The shape of the lesson
By the end of the lesson, learners can
- Follow the single loop: Add resistors in series and watch the current stay equal everywhere while the voltage shares out.
- Same current, shared voltage: The one rule that current is equal everywhere, and the two that voltage shares and resistances add.
- Current: the same at every point.
- Voltage: the p.d.s add up to the e.m.f.
- Resistance: R = R₁ + R₂ + ... ; then I = V ÷ R.
One picture every idea returns to
In a series circuit there is a single path, so the current is the same at every point. The supply voltage is shared as potential difference across the components, and the total resistance is the sum of the individual resistances.
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, One loop, one current.. String the components in a single loop and the current has only one road to travel, so it is the same everywhere. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Follow the single loop | Add resistors in series and watch the current stay equal everywhere while the voltage shares out. Project tab 2 of the lab (Series) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: Same current, shared voltage | The one rule that current is equal everywhere, and the two that voltage shares and resistances add. Model the worked example Worked example on the board, one line at a time. | Numbered Heads Together |
| 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 Same current, shared voltage, 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.
Numbered Heads Together
Teams of four number off. Put up a question, give the team time to make sure every member can answer, then call a number: that student answers for the team, so nobody can leave the thinking to someone else.
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 |
|---|---|
| In series the current is the same all the way round, it is not shared. | It is the voltage that is shared between the components, and the resistances add up. |
Series circuits: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
State what happens to the size of the current at different points in a series circuit, and what happens to the p.d.s across the components.
Three resistors of 2 Ω, 3 Ω and 5 Ω are connected in series. Calculate their combined resistance.
A 12 V battery is connected in series with a 2 Ω resistor and a 4 Ω resistor. Calculate the total resistance and the current in the circuit.
For the circuit in Question 3, calculate the p.d. across each resistor, and show that they add up to the supply voltage.
A third resistor is now added in series to the circuit in Question 3. State and explain what happens to the current.
Two lamps are connected in series with a cell. One lamp is unscrewed and removed. State what happens to the other lamp, and explain in terms of the circuit.
Answer key and marking notes
- The current is the same at every point. ✓
- The p.d.s across the components add up to the e.m.f. of the supply. ✓
- R = 2 + 3 + 5
R = 10 Ω
- R = 2 + 4 = 6 Ω
I = V ÷ R = 12 ÷ 6
I = 2.0 A
- V₁ = IR₁ = 2.0 × 2 = 4 V
V₂ = IR₂ = 2.0 × 4 = 8 V
4 V + 8 V = 12 V ✓
- The current decreases. ✓
- Adding a resistor in series increases the total resistance, and since I = V ÷ R with V fixed, the current falls. ✓
- The other lamp goes out. ✓
- There is only one path, so removing one lamp breaks the circuit and the current stops everywhere. ✓
2 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.
Two lamps are connected in series with a cell. What is true about the current through them?
- AThe current is the same through both lamps.
- BThe current is larger through the first lamp.
- CThe current splits between the two lamps.
- DNo current flows because there are two lamps.
A 3.0 V cell is connected to two identical resistors in series. What is the potential difference across each resistor?
- A1.5 V
- B3.0 V
- C6.0 V
- D0.75 V
Quiz key
In a series circuit there is one path, so the same current flows through every component. Current splitting happens in parallel circuits, not series.
The supply p.d. is shared in series, so two identical resistors each take half: 3.0 ÷ 2 = 1.5 V. Each resistor does not get the full 3.0 V.
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.