The shape of the lesson
By the end of the lesson, learners can
- Tap off a fraction: Change the two resistances and watch how the output voltage tracks their ratio.
- From divider to sensor: Make one resistor a light or heat sensor and the output becomes a signal that responds to the surroundings.
- Equation: V_out = V × R₂ ÷ (R₁ + R₂), output across R₂.
- Shares add: the p.d.s across R₁ and R₂ add up to the supply.
- Sensing: an LDR or thermistor changes one resistance, so the output tracks light or temperature.
One picture every idea returns to
A potential divider is two resistors in series that share the supply voltage in proportion to their resistances, with the output taken across one of them. Replacing one resistor with an LDR or thermistor makes the output voltage respond to light or temperature.
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, Splitting the voltage.. Two resistors in a row do more than slow the current, they carve up the voltage between them. | Whole class, then pairs |
| 5 to 17 min | Teach and check: Tap off a fraction | Change the two resistances and watch how the output voltage tracks their ratio. Project tab 4 of the lab (Potential divider) and let a student drive it while the class predicts. | Whole class, then pairs |
| 17 to 30 min | Teach and check: From divider to sensor | Make one resistor a light or heat sensor and the output becomes a signal that responds to the surroundings. Model the worked example Worked example on the board, one line at a time. | Anticipation / Reaction Guide |
| 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 divider to sensor, 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.
Anticipation / Reaction Guide
Before the lesson, students mark five statements true or false. After the teaching they mark them again and justify any change of mind in one sentence each. The statements come from the examiner traps below.
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 larger share of the voltage appears across the larger resistance. | The output across one resistor equals the supply voltage multiplied by that resistance divided by the total resistance. |
Potential dividers: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
State what a potential divider is, and write the equation for the output voltage taken across the lower resistor R₂.
A 12 V supply is connected across a 6 Ω resistor and a 6 Ω resistor in series. Calculate the output voltage across one of them.
A 9.0 V supply is connected across R₁ = 200 Ω and R₂ = 100 Ω in series. Calculate the output voltage across R₂, and the p.d. across R₁.
In a divider, both resistors are doubled in value. State what happens to the output voltage, and explain.
An LDR is used as the lower resistor R₂ in a divider, with the output taken across the LDR. Describe how the output voltage changes as it gets dark, and explain in terms of resistance.
A thermistor is used in a potential divider to help switch on a cooling fan when a room gets hot. State how the thermistor's resistance changes as the room warms, and hence how the output it controls changes.
Answer key and marking notes
- Two (or more) resistors in series across a supply, with an output taken across one of them. ✓
- V_out = V × R₂ ÷ (R₁ + R₂). ✓
- V_out = 12 × 6 ÷ (6 + 6)
V_out = 6.0 V
equal resistors share the voltage equally
- V_out = 9.0 × 100 ÷ (200 + 100) = 9.0 × 100 ÷ 300
V_out = 3.0 V
p.d. across R₁ = 9.0 − 3.0 = 6.0 V. ✓
- The output voltage is unchanged. ✓
- The output depends on the ratio R₂ ÷ (R₁ + R₂), which is the same when both are doubled. ✓
- As it gets dark, the output voltage rises. ✓
- Darkness increases the LDR resistance. ✓
- So R₂ takes a larger share of the supply, raising V_out. ✓
- As the room warms, the thermistor resistance falls. ✓
- So its share of the supply falls, while the p.d. across the other resistor rises, and that rising output can switch the fan on. ✓
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.
A potential divider uses two equal resistors across a 12 V supply. What is the output voltage taken across one of the resistors?
- A6.0 V
- B12 V
- C3.0 V
- D24 V
In a potential divider, a 2.0 kΩ resistor (R₁) and a 4.0 kΩ resistor (R₂) are in series across a 12 V supply. What is the voltage across R₂?
- A8.0 V
- B4.0 V
- C6.0 V
- D2.0 V
An LDR is used in a potential divider as a light sensor. When bright light shines on the LDR, its resistance falls. What happens to the voltage across the LDR?
- AThe voltage across the LDR decreases.
- BThe voltage across the LDR increases.
- CThe voltage across the LDR stays the same.
- DThe circuit stops working.
A thermistor and a fixed resistor form a potential divider, with the output taken across the fixed resistor. As the thermistor gets hotter, its resistance falls. What happens to the output voltage?
- AIt increases.
- BIt decreases.
- CIt stays constant.
- DIt drops to zero.
Quiz key
With two equal resistors, the supply divides equally, so the output across one resistor is 12 ÷ 2 = 6.0 V. The output equals the full supply only if it is taken across both resistors.
V₂ = R₂ ÷ (R₁ + R₂) × V = 4.0 ÷ 6.0 × 12 = 8.0 V. Option B gives the voltage across R₁ instead.
A smaller resistance takes a smaller share of the supply voltage, so as the LDR's resistance falls in bright light, the voltage across it decreases. The voltage across the other resistor rises to compensate.
As the thermistor's resistance falls, it takes a smaller share of the supply, so the fixed resistor (and hence the output) takes a larger share: the output voltage increases. This is the basis of a temperature-controlled switch.
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.