Physics 9702 · syllabus 9
Cambridge A Level Physics 9702 — Electricity topical past papers
Electric current as a flow of charge, potential difference and power, resistance and resistivity, and how lamps, diodes, LDRs and thermistors behave.
- Questions
- 514
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 184
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Electricity covers
Electric current is a flow of charge carriers, and charge comes in whole multiples of the elementary charge e. Current is the rate of flow of charge, I = Q/t, and in a conductor it depends on the number density of carriers, their charge, their drift speed and the cross-sectional area: I = Anvq. Potential difference is the energy transferred per unit charge as charge passes through a component, which leads to the three forms of the power equation.
Resistance is the ratio V/I. A metal wire at constant temperature obeys Ohm's law, but a filament lamp's resistance rises as it heats up, and a semiconductor diode conducts in one direction only, once a small forward voltage is reached. Resistivity is a property of the material, and R = ρL/A turns it into the resistance of a particular wire. A light-dependent resistor's resistance falls as light intensity rises; a thermistor's falls as its temperature rises.
This topic is AS content, examined on Papers 1 and 2. Its definitions of the coulomb, the volt and the ohm are the vocabulary for D.C. circuits, which follows directly, and for capacitance, magnetic forces on currents and alternating currents at A Level. The equation I = Anvq returns in the Hall voltage, where the same carriers are pushed sideways by a magnetic field.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 9 Electricity
- 9.1 Electric current
- 9.2 Potential difference and power
- 9.3 Resistance and resistivity
What you should be able to do
- Use Q = It and I = Anvq, including comparing drift speeds in wires of different thickness
- Define the volt and use V = W/Q
- Use P = VI, P = I²R and P = V²/R, choosing the form that fits the data
- Sketch and explain the I–V characteristics of a metal wire, a filament lamp and a semiconductor diode
- Use R = ρL/A to find a resistance or a resistivity from a wire's dimensions
- Describe how the resistance of an LDR and of a thermistor changes with light intensity and temperature
Key ideas
- Q = It, with 1 C = 1 A s; charge is quantised in multiples of e = 1.60 × 10⁻¹⁹ C
- I = Anvq, where n is the number density of charge carriers (m⁻³) and v their drift speed
- V = W/Q (1 V = 1 J C⁻¹); P = VI = I²R = V²/R
- R = V/I (Ω); resistivity from R = ρL/A, with ρ in Ω m
How Electricity has been examined
From 2022 to 2026, Electricity was on Paper 2 · AS Level Structured Questions in 11 of 11 exam series, worth about 19% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Multiple Choice | 146 | 11% | 11 of 11 |
| Paper 2 · AS Level Structured Questions | 36 | 19% | 11 of 11 |
| Paper 4 · A Level Structured Questions | 2 | 1% | 2 of 11 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 1.
Filed in Momentum under: Current of Electricity.
A worked example
A nichrome wire is 1.50 m long with a diameter of 0.30 mm, and the resistivity of nichrome is 1.1 × 10⁻⁶ Ω m. Find the wire's resistance and the power it dissipates across a 6.0 V supply of negligible internal resistance.
- Radius r = 0.15 mm = 1.5 × 10⁻⁴ m, so A = πr² = 7.07 × 10⁻⁸ m²
- R = ρL/A = 1.1 × 10⁻⁶ × 1.50 / 7.07 × 10⁻⁸ = 23.3 Ω
- P = V²/R = 6.0² / 23.3 = 1.54 W
Answer: Resistance 23 Ω; power 1.5 W
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Using the diameter as the radius in A = πr² — the area is four times too large, so a resistance comes out four times too small or a resistivity four times too large
- Saying a filament lamp fails Ohm's law because its graph misses the origin — the graph does pass through the origin; the lamp fails because the graph is curved, as its resistance rises with temperature
- Finding the resistance at a point on a curved I–V graph from the tangent's gradient — resistance is V/I at that point, read from the coordinates
- Giving resistivity in Ω m⁻¹ — rearranging R = ρL/A gives ρ = RA/L, so the unit is Ω m
Exam technique
- When explaining why a lamp's resistance rises, give the chain: greater current, higher temperature, more vibration of the lattice ions, more frequent collisions with the charge carriers
- Convert millimetre diameters and mm² areas to metres before substituting, writing the converted value on a line of its own
- On I–V sketches, label the axes, draw both directions of current where relevant (the lamp is symmetrical; the diode passes almost nothing in reverse) and show the diode's forward voltage
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.