# Potential divider

> CIE A-Level Physics · D.C. circuits
> Source: https://www.owlsprep.com/study/cie-9702-u10-potential-divider/

This module covers the theory and application of resistive potential dividers in DC circuits. You will learn to calculate output voltage, analyze variable and sensor-based dividers, and solve common CIE exam problems.

**Prerequisites:** [Ohm's Law for series resistors](https://www.owlsprep.com/study/cie-9702-u10-ohms-law-series-circuits/); [Kirchhoff's Voltage Law](https://www.owlsprep.com/study/cie-9702-u10-kirchhoffs-laws/)

## Learning objectives

- Derive the potential divider equation from Ohm's and Kirchhoff's laws
- Calculate output voltage for resistive potential dividers
- Analyze circuits with variable resistors and sensors as dividers
- Explain the loading effect of connected loads on output voltage

## Principle and Equation Derivation

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

*Notation:* $V_{in}$ = input voltage, $V_{out}$ = output voltage, $R_1, R_2$ = series resistors

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

**Derivation:** 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 $V_{out}$ 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. $$V_{out} = \frac{R_2}{R_1 + R_2} V_{in}$$

*Conclusion:* 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: $R_1 = 2.0\ \text{kΩ}$ and $R_2 = 4.0\ \text{kΩ}$. Calculate the output voltage across $R_1$.

1. Confirm that output is across $R_1$, so $R_1$ goes in the numerator. $V_{in} = 9.0\ \text{V}$.
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. $$V_{out} = 3.0\ \text{V}$$

## Variable Potential Dividers (Potentiometers)

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.

**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.

## Potential Dividers with Sensors

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: $V_{in} = 5.0\ \text{V}$, $R_{fixed} = 2.0\ \text{kΩ}$, $R_{LDR} = 0.5\ \text{kΩ}$, output across $R_{fixed}$.
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. $$V_{out} = 4.0\ \text{V}$$

## Loading Effect of Connected Loads

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.

> **info**
>
> A high-impedance voltmeter draws very little current, so the loading effect is negligible for most practical school experiments. Low-impedance loads cause large, measurable changes to output voltage.

**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.

## Common pitfalls

- **Wrong:** Swapping the numerator resistor in the potential divider equation
  - Why it fails: Output voltage depends on which resistor the output is tapped across; swapping gives an incorrect value
  - Correct: Always double check which resistor the output is measured across, and put that resistor's value in the numerator
- **Wrong:** Claiming rheostats and potential dividers have the same output voltage range
  - Why it fails: Candidates often forget that rheostats cannot reach 0V output
  - Correct: Remember that only potential dividers can output voltage across the full 0V to $V_{in}$ range
- **Wrong:** Ignoring the loading effect when a load is connected across output
  - Why it fails: Candidates use the unloaded equation even when a load is present, leading to wrong results
  - Correct: Always calculate the effective parallel resistance of the output resistor and load before applying the divider equation
- **Wrong:** Assuming NTC thermistor resistance increases with temperature
  - Why it fails: Confusion between negative (NTC) and positive (PTC) temperature coefficient sensors
  - Correct: Remember NTC = Negative: resistance goes down when temperature goes up

## Cheatsheet

| Scenario | Formula / Rule | Key Note |
| --- | --- | --- |
| Unloaded two-resistor divider | $V_{out} = \frac{R_{out}}{R_1 + R_2} V_{in}$ | $R_{out}$ = resistance across output |
| Uniform potentiometer | $V_{out} = \frac{l}{L} V_{in}$ | $l$ = wiper distance from terminal |
| Loaded output | Calculate $R_{eff} = R_{out} \|\| R_{load}$ | Use $R_{eff}$ instead of $R_{out}$ |
| LDR divider | $R_{LDR} \downarrow$ as light intensity $\uparrow$ | Output across R: $V_{out} \uparrow$ when light increases |
| NTC thermistor | $R_{therm} \downarrow$ as temperature $\uparrow$ | Output across R: $V_{out} \uparrow$ when temperature increases |

## 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.

- [Internal resistance](https://www.owlsprep.com/study/cie-9702-u10-internal-resistance/)
- [Particle physics (AS)](https://www.owlsprep.com/study/cie-9702-u11-overview/)
- [Atomic structure](https://www.owlsprep.com/study/cie-9702-u11-atomic-structure/)

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