Potential Dividers (Cambridge (CIE) IGCSE Physics)

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Written by: Katie M

Reviewed by: Caroline Carroll

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Variable potential dividers

Extended Tier Only

When two resistors are connected in series, the potential difference across the power source is shared between them
A potential divider is a circuit which splits potential difference from a power source, so only a fraction goes to a component (a voltmeter, in the diagram below)

Potential divider diagram

Potential divider, IGCSE & GCSE Physics revision notes

A potential divider splits the potential difference of a power source between two components

The potential difference across each resistor depends upon its resistance:
- The resistor with the largest resistance will have a greater potential difference than the other one
- 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
If one resistor is a variable resistor, the potential difference across the other resistor can be altered
- This means the potential difference across any component in parallel with that resistor can also be altered

Resistors as Potential Dividers

Extended Tier Only

When two resistors are connected in series, the power source's e.m.f. is split between the resistors
- This potential difference splits in the same ratio as the resistance of the two resistors
The ratio of potential differences across each resistor can be found using the following equation:

$\frac{R_{1}}{R_{2}} = \frac{V_{1}}{V_{2}}$

Where:
- $R_{1}$ is the resistance of resistor 1 in ohms, Ω
- $R_{2}$ is the resistance of resistor 2 in ohms, Ω
- $V_{1}$ is the potential difference across resistor 1 in volts, V
- $V_{2}$ is the potential difference across resistor 2 in volts, V
Recall that the e.m.f. of the power source will be equal to $V_{1} + V_{2}$

Worked Example

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

When the lamp lights up, $V_{o u t}$ is 5.3 V.

Calculate the e.m.f. of the power source required to illuminate the lamp.

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

Answer:

Step 1: List the known quantities

Resistance of resistor 1, $R_{1}$ = 12 kΩ = 12 000 Ω
Resistance of resistor 2, $R_{2}$ = 20 kΩ = 20 000 Ω
Potential difference across resistor 2, $V_{2} = V_{o u t}$ = 5.3 V

Step 2: Recall the equation for a potential divider

$\frac{R_{1}}{R_{2}} = \frac{V_{1}}{V_{2}}$

Step 3: Substitute the known quantities and determine the potential difference across resistor 1

$\frac{12 000}{20 000} = \frac{V_{1}}{5.3}$

$V_{1} = 3.18 V$

Step 4: Determine the e.m.f. of the power source, $V_{i n}$

$V_{i n} = V_{1} + V_{2}$

$V_{i n} = 3.18 + 5.3 = 8.48 V$

The e.m.f. of the power source when the lamp illuminates is 8.5 V to 2 significant figures

Examiner Tips and Tricks

When thinking about potential dividers, remember that the higher the resistance the more energy it will take to 'push the current through' and therefore the higher the potential difference.

This means that if a component (often shown as a voltmeter in questions) needs to be switched on by a change such as increased light or temperature, then the resistor it is in parallel with needs to become larger compared to the other resistor.

Last updated: 30 September 2024

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