Potential divider

CIE A-Level Physics· 10 min read

1. Principle and Equation Derivation★★☆☆☆⏱ 3 min

📘 Definition

Potential Divider

$V_{in}$ = input voltage, $V_{o u t}$ = output voltage, $R_{1}, R_{2}$ = series resistors

A series circuit of two or more resistors connected across a fixed input voltage, with output voltage tapped across one of the resistors.

Example:

12V input across two equal 1kΩ resistors gives 6V output across the lower resistor.

🔬 Derivation

Goal:

Derive the output voltage equation for a two-resistor potential divider

Starting from:

Ohm's Law and Kirchhoff's Voltage Law for series circuits

1
The same current $I$ flows through both series resistors. Total resistance of the circuit is $R_{1} + R_{2}$ , so by Ohm's law:
2
$I = \frac{V_{in}}{R_1 + R_2}$
3
Output voltage $V_{o u t}$ is tapped across $R_{2}$ , so by Ohm's law for $R_{2}$ :
4
$V_{out} = I R_2$
5
Substitute the expression for current to get the final equation:
6
$V_{out} = \frac{R_2}{R_1 + R_2} V_{in}$

Result:

If output is tapped across $R_{1}$ , replace $R_{2}$ with $R_{1}$ in the numerator of the equation.

📐 Worked Example

A 9.0V battery is connected across two series resistors: $R_{1} = 2.0 kΩ$ and $R_{2} = 4.0 kΩ$ . Calculate the output voltage across $R_{1}$ .

1
Confirm that output is across $R_{1}$ , so $R_{1}$ goes in the numerator. $V_{in} = 9.0 V$ .
2
Substitute values into the potential divider equation:
3
$V_{out} = \frac{R_1}{R_1 + R_2} V_{in} = \frac{2.0}{2.0 + 4.0} \times 9.0$
4
Calculate the final result:
5
$V_{out} = 3.0\ \text{V}$

2. Variable Potential Dividers (Potentiometers)★★★☆☆⏱ 4 min

A variable potential divider uses a sliding contact (called a wiper) to change the fraction of output voltage taken from the total input. Unlike rheostats, which only vary current, potential dividers can adjust output voltage from 0V all the way up to the full input voltage.

📘 Definition

Potentiometer

A three-terminal variable resistor that acts as a continuously variable potential divider, with output voltage adjusted by moving the sliding wiper.

📐 Worked Example

A 10cm long 10kΩ uniform potentiometer is connected across a 12V input. What output voltage is measured when the wiper is 3cm from the bottom output terminal?

1
Resistance is proportional to length for a uniform potentiometer, so the resistance of the 3cm output section is:
2
$R_{out} = \frac{3}{10} \times 10\ \text{kΩ} = 3\ \text{kΩ}$
3
Total resistance of the potentiometer is 10kΩ, substitute into the divider equation:
4
$V_{out} = \frac{3}{10} \times 12\ \text{V} = 3.6\ \text{V}$
5
Output voltage is directly proportional to the wiper position for a uniform potentiometer.

Exam tip:

CIE often asks to compare potential dividers to rheostats for voltage control. Remember that only potential dividers can achieve 0V output.

3. Potential Dividers with Sensors★★★☆☆⏱ 4 min

Potential dividers are widely used with resistive input sensors that change resistance with a physical property (light, temperature, etc.). The output voltage of the divider changes with the physical quantity, which can then be measured and calibrated.

LDR (light-dependent resistor): resistance decreases as light intensity increases
NTC thermistor: resistance decreases as temperature increases
PTC thermistor: resistance increases as temperature increases

📐 Worked Example

An LDR is connected in series with a 2.0kΩ fixed resistor across a 5.0V input, forming a potential divider with output across the fixed resistor. In bright light, the LDR resistance is 0.5kΩ. Calculate the output voltage.

1
Label values: $V_{in} = 5.0 V$ , $R_{f i x e d} = 2.0 kΩ$ , $R_{L D R} = 0.5 kΩ$ , output across $R_{f i x e d}$ .
2
Substitute into the potential divider equation:
3
$V_{out} = \frac{R_{fixed}}{R_{fixed} + R_{LDR}} V_{in} = \frac{2.0}{2.0 + 0.5} \times 5.0$
4
Calculate the result:
5
$V_{out} = 4.0\ \text{V}$

4. Loading Effect of Connected Loads★★★★☆⏱ 3 min

When a load (such as a voltmeter, bulb or other component) is connected across the output of a potential divider, it draws current and forms a parallel combination with the output resistor. This changes the effective resistance of the output section, which changes the output voltage from the unloaded prediction. This effect is called loading.

📐 Worked Example

Two 10kΩ resistors are in series across a 10V input, so unloaded output across one resistor is expected to be 5V. A voltmeter of internal resistance 10kΩ is connected across the output. Calculate the actual measured output voltage.

1
The voltmeter is in parallel with the 10kΩ output resistor. Calculate the effective combined resistance:
2
$R_{eff} = \frac{10 \times 10}{10 + 10} = 5\ \text{kΩ}$
3
The circuit now has the upper 10kΩ in series with 5kΩ effective output resistance. Apply the divider equation:
4
$V_{out} = \frac{5}{10 + 5} \times 10 = 3.3\ \text{V (2 s.f.)}$
5
The loading effect here is very large, dropping output by 1.7V from the unloaded prediction.

5. Common Pitfalls

Wrong move:

Swapping the numerator resistor in the potential divider equation

Why:

Output voltage depends on which resistor the output is tapped across; swapping gives an incorrect value

Correct move:

Always double check which resistor the output is measured across, and put that resistor's value in the numerator

Wrong move:

Claiming rheostats and potential dividers have the same output voltage range

Why:

Candidates often forget that rheostats cannot reach 0V output

Correct move:

Remember that only potential dividers can output voltage across the full 0V to $V_{in}$ range

Wrong move:

Ignoring the loading effect when a load is connected across output

Why:

Candidates use the unloaded equation even when a load is present, leading to wrong results

Correct move:

Always calculate the effective parallel resistance of the output resistor and load before applying the divider equation

Wrong move:

Assuming NTC thermistor resistance increases with temperature

Why:

Confusion between negative (NTC) and positive (PTC) temperature coefficient sensors

Correct move:

Remember NTC = Negative: resistance goes down when temperature goes up

6. Quick Reference Cheatsheet

Scenario	Formula / Rule	Key Note
Unloaded two-resistor divider	$V_{o u t} = \frac{R _{o u t}}{R _{1} + R _{2}} V_{in}$	$R_{o u t}$ = resistance across output
Uniform potentiometer	$V_{o u t} = \frac{l}{L} V_{in}$	$l$ = wiper distance from terminal
Loaded output	Calculate $R_{e f f} = R_{o u t} ∣∣ R_{l o a d}$	Use $R_{e f f}$ instead of $R_{o u t}$
LDR divider	$R_{L D R} ↓$ as light intensity $↑$	Output across R: $V_{o u t} ↑$ when light increases
NTC thermistor	$R_{t h er m} ↓$ as temperature $↑$	Output across R: $V_{o u t} ↑$ when temperature increases

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · 12
Calculate sensor divider output voltage
2023 · 22
Derive potential divider equation
2024 · 11
Analyze LDR potential divider circuit

What's Next

Potential dividers are a core building block of many sensor and measurement circuits that you will encounter in further CIE A-Level Physics topics, including input transducers, analogue electronics and practical measurement experiments. Understanding how loading effects change output voltage is also critical for solving more complex circuit problems, including those involving null measurement methods with potentiometers to find unknown e.m.f. or resistance. This topic forms the foundation for voltage division in both DC and AC circuits, which you will encounter in multiple later sections of the CIE 9702 syllabus. Next, you will build on this knowledge to analyze more complex circuits and learn about sensor behaviour in practical systems.