AQA A-Level Physics: Electricity (7408)

This study guide teaches section 3.5 Electricity of the AQA AS and A-level Physics specification (7407/7408, version 1.4, AS and A-level exams June 2016 onwards). It covers every sub-section of 3.5.1 Current electricity, from 3.5.1.1 to 3.5.1.6, including Required practicals 5 and 6. Electricity is AS and A-level content, not A-level only: at A-level it is assessed in Paper 1 (sections 1 to 5 and 6.1) and is assumed knowledge in Paper 2. Sections 3.9 to 3.13 are the options, of which you study one for Paper 3 Section B; Electricity is not an option.

Follow it with the revision notes and the practice questions. The course hub lists every topic, the printable checklist tracks each outcome, and the free diagnostics show where to start.

What this topic covers

Specification	What you must be able to do
3.5.1.1 Basics of electricity	Current as rate of flow of charge, I = ΔQ/Δt; pd as work done per unit charge, V = W/Q; R = V/I
3.5.1.2 Current–voltage characteristics	I–V graphs for an ohmic conductor, semiconductor diode and filament lamp; Ohm’s law; ideal meters
3.5.1.3 Resistivity	ρ = RA/L; effect of temperature on metals and ntc thermistors; superconductivity; Required practical 5
3.5.1.4 Circuits	Series and parallel resistors; E = IVt; P = IV = I²R = V²/R; cells in series and identical cells in parallel; conservation of charge and energy
3.5.1.5 Potential divider	Constant or variable pd from a supply; with variable resistors, thermistors and LDRs
3.5.1.6 Emf and internal resistance	ε = E/Q, ε = I(R + r); terminal pd; Required practical 6

Not required: the potentiometer as a measuring instrument, positive temperature coefficient thermistors, and the critical magnetic field of a superconductor.

3.5.1.1 Basics of electricity

Current is the rate of flow of charge: I = ΔQ/Δt. One ampere is one coulomb per second. In a metal the charge carriers are free electrons, each of charge magnitude 1.60 × 10⁻¹⁹ C.

Potential difference is the work done (energy transferred) per unit charge: V = W/Q. One volt is one joule per coulomb.

Resistance is defined as R = V/I. One ohm is one volt per ampere. This definition holds for any component at any instant, even when R is not constant.

Worked example 1

A charge of 36 C flows through a resistor in 2.0 minutes. The pd across it is 12 V. Find the current, the number of electrons passing per second, the energy transferred and the resistance.

I = ΔQ/Δt = 36 / 120 = 0.30 A
electrons per second = 0.30 / 1.60 × 10⁻¹⁹ = 1.9 × 10¹⁸
W = QV = 36 × 12 = 432 J
R = V/I = 12 / 0.30 = 40 Ω

3.5.1.2 Current–voltage characteristics

Questions can put either I or V on the horizontal axis, so check the axes first.

• Ohmic conductor (for example a metal wire at constant temperature): a straight line through the origin. I ∝ V, so R is constant.
• Filament lamp: a straight line near the origin, then a curve. With I on the vertical axis, the gradient decreases as V rises. A larger current heats the filament, the metal ions vibrate more, the charge carriers collide with them more often, and the resistance rises. The graph is symmetrical for reversed pd.
• Semiconductor diode: in forward bias almost no current flows until the pd reaches a small threshold pd; above it the current rises steeply. In reverse bias the resistance is very high and the current is close to zero.

Ohm’s law is a special case: I ∝ V under constant physical conditions, such as constant temperature. A filament lamp does not obey it because its temperature changes.

Unless a question says otherwise, treat ammeters as ideal (zero resistance) and voltmeters as ideal (infinite resistance).

Worked example 2

A filament lamp carries 0.20 A at 1.0 V and 0.50 A at 6.0 V. Find its resistance at each point.

at 1.0 V: R = 1.0 / 0.20 = 5.0 Ω
at 6.0 V: R = 6.0 / 0.50 = 12 Ω

The hotter filament has more than double the resistance. Use R = V/I at the point, not the gradient.

3.5.1.3 Resistivity

Resistance depends on the material and the shape: R = ρL/A. Resistivity is

ρ = RA/L, unit Ω m

where L is the length and A the cross-sectional area. For a wire of diameter d, A = πd²/4.

Worked example 3

A wire 0.800 m long with diameter 0.32 mm has a resistance of 4.9 Ω. Find the resistivity of the material.

A = π × (0.32 × 10⁻³)² / 4 = 8.04 × 10⁻⁸ m²
ρ = RA/L = 4.9 × 8.04 × 10⁻⁸ / 0.800 = 4.9 × 10⁻⁷ Ω m

A quick ratio check: a wire of the same material with twice the length and half the diameter has a quarter of the area, so R rises by a factor of 2 × 4 = 8.

Temperature effects

• Metal conductors: resistance increases as temperature rises, because the ions vibrate more and impede the charge carriers.
• ntc thermistors (only negative temperature coefficient thermistors are assessed): resistance decreases as temperature rises, because more charge carriers are released.

Thermistors are used as temperature sensors. A resistance–temperature graph for an ntc thermistor is a curve falling steeply at low temperature and levelling off at high temperature. You can investigate it by heating the thermistor in a water bath and measuring R at a range of temperatures.

Superconductivity

A superconductor is a material that has zero resistivity at and below a critical temperature, which depends on the material. Applications:

• producing strong magnetic fields (large currents in superconducting coils)
• reducing energy loss in the transmission of electric power.

Required practical 5: resistivity of a wire

1. Measure the diameter with a micrometer at several points and in different orientations; take the mean and find A.
2. Clamp the wire along a metre rule. Connect it in series with an ammeter and connect a voltmeter across the section between two crocodile clips.
3. For a range of lengths L, record V and I and calculate R = V/I.
4. Plot R against L. The gradient is ρ/A, so ρ = gradient × A.
5. Use small currents and switch off between readings so the wire does not heat up. A non-zero intercept suggests contact resistance; the gradient method is unaffected.

Because d is small its percentage uncertainty is often the largest, and it doubles in A. See the measurements revision notes for combining uncertainties.

3.5.1.4 Circuits

Arrangement	Current	Pd	Resistance
Series	Same through each	Shared: V = V₁ + V₂ + …	R_T = R₁ + R₂ + R₃ + …
Parallel	Shared: I = I₁ + I₂ + …	Same across each branch	1/R_T = 1/R₁ + 1/R₂ + 1/R₃ + …

These rules follow from:

• Conservation of charge: the total current into a junction equals the total current out.
• Conservation of energy: around any closed loop, the sum of the emfs equals the sum of the pds.

Energy and power: E = IVt, and P = IV = I²R = V²/R.

Cells: in series, emfs add (taking direction into account) and internal resistances add. For n identical cells in parallel, the emf equals that of one cell and the internal resistance is r/n. Three cells of emf 1.5 V and internal resistance 0.30 Ω give 4.5 V and 0.90 Ω in series, or 1.5 V and 0.10 Ω in parallel.

Worked example 4

An 18 V supply with negligible internal resistance is connected to a 6.0 Ω resistor in series with a parallel pair of 12 Ω and 4.0 Ω. Find the current from the supply, the current in each parallel resistor, and the power in the 4.0 Ω resistor.

parallel pair: 1/R = 1/12 + 1/4.0  →  R = 3.0 Ω
total: R_T = 6.0 + 3.0 = 9.0 Ω
I = 18 / 9.0 = 2.0 A
pd across 6.0 Ω = 2.0 × 6.0 = 12 V, so pd across pair = 18 − 12 = 6.0 V
I(12 Ω) = 6.0 / 12 = 0.50 A ;  I(4.0 Ω) = 6.0 / 4.0 = 1.5 A   (sum 2.0 A ✓)
P(4.0 Ω) = V²/R = 6.0² / 4.0 = 9.0 W

The energy transferred in the 6.0 Ω resistor in 5.0 minutes is E = IVt = 2.0 × 12 × 300 = 7200 J.

3.5.1.5 Potential divider

Resistors in series across a supply share its pd in the ratio of their resistances:

V_out = V_in × R₂ / (R₁ + R₂), where V_out is taken across R₂.

Fixed resistors give a constant output. Replacing one with a variable resistor gives a variable output. Replacing one with a thermistor or an LDR (whose resistance falls as light intensity rises) makes the output depend on temperature or light, which is how simple sensing circuits work.

Worked example 5

A 9.0 V supply is connected across a 2.2 kΩ fixed resistor in series with an ntc thermistor. V_out is taken across the fixed resistor. The thermistor’s resistance is 1.8 kΩ at 20 °C and 0.60 kΩ at 60 °C. Find V_out at each temperature.

20 °C: V_out = 9.0 × 2.2 / (2.2 + 1.8) = 4.95 V ≈ 5.0 V
60 °C: V_out = 9.0 × 2.2 / (2.2 + 0.60) = 7.1 V

V_out rises as it warms, because the thermistor takes a smaller share of the pd. Taking V_out across the thermistor reverses this. A load across the output is in parallel with that resistor, lowering the combined resistance and so V_out.

3.5.1.6 Electromotive force and internal resistance

Emf is the energy transferred to electrical form per unit charge passing through the source: ε = E/Q. Real sources have internal resistance r, which transfers some energy inside the source.

ε = I(R + r), or ε = V + Ir

where V = IR is the terminal pd and Ir is the pd across the internal resistance (often called the “lost volts”). The terminal pd falls as the current increases.

Worked example 6

A cell of emf 1.50 V and internal resistance 0.50 Ω is connected to a 2.5 Ω resistor.

I = ε / (R + r) = 1.50 / 3.0 = 0.50 A
terminal pd V = IR = 0.50 × 2.5 = 1.25 V
lost volts = Ir = 0.25 V   (1.25 + 0.25 = 1.50 ✓)
power wasted in the cell = I²r = 0.125 W

Required practical 6: emf and internal resistance

Connect the cell in series with an ammeter and a variable resistor, with a voltmeter across the cell. Vary the resistance and record V and I. Rearranging ε = V + Ir gives

V = −rI + ε

so a graph of V against I is a straight line with gradient −r and y-intercept ε. Keep currents small and switch off between readings so r does not change as the cell warms.

With only two readings, for example V = 1.40 V at 0.20 A and V = 1.10 V at 0.80 A: r = (1.40 − 1.10)/(0.80 − 0.20) = 0.50 Ω and ε = 1.40 + 0.20 × 0.50 = 1.50 V.

Common errors

• Taking resistance from the gradient of a curved I–V graph.
• Forgetting mm → m before squaring in A = πd²/4.
• Adding resistances in parallel, or forgetting to take the reciprocal at the end.
• Saying an ntc thermistor’s resistance rises with temperature.
• Taking ε as the reading on a voltmeter across a cell that is supplying current.
• Giving the gradient of V against I as +r rather than −r.

Next steps

Use the revision notes, then the practice questions. See the exam preparation guide and the free diagnostics.

Official syllabus

AQA AS and A-level Physics specification (7407/7408), version 1.4, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.5 Electricity.