Potential Divider Principle
Key idea: Use the potential divider relationship to find output voltages in series resistor networks, including thermistor and LDR sensing circuits (A Level Physics).
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The core idea
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Learning objectives
- Analyse series, parallel and potential-divider resistor networks.
1. Definitions (Must Know)
A. Potential divider
A potential divider is a series combination of resistors used to produce a fraction of the input potential difference.
B. Output voltage, Vₒᵤₜ
Vₒᵤₜ is the potential difference across a chosen component (often across R₂).
2. Key Ideas (What Earns Marks)
- For an unloaded two-resistor divider with fixed input Vᵢₙ: Vₒᵤₜ = VᵢₙR₂/(R₁ + R₂) if the output is taken across R₂.
- Voltage divides in proportion to resistance because the same current flows through series components: V₁:V₂ = R₁:R₂
- Sensors are often used as one resistor:
- NTC thermistor: resistance decreases when temperature increases
- LDR: resistance decreases when light intensity increases
Write “Vₒᵤₜ across R₂” (or across R₁) before using the formula.
3. Detailed Explanations
A. Derivation of the divider formula
Fixed and variable potential-divider circuits
The upper circuit has resistors R1 and R2 in series across a supply, with output voltage measured across the lower resistor R2. The lower circuit has a three-terminal potentiometer across a supply, with output measured from its slider to the zero-volt end.
View figure data
| Circuit | Connection |
|---|---|
| Fixed divider | R₁ and R₂ are in series; output is across R₂ |
| Variable divider | The potentiometer track is across the supply; output is taken from its wiper |
For series resistors:
I = Vᵢₙ/(R₁ + R₂)
If Vₒᵤₜ is across R₂:
B. How changing resistance changes Vₒᵤₜ
For output across R₂:
- increasing R₂ increases Vₒᵤₜ
- increasing R₁ decreases Vₒᵤₜ
Potential divider: choosing sensor position (scaled)
Two divider curves showing how Vout/Vin changes with sensor resistance depending on whether the sensor is placed at R1 (top) or R2 (bottom).
Scroll across the graph to read all labels.
View figure data
| Sensor resistance (relative to fixed resistor) | Sensor as R2 (bottom): Vout/Vin = R2/(1+R2) | Sensor as R1 (top): Vout/Vin = 1/(R1+1) |
|---|---|---|
| 0.25 | 0.2 | 0.8 |
| 0.5 | 0.333 | 0.667 |
| 1 | 0.5 | 0.5 |
| 2 | 0.667 | 0.333 |
| 4 | 0.8 | 0.2 |
C. Sensor dividers (NTC thermistor / LDR)
If you replace one resistor with a sensor, Vₒᵤₜ becomes a function of temperature or light.
Typical exam task: choose whether to place the sensor as R₁ or R₂ so that Vₒᵤₜ increases when temperature/light increases.
D. Loading the output
The simple divider formula assumes that the output is unloaded, or that the connected device has a resistance much larger than the divider resistances. If a load R_L is connected across R₂, it is in parallel with R₂:
R_eq = R₂∥ R_L = R₂R_L/(R₂ + R_L)
Replace R₂ by R_eq before applying the divider relationship:
Vₒᵤₜ = VᵢₙR_eq/(R₁ + R_eq)
For a finite passive load, R_eq < R₂, so the loaded output is lower than the unloaded output. Current through R₁ now splits between R₂ and R_L; it is not correct to treat the same current as flowing through all three resistors.
If R_L is very large, R₂∥ R_L ≈ R₂ and the unloaded result is recovered. This model assumes a fixed ideal input voltage and negligible wire resistance.
4. Common Mistakes
- Using the formula but taking Vₒᵤₜ across the wrong resistor.
- Forgetting the divider only holds in this simple form when the output is not significantly loaded by another component.
- Not stating the thermistor/LDR trend (NTC thermistor and LDR both decrease resistance when the stimulus increases).
5. Exam Tips
- If the question is qualitative, use ratios: Vₒᵤₜ ∝ R₂/(R₁ + R₂) and discuss what happens as the sensor resistance changes.
- If the question is quantitative, compute Vₒᵤₜ from the formula after converting units.
6. Worked Examples
Modelled example 1
Simple divider calculation
Problem
Study the worked solution
Identify the output resistor
Method
Place R₂ in the numerator.Reason
The output p.d. is measured across R₂.Working
Vₒᵤₜ = VᵢₙR₂/(R₁ + R₂)Calculate the divider fraction
Method
Use one-quarter of the input.Reason
Resistance units cancel in the ratio.Working
Vₒᵤₜ = 121.0/(3.0 + 1.0) = 3.0 V
Guided practice 2
Thermistor placement (qualitative)
Problem
Try this before viewing the solution
Hints
Hint 1: state the sensor trend
Hint 2: track the divider fraction
View solution step by step
Place the NTC in the top arm
Method
Use the thermistor as R₁.Reason
Its falling resistance reduces the denominator without reducing the output-arm numerator.Working
Vₒᵤₜ = VᵢₙR₂/(R₁ + R₂)Track the temperature change
Method
Conclude that output rises.Reason
T↑ makes NTC R₁↓, increasing the fraction across fixed R₂.Working
T↑ ⇒ R₁↓ ⇒ Vₒᵤₜ↑.
Common misconception 3
Output taken across the top resistor
Learner claim
Try this before viewing the solution
View solution step by step
Match numerator to output terminals
Method
Use R₁ in the numerator.Reason
The requested p.d. is across the top resistor.Working
Vₒᵤₜ = VᵢₙR₁/(R₁ + R₂)Evaluate
Method
Obtain 6.0 V.Reason
R₁ is two-thirds of the series total.Working
Vₒᵤₜ = 9.02.0/(2.0 + 1.0) = 6.0 V
Examiner practice 4
Sensor value changes (numerical)
Examination question
Try this before viewing the solution
View solution step by step
Use the output-arm divider form
1 markMethod
Place LDR resistance in the numerator.Reason
Output is across R₂.Working
Vₒᵤₜ = 5.0R₂/(1.0 + R₂)Calculate the dark output
1 markMethod
Use R₂ = 4.0 kΩ.Reason
The resistance units cancel within the ratio.Working
V_dark = 5.04.0/(1.0 + 4.0) = 4.0 VCalculate the bright output
1 markMethod
Use R₂ = 1.0 kΩ.Reason
Light lowers LDR resistance.Working
V_bright = 5.01.0/(1.0 + 1.0) = 2.5 VState the trend
1 markMethod
Output falls as light intensity rises.Reason
The output is measured across the decreasing LDR resistance.Working
4.0 V → 2.5 V.
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark relation, two values and trend.
Challenge 5
LDR placement for “Vout increases with light”
Independent transfer
Try this before viewing the solution
Hints
Hint 1: work backwards from the required output
View solution step by step
Choose the top arm
Method
Place the LDR as R₁.Reason
Then the output-arm resistance R₂ remains fixed.Working
Vₒᵤₜ = VᵢₙR₂/(R_LDR + R₂)Follow the stimulus
Method
Conclude that output rises with illumination.Reason
Light lowers R_LDR, reducing the denominator.Working
Light ↑ ⇒ R₁↓ ⇒ Vₒᵤₜ↑.
Challenge 6
Loaded divider calculation
Independent boundary transfer
Try this before viewing the solution
Hints
Hint 1: replace the loaded lower arm
View solution step by step
Find the unloaded output
Method
Use R₂ as the lower arm.Reason
No load yet changes the divider.Working
V_unloaded = 124.0/(2.0 + 4.0) = 8.0 VReduce the loaded lower arm
Method
Combine R₂ and R_L in parallel.Reason
The load connects across the same output nodes.Working
R_eq = 4.0∥4.0 = 2.0 kΩFind the loaded output
Method
Use R_eq in the divider.Reason
It is the effective resistance across the output terminals.Working
V_loaded = 122.0/(2.0 + 2.0) = 6.0 V
7. Mind Stretchers
Mind stretcher 1: Voltmeter loadingExtension
A 9.0 V divider uses R₁ = R₂ = 10 kΩ. A voltmeter with resistance 10 kΩ is connected across R₂. Compare the ideal unloaded output with the voltmeter reading.
Show Answer
Without the meter:
Vₒᵤₜ = 9.010/(10 + 10) = 4.5 V
The meter is in parallel with R₂:
R_eq = 10∥10 = 5.0 kΩ
So the reading is:
Vₒᵤₜ = 9.05.0/(10 + 5.0) = 3.0 V
The meter resistance is not large compared with the divider resistances, so it significantly loads the output.
Mind stretcher 2: When does the simple divider formula fail?Extension
Give one practical condition under which Vₒᵤₜ = VᵢₙR₂/(R₁ + R₂) does not predict the measured output accurately, and explain why.
Show Answer
If the output is connected to a device with comparable or low resistance (a heavy load), that device is effectively in parallel with R₂.
This changes the effective resistance across the output and therefore changes the divider ratio, so the simple two-resistor formula no longer applies.
Mind stretcher 3: Variable-divider terminologyExtension
A three-terminal potentiometer used as a variable divider has a resistive track connected across the supply and a movable slider that selects Vₒᵤₜ. Do not confuse this with a slide-wire null-method potentiometer, which is a different measurement circuit.
Mind stretcher 4: Simulation Bridge: Potential Divider LabExtension
Concept Explorer: Potential Divider Lab
Vary supply, resistor values, sensor placement, and output loading to track voltage trends and checkpoint your divider reasoning.
- Divider Ratio
- Sensor Placement
- Trend Analysis
- Source and Branch Current
Apply the principle in the Potential Divider Lab. Open More controls, enable Output loading, and compare the unloaded output with different load resistances.
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Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027