An LDR (top arm) and a 1.0\,k resistor (bottom, output) are across 6.0\,V. In bright light the LDR is 1.0\,k. State the output voltage.

WJEC Eduqas style-style practice: A thermistor and a 1.0\,k resistor form a divider across a 6.0\,V supply, output across the thermistor. At a certain temperature the thermistor's resistance is 2.0\,k. Calculate the output voltage.

A Component 1 Calculate question. The output is across the thermistor, so it is R_2 = 2.0\,k and the fixed resistor is R_1 = 1.0\,k (1 mark for identifying the resistors). Use V_out = V_in × R_2R_1 + R_2 = 6.0 × 2.01.0 + 2.0 = 6.0 × 2.03.0 = 4.0\,V (2 marks for the substitution and the answer). Markers reward the correct resistor on top and the value. A common error is to use the fixed resistor as R_2.

WalesElectronicsSyllabus dot point

How do an LDR and a thermistor turn light and temperature into a changing voltage?

Input sensors as variable resistors: the light-dependent resistor (LDR) and the NTC thermistor, how their resistance varies with light and temperature, and using them in a potential divider so the output voltage responds to the physical quantity.

A focused answer to WJEC Eduqas GCSE Electronics on input sensors, covering how an LDR and an NTC thermistor change resistance with light and temperature, and how a sensor in a potential divider produces a voltage that responds to the physical quantity.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-15

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

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What this topic is asking
The light-dependent resistor (LDR)
The NTC thermistor
Using a sensor in a potential divider
Try this

What this topic is asking

WJEC Eduqas wants you to know two key input sensors that behave as variable resistors: the light-dependent resistor (LDR) and the NTC thermistor. You should know how their resistance changes with the physical quantity, and how putting one in a potential divider turns that change in resistance into a change in voltage that the rest of the system can use.

The light-dependent resistor (LDR)

The LDR is made from a semiconductor whose conduction improves when light falls on it, freeing more charge carriers, so the resistance drops. The change is large, which makes the LDR useful, but it is not linear and it responds fairly slowly. On a circuit diagram the LDR is a resistor symbol inside a circle with two arrows pointing in, showing light arriving.

The NTC thermistor

In an NTC thermistor, heating frees more charge carriers in the semiconductor, so the resistance falls as the temperature rises (the opposite of a metal wire, whose resistance rises with temperature). Like the LDR, the thermistor only changes its resistance, so it must be combined with a fixed resistor to produce a usable voltage. Its symbol is a resistor rectangle with a temperature marker.

Using a sensor in a potential divider

This is the bridge between the physical world and the rest of the electronics. Use the potential divider equation $V_{\text{out}} = V_{\text{in}} \times \dfrac{R_2}{R_1 + R_2}$ . If the output is across the fixed resistor and an LDR is the other arm, then in the dark the LDR's high resistance takes most of the voltage, so the output is low; in bright light the output rises. If the output is across the sensor, the behaviour reverses. The fixed resistor is chosen so that the output passes through the threshold you need (often half the supply) at the light or temperature where the system should switch.

Designing a dark sensor

An LDR and a fixed resistor form a divider across a $9.0\,\text{V}$ supply. The output is taken across the fixed resistor and must rise above $4.5\,\text{V}$ in bright light. In bright light the LDR is $500\,\Omega$ . Choose a fixed resistor that gives $4.5\,\text{V}$ in bright light, and say what the output does in the dark.

step 1: Set the bright-light condition

We want the output (across the fixed resistor $R_2$ ) to be half the supply, $4.5\,\text{V}$ , in bright light. Half the supply means the two arms are equal, so $R_2 = R_{\text{LDR}} = 500\,\Omega$ .

step 2: Check with the equation

$V_{\text{out}} = 9.0 \times \dfrac{500}{500 + 500} = 9.0 \times \dfrac{1}{2} = 4.5\,\text{V}$ . Correct.

step 3: Consider the dark

In the dark the LDR's resistance rises to a high value (say tens of kilohms), so the LDR takes nearly all the voltage and the output across the $500\,\Omega$ fixed resistor becomes small.

step 4: State the result

A $500\,\Omega$ fixed resistor gives $4.5\,\text{V}$ in bright light; in the dark the output falls well below this. So the output is high in light and low in dark, which a later comparator can use to switch a lamp on when it gets dark.

Try this

Q1. State what happens to the resistance of an NTC thermistor as it is cooled. [1 mark]

Cue. Its resistance increases as it gets colder.

Q2. An LDR (top arm) and a $1.0\,\text{k}\Omega$ resistor (bottom, output) are across $6.0\,\text{V}$ . In bright light the LDR is $1.0\,\text{k}\Omega$ . State the output voltage. [2 marks]

Cue. Equal arms give half: $V_{\text{out}} = 6.0 \times \dfrac{1.0}{1.0 + 1.0} = 3.0\,\text{V}$ .

Exam-style practice questions

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

Eduqas style4 marksAn LDR is connected in series with a fixed resistor across a

9.0\,\text{V}

supply, with the output taken across the fixed resistor. Explain how the output voltage changes as it gets darker.

Show worked answer →

A Component 1 Explain question on a sensing divider. As it gets darker, the LDR's resistance increases (1 mark). The LDR now takes a larger share of the supply voltage, so the share across the fixed resistor (the output) falls (2 marks for the reasoning using the potential divider). Therefore the output voltage decreases as it gets darker (1 mark for the overall direction). Markers reward the rise in LDR resistance and the fall in the output across the fixed resistor. A common error is to forget which resistor the output is taken across.

Eduqas style3 marksA thermistor and a

1.0\,\text{k}\Omega

resistor form a divider across a

6.0\,\text{V}

supply, output across the thermistor. At a certain temperature the thermistor's resistance is

2.0\,\text{k}\Omega

. Calculate the output voltage.

Show worked answer →

A Component 1 Calculate question. The output is across the thermistor, so it is $R_2 = 2.0\,\text{k}\Omega$ and the fixed resistor is $R_1 = 1.0\,\text{k}\Omega$ (1 mark for identifying the resistors). Use $V_{\text{out}} = V_{\text{in}} \times \dfrac{R_2}{R_1 + R_2} = 6.0 \times \dfrac{2.0}{1.0 + 2.0} = 6.0 \times \dfrac{2.0}{3.0} = 4.0\,\text{V}$ (2 marks for the substitution and the answer). Markers reward the correct resistor on top and the value. A common error is to use the fixed resistor as $R_2$ .

Related dot points

Sources & how we know this

WJEC Eduqas GCSE Electronics specification (from 2017) — WJEC Eduqas (2017)