe the resistance of an object using physical properties of that object, with essential knowledge 11.3.A.1, 11.3.A.2 and sub-statements i to iii; and 11.3.B, describe the electrical characteristics of elements of a circuit, with essential knowledge 11.3.B.1 and sub-statements i to iv. The topic prints no boundary statement. Relevant equations are R = rho times length over area (11.3.A.2) and I = the potential difference divided by R (11.3.B.1), both printed on the equation sheet. Essential knowledge 11.3.A.2.iii gives the integral form for a resistor of uniform geometry whose resistivity varies along its length; that statement is unique to AP Physics C and is not printed on the sheet. The other nine statements appear word for word in the AP Physics 2 framework, which lists the same five suggested skills: 1.B, 2.B, 2.D, 3.A and 3.B. Learning objective 11.3.B is aligned to two of the four sample free-response questions in the course description, Question 3 and Question 4, more than any other Unit 11 objective.

What Topic 11.3 requires

Topic 11.3 of AP Physics C: Electricity and Magnetism Unit 11 carries two learning objectives, which is the CED signalling that geometry and circuit behaviour are separate ideas. It prints no boundary statement.

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. Relevant equation $R = \dfrac{\rho \ell}{A}$ .
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.A.2.iii The total resistance of a resistor with uniform geometry, but that is made of a material whose resistivity varies along the length of the resistor, is given by $R = \displaystyle\int \frac{\rho(\ell)\, d\ell}{A}$ .

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. Relevant equation $I = \dfrac{\Delta V}{R}$ .
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.

The suggested skills are 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; 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.

Five suggested skills is the joint largest count in Unit 11, shared with Topic 11.8. The pairing of 1.B with 3.A is the fingerprint of a lab question, and the unit's exam-preparation note says the third free-response question is the Experimental Design and Analysis question, where students will be required to derive relevant equations, linearize and analyze data.

What calculus changes here, and it is exactly one statement

Nine of this topic's ten essential-knowledge statements appear word for word in the AP Physics 2 framework. One does not, and it is 11.3.A.2.iii, the resistivity integral. That is the whole of the calculus in Topic 11.3, and it is worth being blunt about that rather than dressing up the rest.

What this means practically:

The definitions are the same. 11.3.A.1, 11.3.A.2, 11.3.A.2.i and 11.3.A.2.ii are identical across both courses.
All four of the 11.3.B statements are identical. Ohm's law, ohmic materials, the temperature statement and the graph-slope statement are the same sentences in both frameworks.
Both courses print the same two equations on their respective sheets: $R = \rho\ell/A$ and $I = \Delta V / R$ .
The suggested skills are identical too: 1.B, 2.B, 2.D, 3.A and 3.B in both.

So if you want the underlying ideas explained at length, the AP Physics 2 Topic 11.3 page covers nine tenths of this material and is written for the algebra-based exam. That page is for AP Physics 2 students; this one is for AP Physics C students, and it spends its space on the tenth statement, on the microscopic form $\vec{E} = \rho\vec{J}$ that only the C course carries, and on the symbolic style the C free-response section expects.

One genuine difference in emphasis rather than content: on the C exam, skill 2.A alone carries 25 to 30% of the multiple-choice section, and Science Practice 2 carries 40 to 45% of the free-response section. A Physics C resistivity question is far more likely to want $R = 3\rho_0 \ell / (2A)$ than 24 ohms.

Resistance belongs to the object, resistivity to the material

Statement 11.3.A.1 defines resistance as a measure of the degree to which an object opposes the movement of electric charge. Statement 11.3.A.2.i defines resistivity as a fundamental property of a material that depends on its atomic and molecular structure. The two nouns are the whole distinction, and the CED chose them carefully.

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

Read the equation as a translation between the two. Resistivity $\rho$ is what the substance is; length and cross-sectional area are what you did to it; resistance $R$ is what the resulting object does in a circuit. Cut a wire in half and its resistance halves while its resistivity does not move. Melt two identical wires into one of double the cross-section and the resistance halves again, still at unchanged resistivity.

Three consequences that carry marks:

Doubling the diameter quarters the resistance, because $A = \pi r^2$ and area goes as the square of a linear dimension. This is the single most common factor-of-change question in the topic, and it is skill 2.D.
Stretching a wire raises its resistance faster than the length ratio suggests. Stretching at constant volume to double the length halves the area, so $R$ goes up by a factor of four, not two.
The units follow. Resistivity is in ohm metres, because $\rho = RA/\ell$ has units $\Omega \cdot \mathrm{m}^2 / \mathrm{m}$ .

Topic 11.1 gives the same physics one level down. Statement 11.1.A.2.iii says a potential difference across a conductor creates a field inside it, $\vec{E} = \rho\vec{J}$ . Multiply that by a length and divide by a current and you have $R = \rho\ell/A$ again: $\Delta V = E\ell = \rho J \ell = \rho (I/A) \ell$ , so $\Delta V / I = \rho\ell/A$ . The point-by-point statement and the whole-object statement are the same law, and only AP Physics C carries both. The resistance against resistivity comparison is the short version of this section.

The resistivity integral (11.3.A.2.iii), and what it does not say

Statement 11.3.A.2.iii is the C-only line:

R = \int \frac{\rho(\ell)\, d\ell}{A}

Read the sentence attached to it, because every clause is doing work: the total resistance of a resistor with uniform geometry, but that is made of a material whose resistivity varies along the length of the resistor.

Where the integral comes from. Chop the resistor into slabs of thickness $d\ell$ , each with the same area $A$ but its own local resistivity $\rho(\ell)$ . Each slab is a tiny resistor of resistance $dR = \rho(\ell)\, d\ell / A$ . The slabs are in series, because every charge that crosses one must cross them all, which is the 11.5.A.1.i definition. Series resistances add, so the total is the integral. That derivation is worth being able to state in one sentence, because skill 2.A questions ask you to justify the setup rather than just evaluate it.

Uniform geometry means $A$ comes out. Because the area is constant, the integral is $\dfrac{1}{A}\displaystyle\int \rho(\ell)\, d\ell$ , and the integral of a resistivity over a length divided by that length is just the average resistivity. So for a uniform-area resistor,

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

and if $\rho$ varies linearly you can write the answer down by inspection using the mean of the two end values. That check catches most algebra errors in one line, and it explains why a linear profile is a favourite: the calculus and the shortcut agree.

What the statement does not cover. It says uniform geometry. A resistor whose cross-section varies, a cone or a tapered wire, is a different integral, $R = \int \rho\, d\ell / A(\ell)$ , with the area inside. The framework does not print that form anywhere, and Topic 11.3 prints no boundary statement that would rule it in or out. The honest position is that the statement you can cite covers varying resistivity at constant area, so if a question hands you a varying area, build the integral from the series-of-slabs argument rather than from a remembered formula.

It also says along the length. A current density that varies across the radius, which is the situation 11.1.A.3 sets up, is a different problem again and is handled there by integrating $\vec{J}$ over the area rather than $\rho$ over the length.

Ohm's law, printed with the delta

Statement 11.3.B.1 says Ohm's law relates current, resistance, and potential difference across a conductive element of a circuit, and the sheet prints it as

I = \frac{\Delta V}{R}

Not $V = IR$ . The delta is not decoration, and dropping it is how a correct formula produces a wrong answer. $\Delta V$ is the potential difference across that particular element, the difference between the potentials at its two ends. It is not the battery emf, not the potential at a point, and not the potential difference across the whole network unless that element is the whole network.

The habit that prevents the error: whenever you write Ohm's law, name the element in the same breath. $I_2 = \Delta V_2 / R_2$ is safe; $I = V/R$ is an invitation to use the wrong voltage.

Statement 11.3.B.1.i defines the class of materials the law applies to: materials that obey Ohm's law have constant resistance for all currents and are called ohmic materials. That is the real content of the law. Every element has a ratio $\Delta V / I$ at any instant; being ohmic means that ratio does not depend on the current.

The exam then hands you the assumption for free. The reference information for AP Physics C: Electricity and Magnetism lists seven standing conventions used unless otherwise stated, and one of them is: resistors and lightbulbs are ohmic. Two others matter for this topic, that strings, springs, batteries, wires and meters are ideal, and that the direction of current is the direction in which positive charges would drift.

So unless a question says otherwise, every resistor and every bulb on the exam has a resistance that does not change. A question describing a filament that warms up has switched that default off deliberately, and that is your signal to reach for 11.3.A.2.ii instead. The Ohm's law guide and the Ohm's law calculator cover the arithmetic and the rearrangements.

Two temperature statements that look opposed

The CED prints these two statements a page apart, and both are in the required content:

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 are not a contradiction, and being able to say why is a skill 3.B claim.

11.3.A.2.ii is about real conductors. In a metal, raising the temperature makes the lattice vibrate more, carriers scatter more often, and the resistivity rises. The word "typically" is there because it is not universal.

11.3.B.1.ii is the definition of an idealisation. Ohmic is a model, not a substance. An ohmic material is by construction one whose resistivity does not vary, and 11.3.B.1.i says the same thing in terms of current rather than temperature: constant resistance for all currents.

The reference information decides which applies. Resistors and lightbulbs are ohmic unless otherwise stated, so the default on the exam is 11.3.B.1.ii. When a question describes a warming filament, a bulb that takes several minutes to reach full brightness, or a resistor whose temperature is rising, it has turned the default off and 11.3.A.2.ii is now the operative statement.

Statement 11.3.B.1.iii is the bridge between them: resistors can also convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment. That is the mechanism by which a circuit heats its own resistors and so moves their resistivity.

The CED signposts this in the unit's essential questions, one of which asks why warming bulbs take several minutes to shine brightly, and again in a Topic 11.3 sample activity: give students water, modeling clay, or a related substance and ask them to determine whether the substance is ohmic by applying various voltages and measuring the resulting current. The Topic 11.4 boundary statement then draws the line on how far you take the thermal side: the course only expects students to analyze the transfer of mechanical and electrical energy, although students should be aware that electrical energy can also be dissipated in the form of thermal energy.

Reading a current-against-potential-difference graph (11.3.B.1.iv)

Statement 11.3.B.1.iv fixes the axes and it is easy to get backwards: 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 on the vertical axis, potential difference on the horizontal.

With those axes, $I = \Delta V / R$ is a straight line through the origin with

\text{slope} = \frac{1}{R}

so the resistance is the reciprocal of the slope, not the slope. A steeper line means a smaller resistance. Two safety checks: the units of the slope are amps per volt, which are inverse ohms, and a line through the origin is what an ohmic element must give, because zero potential difference has to mean zero current.

Three readings that come up:

A straight line through the origin: ohmic. The resistance is $1/\text{slope}$ , and the chord from the origin to any data point gives the same value.
A curve that bends toward the horizontal axis as $\Delta V$ grows: resistance rising. A filament lamp does this, because it heats up and 11.3.A.2.ii takes over.
A straight line that misses the origin: something else is in the circuit. In a lab that is usually contact resistance or a nonideal supply, and the intercept is your evidence for it.

One precision worth carrying into a free-response answer. The CED licenses the slope method for an ohmic element only. For a nonohmic element, the ratio $\Delta V / I$ at a chosen point and the local slope $dI / d(\Delta V)$ are different numbers, and the CED does not tell you to call either of them "the resistance". If a question hands you a curved graph, answer with the ratio at the stated operating point and say that is what you are using.

The lab framing follows from skills 1.B and 3.A. To find a material's resistivity, measure the resistance of several lengths of the same wire and plot $R$ against $\ell$ : the slope is $\rho / A$ , and multiplying by the measured cross-sectional area gives $\rho$ . That is the linearisation the Experimental Design and Analysis question asks for, and the CED's Instructional Strategies section gives the matching activity of scaffolding a design to determine whether a lightbulb can be considered an ohmic resistor.

How Topic 11.3 is tested

Learning objective 11.3.B carries an unusual amount of the CED's own sample assessment. It appears on two of the four sample free-response questions: Question 3, the Experimental Design and Analysis question, aligns to 11.3.B, 12.3.B and 12.4.A, and Question 4, the Qualitative/Quantitative Translation question, aligns to 10.3.A, 11.1.A, 11.7.A, 11.3.B and 11.8.B. No other Unit 11 objective appears twice in that set.

That is the shape of it. Topic 11.3 is rarely the subject of a question and almost always the tool inside one: the step where a resistance becomes a current, or a current becomes a potential difference, in the middle of a circuit, a lab analysis, or an RC derivation.

What that means for preparation:

Know the two equations cold, because you will be using them under time pressure inside longer problems. Both are printed on the C: E&M sheet.
Expect symbolic answers. The 25 to 30% multiple-choice weighting on skill 2.A and the 40 to 45% free-response weighting on Science Practice 2 both push that way.
Expect factor-of-change questions. Skill 2.D is listed here, and $R = \rho\ell/A$ with $A = \pi r^2$ is the cleanest functional-dependence relationship in the unit.
Expect a graph with the axes in the CED's order, current against potential difference, and remember the resistance is the reciprocal of the slope.

From here the topic feeds directly into electric power, where $P = I \Delta V$ combines with Ohm's law to give the two derived forms, and into compound DC circuits, where equivalent resistance turns a network into a single $R$ you can put back into $I = \Delta V / R$ .

A resistor whose resistivity varies along its length

A resistor of uniform cross-sectional area $A = 5.0 \times 10^{-7}\ \mathrm{m^2}$ and length $L = 0.20$ m is made of a material whose resistivity varies along its length as $\rho(x) = \rho_0\left(1 + \dfrac{x}{L}\right)$ , with $\rho_0 = 4.0 \times 10^{-5}\ \Omega \cdot \mathrm{m}$ at $x = 0$ . Find (a) the resistance, symbolically then numerically, (b) the current when 12 V is applied across it, and (c) the electric field inside the material at each end.

(a) Statement 11.3.A.2.iii applies: the geometry is uniform and the resistivity varies along the length. $R = \displaystyle\int_0^L \frac{\rho(x)\, dx}{A} = \frac{\rho_0}{A}\int_0^L \left(1 + \frac{x}{L}\right) dx$ .
The sheet's calculus table prints $\int x^n dx = \dfrac{1}{n+1}x^{n+1}$ for $n \neq -1$ , so $\displaystyle\int_0^L \left(1 + \frac{x}{L}\right) dx = L + \frac{L^2}{2L} = \frac{3L}{2}$ , giving $R = \dfrac{3\rho_0 L}{2A}$ .
Check it against the average-resistivity shortcut. Because $A$ is constant, $R = \bar{\rho} L / A$ , and a linear profile has mean value $\bar{\rho} = \tfrac{1}{2}(\rho_0 + 2\rho_0) = 1.5\rho_0$ . That gives $R = 1.5\rho_0 L / A = 3\rho_0 L / (2A)$ , the same expression.
Numerically, $R = \dfrac{3(4.0 \times 10^{-5})(0.20)}{2(5.0 \times 10^{-7})} = \dfrac{2.4 \times 10^{-5}}{1.0 \times 10^{-6}} = 24\ \Omega$ . For comparison, a uniform $\rho_0$ would give $\rho_0 L / A = 16\ \Omega$ , and the ratio $24/16 = 1.5$ is the average-resistivity factor showing up as expected.
(b) Ohm's law in the printed form: $I = \dfrac{\Delta V}{R} = \dfrac{12\ \mathrm{V}}{24\ \Omega} = 0.50$ A. The current is the same everywhere along the resistor, because the slabs are in series and 11.5.A.1.i requires it.
(c) The current density is also the same everywhere, since $A$ is constant: $J = I/A = \dfrac{0.50}{5.0 \times 10^{-7}} = 1.0 \times 10^{6}\ \mathrm{A/m^2}$ . Statement 11.1.A.2.iii gives $E = \rho J$ , so at $x = 0$ , $E = (4.0 \times 10^{-5})(1.0 \times 10^{6}) = 40$ V/m, and at $x = L$ , where $\rho = 2\rho_0$ , $E = 80$ V/m.
Final check: the potential difference is the field integrated along the length, $\Delta V = \int_0^L E\, dx = J\rho_0 (3L/2) = (1.0 \times 10^{6})(4.0 \times 10^{-5})(0.30) = 12$ V. It returns the applied 12 V, so parts (a) and (c) are consistent.

(a) $R = \dfrac{3\rho_0 L}{2A} = 24\ \Omega$ , which is 1.5 times the uniform- $\rho_0$ value because the mean resistivity is $1.5\rho_0$ . (b) $I = 0.50$ A. (c) The internal field rises from 40 V/m at the low-resistivity end to 80 V/m at the other, and integrating it along the length returns the applied 12 V.

Is it ohmic? Reading the graph the CED specifies

Two circuit elements are tested by applying several potential differences and recording the current. Element X gives 0.400 A at 1.0 V, 0.800 A at 2.0 V, 1.200 A at 3.0 V and 1.600 A at 4.0 V. Element Y gives 0.500 A at 1.0 V, 0.800 A at 2.0 V, 1.000 A at 3.0 V and 1.143 A at 4.0 V. Decide which is ohmic, give its resistance, and say what can and cannot be claimed about the other.

Plot as statement 11.3.B.1.iv specifies: current on the vertical axis as a function of the potential difference on the horizontal axis. Then a straight line through the origin means ohmic, and the slope is $1/R$ .
Element X: the ratios $I / \Delta V$ are $0.400$ , $0.400$ , $0.400$ and $0.400$ A/V. Constant, so the graph is a straight line through the origin and the element is ohmic by 11.3.B.1.i, which requires constant resistance for all currents.
Its resistance is the reciprocal of the slope, not the slope: $R = \dfrac{1}{0.400\ \mathrm{A/V}} = 2.50\ \Omega$ . Units confirm it, since amps per volt are inverse ohms.
Element Y: the ratios $\Delta V / I$ are $1.0/0.500 = 2.00\ \Omega$ , $2.0/0.800 = 2.50\ \Omega$ , $3.0/1.000 = 3.00\ \Omega$ and $4.0/1.143 = 3.50\ \Omega$ . The ratio rises steadily, so the graph curves toward the horizontal axis and Y is not ohmic.
Statement 11.3.A.2.ii supplies the likely reason: the resistivity of a conductor typically increases with temperature, and 11.3.B.1.iii says resistors convert electrical energy to thermal energy, which may change the temperature of the resistor. Element Y behaves like a filament lamp warming up.
State the limit of the claim. For Y, the CED does not license calling any single number "the resistance": 11.3.B.1.iv is written for an ohmic element. Quote a value at a stated operating point instead, for example $3.00\ \Omega$ at 3.0 V, and say so explicitly.
Note also that the exam's reference information declares resistors and lightbulbs ohmic unless otherwise stated. Data like Element Y's is the question telling you it has switched that default off.

Element X is ohmic with $R = 2.50\ \Omega$ , the reciprocal of the 0.400 A/V slope. Element Y is not ohmic: its ratio of potential difference to current rises from 2.00 to 3.50 ohms across the range, consistent with a conductor whose resistivity increases as it heats. For Y you can only quote a resistance at a stated operating point.

Measuring resistivity by linearisation, then predicting a new value

Students measure the resistance of several lengths cut from the same spool of wire and plot $R$ against length $\ell$ . The best-fit line has slope $8.6\ \Omega/\mathrm{m}$ and passes essentially through the origin. The wire's diameter is measured as $0.40$ mm. (a) Find the resistivity. (b) Predict the resistance of a 2.5 m length of wire of the same material with half the diameter. (c) What would a nonzero vertical intercept have meant?

(a) Rearrange 11.3.A.2 to expose the plotted variables: $R = \dfrac{\rho}{A}\ell$ , so a graph of $R$ against $\ell$ is a straight line through the origin with slope $\rho / A$ . That rearrangement is the linearisation the Experimental Design and Analysis question asks for.
Cross-sectional area from the diameter: $r = 0.20\ \mathrm{mm} = 2.0 \times 10^{-4}$ m, so $A = \pi r^2 = \pi (4.0 \times 10^{-8}) = 1.257 \times 10^{-7}\ \mathrm{m^2}$ .
$\rho = \text{slope} \times A = (8.6)(1.257 \times 10^{-7}) = 1.081 \times 10^{-6}\ \Omega \cdot \mathrm{m}$ , or $1.1 \times 10^{-6}\ \Omega \cdot \mathrm{m}$ to two significant figures, which is the limiting precision of the two-figure slope and diameter.
(b) Skill 2.D, functional dependence. Halving the diameter quarters the area, and $R$ is inversely proportional to $A$ , so the resistance per metre becomes $4 \times 8.6 = 34.4\ \Omega/\mathrm{m}$ . For 2.5 m, $R = (34.4)(2.5) = 86\ \Omega$ .
Confirm the long way: $A' = A/4 = 3.14 \times 10^{-8}\ \mathrm{m^2}$ , so $R = \dfrac{(1.081 \times 10^{-6})(2.5)}{3.14 \times 10^{-8}} = 86\ \Omega$ . The two routes agree, and the factor-of-change route is the one the exam rewards for speed.
(c) A nonzero intercept would mean a resistance present at zero wire length, so something other than the wire is being measured: contact or lead resistance, or an ohmmeter offset. The physical model predicts an intercept of zero, and reporting the discrepancy is part of what the lab question scores.

(a) $\rho = 1.1 \times 10^{-6}\ \Omega \cdot \mathrm{m}$ , from slope times cross-sectional area. (b) $86\ \Omega$ , because halving the diameter quarters the area and so quadruples the resistance per metre. (c) A nonzero intercept means resistance that does not belong to the wire, such as contact or lead resistance.

Frequently asked questions

What is the difference between resistance and resistivity?

Resistance is a property of an object and resistivity is a property of the material it is made from. The AP Physics C course description makes the distinction in its wording: essential knowledge 11.3.A.1 says resistance is a measure of the degree to which an object opposes the movement of electric charge, while 11.3.A.2.i says resistivity is a fundamental property of a material that depends on its atomic and molecular structure. The two are linked by R equals rho times length divided by cross-sectional area, which is printed on the equation sheet. Cutting a wire in half halves its resistance and leaves its resistivity unchanged, because you changed the object and not the substance.

How do you find the resistance when the resistivity varies along the length?

Integrate. Essential knowledge 11.3.A.2.iii in the AP Physics C course description states that the total resistance of a resistor with uniform geometry, but that is made of a material whose resistivity varies along the length of the resistor, is given by the integral of rho of ell d ell divided by the cross-sectional area. The derivation is a series argument: each thin slab has resistance rho times d ell over A, every charge that crosses one slab must cross them all, and series resistances add. Because the area is constant it comes outside the integral, so the answer is always the average resistivity times length over area. This statement does not appear in the AP Physics 2 framework.

Is Ohm's law on the AP Physics C E&M equation sheet?

Yes, and it is printed as I equals delta V divided by R, with the delta on the potential difference, rather than as V equals I R. The delta carries the meaning: the numerator is the potential difference across that particular element, not a potential at a point and not the battery emf. Essential knowledge 11.3.B.1 says the same in words, that Ohm's law relates current, resistance, and potential difference across a conductive element of a circuit. The AP Physics 2 equation sheet prints the identical form. Also printed on the C: E&M sheet is R equals rho times length over area.

Does resistivity change with temperature in AP Physics C?

The course description prints two statements that appear to disagree, and both are correct. Essential knowledge 11.3.A.2.ii says the resistivity of a conductor typically increases with temperature, which describes real metals. Essential knowledge 11.3.B.1.ii says the resistivity of an ohmic material is constant regardless of temperature, which is the definition of the idealisation. The exam's reference information settles which one applies by default, listing among its standing conventions that resistors and lightbulbs are ohmic. So resistance is constant unless a question tells you otherwise, and a question that describes a warming filament or a bulb that takes time to reach full brightness has deliberately switched that default off.

How do you read resistance off a current versus voltage graph?

Take the reciprocal of the slope, not the slope. Essential knowledge 11.3.B.1.iv specifies the axes: 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. With current on the vertical axis, I equals delta V over R is a straight line through the origin whose slope is one over R, so a steeper line means a smaller resistance. The units confirm it, since the slope is in amps per volt, which are inverse ohms. The statement is written for an ohmic element only, so a curved graph does not have a single resistance to read.

What is different about AP Physics C Topic 11.3 compared with AP Physics 2?

One essential knowledge statement. Nine of the ten statements in AP Physics C Topic 11.3 appear word for word in the AP Physics 2 framework, both courses print the same two equations on their equation sheets, and both list the same five suggested skills. The addition is essential knowledge 11.3.A.2.iii, the integral for a resistor of uniform geometry whose resistivity varies along its length. AP Physics C also carries the microscopic version of the same law in Topic 11.1, that the electric field inside a conductor equals the resistivity times the current density, which AP Physics 2 does not have, and it expects symbolic rather than numerical answers far more often.

How would you measure the resistivity of a wire in an AP Physics C lab?

Measure the resistance of several different lengths cut from the same wire, plot resistance against length, and take the slope. Rearranging R equals rho times length over area shows that this graph is a straight line through the origin with slope equal to resistivity divided by cross-sectional area, so multiplying the slope by the measured cross-sectional area gives the resistivity. That linearisation is what the Experimental Design and Analysis free-response question asks for, and skills 1.B and 3.A are both listed on this topic for it. A nonzero vertical intercept is evidence of resistance that does not belong to the wire, such as contact or lead resistance.