Hysteresis in a transducer means that the output for a given input depends on
Answer: A
The two branches of the characteristic do not coincide, so a reading taken while loading differs from one taken while unloading. Mechanical friction and magnetic or elastic memory in the materials are the usual causes, and it is quoted as a percentage of full scale.
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The linearity of a transducer is usually expressed as
Answer: B
A straight line is fitted to the calibration data and the greatest deviation from it is quoted. Which line is fitted matters, since a line through the end points and a least squares line give different figures for the same data.
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A strain gauge of 120 \( \Omega \) and gauge factor 2 undergoes a strain of 0.001. The change of resistance is
Answer: C
From \( \Delta R = R \times G \times \epsilon \) the change is \( 120 \times 2 \times 0.001 = 0.24\ \Omega \). A fifth of one per cent on a hundred and twenty ohms is why a bridge and an amplifier are needed to see it at all.
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A quarter bridge with a 120 \( \Omega \) gauge of factor 2 is excited at 5 V and strained by 0.0005. The output is about
Answer: D
The output of a quarter bridge is \( V G \epsilon /4 \), that is \( 5 \times 2 \times 0.0005/4 = 1.25 \times 10^{-3} \) V. A half bridge would give twice this and a full bridge four times, which is the reason for using more gauges.
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A platinum resistance thermometer of 100 \( \Omega \) at 0 \( ^\circ \)C has \( \alpha = 0.00385 \) per \( ^\circ \)C. At 100 \( ^\circ \)C its resistance is
Answer: A
Correct answer: 138.5 \( \Omega \)
\[ R = R_0(1+\alpha T) = 100\,(1 + 0.00385\times100) = 100 \times 1.385 = 138.5\ \Omega. \] This is the standard Pt100 characteristic, and 138.5 ohms at 100 degrees is the figure to recognise. Applying the coefficient once rather than over the hundred degrees gives the 103.85 offered.
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An LVDT of sensitivity 20 mV per mm shows an output of 250 mV. The core displacement is
Answer: B
Correct answer: 12.5 mm
\[ \frac{250\ \mathrm{mV}}{20\ \mathrm{mV/mm}} = 12.5\ \mathrm{mm}. \] Dividing by the sensitivity converts output back to input; multiplying instead gives 5000 mm, which is offered here. The sign of the displacement is not recoverable from magnitude alone, which is why an LVDT is used with phase-sensitive detection.
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A thermocouple of 41 \( \mu \)V per \( ^\circ \)C has its reference junction at 25 \( ^\circ \)C and gives 4.10 mV. The measuring junction is at
Answer: C
The emf corresponds to a difference of \( 4100/41 = 100 \) degrees, and adding the reference temperature of twenty-five gives a measuring junction at 125 degrees. Forgetting the cold junction is the commonest error in thermocouple work.
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A capacitive transducer has plates of 100 mm\( ^2 \) separated by 1 mm of air. Taking \( \epsilon_0 = 8.85 \) pF/m, its capacitance is about
Answer: D
From \( C = \epsilon_0 A/d \), with \( A = 10^{-4} \) m² and \( d = 10^{-3} \) m, the capacitance is \( 8.85 \times 10^{-12} \times 10^{-1} = 0.885 \) pF. Such a small value shows why stray capacitance is the dominant practical difficulty.
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Halving the spacing of that capacitive transducer changes its capacitance to
Answer: A
Correct answer: 1.77 pF
Parallel plate capacitance is inversely proportional to the spacing, \[ C = \frac{\epsilon A}{d}, \] so halving \( d \) doubles \( C \): \( 2 \times 0.885 = 1.77 \) pF. That inverse relation makes the variable-gap capacitor very sensitive but distinctly non-linear, which is why such transducers are used over small gaps or in a differential arrangement.
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A semiconductor strain gauge of factor 120 and resistance 350 \( \Omega \) is strained by 0.0001. The resistance change is
Answer: B
The product is \( 350 \times 120 \times 0.0001 = 4.2\ \Omega \), against 0.07 ohm for a metallic gauge of factor two on the same strain. The sixty-fold advantage is real but comes with poor linearity and heavy temperature dependence.
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