WJEC Eduqas Eduqas 2021-style practice: A Wheatstone bridge has three 1.0\ k resistors and a strain gauge whose resistance rises from 1.0\ k to 1.02\ k under load. The bridge is supplied with 6.0\ V. Calculate the output (the difference between the two midpoint voltages) when balanced and when strained.

Balanced (up to 2 marks): with all four arms equal, each midpoint sits at half the supply, 3.0\ V, so the output difference is 0\ V (the bridge is balanced). Strained (up to 4 marks): one divider stays at 3.0\ V. The other has the strain gauge (1.02\ k) and a 1.0\ k resistor. Its midpoint voltage is 6.0 × R_fixedR_fixed + R_gauge or 6.0 × R_gauge depending on arrangement; taking the gauge in the lower arm, V = 6.0 × 1.021.0 + 1.02 = 6.0 × 0.5050 = 3.030\ V. The output is the difference 3.030 - 3.000 = 0.030\ V = 30\ mV. Markers reward the 0\ V balanced output, the strained midpoint near 3.03\ V, and the small difference of order 30\ mV that an instrumentation amplifier then amplifies.

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How do we extract a tiny sensor signal from noise and condition it for the next stage?

Instrumentation systems: sensors and transducers, the Wheatstone bridge, the instrumentation (difference) amplifier, common-mode rejection, and signal conditioning.

An Eduqas A-Level Electronics answer on instrumentation systems: input transducers and sensors, the Wheatstone bridge for small resistance changes, the instrumentation (difference) amplifier with high common-mode rejection, and the signal conditioning that turns a tiny noisy sensor signal into a clean usable voltage.

Generated by Claude Opus 4.813 min answerUpdated 2026-06-02

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

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Quick answer

An input transducer (sensor) converts a physical quantity into an electrical signal: a thermistor (temperature), an LDR (light), a strain gauge (force), a microphone (sound). Small resistance changes are best read with a Wheatstone bridge, two potential dividers compared against each other, which gives zero output when balanced and a small difference voltage as one arm changes. That difference sits on a large common-mode level and is usually tiny and noisy, so it is amplified by an instrumentation (difference) amplifier, which amplifies the difference between its inputs while strongly rejecting any signal common to both (a high common-mode rejection ratio). Signal conditioning then scales, filters and offsets the result so it suits the next stage (a comparator or analogue-to-digital converter).

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What this dot point is asking

Eduqas wants you to describe sensors and transducers, the Wheatstone bridge for small resistance changes, the instrumentation (difference) amplifier with high common-mode rejection, and signal conditioning. This is how a system reads the physical world accurately.

The answer

Sensors and transducers

The Wheatstone bridge

The instrumentation amplifier and common-mode rejection

Signal conditioning

Finding the bridge output for a small sensor change

A Wheatstone bridge has three $2.0\ \text{k}\Omega$ resistors and a sensor that changes from $2.0\ \text{k}\Omega$ to $2.10\ \text{k}\Omega$ . The supply is $5.0\ \text{V}$ and the sensor is the lower arm of one divider. Find the output difference voltage.

step 1: Reference divider midpoint

The reference divider (two $2.0\ \text{k}\Omega$ arms) sits at half the supply: $V_\text{ref} = 2.5\ \text{V}$ .

step 2: Sensor divider midpoint

With the sensor ( $2.10\ \text{k}\Omega$ ) in the lower arm and $2.0\ \text{k}\Omega$ above: $V_\text{sensor} = 5.0 \times \frac{2.10}{2.0 + 2.10} = 5.0 \times \frac{2.10}{4.10} = 5.0 \times 0.5122 = 2.561\ \text{V}$ .

step 3: Output difference

$V_\text{out} = V_\text{sensor} - V_\text{ref} = 2.561 - 2.500 = 0.061\ \text{V} = 61\ \text{mV}$ , which an instrumentation amplifier would then boost to a usable level.

Examples in context

Instrumentation systems are how electronic devices measure the world accurately: electronic scales and load cells use a strain-gauge bridge and an instrumentation amplifier, medical monitors extract tiny biopotentials from large common-mode interference, and industrial sensors feed conditioned signals to controllers. The bridge, the difference amplifier and signal conditioning are exactly the input subsystem a non-exam assessment project needs when it must measure something precisely.

Try this

Q1. State the output of a Wheatstone bridge when it is balanced. [1 mark]

Cue. Zero (the two midpoint voltages are equal).

Q2. State what an instrumentation amplifier rejects. [1 mark]

Cue. The common-mode signal (any voltage common to both inputs, including noise).

Q3. Name one example of signal conditioning applied after the amplifier. [1 mark]

Cue. Low-pass filtering to remove noise (or offsetting or scaling the signal).

Exam-style practice questions

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

Eduqas 20216 marksA Wheatstone bridge has three

1.0\ \text{k}\Omega

resistors and a strain gauge whose resistance rises from

1.0\ \text{k}\Omega

1.02\ \text{k}\Omega

under load. The bridge is supplied with

6.0\ \text{V}

. Calculate the output (the difference between the two midpoint voltages) when balanced and when strained.

Show worked answer →

Balanced (up to 2 marks): with all four arms equal, each midpoint sits at half the supply, $3.0\ \text{V}$ , so the output difference is $0\ \text{V}$ (the bridge is balanced).

Strained (up to 4 marks): one divider stays at $3.0\ \text{V}$ . The other has the strain gauge ( $1.02\ \text{k}\Omega$ ) and a $1.0\ \text{k}\Omega$ resistor. Its midpoint voltage is $6.0 \times \dfrac{R_\text{fixed}}{R_\text{fixed} + R_\text{gauge}}$ or $6.0 \times \dfrac{R_\text{gauge}}{\dots}$ depending on arrangement; taking the gauge in the lower arm, $V = 6.0 \times \dfrac{1.02}{1.0 + 1.02} = 6.0 \times 0.5050 = 3.030\ \text{V}$ . The output is the difference $3.030 - 3.000 = 0.030\ \text{V} = 30\ \text{mV}$ .

Markers reward the $0\ \text{V}$ balanced output, the strained midpoint near $3.03\ \text{V}$ , and the small difference of order $30\ \text{mV}$ that an instrumentation amplifier then amplifies.

Eduqas 20195 marksExplain why an instrumentation amplifier is used to amplify the output of a sensor bridge, referring to common-mode rejection.

Show worked answer →

Why an instrumentation amplifier (up to 3 marks): the bridge output is a small difference voltage sitting on top of a large common-mode voltage (both midpoints are near half the supply) and is often contaminated by noise picked up equally on both wires. An instrumentation amplifier amplifies the difference between its two inputs while rejecting any signal common to both, so it extracts the tiny wanted signal and ignores the common-mode level and noise.

Common-mode rejection (up to 2 marks): the common-mode rejection ratio (CMRR) measures how well the amplifier rejects common signals; a high CMRR means interference and the bias level are strongly suppressed while the difference signal is amplified. This gives a clean, accurate output from a noisy environment.

Markers reward the small difference on a large common-mode level, amplifying the difference while rejecting the common-mode, and a high CMRR giving noise immunity.

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Sources & how we know this

Eduqas GCE AS/A Level Electronics specification (A410QS) — WJEC Eduqas (2017)