The shape of the lesson
By the end of the lesson, learners can
- Current and power: The top trace is the current, i = I₀ sinωt.
- Why the √2 appears: The r.m.s. value is built from the power, not the current.
- Sinusoidal a.c.: i = I₀ sinωt, ω = 2πf, T = 1/f.
- R.M.S.: divide peak by √2; mean power = Iₕₘₛ²R = ½ peak power.
- Rectification: half-wave (1 diode), full-wave (4-diode bridge); smoothing capacitor in parallel with the load reduces ripple, full-wave ripple at twice the supply frequency.
One picture every idea returns to
A sinusoidal current is i = I₀ sin ωt, with peak value I₀, angular frequency ω = 2πf and period T = 1/f. The power i²R is always positive and averages to half its peak, so the mean power is ½ the peak power. The r.m.s. value is then I₀/√2, and likewise V₀/√2 for voltage.
Forty-five minutes, phase by phase
| Time | Phase | What happens in the room | Grouping |
|---|---|---|---|
| 0 to 5 min | Hook: predict | Open with the question in the title, The current that turns around.. Alternating current rises and falls and reverses many times a second. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Current and power | The top trace is the current, i = I₀ sinωt. Project tab 1 of the lab (Characteristics of a.c.) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Why the √2 appears | The r.m.s. value is built from the power, not the current. | Mix-Pair-Share |
| 30 to 43 min | Check and plenary | The quick-check questions from the student lesson, answered on mini whiteboards one at a time; re-teach the one most get wrong, then a random call on the last. | Individual, then whole class |
| 43 to 45 min | Exit and homework | Set the worksheet below (7 questions, 65 marks) from the topic practice set. Learners self-assess against the objectives before they leave. | Individual |
Protect the parts that carry the learning
The sequence carries about fifty minutes of material, so the lab is driven by one student while the class predicts rather than explored freely. Times are a guide.
Protected: the check at 30 minutes. It is the only individual read on whether the key idea landed, and it is not cut.
Cut first if the build overruns: the last check in Why the √2 appears, never the check or the plenary.
In a 60 minute block: run worksheet questions 1 to 3 in class as a Rally Coach set and model both the hardest worked example and its check live.
Mix-Pair-Share
Students mix around the room until you call pair; they pair with the nearest classmate, discuss the prompt, then mix again. Three prompts of rising difficulty, three different partners.
The full structure, with its variations and the other twelve, is on the cooperative-learning page.
What to head off, and how
| Trap learners fall into | Teaching move that pre-empts it |
|---|---|
| The simple average of a sinusoidal current over a whole cycle is zero, which is why we use the r.m.s. value instead. | The factor 1/√2 applies only to sinusoidal signals. The r.m.s. current is defined through power: it is the steady d.c. that dissipates the same mean power, so mean power = ½ peak power, not the peak power. |
Characteristics of alternating current: topic worksheet
7 questions, 65 marks. Written in the style of the papers; no past paper question is reproduced. These questions cover the whole topic, so sibling lessons share them.
A sinusoidal current is described by i = I₀ sinωt, with a peak value of 8.0 A and a frequency of 50 Hz.
(a) State what is meant by the peak value. [1]
(b) Calculate the r.m.s. value of the current. [2]
(c) Calculate the period of the current. [1]
(d) Calculate the angular frequency ω. [2]
(e) Write the equation for the current with these numerical values. [1]
(f) Calculate the current at t = 5.0 ms. [2]
(g) State the time at which the current first returns to zero after t = 0. [1]
A heating element of resistance 46 Ω is connected to the 230 V (r.m.s.), 50 Hz mains supply.
(a) Calculate the peak voltage of the supply. [2]
(b) Calculate the r.m.s. current in the element. [2]
(c) Calculate the mean power dissipated. [2]
(d) Calculate the peak power dissipated. [2]
(e) State the ratio of mean power to peak power. [1]
(f) State why r.m.s. values, rather than the simple average, are quoted for a.c. [1]
A resistor of resistance 12 Ω carries a sinusoidal current of peak value 3.0 A.
(a) Calculate the r.m.s. current. [2]
(b) Calculate the mean power dissipated in the resistor. [2]
(c) Calculate the peak (maximum instantaneous) power. [2]
(d) State the relationship between the mean power and the peak power. [1]
(e) State the value of the steady direct current that would dissipate the same mean power. [2]
Fig. 4.1 shows how an alternating current varies with time.
(a) State the peak value of the current. [1]
(b) Read the period from the graph and hence calculate the frequency. [2]
(c) Calculate the angular frequency ω. [2]
(d) Write the equation of the current in the form i = I₀ sinωt. [1]
(e) Calculate the r.m.s. value of the current. [2]
(f) State the time at which the current first reaches its peak. [1]
An a.c. supply of r.m.s. value 12 V and frequency 50 Hz is connected to a single-diode (half-wave) rectifier and a load. Parts (a) and (b) use the a.c. characteristics; the rest use rectification.
(a) Calculate the peak voltage of the supply. [2]
(b) Stating any assumption, give the peak output voltage across the load. [1]
(c) For how long in each cycle does the diode conduct? [2]
(d) A smoothing capacitor is now connected across the load. Describe the output. [2]
(e) State the ripple frequency of this smoothed half-wave output. [1]
A full-wave rectifier supplied from 50 Hz a.c. feeds a load of resistance 2.2 kΩ with a smoothing capacitor of 100 μF in parallel.
(a) Calculate the time constant of the load and capacitor. [2]
(b) State the frequency and period of the ripple before smoothing. [2]
(c) By comparing the time constant with the ripple period, explain whether the smoothing is effective. [2]
(d) State and explain the effect of replacing the capacitor with a 470 μF one. [2]
This question follows a power supply from end to end. A transformer secondary provides 9.0 V (r.m.s.) at 50 Hz to a bridge rectifier, whose output is smoothed and connected to a load.
(a) Calculate the peak voltage at the secondary. [2]
(b) State the type of rectification and the number of diodes used. [2]
(c) State the ripple frequency of the rectified output. [2]
(d) A smoothing capacitor is added across the load. Describe the output voltage. [2]
(e) The load is changed to draw a larger current (a smaller resistance). State and explain the effect on the ripple. [3]
Answer key and marking notes
- (a)The maximum value of the current (the amplitude) ✓
- (b)Iₕₘₛ = I₀/√2 = 8.0/1.414 ✓ = 5.7 A ✓
- (c)T = 1/f = 1/50 = 0.020 s (20 ms) ✓
- (d)ω = 2πf = 2π × 50 ✓ = 314 rad s⁻¹ ✓
- (e)i = 8.0 sin(314t) ✓
- (f)ωt = 314 × 0.0050 = 1.57 rad (½π) ✓; i = 8.0 sin(½π) = 8.0 A ✓
- (g)At ωt = π, t = T/2 = 10 ms ✓
- (a)V₀ = Vₕₘₛ × √2 = 230 × 1.414 ✓ = 325 V ✓
- (b)Iₕₘₛ = Vₕₘₛ/R = 230/46 ✓ = 5.0 A ✓
- (c)Mean power = Vₕₘₛ²/R = 230²/46 ✓ = 1150 W (1.15 kW) ✓
- (d)Peak power = V₀²/R = 325²/46 ✓ = 2300 W ✓
- (e)Mean : peak = 1 : 2 (mean is half the peak) ✓
- (f)The simple average of a sinusoid over a cycle is zero, so it is useless; the r.m.s. gives the equivalent steady (d.c.) value for power ✓
- (a)Iₕₘₛ = 3.0/√2 ✓ = 2.1 A ✓
- (b)Mean power = Iₕₘₛ²R = 2.12² × 12 ✓ = 54 W (= ½ × 3.0² × 12) ✓
- (c)Peak power = I₀²R = 3.0² × 12 ✓ = 108 W ✓
- (d)The mean power is half the peak power ✓
- (e)2.1 A ✓, which is the r.m.s. current (the d.c. that gives the same mean power) ✓
- (a)2.0 A ✓
- (b)T = 40 ms = 0.040 s ✓; f = 1/T = 25 Hz ✓
- (c)ω = 2π/T = 2π/0.040 ✓ = 157 rad s⁻¹ ✓
- (d)i = 2.0 sin(157t) ✓
- (e)Iₕₘₛ = 2.0/√2 ✓ = 1.4 A ✓
- (f)At a quarter period, t = 10 ms ✓
- (a)V₀ = 12 × √2 ✓ = 17 V ✓
- (b)About 17 V ✓, assuming the diode has no voltage drop (ideal diode)
- (c)One half of the 20 ms cycle, so 10 ms ✓✓
- (d)A nearly steady d.c. near the peak voltage, with a small ripple as the capacitor discharges between peaks ✓✓
- (e)50 Hz (one charging pulse per cycle for half-wave) ✓
- (a)τ = RC = 2200 × 100 × 10⁻⁶ ✓ = 0.22 s ✓
- (b)Full-wave ripple frequency = 100 Hz ✓; period = 1/100 = 10 ms ✓
- (c)τ (220 ms) is much greater than the ripple period (10 ms) ✓, so the capacitor barely discharges between peaks: the smoothing is effective and the ripple is small ✓
- (d)The ripple is smaller still ✓; a larger C gives a longer time constant, so even less discharge between peaks ✓
- (a)V₀ = 9.0 × √2 ✓ = 12.7 V ✓
- (b)Full-wave rectification ✓, using four diodes (a bridge) ✓
- (c)Twice the supply frequency = 100 Hz ✓✓
- (d)A nearly steady d.c. close to the peak (about 12.7 V) with a small ripple ✓✓
- (e)The ripple becomes larger ✓; the smaller load resistance gives a shorter time constant RC ✓, so the capacitor discharges more between peaks ✓
The lesson's own checks as a quiz
The quick-check and exam-style questions from the student lesson, printable with the key.
The r.m.s. value of a sinusoidal current of peak value I₀ is:
- AI₀ ÷ √2
- BI₀ ÷ 2
- C√2 × I₀
- Dequal to I₀
For a sinusoidal supply, the mean power delivered to a resistor is:
- Ahalf the peak power
- Bequal to the peak power
- Ctwice the peak power
- Dzero on average
The UK mains has an r.m.s. voltage of 230 V. The peak voltage is closest to:
- A325 V
- B230 V
- C163 V
- D460 V
A sinusoidal supply has a frequency of 50 Hz. Its period is:
- A20 ms
- B50 ms
- C2.0 ms
- D100 ms
Quiz key
For a sinusoidal current the r.m.s. value is the peak divided by the square root of two, I₀/√2.
The power i²R varies as sin²ωt, whose mean over a cycle is ½, so the mean power is half the peak power.
V₀ = Vₕₘₛ × √2 = 230 × 1.41 ≈ 325 V.
T = 1/f = 1/50 = 0.020 s = 20 ms.
What to project
The student lesson embeds these tabs. Open them full screen on the projector; every slider has a drag handle on the stage.
Slides for this lesson
A projectable deck for every pack is on the way. Until then, present the student lesson full screen: its sections, figure and lab tabs follow the same order as the plan above.