The shape of the lesson
By the end of the lesson, learners can
- Read off the power: Set the current and voltage and watch the power, then the energy build up over time.
- From power to energy: Multiply power by time to get the energy transferred.
- Power: P = IV, in watts (one watt is one joule per second).
- Energy: E = IVt = Pt, in joules, with t in seconds.
- Rearrange: I = P ÷ V and V = P ÷ I when the power is given.
One picture every idea returns to
Electrical power is the rate of transferring energy, P = I multiplied by V, measured in watts. Over a time t the energy transferred is E = I multiplied by V multiplied by t.
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, Energy every second.. A bright heater and a dim phone charger both run on electricity, but one pours out energy far faster. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Read off the power | Set the current and voltage and watch the power, then the energy build up over time. Project tab 8 of the lab (Power and energy) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: From power to energy | Multiply power by time to get the energy transferred. Model the worked example Worked example on the board, one line at a time. | Round Table |
| 30 to 36 min | Check | The four Stage 1 questions from the student lesson, answered on mini whiteboards or by show of fingers. Re-teach the one most get wrong. | Individual, then whole class |
| 36 to 43 min | Plenary | The exam-style questions from Stage 2, one at a time: thirty seconds alone, then a random call. The feedback lines in the key give the examiner's wording. | Individual, random call |
| 43 to 45 min | Exit and homework | Set the worksheet below (6 questions, 14 marks). 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 From power to energy, 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.
Round Table
One sheet travels round the team. Each student adds one line of working or one idea and passes it on, so the solution is built four hands at a time and every contribution is visible.
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 |
|---|---|
| Power is current multiplied by voltage, P = I V, in watts. | To get the energy, multiply the power by the time, E = P t = I V t, in joules. |
Electrical power and energy: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
A heater draws a current of 2.0 A from the 230 V mains. Calculate its power.
The same 460 W heater runs for 10 s. Calculate the energy it transfers.
A 12 V motor has a power of 36 W. Calculate the current it draws.
A 9.0 V lamp draws a current of 0.40 A. Calculate the energy it transfers in 5.0 minutes, giving your answer in joules.
A device transfers 6000 J of energy in 30 s. Calculate its power.
Lamp X has a power of 60 W. Lamp Y has a power of 100 W. Lamp X is left on for 5.0 minutes and lamp Y for 3.0 minutes. Calculate the energy transferred by each, and state which uses more.
Answer key and marking notes
- P = IV = 2.0 × 230
P = 460 W
- E = Pt = 460 × 10
E = 4600 J
- I = P ÷ V = 36 ÷ 12
I = 3.0 A
- t = 5.0 × 60 = 300 s
E = IVt = 0.40 × 9.0 × 300
E = 1080 J
convert minutes to seconds first
- P = E ÷ t = 6000 ÷ 30
P = 200 W
- X: E = Pt = 60 × 300 = 18 000 J
Y: E = Pt = 100 × 180 = 18 000 J
Both transfer the same energy, 18 000 J. ✓
higher power but less time can give the same energy
4 multiple-choice questions on this lesson
From the topic question bank. Assign them in the app for a marked score in your gradebook, or run the topic as a class game.
An electric heater operates at 230 V and draws a current of 4.0 A. What is its power?
- A920 W
- B57.5 W
- C234 W
- D0.017 W
A 60 W lamp is left on for 2.0 hours. How much energy does it transfer? (1 hour = 3600 s)
- A432 000 J
- B120 J
- C216 000 J
- D30 J
A current of 3.0 A flows through a resistor of resistance 4.0 Ω. What power is dissipated? (Use P = I²R)
- A36 W
- B12 W
- C48 W
- D1.3 W
A resistor of resistance 6.0 Ω has a potential difference of 12 V across it. What power is dissipated? (Use P = V² ÷ R)
- A24 W
- B2.0 W
- C72 W
- D144 W
Quiz key
P = IV = 4.0 × 230 = 920 W. Option B divides the values; C adds them.
E = Pt = 60 × (2.0 × 3600) = 60 × 7200 = 432 000 J. Option C uses only 1 hour; B forgets to convert hours to seconds.
P = I²R = 3.0² × 4.0 = 9.0 × 4.0 = 36 W. Option B forgets to square the current (3 × 4); C squares the wrong way.
P = V² ÷ R = 12² ÷ 6.0 = 144 ÷ 6.0 = 24 W. Option B divides without squaring (12 ÷ 6); D forgets to divide by R.
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.