WJEC WJEC Eduqas 2020-style practice: A Wheatstone bridge has three fixed 1.0\,k resistors and a strain gauge whose resistance is 1.0\,k when unstrained. The bridge is supplied with 6.0\,V. State the output when the bridge is balanced, and find the output voltage when strain raises the gauge to 1.02\,k.

A Wheatstone bridge is two potential dividers in parallel; the output is the difference between their midpoints. Balanced: with all four arms equal, both dividers give 3.0\,V, so the output difference is 0\,V. Strained, the gauge arm now gives V = 6.0 × 1.001.00 + 1.02 = 6.0 × 0.495 = 2.97\,V (taking the output across the fixed 1.0\,k partner), while the reference divider still gives 3.0\,V. Output: 3.0 - 2.97 = 0.03\,V = 30\,mV. Markers reward the zero balanced output, the changed divider value, and the small differential output of about 30\,mV.

WalesElectronicsSyllabus dot point

How is a small sensor signal conditioned, amplified and calibrated into a usable measurement?

Instrumentation systems: sensors and transducers, the Wheatstone bridge, signal conditioning and amplification, calibration, and the use of the instrumentation amplifier.

A focused answer to WJEC A-Level Electronics instrumentation systems, covering sensors and transducers, the Wheatstone bridge for resistive sensors, signal conditioning and amplification, calibration, and the role of the instrumentation amplifier.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-16

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

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What this dot point is asking

Instrumentation systems turn a physical measurement into a clean, calibrated electrical reading. WJEC expects you to describe sensors and transducers, use the Wheatstone bridge for resistive sensors, explain signal conditioning and amplification, discuss calibration, and link to the instrumentation amplifier. The Wheatstone-bridge calculation and the case for an instrumentation amplifier are reliable, high-mark exam content.

The answer

Sensors and the measurement chain

The Wheatstone bridge

The bridge is ideal for resistive sensors because it gives zero output at the reference condition, so only the change is measured, and it cancels effects (like temperature) common to its arms.

Signal conditioning and the instrumentation amplifier

Calibration

Calibration relates the conditioned output to the true value of the measured quantity, by comparing the system's reading against known standards and adjusting the gain and offset so the scale reads correctly.

Why a bridge is better than a single divider for a sensor

A designer compares a single potential divider and a Wheatstone bridge for a strain gauge. Explain the advantage of the bridge.

step State what a single divider gives

A single divider with the gauge gives an output voltage that includes a large standing value plus a tiny change due to strain, so the change is a small fraction of a large number, hard to amplify cleanly.

step State what the bridge gives

A bridge compares the gauge divider against a matched reference divider, so at zero strain the output is zero; only the change appears at the output.

step Draw the conclusion

The bridge output is just the small change, with no large standing offset, so it can be amplified by a high-gain instrumentation amplifier without the amplifier saturating. The bridge also cancels temperature effects common to both arms, giving a cleaner, more stable measurement.

Examples in context

Example 1. An electronic weighing scale: Strain gauges in a Wheatstone bridge under the platform give a tiny voltage proportional to the load, which an instrumentation amplifier raises and an ADC reads. Calibration with known masses sets the scale so the display reads the correct weight, showing the full sensor-to-reading chain.
Example 2. A temperature logger: A thermistor in a bridge produces a small differential signal as temperature changes; signal conditioning amplifies and filters it before logging. The bridge cancels supply-voltage drift common to both arms, giving a stable reading that a single divider could not.
Example 3. A blood-pressure monitor: A pressure sensor bridge gives a small signal that is amplified, filtered to remove movement noise, and calibrated against a known pressure. The instrumentation amplifier's common-mode rejection is essential here to strip out electrical interference from the tiny medical signal.

Try this

Q1. State the output of a Wheatstone bridge when it is balanced, and what causes a non-zero output. [2 marks]

Cue. Zero output when balanced; a change in the sensor arm's resistance unbalances the bridge and produces a small differential output.

Q2. Give two functions performed by signal conditioning in an instrumentation system. [2 marks]

Cue. Any two of: amplifying a small signal, filtering out noise, adding or removing an offset, converting the signal to a suitable form.

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.

WJEC Eduqas 20205 marksA Wheatstone bridge has three fixed

1.0\,\text{k}\Omega

resistors and a strain gauge whose resistance is

1.0\,\text{k}\Omega

when unstrained. The bridge is supplied with

6.0\,\text{V}

. State the output when the bridge is balanced, and find the output voltage when strain raises the gauge to

1.02\,\text{k}\Omega

Show worked answer →

A Wheatstone bridge is two potential dividers in parallel; the output is the difference between their midpoints.

Balanced: with all four arms equal, both dividers give $3.0\,\text{V}$ , so the output difference is $0\,\text{V}$ .

Strained, the gauge arm now gives $V = 6.0 \times \dfrac{1.00}{1.00 + 1.02} = 6.0 \times 0.495 = 2.97\,\text{V}$ (taking the output across the fixed $1.0\,\text{k}\Omega$ partner), while the reference divider still gives $3.0\,\text{V}$ .

Output: $3.0 - 2.97 = 0.03\,\text{V} = 30\,\text{mV}$ .

Markers reward the zero balanced output, the changed divider value, and the small differential output of about $30\,\text{mV}$ .

WJEC Eduqas 20184 marksExplain the purpose of signal conditioning in an instrumentation system, and why the small differential output of a sensor bridge is usually amplified by an instrumentation amplifier.

Show worked answer →

Signal conditioning prepares a raw sensor signal so the rest of the system can use it: this includes amplifying a small signal, filtering out noise, removing or adding offsets, and converting it into a suitable form.

A sensor bridge produces a very small differential voltage (often only millivolts) sitting on a larger common voltage, from relatively high-impedance outputs. An instrumentation amplifier is used because it has a very high input resistance (so it does not load and unbalance the bridge), a high and well-defined differential gain (to lift the tiny signal), and strong common-mode rejection (to remove the common voltage and pick-up noise).

Markers reward the role of signal conditioning (amplify, filter, offset) and the instrumentation amplifier's high input resistance, gain and common-mode rejection.

Related dot points

Sources & how we know this

WJEC Eduqas GCE A-level Electronics specification — WJEC Eduqas (2017)