Resistance, Resistivity and Ohm's Law: Study Guide (Physics 2 11.3)

What resistance measures

In Topic 11.2 you built circuits from loops. Connect the same battery to a short thick copper wire and then to a thin bulb filament, and you get very different currents. The difference comes from resistance.

Resistance R is a measure of how strongly an object opposes the movement of electric charge through it. For any element in a circuit it is defined by

R = ΔV / I

where ΔV is the potential difference across the element and I is the current through it. The unit is the ohm (Ω), and 1 Ω = 1 V/A. A 15 Ω resistor with 6.0 V across it carries a current of 6.0 V ÷ 15 Ω = 0.40 A.

Why do objects resist at all? In a metal, the free electrons drift through a lattice of atoms that vibrate. The electrons keep colliding with them, which limits the drift. Those collisions also pass energy to the atoms, so the material warms up. A resistor converts electrical energy into thermal energy, which can raise the temperature of the resistor and of whatever is around it. Topic 11.4 deals with the rate of that transfer.

Resistance from shape and material

For a resistor with uniform geometry (the same cross-section all along, like a wire or a rod), the resistance depends on three things:

R = ρL / A

• L is the length in metres. A longer conductor has more resistance, in direct proportion. Charge has further to travel through the material.
• A is the cross-sectional area in m². A wider conductor has less resistance, in inverse proportion. There are more paths side by side for the charge.
• ρ (rho) is the resistivity of the material, in Ω·m.

Resistivity is a fundamental property of a material. It comes from the material’s atomic and molecular structure: how many free charge carriers there are and how easily they move. It quantifies how strongly the material itself opposes the motion of charge. Any sample of copper at a given temperature has the same resistivity, whatever its size or shape. Resistance, in contrast, belongs to one particular object.

Typical values at 20 °C:

Material	Resistivity ρ (Ω·m)	Typical use
Copper	1.68 × 10⁻⁸	connecting wires
Aluminium	2.82 × 10⁻⁸	power lines
Nichrome (nickel–chromium alloy)	1.10 × 10⁻⁶	heating elements

Nichrome’s resistivity is about 65 times that of copper. That is why a toaster element is made of nichrome and the cable that feeds it is copper. A 1.0 m length of copper with a cross-section of 1.0 mm² has R = (1.68 × 10⁻⁸ Ω·m)(1.0 m) ÷ (1.0 × 10⁻⁶ m²) = 0.0168 Ω. That is why connecting wires are usually treated as having no resistance.

Wires are circles. If you are given a diameter d, the cross-sectional area is A = π(d/2)². Doubling the diameter makes the area four times larger, so it cuts R to a quarter.

Resistivity and temperature

The resistivity of a conductor typically increases with temperature. In a hotter metal the atoms vibrate more strongly, so drifting electrons collide with them more often. A tungsten lamp filament has a much larger resistance when it glows than when it is cold.

Ohm’s law and ohmic elements

Ohm’s law relates the current in an element to the potential difference across it and its resistance:

I = ΔV / R

Some materials keep the same resistance whatever the current. These are called ohmic materials. For an ohmic element, doubling ΔV doubles I, so I ∝ ΔV. In the model used in this course, the resistivity of an ohmic material stays constant regardless of temperature. Treat a “resistor” in a question as ohmic unless you are told otherwise.

Elements whose resistance changes with current are non-ohmic. The filament bulb is the standard example: a larger current heats the filament, the resistivity rises and so R rises. R = ΔV/I still gives the resistance at each moment. It just is not a constant.

Reading resistance from a graph

The best way to test whether an element is ohmic is to measure I for several values of ΔV and plot a graph of current I (vertical) against potential difference ΔV (horizontal).

• An ohmic element gives a straight line through the origin.
• Since I = (1/R)ΔV, the slope of the I–ΔV graph is 1/R, so R = 1/slope. A steeper line means a smaller resistance.
• If the axes are swapped (ΔV vertical, I horizontal), the slope is R itself. Always check which quantity is on which axis.
• For a non-ohmic element the graph curves. At any point, R = ΔV/I uses the coordinates of that point (the slope of the line from the origin to the point), not the slope of the curve there.

Measuring resistance in the lab

To find the resistance of an element, put it in a loop with a source and a way of changing ΔV (a variable resistor or an adjustable supply). Place an ammeter in the loop to measure I and a voltmeter across the element to measure ΔV, as in Topic 11.2. Take readings over a range of values, then plot I against ΔV.

To find the resistivity of a material, use a wire of that material with a known diameter. Measure the resistance of different lengths of it (move one contact along the wire). Since R = (ρ/A)L, a graph of R against L should be a straight line through the origin with slope ρ/A. Then ρ = slope × A. Good practice: measure the diameter at several points with a micrometer and average; keep the current small so the wire does not heat up and change its resistivity.

Worked example 1: a heating wire

Question. A heating element is made of 2.4 m of nichrome wire (ρ = 1.10 × 10⁻⁶ Ω·m) with a diameter of 0.50 mm. (a) Find its resistance. (b) Find the current when it is connected across 12 V. Assume the resistivity stays constant.

1. Convert the diameter: d = 0.50 mm = 0.50 × 10⁻³ m, so the radius is 0.25 × 10⁻³ m.
2. Cross-sectional area: A = π(0.25 × 10⁻³ m)² = 1.963 × 10⁻⁷ m².
3. (a) R = ρL/A = (1.10 × 10⁻⁶ Ω·m)(2.4 m) ÷ (1.963 × 10⁻⁷ m²) = 13.45 Ω.
4. (b) I = ΔV/R = 12 V ÷ 13.45 Ω = 0.892 A.

Answer. (a) R = 13 Ω (13.4 Ω). (b) I = 0.89 A.

Check. Units: (Ω·m)(m)/m² = Ω. A common slip is to use the diameter in place of the radius, which makes A four times too big and gives 3.4 Ω. A copper wire of the same size would have only 0.21 Ω. It would barely warm up, which is why heating elements use a high-resistivity alloy.

Worked example 2: predicting a factor of change

Question. Wire A has a resistance of 16 Ω. Wire B is made of the same material. It is half as long as A and has twice the diameter. (a) Find the resistance of B. (b) Each wire in turn is connected across the same battery. How does the current in B compare with the current in A?

1. Write the ratio so that ρ cancels: R_B / R_A = (L_B / L_A) × (A_A / A_B).
2. Length factor: L_B / L_A = 1/2.
3. Area factor: area depends on diameter squared, so A_B / A_A = 2² = 4, and A_A / A_B = 1/4.
4. R_B / R_A = (1/2)(1/4) = 1/8, so R_B = 16 Ω ÷ 8 = 2.0 Ω.
5. (b) Same ΔV, so I = ΔV/R and I_B / I_A = R_A / R_B = 8.

Answer. (a) 2.0 Ω. (b) The current in B is 8 times the current in A.

Check. Both changes (shorter and thicker) lower the resistance, so R_B must be smaller than R_A. The usual error is to treat the diameter like the area and get a factor of 1/4 instead of 1/8.

Worked example 3: resistivity from a graph

Question. A student measures the resistance of different lengths of a wire of unknown alloy. The diameter is 0.30 mm. The results are below. Plot a suitable graph and use it to find the resistivity of the alloy.

L (m)	0.20	0.40	0.60	0.80	1.00
R (Ω)	2.9	5.6	8.6	11.3	14.3

1. R = (ρ/A)L, so plot R (vertical) against L (horizontal). Expect a straight line through the origin with slope ρ/A. Figure 2 shows the plot.
2. Draw the best-fit line. It passes very close to the origin and through about (1.00 m, 14.2 Ω). Slope = 14.2 Ω ÷ 1.00 m = 14.2 Ω/m. (A least-squares fit gives 14.25 Ω/m.)
3. Area: A = π(0.15 × 10⁻³ m)² = 7.07 × 10⁻⁸ m².
4. ρ = slope × A = (14.2 Ω/m)(7.07 × 10⁻⁸ m²) = 1.00 × 10⁻⁶ Ω·m.

Answer. ρ ≈ 1.0 × 10⁻⁶ Ω·m.

Interpretation and check. The result is of the same order as nichrome (1.10 × 10⁻⁶ Ω·m), which is reasonable for a resistance alloy. Using the slope of a best-fit line is better than using one reading, because it averages out random errors in individual measurements. The line passes through the origin, so there is no sign of a systematic error such as contact resistance.

Common misconceptions

• “R = ΔV/I means a bigger voltage makes a bigger resistance.” For an ohmic element, raising ΔV raises I in proportion, and R stays the same. R depends on the material and shape.
• “A thicker wire has more resistance because there is more metal.” More cross-section gives more paths for charge, so R goes down.
• Using the diameter as the radius, or forgetting to square it. A = π(d/2)². Doubling d divides R by 4.
• Treating mm² like mm. 1 mm² = 10⁻⁶ m², not 10⁻³ m².
• “Resistivity and resistance are the same thing.” Resistivity belongs to the material; resistance belongs to the object.
• “The slope of an I–ΔV graph is the resistance.” It is 1/R. A steeper line means less resistance.
• “Every element obeys Ohm’s law.” You can always work out R = ΔV/I at one moment, but only ohmic elements keep that R constant, so only they have I ∝ ΔV. A filament bulb does not.
• “Current is used up in a resistor.” The same current enters and leaves it. What the resistor takes is energy, which becomes thermal energy.

Where this leads

Resistance lets you calculate currents. Topic 11.4 uses it to find the rate at which a resistor converts electrical energy: continue with Topic 11.4, Electric Power. Later in the unit you will combine resistors into compound circuits. First test yourself with the practice questions, then use the revision notes and the topic checklist to consolidate.