Resistance, Resistivity and Ohm's Law: Study Guide (Physics C: E&M 11.3)

Course note. This guide is for the calculus-based Physics C: Electricity and Magnetism course, Topic 11.3. You will derive the resistance of a wire from the field inside it, and integrate when the material changes along the wire. You will also plot and read experimental data, which the course expects you to do with care.

Constant used throughout: copper has resistivity 1.7 × 10⁻⁸ Ω·m at room temperature. Every other material in this guide is given its own value.

What resistance measures

In Topic 11.2 you saw that a bulb lights only if there is a potential difference across it. The same potential difference does not drive the same current through every object. A short thick copper wire lets a large current through; a thin wire of a heating alloy lets much less through. Resistance measures how strongly an object opposes the movement of charge through it:

R = ΔV / I

• ΔV is the potential difference across the object and I is the current in it.
• The SI unit is the ohm: 1 Ω = 1 V/A.
• A large R means a large potential difference is needed for a given current.

Resistance belongs to an object: this wire, this lamp, this resistor. It depends on what the object is made of and on its shape. The next section separates those two parts.

From E = ρJ to R = ρℓ/A

In Topic 11.1 you met the link between the field inside a conductor and the current density:

E = ρJ

Here ρ is the resistivity of the material, in Ω·m. Now apply it to a straight wire of length ℓ and uniform cross-sectional area A, made of one material, carrying a steady current I.

1. The current spreads evenly over the cross-section, so J = I/A.
2. The field inside is uniform and along the wire, so the potential difference between the ends is ΔV = Eℓ.
3. Combine: ΔV = Eℓ = ρJℓ = ρ(I/A)ℓ.
4. Divide by I: R = ΔV/I = ρℓ/A.

This shows how each factor matters:

• Length. R ∝ ℓ. Double the length and the charge has twice as far to go against the same opposition, so R doubles.
• Area. R ∝ 1/A. Double the area and there are twice as many parallel paths, so R halves. For a round wire, A = πr² = πd²/4, so doubling the diameter makes R four times smaller.
• Material. R ∝ ρ. Copper’s ρ is about 1.7 × 10⁻⁸ Ω·m; the alloy in Worked example 1 is about 65 times larger, and insulators are larger by many powers of ten.

Stretching a wire. If a wire is pulled to n times its length, its volume stays the same, so its area falls to A/n. Then R = ρ(nℓ)/(A/n) = n²(ρℓ/A). Stretching to twice the length makes R four times larger.

Resistivity is a property of the material

Resistivity does not depend on the size or shape of a sample. It depends on the atomic and molecular structure of the material: how many charge carriers it has and how often they are scattered as they drift. A copper coin and a copper cable have the same resistivity but very different resistances.

Temperature. In a metal, the charge carriers scatter off the vibrating ions. When the metal gets hotter, the ions vibrate more, the carriers scatter more often, and the resistivity increases. So the resistivity of a conductor typically rises with temperature.

The ohmic model. In many problems you treat a resistor as ohmic: its resistivity, and so its resistance, stays constant regardless of temperature. Real resistors are designed to come close to this over their working range. When a question says an element is ohmic, or simply gives you a fixed R, use a constant value.

Resistors and thermal energy. As charge drifts through a resistor, electrical energy is converted to thermal energy. This can raise the temperature of the resistor and of its surroundings. Topic 11.4 shows how to calculate the rate. For a component whose resistivity depends on temperature, such as the thin metal filament of a lamp, this heating changes the resistance as the current rises.

When the resistivity changes along the wire

Suppose the wire has a uniform area A, but its material changes gradually along its length, so the resistivity is a function ρ(x) of the distance x from one end. You cannot use a single ρ in ρℓ/A.

Split the wire into thin slices of thickness dx. Each slice is short enough that ρ is constant across it, so its resistance is

dR = ρ(x) dx / A

The same current passes through every slice in turn, and the potential differences across them add. So the resistances of the slices add too:

R = (1/A) ∫₀ˡ ρ(x) dx

Check: if ρ is constant, the integral gives ρℓ and you get back R = ρℓ/A. Because the current is the same in every slice, the potential difference is not shared evenly: more of it falls across the slices with higher resistivity. The field is larger there too, since E = ρJ and J is the same everywhere.

Ohm’s law

The definition R = ΔV/I works at any single moment for any element. Ohm’s law is a stronger statement about some materials:

ΔV = IR, with R constant for all currents

• Elements that obey it are called ohmic. Double the potential difference and the current doubles.
• Elements that do not are non-ohmic. A filament lamp is the usual example. As the current rises, the filament heats up, its resistivity rises, and so does its resistance.

Reading resistance from a graph. For an ohmic element, a graph of current I (vertical) against potential difference ΔV (horizontal) is a straight line through the origin. Its slope is ΔI/ΔV = 1/R, so R = 1/slope. For a non-ohmic element, R at a given point is ΔV/I at that point, read from the coordinates; it is not 1 ÷ (the slope of the tangent).

Worked example 1: designing a heating coil

Question. A heating coil is made from 2.5 m of alloy wire with diameter 0.50 mm. The alloy has resistivity 1.1 × 10⁻⁶ Ω·m. Treat the coil as ohmic. (a) Find its resistance. (b) Find the current when 7.0 V is applied, and check the field in the wire using E = ρJ. (c) The same piece of wire is stretched evenly to 5.0 m. Find the new resistance. (d) Compare with a copper wire of the same size.

1. Convert the diameter and find the area: r = 0.25 mm = 2.5 × 10⁻⁴ m, so A = π(2.5 × 10⁻⁴ m)² = 1.96 × 10⁻⁷ m².
2. (a) R = ρℓ/A = (1.1 × 10⁻⁶ Ω·m)(2.5 m) ÷ (1.96 × 10⁻⁷ m²) = 14 Ω (14.0 Ω).
3. (b) I = ΔV/R = 7.0 V ÷ 14.0 Ω = 0.50 A. Field from the potential difference: E = ΔV/ℓ = 7.0 V ÷ 2.5 m = 2.8 V/m. Field from the current density: J = I/A = 2.55 × 10⁶ A/m², so ρJ = (1.1 × 10⁻⁶)(2.55 × 10⁶) = 2.8 V/m. The two agree.
4. (c) Twice the length at constant volume means half the area, so R increases by 2 × 2 = 4: 56 Ω.
5. (d) Copper: R = (1.7 × 10⁻⁸)(2.5) ÷ (1.96 × 10⁻⁷) = 0.22 Ω, about 65 times smaller. The ratio is just the ratio of the resistivities, because the shape is the same.

Check. Units: Ω·m × m ÷ m² = Ω. Copper connecting wires have much less resistance than the coil, which is why we usually treat them as ideal (Topic 11.2).

Worked example 2: resistance from plotted data

Question. A student connects a coil of wire to a variable supply and records the potential difference across it and the current in it. The wire is 1.20 m long with diameter 0.30 mm. (a) Plot I against ΔV and decide whether the coil is ohmic over this range. (b) Find its resistance from the graph. (c) Find the resistivity of the wire.

ΔV (V)	0.50	1.00	1.50	2.00	2.50	3.00
I (A)	0.082	0.158	0.247	0.322	0.405	0.480

(a) Plot and decide. Choose scales that use most of the grid: ΔV from 0 to 3.5 V and I from 0 to 0.6 A, with units on each axis (Figure 2). The points lie close to a straight line through the origin, so the coil behaves as ohmic over this range. The small scatter is measurement uncertainty. Calculating ΔV/I for each row gives values between 6.07 Ω and 6.33 Ω with no steady trend, which supports this.

(b) Resistance from the slope. Draw one best-fit line, not a line through the first and last points. Read the slope from a large triangle: the line passes through (0, 0) and (3.00 V, 0.483 A), so slope = 0.483 A ÷ 3.00 V = 0.161 A/V. Then R = 1/slope = 1 ÷ 0.161 A/V = 6.2 Ω.

(c) Resistivity. A = π(1.5 × 10⁻⁴ m)² = 7.07 × 10⁻⁸ m². ρ = RA/ℓ = (6.21 Ω)(7.07 × 10⁻⁸ m²) ÷ 1.20 m = 3.7 × 10⁻⁷ Ω·m.

Interpretation. A best-fit line uses all six readings at once, so random errors partly cancel. If the points had curved away from the line at higher potential differences, that would suggest the wire was warming up and its resistance rising.

Worked example 3: a wire with graded resistivity

Question. A rod of length L = 0.80 m and uniform area A = 2.0 × 10⁻⁷ m² is made of a material whose resistivity rises steadily from one end: ρ(x) = ρ₀(1 + 2x/L), with ρ₀ = 4.0 × 10⁻⁷ Ω·m. (a) Derive its resistance. (b) A potential difference of 1.6 V is applied across it. How is this shared between the two halves? (c) Compare the field at the two ends.

(a) Resistance. Add slices in series:

R = (1/A) ∫₀ᴸ ρ₀(1 + 2x/L) dx = (ρ₀/A)[x + x²/L]₀ᴸ = (ρ₀/A)(L + L) = 2ρ₀L/A

With numbers: R = 2(4.0 × 10⁻⁷ Ω·m)(0.80 m) ÷ (2.0 × 10⁻⁷ m²) = 3.2 Ω.

(b) Sharing the potential difference. The current is I = 1.6 V ÷ 3.2 Ω = 0.50 A in every slice. The first half has resistance (ρ₀/A)[x + x²/L] from 0 to L/2, which is (ρ₀/A)(L/2 + L/4) = (3/4)(ρ₀L/A). That is 3/8 of the total. So the first half takes (3/8)(1.6 V) = 0.60 V and the second half takes 1.0 V.

(c) Fields. J = I/A = 2.5 × 10⁶ A/m² everywhere. At x = 0, E = ρ₀J = 1.0 V/m. At x = L, ρ = 3ρ₀, so E = 3.0 V/m.

Check. If ρ were constant at ρ₀, R would be ρ₀L/A = 1.6 Ω; the graded rod has twice that, which makes sense because its average resistivity is 2ρ₀. The average field is 1.6 V ÷ 0.80 m = 2.0 V/m, between the two end values.

Common misconceptions

• “R = ΔV/I is Ohm’s law.” It is the definition of resistance. Ohm’s law says R stays constant as the current changes.
• “Resistance and resistivity are the same thing.” Resistivity belongs to the material; resistance belongs to the object and also depends on ℓ and A.
• Using diameter in place of radius. A = πr² = πd²/4. Forgetting to halve d makes A four times too big and R four times too small.
• “Stretching to twice the length doubles R.” The area also halves, so R becomes four times larger.
• Taking the slope of an I–ΔV graph as R. With I on the vertical axis, the slope is 1/R.
• Using a tangent slope for a non-ohmic element. R at a point is ΔV/I at that point.
• “A thicker wire has more resistance because there is more metal.” More area means more paths for charge, so less resistance.
• Using one value of ρ when ρ varies. Integrate ρ(x) dx and divide by A.

Where this leads

Next, Topic 11.4, Electric Power, uses ΔV = IR to find how fast resistors convert electrical energy, and why some bulbs glow more brightly than others. Topic 11.5 combines resistors in series and parallel. Go back to Topic 11.2, Simple Circuits if loops and symbols feel shaky. Practise now with the practice questions, then use the revision notes and the topic checklist.