What is a potential divider in the context of D.C. Circuits?

A potential divider, also known as a voltage divider, is a simple circuit used to produce a specific voltage output from a higher voltage source. It consists of two or more resistors in series across a voltage supply. The output voltage is taken from the junction of the resistors, allowing for the division of the input voltage in proportion to the resistance values. This concept is crucial in Cambridge International A Levels Physics for understanding how voltage can be controlled and distributed in circuits.

How do you calculate the output voltage of a potential divider?

To calculate the output voltage of a potential divider, you can use the formula V_out = (R2 / (R1 + R2)) * V_in, where V_out is the output voltage, R1 and R2 are the resistances of the two resistors, and V_in is the input voltage. This formula is derived from Ohm's Law and the principle of proportionality in series circuits, which is a key concept in D.C. Circuits for Cambridge International A Levels Physics.

Why are potential dividers important in electronic circuits?

Potential dividers are important in electronic circuits because they allow for the control and adjustment of voltage levels. This is essential for powering different components that require specific voltage levels, such as sensors and microcontrollers. Understanding potential dividers is crucial for students studying D.C. Circuits in Cambridge International A Levels Physics, as it helps in designing circuits that can efficiently manage power distribution.

What are some practical applications of potential dividers?

Potential dividers are widely used in various applications, such as adjusting volume controls in audio equipment, setting reference voltages in analog-to-digital converters, and controlling the brightness of lights. In the context of Cambridge International A Levels Physics, understanding these applications helps students appreciate the practical significance of potential dividers in everyday electronic devices and systems.

How does the choice of resistors affect the performance of a potential divider?

The choice of resistors in a potential divider affects both the output voltage and the power efficiency of the circuit. Using resistors with higher resistance values can reduce power consumption but may also limit the current available to the load. Conversely, lower resistance values can provide more current but at the cost of increased power dissipation. In Cambridge International A Levels Physics, students learn to balance these factors to optimize circuit performance in D.C. Circuits.

Why is "Putting $R_1$ on top" a common mistake in Potential dividers?

Why it happens: Memorisation slip. How to avoid it: PD across R_2 has R_2 ON TOP of fraction. Confirmation: if R_2 , V_2 V_total (all PD across R_2). Source: 9702 Examiner Reports 2022-2024

D.C. CircuitsCambridge International A Levels Physics (9702)

Potential dividers

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Potential Dividers Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Series resistors split voltage. Sensors via potential divider.

What you’ll learn

Mapped to the Cambridge IGCSE 9702 syllabus (2025-2027).

10.3.1 — Apply potential divider rule.
10.3.2 — Design sensor circuits.
10.3.3 — Account for loading.

Potential divider rule

Voltage splits with resistance.

Rule. PD across $R_2$ (in series with $R_1$ , across total $V$ ): $V_2 = V \frac{R_2}{R_1 + R_2}$

Same current through both $R_1$ and $R_2$ : $I = V/(R_1 + R_2)$ . So $V_2 = IR_2 = VR_2/(R_1 + R_2)$ .

Special cases.

$R_1 = R_2$ : $V_2 = V/2$ .
$R_2 \gg R_1$ : $V_2 \approx V$ (most across larger).
$R_2 \ll R_1$ : $V_2 \approx 0$ .

Example. 12 V across 4 Ω and 8 Ω: $V_8 = 12 \times 8/12 = 8$ V; $V_4 = 4$ V.

Cambridge tip. Memorise formula — saves time.

$V_2 = V R_2/(R_1 + R_2)$ .
Same current through both.
Linear in $R_2$ if $R_1$ fixed.

See the full worked example for potential dividers →

Sensor circuits

Variable $R$ → variable output.

Thermistor (NTC). Resistance DECREASES with temperature.

In divider with fixed $R$ :

Output across THERMISTOR: decreases with $T$ .
Output across FIXED resistor: increases with $T$ .

LDR (Light Dependent Resistor). Resistance DECREASES with light intensity. Similar circuit logic.

Cambridge tip. Identify which device is across the output — output behaviour follows.

Thermistor/LDR sense temperature/light.
Output depends on which $R$ is across it.
Useful for control circuits.

See the full worked example for potential dividers →

Loading effect

Voltmeter changes output.

Problem. When a load (e.g. voltmeter) is connected across part of a divider, it draws current and DECREASES the PD.

Solution. Voltmeter resistance should be MUCH GREATER than the resistance it's connected across. Ideal voltmeter: infinite resistance.

Effect. Combine voltmeter and resistor in PARALLEL; effective resistance LOWER. Output PD then lower than 'no load' case.

Example. 10 V across two 10 kΩ; voltmeter 50 kΩ reads across one.

Effective parallel: $R_p = (10 \times 50)/(60) \approx 8.33$ kΩ.
$V_{\text{read}} = 10 \times 8.33/18.33 \approx 4.55$ V (not 5 V).

Cambridge tip. Voltmeter must be 'high resistance' to give true reading.

Load reduces output.
Voltmeter must be high R.
Combine voltmeter || resistor in parallel.

See the full worked example for potential dividers →

How it’s examined

Potential dividers very common on Paper 2. Most-tested: rule (5 marks), sensor circuit (7-10 marks), loading (7 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

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Step-by-step worked examples — Potential dividers

Step-by-step solutions to past-paper-style questions on potential dividers, written exactly the way a tutor would explain them at the board.

1Potential divider (5 marks)
Extended• Adapted from 9702/22 May/Jun 2024• potential divider
Question
A 12 V supply is connected across two resistors in series: 4 Ω and 8 Ω. Find the PD across the 8 Ω resistor. (5 marks)
Step-by-step solution
1. Step 1
  Potential divider rule. $V_2 = V \times \dfrac{R_2}{R_1 + R_2}$ .
  $V_2 = 12 \times \dfrac{8}{4 + 8} = 12 \times \dfrac{8}{12} = 8 \text{ V}$
Answer
$V = 8$ V.
2Thermistor sensor (7 marks)
Extended• thermistor
Question
A potential divider has 12 V supply, fixed 10 kΩ resistor, and thermistor in series. Thermistor at room temp = 5 kΩ. Output is the PD across the thermistor. Find output. What happens if temperature rises? (7 marks)
Step-by-step solution
1. Step 1
  Output across thermistor.
  $V_T = 12 \times \dfrac{5}{10 + 5} = 4.0 \text{ V}$
2. Step 2
  Thermistor (NTC) resistance DECREASES with temperature.
3. Step 3
  Output PD across thermistor decreases. Output proportional to $R_T/(R_1 + R_T)$ . As $R_T$ falls, this fraction falls.
Answer
Output = 4.0 V at room T. Output DECREASES as T rises.
3Loaded potential divider (8 marks)
Extended• loading
Question
Potential divider has 10 V supply and two 10 kΩ resistors. A voltmeter of 50 kΩ resistance reads across one of them. Find the actual reading. (8 marks)
Step-by-step solution
1. Step 1
  Voltmeter in parallel with 10 kΩ. Combined resistance:
  $R_p = \dfrac{10 \times 50}{10 + 50} = \dfrac{500}{60} \approx 8.33 \text{ k}\Omega$
2. Step 2
  Now divider: 10 kΩ in series with $R_p$ .
  $V_p = 10 \times \dfrac{8.33}{10 + 8.33} \approx 4.55 \text{ V}$
3. Step 3
  Voltmeter reads ~4.55 V (lower than ideal 5 V because of loading).
Answer
$V \approx 4.55$ V.

Key Formulae — Potential dividers

The formulae you need to memorise for potential dividers on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Potential divider rule
$V_2 = V_{\text{total}} \times \dfrac{R_2}{R_1 + R_2}$
When to use
PD across $R_2$ in series with $R_1$ , both across $V_{\text{total}}$ .

Key Definitions and Keywords — Potential dividers

Definitions to memorise and the exact keywords mark schemes credit for potential dividers answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Potential divider
Examiner keyword
Series resistors that split a voltage in proportion to their resistance.
Loading effect
Voltmeter (or load) drawing current from a potential divider reduces the output PD.

Common Mistakes and Misconceptions — Potential dividers

The traps other students keep falling into on potential dividers questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Putting $R_1$ on top
9702 Examiner Reports 2022-2024
Why it happens
Memorisation slip.
How to avoid it
PD across $R_2$ has $R_2$ ON TOP of fraction. Confirmation: if $R_2 \to \infty$ , $V_2 \to V_{\text{total}}$ (all PD across $R_2$ ).

Practice questions

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Past paper style quiz

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Video lesson

Short walkthrough of the concepts students most often get stuck on.

Potential dividers — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Potential dividers

Short Study Notes in the form of Slides

Detailed Notes

What you’ll learn

How it’s examined

1Potential divider (5 marks)

2Thermistor sensor (7 marks)

3Loaded potential divider (8 marks)

Potential divider rule

Potential divider

Loading effect

✕Putting R1R_1R1​ on top

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Potential dividers — frequently asked questions

✕Putting $R_1$ on top