The shape of the lesson
By the end of the lesson, learners can
- Slide the tap: Move the divider tap, or swap in a thermistor or LDR, and watch the output voltage follow V × R₂ / (R₁ + R₂) as the resistances change.
- The divider rule: The output is the supply scaled by the resistance ratio.
- Building a sensor: Sensor dividers turn a physical change into a voltage.
- Output: V R₂/(R₁+R₂).
- Sensor: thermistor or LDR.
- Null: potentiometer reads zero.
One picture every idea returns to
A potential divider uses two resistors in series to split a supply voltage. The output across R₂ is Vₒᵤₜ = Vᵢₙ × R₂ / (R₁ + R₂). A potentiometer is a continuously variable divider, used in null methods to compare potential differences. Replacing one resistor with a thermistor or LDR makes a sensor circuit whose output voltage changes with temperature or light.
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, Tapping off a voltage.. Two resistors in series share the supply voltage in proportion to their resistances. Put the figure on the board: it is the one picture every idea in the lesson returns to. | Whole class, then pairs |
| 5 to 13 min | Teach and check: Slide the tap | Move the divider tap, or swap in a thermistor or LDR, and watch the output voltage follow V × R₂ / (R₁ + R₂) as the resistances change. Project tab 7 of the lab (Potential divider and potentiometer) and let a student drive it while the class predicts. | Whole class, then pairs |
| 13 to 21 min | Teach and check: The divider rule | The output is the supply scaled by the resistance ratio. | Pairs |
| 21 to 30 min | Teach and check: Building a sensor | Sensor dividers turn a physical change into a voltage. | Frayer Model |
| 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 Building a 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.
Frayer Model
Pairs fill a four-box grid for the key term: definition, characteristics, examples and non-examples. The non-examples box is where the misconceptions surface, so give it the most time.
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 |
|---|---|
| Use the resistance of the component you take the output across in the numerator: Vₒᵤₜ = V R₂/(R₁ + R₂). | When a sensor replaces one resistor, decide carefully which resistor the output is measured across, because a falling sensor resistance can make the output rise or fall depending on the arrangement. The unloaded divider equation assumes negligible current is drawn from the output. |
Potential dividers: worksheet
6 questions, 14 marks. Written in the style of the papers; no past paper question is reproduced.
State the potential divider equation for the output across the lower resistor R₂.
A 12 V supply is divided by two equal resistors. Find the output across one of them.
A 9.0 V supply is connected across R₁ = 2.0 kΩ and R₂ = 1.0 kΩ in series. Find the output across R₂.
A potential divider must give 5.0 V across R₂ = 1.0 kΩ from a 15 V supply. Find the required value of R₁.
In a potential divider the output is taken across a thermistor. As the temperature rises, explain what happens to the output voltage.
State the purpose of the null method used with a potentiometer.
Answer key and marking notes
- Vₒᵤₜ = V × R₂ / (R₁ + R₂) ✓
- where V is the supply voltage ✓
- Equal resistors share the supply equally ✓
- Vₒᵤₜ = 12 × ½ = 6.0 V ✓
- Vₒᵤₜ = V R₂/(R₁ + R₂) = 9.0 × 1.0/(2.0 + 1.0) ✓
- Vₒᵤₜ = 9.0 × 1/3 = 3.0 V ✓
- 5.0 = 15 × 1.0/(R₁ + 1.0), so R₁ + 1.0 = 3.0 ✓
- R₁ = 2.0 kΩ ✓
- The thermistor resistance falls as it warms ✓
- so its share of the total resistance falls, and the output voltage across it decreases ✓
- To compare two potential differences without drawing current at balance ✓
- the potentiometer is adjusted until the galvanometer reads zero ✓
14 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.
The output of a potential divider across R₂ is:
- AV R₂/(R₁+R₂)
- BV R₁/(R₁+R₂)
- CV(R₁+R₂)/R₂
- DV R₁R₂
A 12 V supply is divided by two equal resistors. The output across one is:
- A12 V
- B3.0 V
- C6.0 V
- D24 V
A 10 V supply has R₁ = 1.0 kΩ and R₂ = 4.0 kΩ. The output across R₂ is:
- A2.0 V
- B8.0 V
- C5.0 V
- D10 V
A potentiometer provides a:
- Afixed current
- Bstore of charge
- Clarger e.m.f.
- Dcontinuously variable potential difference
In a divider with the output across a thermistor, raising the temperature makes the output:
- Adecrease
- Bincrease
- Cstay the same
- Dequal the supply
In a null method with a potentiometer, the galvanometer reads:
- Aa maximum
- Bthe supply voltage
- Czero at balance
- Dhalf the supply
Increasing R₂ (the output resistor) makes the output voltage:
- Adecrease
- Bincrease
- Cstay the same
- Dbecome zero
In a divider with the output across an LDR, brighter light makes the output:
- Aincrease
- Bstay constant
- Cequal the supply
- Ddecrease
A 5.0 V output across R₂ = 2.0 kΩ is wanted from a 12 V supply. R₁ should be about:
- A2.8 kΩ
- B1.0 kΩ
- C4.8 kΩ
- D5.0 kΩ
The output of a potential divider taken across R₁ is:
- AV R₂/(R₁+R₂)
- BV/R₁
- CV R₁/(R₁+R₂)
- DV R₁R₂
A 9.0 V supply has R₁ = R₂ = 3.0 kΩ. The output across R₂ is:
- A9.0 V
- B4.5 V
- C3.0 V
- D1.5 V
A sensor potential divider is useful because its output voltage:
- Ais always constant
- Bequals the supply
- Cis independent of the sensor
- Dvaries with a physical condition such as temperature or light
Two equal resistors in a potential divider share the supply voltage:
- Aequally
- Bin the ratio 2:1
- Call across one
- Dnot at all
A 6.0 V supply has R₁ = 2.0 kΩ and R₂ = 1.0 kΩ. The output across R₂ is:
- A4.0 V
- B3.0 V
- C2.0 V
- D6.0 V
Quiz key
Output across R₂ is V R₂/(R₁+R₂).
Equal resistors share equally: 6.0 V.
Vₒᵤₜ = 10 × 4.0/(1.0+4.0) = 8.0 V.
A potentiometer is a variable potential divider.
The thermistor resistance falls, so its share of the voltage falls.
Balance is when no current flows through the galvanometer.
A larger R₂ takes a larger share of the supply.
The LDR resistance falls in brighter light, so its share of the voltage falls.
5.0 = 12 × 2.0/(R₁+2.0) gives R₁+2.0 = 4.8, so R₁ = 2.8 kΩ.
Output across R₁ is V R₁/(R₁+R₂).
Equal resistors: 9.0 × ½ = 4.5 V.
Replacing a resistor with a sensor makes the output respond to the condition.
Equal resistances take equal shares of the supply.
Vₒᵤₜ = 6.0 × 1.0/(2.0+1.0) = 2.0 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.