Potential Dividers | CIE AS Physics Revision Notes 2025

Potential Divider Circuit

When two resistors are connected in series, through Kirchhoff’s Second Law, the potential difference across the power source is divided between them
Potential dividers are circuits which produce an output voltage as a fraction of its input voltage
Potential dividers have three main purposes:
- To provide a variable potential difference
- To enable a specific potential difference to be chosen
- To split the potential difference of a power source between two or more components
Potential dividers are used widely in volume controls and sensory circuits using LDRs and thermistors
Potential divider circuits are based on the ratio of voltage between components. This is equal to the ratio of the resistances of the resistors in the diagram below, giving the following equation:

$V_{o u t} = \frac{R_{2}}{R_{1} + R_{2}} V_{i n}$

The resistance R₂ on the numerator of the fraction is always the resistance of the component that V_outis connected to

Potential divider circuit diagram

10-2-1-potential-divider-diagram-and-equation

A potential divider circuit is a combination of resistors, voltage in from a source and voltage out

The input voltage V_in is applied to the top and bottom of the series resistors
The output voltage V_out is measured from the centre to the bottom of resistor R₂
The potential difference V across each resistor depends upon its resistance R:
- The resistor with the largest resistance will have a greater potential difference than the other one from V = IR
- If the resistance of one of the resistors is increased, it will get a greater share of the potential difference, whilst the other resistor will get a smaller share
In potential divider circuits, the p.d across a component is proportional to its resistance from V = IR

Worked example

The circuit is designed to light up a lamp when the input voltage exceed a preset value.

It does this by comparing V_out with a fixed reference voltage of 5.3 V.

WE - potential divider question image, downloadable AS & A Level Physics revision notes

V_out is equal to 5.3

Calculate the input voltage V_in.

Answer:

Step 1: List the known quantities

Resistance of first resistor, R₁ = 20 kΩ
Resistance of the second resistor, R₂ = 12 kΩ
V_out= 5.3 V

Step 2: State the potential divider equation

The R₁ is on the numerator as this is the resistor that V_out is from

$V_{o u t} = (\frac{R_{1}}{R_{1} + R_{2}}) V_{i n}$

Step 3: Rearrange for the input voltage, V_in

$V_{i n} = \frac{V_{o u t}}{(\frac{R_{1}}{R_{1} + R_{2}})} = (\frac{R_{1} + R_{2}}{R_{1}}) V_{o u t}$

Step 4: Substitute in values

$V_{i n} = (\frac{12 + 20}{20}) \times 5.3 = 8.5 V (2 s . f .)$

Exam Tip

Always make sure the correct resistance is in the numerator of the potential divider equation. This will be the resistance of the component you want to find the output voltage of.

Potential Dividers (CIE AS Physics)

Revision Note

Potential Divider Circuit

Potential divider circuit diagram

Worked example

Exam Tip

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Author: Ashika