Resistance vs Resistivity: What Is the Difference?

The distinction, stated once

Resistance belongs to an object. Essential knowledge 11.3.A.1 in AP Physics 2 defines it as a measure of the degree to which an object opposes the movement of electric charge. Change the object, by making it longer or thinner, and the resistance changes. Its unit is the ohm, $\Omega$.

Resistivity belongs to the material. Essential knowledge 11.3.A.2.i calls it 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. Change the shape of the object and the resistivity does not move, because it was never about the shape. Its unit is the ohm metre, $\Omega \cdot \mathrm{m}$.

One equation carries both, and the AP Physics 2 sheet prints it in the electricity block:

Read it as a recipe with two ingredients. The material supplies $\rho$. The workshop supplies the length $\ell$ and the cross-sectional area $A$. Multiply them together the way the equation says and you get the resistance of the finished object.

That is the whole difference, and the test for it takes one sentence. Ask what happens if you cut the object in half. Its resistance halves. Its resistivity is unchanged. If your answer to a question would change when the object is reshaped, you were talking about resistance; if it would not, you were talking about resistivity.

Side by side

The two rows that decide most questions are the geometry rows, and they are not symmetric. Length is in the numerator, so doubling the length doubles $R$. Area is in the denominator, so doubling the diameter quadruples the area and cuts $R$ to a quarter. A question that says "twice as thick" is asking about diameter, and the factor is four, not two.

The last row is the practical one. Resistors and light bulbs are sold and labelled by resistance, because resistance is what a circuit responds to. Resistivity is a material data-sheet number. The AP Physics 2 course never supplies a table of resistivities and the sheet prints no values, so if a problem needs $\rho$ it will give you $\rho$.

What the CED requires, and how the two objectives split

Topic 11.3, Resistance, Resistivity, and Ohm's Law has two learning objectives, and the split between them is exactly the split on this page.

11.3.A asks you to describe the resistance of an object using physical properties of that object. Its essential knowledge is 11.3.A.1, which defines resistance as above, and 11.3.A.2, which states that the resistance of a resistor with uniform geometry is proportional to its resistivity and length and is inversely proportional to its cross-sectional area, with $R = \rho \ell / A$ as the relevant equation. The two sub-points are 11.3.A.2.i, defining resistivity, and 11.3.A.2.ii, which states that the resistivity of a conductor typically increases with temperature.

11.3.B asks you to describe the electrical characteristics of elements of a circuit. Its single essential knowledge, 11.3.B.1, states that Ohm's law relates current, resistance, and potential difference across a conductive element of a circuit, with the relevant equation printed as

and four sub-points: 11.3.B.1.i, that materials which obey Ohm's law have constant resistance for all currents and are called ohmic materials; 11.3.B.1.ii, that the resistivity of an ohmic material is constant regardless of temperature; 11.3.B.1.iii, that resistors can also convert electrical energy to thermal energy, which may change the temperature of both the resistor and the resistor's environment; and 11.3.B.1.iv, that 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.

That is the whole of Topic 11.3, checked statement by statement against the page. It carries no boundary statement.

Two details are worth pulling out. First, the delta on Ohm's law is printed. The sheet writes $I = \Delta V / R$, not $I = V/R$, because the quantity that drives current through a resistor is a potential difference between its two ends, not a potential at a point. Second, 11.3.A.2 says "a resistor with uniform geometry". The proportionality is stated for an object whose cross-section does not vary along its length. A tapered conductor is outside what that equation covers.

The course description also flags this pair explicitly. In the Unit 11 guidance on preparing for the AP exam, the CED says students should know the differences in meaning between "current", "potential difference", "resistance", "resistivity", and "capacitance", and that scaffolded instruction in using the correct vocabulary will help students on the free-response section. The distinction on this page is one the framework names as a vocabulary target.

The case that separates them: one material, three wires

Take a single spool of wire, so the resistivity is fixed at one value, and cut three pieces with different geometry. Nothing about the material changes. Every difference in the numbers comes from $\ell$ and $A$.

Suppose the material has $\rho = 5.0 \times 10^{-7} \ \Omega \cdot \mathrm{m}$, given, and start from a piece that is $2.0 \ \mathrm{m}$ long with a cross-sectional area of $1.0 \times 10^{-6} \ \mathrm{m^2}$.

One column never moves. Three others do. The resistivity column is the material talking, and it says the same thing about all three pieces because they came off the same spool.

Read the third row carefully, because it is where marks are lost. Twice the diameter is four times the area, and area is in the denominator, so the resistance drops by a factor of four rather than two. The current then rises by four, from $6.0$ to $24 \ \mathrm{A}$.

Now run the comparison the other way, which is the version that appears on exams more often. Two wires have the same resistance. What can you say about their materials? Nothing, until you know their dimensions. A wire of $\rho = 1.0 \times 10^{-6} \ \Omega \cdot \mathrm{m}$, length $4.0 \ \mathrm{m}$ and area $1.0 \times 10^{-6} \ \mathrm{m^2}$ has $R = 4.0 \ \Omega$. So does a wire of twice that resistivity, the same area, and half the length. Equal resistance is not evidence of equal resistivity, and equal resistivity is not evidence of equal resistance.

Reading resistance off a graph, and the slope that is not R

Essential knowledge 11.3.B.1.iv is precise about which graph and which 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. Current on the vertical axis, potential difference on the horizontal.

Rearranging the printed form of Ohm's law shows what that slope is:

So the slope of $I$ against $\Delta V$ is $1/R$, and the resistance is the reciprocal of the slope. The CED's wording is careful for exactly this reason: it says determined from the slope, not equal to the slope.

This is the single most productive error on this topic, because it returns a number of the right size for a plausible resistance. A slope of $0.25 \ \mathrm{A/V}$ read as $0.25 \ \Omega$ instead of $4.0 \ \Omega$ is wrong by a factor of sixteen and still looks like a resistance.

Two further readings of the same graph are worth having.

• Straight through the origin means ohmic. By 11.3.B.1.i, a material that obeys Ohm's law has constant resistance for all currents, so its $I$ against $\Delta V$ graph is a straight line. A curved graph means the resistance is not constant, and the element is not ohmic.
• A curve that flattens as $\Delta V$ rises means the resistance is climbing. For a filament bulb this is what 11.3.B.1.iii describes: the resistor converts electrical energy to thermal energy, the temperature rises, and by 11.3.A.2.ii the resistivity of a conductor typically increases with temperature, so $R = \rho \ell / A$ grows. The Ohm's law guide works through the three rearrangements and the power relation that goes with them.

Two CED statements about temperature that pull opposite ways

Both of these are in Topic 11.3, three lines apart, and they look like a contradiction until you notice they describe different things.

• 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 consistent because being ohmic is an idealisation, not a measurement. An ohmic material is defined in 11.3.B.1.i by having constant resistance for all currents, and constant resistance at fixed geometry means constant resistivity, so temperature independence is built into the definition rather than discovered. Real conductors are not perfectly ohmic, and 11.3.A.2.ii is the statement about them.

What to do with each on an exam:

• If a question calls an element ohmic, or gives you a straight-line $I$ against $\Delta V$ graph, treat $\rho$ and therefore $R$ as fixed. Heating does not enter.
• If a question describes a real filament heating up, or gives you a curved graph, use 11.3.A.2.ii: the resistivity rises, so the resistance rises, so the current is lower than the cold-resistance value would predict.

The word "typically" in 11.3.A.2.ii is also doing work. The CED does not claim that every material's resistivity rises with temperature, only that a conductor's typically does. Do not upgrade that into a universal rule.

When it costs a mark

Reading the slope of $I$ against $\Delta V$ as $R$. It is $1/R$. Covered above, and worth checking twice on any graph question.

Doubling the resistance when the wire is made twice as thick. Twice as thick means twice the diameter, which is four times the area, which is one quarter the resistance. Halving is the answer to "twice the area", which is a different sentence.

Saying resistivity depends on the length of the wire. It does not, and this is the error the whole page exists to prevent. Length changes $R$ through the equation; $\rho$ is the material coefficient sitting in front of the geometry.

Answering in the wrong unit. Ohms for resistance, ohm metres for resistivity. A resistivity quoted in ohms, or a resistance quoted in ohm metres, tells a grader you have not separated the two ideas without their needing to check the arithmetic.

Using $R = \rho \ell / A$ on a tapered conductor. Essential knowledge 11.3.A.2 states the relationship for a resistor with uniform geometry. If the cross-section varies, that form does not apply, and AP Physics 2 will not ask you to handle the varying case.

Dropping the delta in Ohm's law. The sheet prints $I = \Delta V / R$. Writing $I = V/R$ and then substituting the potential at a single node, rather than the difference across the element, is how a multi-resistor circuit gets the wrong current.

Assuming equal resistance means the same material. Two objects of different materials can have identical resistance, and two objects of the same material almost always have different resistance. Neither quantity determines the other without the geometry.

Treating resistivity as something that adds. Resistances in series add, by $R_{\mathrm{eq,s}} = \sum_i R_i$ on the sheet. Resistivities never add. Joining a $\rho$ of $1 \times 10^{-6}$ to a $\rho$ of $2 \times 10^{-6}$ does not make $3 \times 10^{-6}$; it makes an object whose two pieces have separate resistances that you add.

When they track together, and why that lulls you

The two move in step whenever the geometry is held fixed, and that covers most of the circuits you meet first.

A single fixed resistor. Its length and area never change, so $R$ is just $\rho$ multiplied by a constant. Every statement about one is a statement about the other, and nothing in the problem forces you to say which you mean. This is the circuit the pair is first taught on, which is exactly why the distinction stays invisible.

A heating filament. Here $\rho$ rises and $R$ rises with it, again because the geometry is fixed. The two rise together, so the sentence "the resistance goes up because the material resists more when hot" is true and gives no reason to separate the ideas.

Identical wires. Compare two pieces cut from the same spool to the same dimensions and both quantities are equal. A comparison built only from identical objects cannot expose the difference.

The distinction turns on exactly three situations, and they are the three that keep appearing on exams.

1. The geometry changes. Cut, stretch, thicken, or coil a wire, and $R$ moves while $\rho$ holds still.
2. The material changes at fixed geometry. Two wires of identical dimensions and different resistance differ only in $\rho$, and the ratio of their resistances is the ratio of their resistivities.
3. You are asked to identify a material. Resistance cannot do it, because any resistance can be built from any material by choosing the dimensions. Resistivity can, because it is the property the material carries.

If you can handle those three, the pair is finished.

Where this sits on the AP exam, and what the calculus course adds

This pair is Topic 11.3 in Unit 11, Electric Circuits, which the CED weights at 15 to 18 percent of the multiple-choice section over a suggested 12 to 20 class periods. The suggested skills listed for the topic 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. Skill 2.D is why so many questions on this topic are of the form "the wire is replaced by one twice as long and half as thick; by what factor does the current change".

AP Physics C: Electricity and Magnetism carries Topic 11.3 under the same number, with the same title, and with essential knowledge 11.3.A.1, 11.3.A.2, 11.3.A.2.i, 11.3.A.2.ii and all five statements of 11.3.B word for word identical to the AP Physics 2 versions. It adds exactly one item, 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

That integral is the honest generalisation of $R = \rho \ell / A$: add up the resistance of each thin slice. It is also a good way to see why the algebra-based form needs uniform geometry and uniform material, since it is the special case where $\rho$ is constant and the integral collapses to a multiplication.

The calculus course adds one more pointer worth knowing about, in its Topic 11.1. Essential knowledge 11.1.A.2.iii there states that a potential difference across a conductor creates an electric field within the conductor that is proportional to the resistivity of the conductor and the current density, with the relevant equation $\vec{E} = \rho \vec{J}$. That is resistivity acting at a point inside the material, with no length or area anywhere in sight, which is the clearest possible statement that resistivity is a local material property and resistance is a whole-object one.

For the two quantities that Ohm's law connects to resistance, see voltage vs current. For what the meters that measure them are doing, see ammeter vs voltmeter. For combining resistances once you have them, the series vs parallel circuits guide carries the reduction procedure, and the Ohm's law calculator handles the arithmetic.