SingaporeElectronicsSyllabus dot point

How can two resistors split a supply voltage into a smaller, chosen value, and what happens when one of them is a sensor?

Apply the potential divider equation to find an output voltage and explain how a sensor in a divider produces a varying voltage

A focused answer to the O-Level Electronics outcome on the potential divider. The divider equation, choosing resistors for a wanted output, and using a sensor in a divider to make a varying voltage.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-06

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
The answer
Examples in context
Try this

What this dot point is asking

SEAB wants you to use the potential divider equation to find the output voltage from two series resistors, and to explain how replacing one resistor with a sensor turns the divider into a circuit whose output voltage changes with light, temperature or position. The central insight is that two resistors in series share the supply voltage in proportion to their resistances, so you can tap off any voltage you want, and a sensor makes that tapped voltage respond to the world.

The answer

How a potential divider works

Two resistors in series carry the same current, so the supply voltage is shared between them in proportion to their resistances. The larger resistor takes the larger share of the voltage. By tapping the connection between them, you obtain a fraction of the supply voltage as an output.

The potential divider equation

If the output is taken across the lower resistor $R_2$ , then:

V_{out} = \frac{R_2}{R_1 + R_2}\,V_{in}

The resistor that the output is measured across goes on top of the fraction. The denominator is the total resistance, $R_1 + R_2$ . This single equation is the most frequently used formula in the whole course.

Choosing resistors for a wanted output

To get a particular output voltage, choose the ratio of the resistors. For half the supply, make the two resistors equal. For a quarter of the supply across $R_2$ , make $R_1$ three times $R_2$ . The actual resistance values are usually chosen in the kilohm range so the divider draws only a small current and wastes little power.

Using a sensor in the divider

Replace one resistor with a sensor whose resistance changes, and the output voltage now varies with the quantity the sensor measures:

A light-dependent resistor (LDR) has a high resistance in the dark and a low resistance in bright light.
A thermistor (the common type) has a high resistance when cold and a low resistance when hot.

Where the sensor sits in the divider decides whether the output rises or falls with the input. If the output is across the sensor, then when the sensor resistance rises, its share of the voltage rises, so the output rises. Swapping the sensor and the fixed resistor reverses the behaviour.

Worked example

A potential divider has a fixed $6.0\ \text{k}\Omega$ resistor on top and a thermistor below it, across a $9.0\ \text{V}$ supply, with the output taken across the thermistor. The thermistor resistance is $6.0\ \text{k}\Omega$ when warm and $18\ \text{k}\Omega$ when cold. Find the output voltage in each case.

Step 1: Write the divider equation for this circuit

The output is across the thermistor (the lower resistor), so $V_{out} = \dfrac{R_{th}}{R_1 + R_{th}} \times 9.0$ , with $R_1 = 6.0\ \text{k}\Omega$ .

Step 2: Find the warm output

When warm, $R_{th} = 6.0\ \text{k}\Omega$ : $V_{out} = \dfrac{6.0}{6.0 + 6.0} \times 9.0 = \dfrac{6.0}{12} \times 9.0 = 4.5\ \text{V}$ .

Step 3: Find the cold output

When cold, $R_{th} = 18\ \text{k}\Omega$ : $V_{out} = \dfrac{18}{6.0 + 18} \times 9.0 = \dfrac{18}{24} \times 9.0 = 6.75\ \text{V}$ .

Step 4: Interpret the change

As the thermistor cools, its resistance rises, so its share of the voltage rises and the output climbs from $4.5\ \text{V}$ to $6.75\ \text{V}$ . This rising voltage in the cold is exactly what a heating-control circuit needs.

Common traps

Putting the wrong resistor on top: The numerator is the resistor the output is measured across, not always $R_2$ or always the bottom one. Check which component the output is taken from.
Forgetting it is the ratio that matters: The output depends on the ratio of resistances, not their absolute size, so $2\ \text{k}\Omega$ with $2\ \text{k}\Omega$ gives the same half-supply output as $20\ \text{k}\Omega$ with $20\ \text{k}\Omega$ .
Getting the sensor direction backwards: Whether the output rises or falls with light or heat depends on which arm holds the sensor. Trace the resistance change through the equation.
Loading the divider heavily: Connecting a low-resistance load across the output draws current and changes the output voltage. The divider works cleanly only when the load draws little current.
Mixing up LDR and thermistor behaviour: An LDR's resistance falls in bright light; a common thermistor's resistance falls when hot. State the change clearly before using it.

Examples in context

Example 1. A dark-detecting circuit. An LDR in a potential divider feeds the next stage of a circuit. Placing the LDR so that the output rises as it gets darker lets the circuit switch on a light automatically at dusk. The divider turns "how bright is it" into a voltage the rest of the electronics can act on.

Example 2. A volume control. A potentiometer is a potential divider with a movable tap. Turning the knob moves the tap, changing the ratio of the two parts of the track and so the fraction of the input passed on to the amplifier. This is why almost every volume and brightness control is a potential divider in disguise.

Try this

Cue. Two equal resistors form a divider across a $6.0\ \text{V}$ supply. Find the output across one of them. Equal resistors split equally, so $V_{out} = 3.0\ \text{V}$ .
Cue. A $1.0\ \text{k}\Omega$ resistor is on top and a $3.0\ \text{k}\Omega$ resistor below, across $8.0\ \text{V}$ , output across the lower one. Find the output. $V_{out} = \dfrac{3.0}{1.0 + 3.0} \times 8.0 = 6.0\ \text{V}$ .
Cue. State what happens to the output of an LDR-on-top divider as the light increases, with the output taken across the lower fixed resistor. Brighter light lowers the LDR resistance, so the fixed resistor takes a larger share and the output rises.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksA potential divider is made from a

4.0\ \text{k}\Omega

resistor and a

2.0\ \text{k}\Omega

resistor in series across a

12\ \text{V}

supply. The output is taken across the

2.0\ \text{k}\Omega

resistor. Calculate the output voltage.

Show worked answer →

The output across the lower resistor is $V_{out} = \dfrac{R_2}{R_1 + R_2}\,V_{in}$ .

$V_{out} = \dfrac{2.0}{4.0 + 2.0} \times 12 = \dfrac{2.0}{6.0} \times 12 = 4.0\ \text{V}$ .

What markers reward: the divider formula with the correct resistor in the numerator (the one the output is taken across), and the answer $4.0\ \text{V}$ . Putting the wrong resistor on top gives $8.0\ \text{V}$ .

Original4 marksIn a potential divider, a fixed

10\ \text{k}\Omega

resistor is in series with a light-dependent resistor (LDR), across a

5.0\ \text{V}

supply, with the output taken across the LDR. Explain how the output voltage changes as the light level falls, given that the LDR resistance rises in the dark.