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Measurements · Module 6 of 10

Analog Electrical Measurements

Most transducers speak in volts and ohms. This module covers the Wheatstone bridge that turns a resistance change into a voltage, and the loading error that a real meter introduces.

01

Readiness check

This module measures electrical quantities. Tick only what you can do closed-notes.

  • Cross-multiply a ratio of resistances.
  • Form a ratio like R/(R + r).
  • Recall Ohm's law and a voltage divider.
  • Recall that meters have an input resistance.
  • Convert a fraction to a percentage.
0 or 1 weak itemsContinue with this module.
2 weak itemsRevisit Thevenin and loading in Electrical Circuits, Module 3.
3 or more weak itemsRevisit calibration in Module 1.
02

The core idea

A Wheatstone bridge turns a small resistance change into a measurable voltage and is balanced when R1/R2 = R3/R4. Any real meter draws some current, so the value it reads is scaled by Rm/(Rm + Rs): loading error, minimized by a high input impedance.

bridge balance: R1/R2 = R3/R4loading: Vread/Vtrue = Rm/(Rm + Rs)loading error = −Rs/(Rs + Rm)

Many mechanical measurands, strain, temperature, displacement, reach the electronics as a change in resistance or a small voltage, so the analog circuit that reads them shapes the result. The workhorse is the Wheatstone bridge: four resistances in a diamond with a source across one diagonal and a detector across the other. When the ratios match, R1/R2 = R3/R4, the bridge is balanced and the detector reads zero; a small change in one arm, as from a strain gauge, unbalances it into a voltage proportional to that change, which is how tiny resistance shifts become usable signals. The second essential idea is loading. An ideal voltmeter would draw no current, but a real one has a finite input resistance Rm, and connecting it across a source with output resistance Rs forms a divider, so the reading is only Rm/(Rm + Rs) of the true value. The resulting loading error is negative and shrinks as Rm grows relative to Rs, which is why instruments are designed with high input impedance and why impedance matching matters. Amplifiers, filters, and careful grounding then condition the bridge output for the digital stages ahead.

The skill works when: you balance the bridge by matching arm ratios and keep Rm far above Rs.
The skill breaks down when: a meter is assumed ideal, or the bridge balance ratio is inverted.
The concept. A Wheatstone bridge nulls when its arm ratios match. Unbalancing one arm produces a voltage proportional to the resistance change.
03

The skills, taught in order

Five skills read a transducer's electrical output without corrupting it.

6.1 Measuring current, voltage, resistance

Ammeters go in series, voltmeters across, and resistance is found by driving a known current and reading the voltage. Each connection perturbs the circuit, which is exactly what loading analysis quantifies.

6.2 The Wheatstone bridge

The bridge compares two voltage dividers. At balance, R1/R2 = R3/R4 and the detector reads zero; a small resistance change unbalances it into a voltage. Null operation is precise; deflection operation is fast.

6.3 Loading error and impedance matching

A meter of input resistance Rm across a source of resistance Rs reads Rm/(Rm + Rs) of the true voltage. The loading error is small only when Rm is much larger than Rs, the reason instruments use high input impedance.

Rm / RsVread/VtrueLoading error
100.909−9.1%
1000.990−1.0%
10000.999−0.1%

Loading error falls as the meter's impedance grows relative to the source. Aim for at least a hundred to one.

6.4 Amplifiers and conditioning

A bridge output is often only millivolts, so an instrumentation amplifier with high input impedance and common-mode rejection boosts it without loading the bridge, readying it for the converter.

6.5 Filters, grounding, and shielding

Real analog signals carry noise. Low-pass filtering, a single clean ground, and shielded cable keep interference, especially at mains frequency, from swamping a small transducer signal.

Engineering connection: a strain-gauge bridge feeding a high-impedance instrumentation amplifier is the standard front end of a load cell, chosen precisely to avoid loading error.

04

Worked example 1: balancing a bridge

A Wheatstone bridge has R1 = 100 Ω, R2 = 200 Ω, and R3 = 150 Ω. What value of R4 balances it?

Figure 1. Balance requires equal arm ratios, so R4 is found by cross-multiplying, giving 300 ohms.
  1. ProblemFind R4 that balances the bridge in Figure 1.
  2. Given / findR1 = 100 Ω, R2 = 200 Ω, R3 = 150 Ω. Find R4.
  3. AssumptionsIdeal resistors; balance means zero detector current.
  4. ModelBalance: R1/R2 = R3/R4, so R4 = R3R2/R1.
  5. EquationsR4 = 150 × 200 / 100
  6. SolveR4 = 300 Ω.
  7. CheckR1/R2 = 0.5 and R3/R4 = 150/300 = 0.5, so the ratios match and the bridge nulls.
  8. ConclusionWith R4 = 300 Ω the bridge is balanced; any deviation from 300 Ω would appear as a detector voltage.
Result. R4 = 300 Ω for balance.
05

Worked example 2: voltmeter loading error

A voltmeter of input resistance Rm = 1 MΩ measures a source whose output resistance is Rs = 100 kΩ. Find the loading error.

Figure 2. The source resistance and the meter form a divider. The meter reads 90.9 percent of the true voltage, a 9.1 percent loading error.
  1. ProblemFind the loading error for the meter in Figure 2.
  2. Given / findRm = 1 MΩ, Rs = 100 kΩ. Find the loading error.
  3. AssumptionsSource modelled as an ideal voltage behind Rs; meter as Rm.
  4. ModelVread/Vtrue = Rm/(Rm + Rs); error = that ratio − 1.
  5. Equationsratio = 1000 / (1000 + 100)error = ratio − 1
  6. Solveratio = 1000/1100 = 0.909; error = −9.1%.
  7. CheckRm/Rs = 10, and the table gives −9.1 percent at a ratio of 10, matching.
  8. ConclusionThe meter reads 9.1 percent low; a 10 MΩ meter would cut the error to about 1 percent.
Result. Loading error −9.1 percent.
06

Misconceptions and diagnostics

MistakeSymptomDiagnostic questionCorrection
Assuming an ideal meterReading trusted despite low impedance"How does Rm compare to Rs?"Account for loading unless Rm ≫ Rs.
Inverting the balance ratioWrong R4 for balance"Do the arm ratios match?"Balance is R1/R2 = R3/R4.
Loading a bridge outputMillivolt signal collapses"Is the amplifier high impedance?"Buffer a bridge with a high-impedance amplifier.
Ignoring groundingMains hum on the signal"One clean ground?"Use a single ground and shielding.
07

Practice ladder

Level 1 · Direct skill

A bridge has R1 = 200 Ω, R2 = 400 Ω, R3 = 100 Ω. Find R4 for balance.

Show answer

R4 = R3R2/R1 = 100 × 400 / 200 = 200 Ω.

Level 2 · Mixed concept

A voltmeter with Rm = 1 MΩ measures a source with Rs = 50 kΩ. Find the loading error.

Show answer

ratio = 1000/1050 = 0.952; error = −4.76%.

Level 3 · Independent problem

A source has Rs = 1 kΩ. What meter input resistance keeps the loading error below 1 percent?

Show answer

Need Rm/(Rm+Rs) ≥ 0.99, so Rm ≥ 99 Rs = 99 kΩ.

Transfer task | Real engineering

A strain gauge changes resistance by a tiny fraction. Explain why a Wheatstone bridge, not a direct ohmmeter, is used to read it.

What good work looks like

A gauge's resistance change is a fraction of a percent, far too small to read directly with useful precision. A bridge nulls out the large baseline resistance so only the change produces an output, which an amplifier then boosts. A good answer notes the small ΔR and the bridge's baseline cancellation.

08

Working with AI, and proving it yourself

Use AI as an examiner, not a solver

"Check that I matched the arm ratios for bridge balance."
"Give me source and meter resistances; I will find the loading error."
"Find R4 for me." Cross-multiply the balance yourself.
"Is this meter fine?" Compute the loading error yourself.

Portfolio task

Analyze one real analog measurement: compute a bridge balance or a loading error and state how to reduce it.

Must include: a bridge balance or a loading ratio, the resulting error, and a mitigation.
09

Retrieval and spaced review

Closed notes. Answer out loud, then reveal.

1. Write the bridge balance condition.

R1/R2 = R3/R4.

2. Write the loading ratio.

Vread/Vtrue = Rm/(Rm + Rs).

3. How to reduce loading error?

Make the meter's input resistance much larger than the source resistance.

4. Why a bridge for a strain gauge?

It cancels the baseline resistance so only the tiny change produces output.

5. Why an instrumentation amplifier?

High input impedance and common-mode rejection boost a bridge output without loading it.

TodayFinish this quiz and Levels 1 and 2 of the ladder.
+1 dayRe-derive a bridge balance and a loading error.
+3 daysAnalyze one real analog front end.
+7 daysMove on to sampling and data acquisition in Module 7.
+30 daysCheck meter impedance against source before trusting a reading.
10

Textbook mapping

This module follows Figliola and Beasley, Theory and Design for Mechanical Measurements, 5th edition. Use these references to read further.

Topic in this moduleWhere to read more
Resistance measurement and the bridgeFigliola and Beasley, Section 6.4, Analog Devices: Resistance Measurements
Loading errors and impedance matchingFigliola and Beasley, Section 6.5, Loading Errors and Impedance Matching
Amplifiers and filtersFigliola and Beasley, Sections 6.6 and 6.8

Section numbers refer to Figliola and Beasley, 5th edition. Any edition with the same chapter titles is equivalent for study.