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).

  • GCE A-Level H2 Physics 2027
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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
Always state where Vout is measured

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.

A fixed two-resistor potential divider beside a three-terminal variable potential dividerA fixed two-resistor potential divider beside a three-terminal variable potential divider
Define the output between two labelled points before writing a ratio. In the fixed divider shown, the output is across the lower resistor R₂.
View figure data
Potential-divider topology
CircuitConnection
Fixed dividerR₁ and R₂ are in series; output is across R₂
Variable dividerThe potentiometer track is across the supply; output is taken from its wiper

For series resistors:

I = Vᵢₙ/(R₁ + R₂)

If Vₒᵤₜ is across R₂:

Vₒᵤₜ = IR₂; = (Vᵢₙ/(R₁ + R₂))R₂; = VᵢₙR₂/(R₁ + 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.

Two divider curves showing how Vout/Vin changes with sensor resistance depending on whether the sensor is placed at R1 (top) or R2 (bottom).Two divider curves showing how Vout/Vin changes with sensor resistance depending on whether the sensor is placed at R1 (top) or R2 (bottom).
If the sensor resistance decreases when the stimulus increases (NTC thermistor, LDR), placing it as R1 makes Vout rise with the stimulus (for Vout across the bottom resistor).
Open full-size graph
View figure data
Values for Potential divider: choosing sensor position (scaled)
Sensor resistance (relative to fixed resistor)Sensor as R2 (bottom): Vout/Vin = R2/(1+R2)Sensor as R1 (top): Vout/Vin = 1/(R1+1)
0.250.20.8
0.50.3330.667
10.50.5
20.6670.333
40.80.2

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)

Unloaded and loaded potential-divider circuitsSide-by-side circuits. The unloaded circuit has R1 and R2 in series, with output across R2. In the loaded circuit, load resistance RL is connected in parallel with R2, reducing the effective resistance of the output branch.Unloaded outputLoaded outputOnly R₁ and R₂ set the divider ratio.Rₗ is connected across the output terminals.Vᵢₙ+0 VR₁R₂VoutVout = Vin × R₂ / (R₁ + R₂)Vᵢₙ+0 VR₁R₂RₗVoutOutput branch: Rₑₚ = R₂ ∥ RₗUse Rₑₚ in place of R₂ in the divider formula.
Scroll diagram horizontally to read all labels.
Connecting a load across R₂ puts the load in parallel with R₂. Replace that branch by Rₑₚ = R₂ ∥ Rₗ before applying the potential-divider relationship.

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.

Loading limit check

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

Core

Problem

Vᵢₙ = 12 V, R₁ = 3.0 kΩ and R₂ = 1.0 kΩ. Find Vₒᵤₜ across R₂.
Study the worked solution
  1. Identify the output resistor

    Method

    Place R₂ in the numerator.

    Reason

    The output p.d. is measured across R₂.

    Working

    Vₒᵤₜ = VᵢₙR₂/(R₁ + R₂)
  2. 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)

About 5 min

Problem

You want Vₒᵤₜ across the bottom resistor to increase as temperature increases. Should an NTC thermistor be R₁ (top) or R₂ (bottom)?

Try this before viewing the solution

NTC position

Hints

Hint 1: state the sensor trend
An NTC thermistor’s resistance decreases when temperature rises.
Hint 2: track the divider fraction
For output across R₂, inspect R₂/(R₁ + R₂) when one arm decreases.
View solution step by step
  1. 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)
  2. 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

Find and correct the mistake

Learner claim

For Vᵢₙ = 9.0 V, R₁ = 2.0 kΩ and R₂ = 1.0 kΩ, a learner uses R₂ in the numerator even though output is across R₁. Diagnose the error and find the output.

Try this before viewing the solution

Resistance in the numerator

View solution step by step
  1. 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)
  2. 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)

4 marks

Examination question

A 5.0 V divider has fixed R₁ = 1.0 kΩ and an LDR as R₂, with output across R₂. Find the output when R₂ = 4.0 kΩ in darkness and 1.0 kΩ in bright light, then state the output trend. [4 marks]

Try this before viewing the solution

View solution step by step
  1. Use the output-arm divider form

    1 mark

    Method

    Place LDR resistance in the numerator.

    Reason

    Output is across R₂.

    Working

    Vₒᵤₜ = 5.0R₂/(1.0 + R₂)
  2. Calculate the dark output

    1 mark

    Method

    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 V
  3. Calculate the bright output

    1 mark

    Method

    Use R₂ = 1.0 kΩ.

    Reason

    Light lowers LDR resistance.

    Working

    V_bright = 5.01.0/(1.0 + 1.0) = 2.5 V
  4. State the trend

    1 mark

    Method

    Output falls as light intensity rises.

    Reason

    The output is measured across the decreasing LDR resistance.

    Working

    4.0 V → 2.5 V.

Challenge 5

LDR placement for “Vout increases with light”

Minimal support

Independent transfer

An LDR’s resistance decreases as light intensity increases. You want output across the bottom resistor to increase with light. Decide whether the LDR should be the top or bottom arm and justify the direction.

Try this before viewing the solution

Hints

Hint 1: work backwards from the required output
Ask which arm can decrease while making the fraction across the fixed bottom arm larger.
View solution step by step
  1. 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)
  2. 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

Minimal support

Independent boundary transfer

A divider has Vᵢₙ = 12 V, R₁ = 2.0 kΩ and R₂ = 4.0 kΩ. A 4.0 kΩ load is connected across R₂. Find unloaded and loaded output voltages.

Try this before viewing the solution

Hints

Hint 1: replace the loaded lower arm
When connected, R₂ and R_L form a parallel equivalent before the divider relation is applied.
View solution step by step
  1. 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 V
  2. Reduce 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 kΩ
  3. Find 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.

BetaO LevelA LevelElectricityBest for: O Level practical electricity
  • Divider Ratio
  • Sensor Placement
  • Trend Analysis
  • Source and Branch Current

Open the full interactive simulation on its own page

Use the standalone simulation page for the live controls, SVG scene, run modes, and scoring flow.

The lesson stays lightweight and links out to the dedicated simulation page.

Apply the principle in the Potential Divider Lab. Open More controls, enable Output loading, and compare the unloaded output with different load resistances.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027