--- title: "Gauge resistance change and quarter-bridge output" subject: "Instrumentation and Measurement" chapter: 6 key: "semester_4/instrumentation_and_measurement:gauge-resistance-change-quarter-bridge" source: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/gauge-resistance-change-quarter-bridge site: BCT Engineering --- # Gauge resistance change and quarter-bridge output Instrumentation and Measurement · Chapter 6: Strain Gage - Asked in: year not given, 6 marks - Video: [Wheatstone Bridge with strain gauge explained](https://www.youtube.com/watch?v=GjmM6KKhgp0) — learnchannel - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_4/instrumentation_and_measurement/gauge-resistance-change-quarter-bridge - Whole chapter: https://www.bctengineer.com/md/semester_4/instrumentation_and_measurement/chapter-6 ## Question A 120 $\Omega$ strain gauge has a gauge factor of 2.1 and is bonded to a steel member that is strained by 1500 microstrain. Calculate the change in resistance and the fractional change in resistance. If the gauge forms one arm of an initially balanced Wheatstone bridge with all arms 120 $\Omega$ and 5 V excitation, find the bridge output voltage. ## Short answer Use dR/R = G epsilon = 2.1 x 1500e-6 = 0.00315, so dR = 0.378 ohm. Quarter bridge output is about Vex/4 times dR/R = 5/4 x 0.00315 = 3.94 mV; exact calculation gives 3.93 mV. ## Answer Given: $R_g = 120\ \Omega$, $G = 2.1$, $\varepsilon = 1500\ \mu\varepsilon = 1.5\times10^{-3}$. #### Change in resistance $$ \frac{\Delta R}{R} = G\varepsilon = 2.1 \times 1.5\times10^{-3} = 3.15\times10^{-3} = 0.315\% $$ $$ \Delta R = 120 \times 3.15\times10^{-3} = 0.378\ \Omega $$ #### Quarter-bridge output All arms are 120 $\Omega$ and $V_{ex} = 5$ V. With one active arm of resistance $R + \Delta R$: $$ V_o = V_{ex}\left[\frac{R + \Delta R}{2R + \Delta R} - \frac{1}{2}\right] = V_{ex}\frac{\Delta R}{2(2R + \Delta R)} $$ $$ V_o = 5\times\frac{0.378}{2(240.378)} = 3.931\times10^{-3}\ \text{V} $$ The usual small-change approximation gives $$ V_o \approx \frac{V_{ex}}{4}\frac{\Delta R}{R} = \frac{5}{4}(3.15\times10^{-3}) = 3.94\ \text{mV} $$ ```text +---- R1 = 120 ----B---- R_g = 120 + dR ----+ | | | 5 V (+) A V_o (between B and D) C (-) | | | +---- R3 = 120 ----D------ R2 = 120 --------+ (5 V excitation applied across A and C) ``` **Answer: $\Delta R = 0.378\ \Omega$ ($0.315\%$); bridge output $\approx 3.93$ mV (3.94 mV by the approximate formula).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For G = 2.1 and 1500 microstrain, what is the fractional change in resistance? A. 3.15 x 10^-3 (0.315%) B. 1.4 x 10^-3 C. 3.15 x 10^-2 D. 0.714 x 10^-3 Answer: A. dR/R = G x strain = 2.1 x 1.5e-3 = 3.15e-3. (Key point: Fractional change) 2. For a 120 ohm gauge, what is the change in resistance? A. 3.78 ohm B. 0.378 ohm C. 0.0378 ohm D. 37.8 ohm Answer: B. dR = 120 x 3.15e-3 = 0.378 ohm. (Key point: Resistance change) 3. What is the approximate output of a quarter bridge with 5 V excitation and dR/R = 3.15e-3? A. 15.75 mV B. 7.9 mV C. 3.94 mV D. 0.39 mV Answer: C. Vo is about (Vex/4)(dR/R) = (5/4)(3.15e-3) = 3.94 mV; the exact formula gives 3.93 mV. 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