12 min readConcept sheet1-mark MCQ, 2-mark definitions and ratio numericals, and a 3-mark derivation of ρ = m/(ne²τ); also a NEET favourite

Ohm's law, resistivity and conductivity

At constant temperature, V = IR. Resistance R = ρl/A depends on the wire's shape, but resistivity ρ = m/(ne²τ) depends only on the material and temperature.

Ohm's lawV = IR, when the temperature and other physical conditions stay the same; R in ohm (Ω), 1 Ω = 1 V A⁻¹

Resistance and resistivityR = ρl/A; l = length, A = area of cross-section, ρ = resistivity in Ω m

Conductivityσ = 1/ρ = ne²τ/m; SI unit S m⁻¹ (siemens per metre), the same as Ω⁻¹ m⁻¹

Ohm's law in vector formj = σE, or E = ρj; j = current density (A m⁻²), E = electric field

Ohm's law and what resistance depends on

Ohm's law: the current through a conductor is directly proportional to the potential difference across it, if the temperature and other physical conditions stay the same. So V ∝ I, or V = IR. The constant R is the resistance. It depends on the material, the dimensions and the temperature of the conductor. Join two identical slabs end to end: the length doubles and R doubles, so R ∝ l. Put them side by side: the area doubles and R halves, so R ∝ 1/A. Together, R = ρl/A. The constant ρ is the resistivity: the resistance of a piece of the material with unit length and unit area of cross-section. Its reciprocal is the conductivity σ = 1/ρ. Resistors in series and parallel are not asked as theory in your syllabus, but almost every numerical uses the rules in the table below.

Derivation: Ohm's law and ρ from the drift of electrons

Start from the results of the previous topic: v_d = eEτ/m (in magnitude) and I = neAv_d.

1
Put v_d into I: I = neA(eEτ/m) = (ne²τ/m)AE.
2
Divide by A. Current density j = I/A = (ne²τ/m)E.
3
Write σ = ne²τ/m. Then j = σE: this is Ohm's law in vector form.
4
For a wire of length l with potential difference V across it: E = V/l and j = I/A. So I/A = σV/l.
5
Rearrange: V = (l/(σA))I = (ρl/A)I = IR, with R = ρl/A.
6★
So, ρ = 1/σ = m/(ne²τ). Resistivity depends on n and τ, which belong to the material and its temperature, not to the size of the wire.

★ key step

Resistivity at room temperature (orders of magnitude)

Class	Examples	Resistivity ρ (Ω m)	Typical use
Metals (conductors)	Silver, copper, aluminium	about 10⁻⁸	Connecting wires, cables
Alloys	Nichrome, manganin, constantan	about 10⁻⁷ to 10⁻⁶	Heater coils, standard resistors
Semiconductors	Carbon (graphite), germanium, silicon	about 10⁻⁵ to 10³	Diodes, chips, sensors
Insulators	Glass, rubber, quartz	10¹⁰ or more	Insulating covers and supports

Resistors in series and parallel (used in every numerical)

Feature	Series	Parallel
Same for each resistor	Current I	Potential difference V
Shared out	V = V₁ + V₂ + ...	I = I₁ + I₂ + ...
Equivalent resistance	Rs = R₁ + R₂ + ...	1/Rp = 1/R₁ + 1/R₂ + ...
Two resistors	R₁ + R₂	R₁R₂/(R₁ + R₂)
n equal resistors R	nR	R/n
Result compared with the parts	More than the largest	Less than the smallest

Worked example: resistivity and a stretched wire

Problem

A wire of length 2.0 m and area of cross-section 0.40 mm² has a resistance of 5.0 Ω. (a) Find the resistivity and conductivity of its material. (b) The wire is stretched uniformly to double its length. Find its new resistance.

Working

Given: l = 2.0 m, A = 0.40 mm² = 0.40 × 10⁻⁶ m², R = 5.0 Ω.
(a) Formula: R = ρl/A, so ρ = RA/l.
ρ = (5.0 Ω × 0.40 × 10⁻⁶ m²) ÷ 2.0 m = 1.0 × 10⁻⁶ Ω m.
σ = 1/ρ = 1 ÷ (1.0 × 10⁻⁶ Ω m) = 1.0 × 10⁶ S m⁻¹.
(b) Stretching keeps the volume Al fixed. The new length is 2l, so the new area is A/2.
R' = ρ(2l)/(A/2) = 4ρl/A = 4R = 4 × 5.0 Ω = 20 Ω.

Answer

(a) ρ = 1.0 × 10⁻⁶ Ω m (close to nichrome), σ = 1.0 × 10⁶ S m⁻¹; (b) 20 Ω. The resistivity does not change on stretching.

Practical: resistivity of a wire from a V-I graph

Aim: to find the resistivity of two or three wires by plotting potential difference against current.

1
Measure a length l of the wire (for example 50 cm) with a metre scale. Measure its diameter d with a screw gauge at several places and take the mean.
2
Join the wire in series with a battery, a plug key, a rheostat and an ammeter. Join a voltmeter across the wire. Draw this circuit, marking + and − on both meters.
3
Close the key. Change the rheostat to get five or six pairs of readings of I and V. Keep the current small so the wire does not heat up.
4
Plot V on the y-axis against I on the x-axis. Observation: the points lie on a straight line through the origin.
5
Find the slope of the line: R = ΔV/ΔI.
6
Calculate ρ = RA/l = πd²R/(4l). Repeat for the other wires. Result: for wires of the same l and d, the steeper line belongs to the material of higher resistivity.

In real life

House wiring uses copper or aluminium because their resistivity is very low. The heating coil of an electric iron or kettle uses nichrome, whose resistivity is about 60 times that of copper.

Exam angle

2 marks: “A wire of resistance R is stretched uniformly to three times its length. Find its new resistance. Does its resistivity change?” The volume is fixed, so the area becomes A/3. R' = ρ(3l)/(A/3) = 9R. The resistivity does not change, as the material and temperature are the same.

3 marks: “Using the idea of drift velocity, derive Ohm's law and show that the resistivity is ρ = m/(ne²τ).” I = neAv_d with v_d = eEτ/m gives j = (ne²τ/m)E = σE. Use E = V/l and j = I/A to get V = (ρl/A)I = IR, with ρ = m/(ne²τ).

1 marks: “Write the SI unit of conductivity.” S m⁻¹ (siemens per metre), the same as Ω⁻¹ m⁻¹.

Remember it

R = ρl/A changes with shape; ρ = m/(ne²τ) changes only with material and temperature. Stretch a wire to n times its length and R becomes n² times.