Physics 9702 · syllabus 21
Cambridge A Level Physics 9702 — Alternating currents topical past papers
Sinusoidal alternating current and voltage: period, peak and r.m.s. values, mean power, rectification with diodes and smoothing with a capacitor.
- Questions
- 59
- Years
- 2009–2026
- With mark scheme
- 100%
- Questions 2022–2026
- 20
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What Alternating currents covers
An alternating current reverses its direction periodically. For a sinusoidal supply the current or voltage is described by x = x₀ sin ωt, with peak value x₀, period T and frequency f = 1/T = ω/2π. Because the power dissipated in a resistor depends on the current squared, it is never negative, and over a cycle it averages to half its peak value.
The root-mean-square value of an alternating current is the steady direct current that would dissipate the same mean power in a resistor; for a sine wave, the r.m.s. current is I₀/√2 and the r.m.s. voltage V₀/√2. Mains voltages are quoted as r.m.s. values, so a supply described as 230 V has a peak of about 325 V.
Rectification turns a.c. into a current in one direction. A single diode gives half-wave rectification, passing alternate half-cycles only; four diodes in a bridge give full-wave rectification, using both halves of every cycle. A capacitor across the load smooths the output by discharging through it between peaks, and the larger the product of capacitance and load resistance, the smaller the ripple. This is A Level content, examined on Paper 4.
In the syllabus
The headings this topic has in the Cambridge International 2025–2027 syllabus, as the board prints them.
- 21 Alternating currents
- 21.1 Characteristics of alternating currents
- 21.2 Rectification and smoothing
What you should be able to do
- Read period, frequency and peak value from a graph or from an equation such as V = V₀ sin ωt
- Explain what an r.m.s. value means and convert between peak and r.m.s. values
- Calculate the mean power in a resistive load as half the peak power
- Sketch the output of half-wave and full-wave rectifiers and explain which diodes conduct
- Explain how a smoothing capacitor works and how capacitance and load resistance affect the ripple
Key ideas
- x = x₀ sin ωt, with ω = 2πf
- For a sinusoidal supply: r.m.s. current = I₀/√2 and r.m.s. voltage = V₀/√2
- Mean power in a resistor = ½ × peak power = (r.m.s. current)² × R = (r.m.s. voltage)²/R
- Smoothing: between peaks the capacitor discharges through the load with time constant RC; a larger RC gives less ripple
How Alternating currents has been examined
From 2022 to 2026, Alternating currents was on Paper 4 · A Level Structured Questions in 10 of 11 exam series, worth about 6% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 4 · A Level Structured Questions | 20 | 6% | 10 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: 10.
Filed in Momentum under: Alternating Currents.
A worked example
A supply p.d. V = 170 sin(120πt), with V in volts and t in seconds, is connected across a 50 Ω resistor. Find the frequency of the supply, the r.m.s. p.d. and the mean power dissipated.
- Comparing with V = V₀ sin ωt: V₀ = 170 V and ω = 120π rad s⁻¹
- f = ω/2π = 120π / 2π = 60 Hz
- r.m.s. p.d. = V₀/√2 = 170 / 1.414 = 120 V
- Mean power = (r.m.s. p.d.)²/R = 120.2² / 50 = 289 W
- Check: peak power = 170² / 50 = 578 W, and the mean is half of this
Answer: Frequency 60 Hz; r.m.s. p.d. 120 V; mean power 290 W
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Putting peak values into P = V²/R and calling the result the mean power — that is the peak power, twice the mean
- Dividing by 2 instead of √2 to get an r.m.s. value — the r.m.s. value of a sine wave is about 0.71 of the peak, not 0.5
- Reading the frequency straight from ω — in V = 170 sin(120πt), ω = 120π rad s⁻¹ and f = 60 Hz, not 120π Hz
- Drawing full-wave rectified output with gaps between the humps — full-wave rectification uses both halves of every cycle, so the humps meet
- Saying a smoothing capacitor stores the a.c. — it charges near each peak and supplies current to the load while the rectified p.d. falls
Exam technique
- When sketching smoothed output, draw the discharge between peaks as a gentle fall and the recharge as a steep rise back to the peak
- For bridge rectifier questions, trace the current through two diodes in each half-cycle and show that it passes through the load in the same direction both times
- When defining an r.m.s. value, include 'steady direct current' and 'same mean power in a resistor'
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.