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Chapter 1 · 3 hours

Fundamentals of Measurement

Practice questions

Practice questions and answers

3 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 5 marks

Explain the fundamental methods of measurement. Differentiate between the deflection method and the null method with one example of each.

Answer

Measurement is the process of comparing an unknown quantity with a standard of the same kind. The methods are grouped by how the comparison is made.

Fundamental methods

  • Direct method: the unknown is compared directly with the standard or read on a calibrated scale. Example: length with a steel rule, mass on a beam balance.
  • Indirect method: the quantity is found by measuring other quantities related to it by a known law. Example: power from P=VIP = VI, density from mass and volume.
  • Comparison method: the unknown is compared with a known quantity of the same kind (a balance with standard weights).
  • Substitution method: the unknown is replaced by a known standard that gives the same effect.
  • Null method: the known quantity is adjusted until the difference is zero.

Deflection and null methods

PointDeflection methodNull method
PrincipleOutput shown as pointer deflectionOpposing known effect is adjusted until detector reads zero
ReadingFrom the scaleFrom the setting of the known quantity
AccuracyLower; depends on instrument calibrationHigher; depends on the standard and detector sensitivity
Loading effectDraws power from sourceDraws almost no power at balance
SpeedFastSlow, needs balancing
ExampleMoving-coil ammeter, bourdon gaugePotentiometer, Wheatstone bridge, deadweight pressure balance

Example of null method: in a Wheatstone bridge the variable arm is adjusted until the galvanometer shows zero, and the unknown resistance is Rx=R2R3/R1R_x = R_2 R_3/R_1.

Example of deflection method: a spring balance shows weight from the spring extension on a marked scale.

  • Practice · 6 marks

Draw the block diagram of a generalized measurement system and explain the function of each stage, taking a thermocouple-based temperature indicator as the example.

Answer

A generalized measurement system converts the measured quantity (measurand) into a form suitable for observation. It has three main stages: sensing/transducing, signal conditioning, and output (readout).

 Measurand    +-----------+   +-------------+   +----------+
 ----------->| Sensor /  |-->|   Signal    |-->| Terminal |
 (temperature)| transducer|   | conditioning|   | (readout)|
              +-----------+   +-------------+   +----------+
   Stage I        Stage II (modify, amplify)      Stage III
                         ^
                    Power supply / calibration

Functions of the stages

  1. Sensor-transducer stage (detector): senses the measurand and gives an output related to it. It should respond only to the measurand and not load the source.
  2. Signal-conditioning stage: modifies the transducer output into a suitable form. It includes amplification, filtering, bridge circuits, analog-to-digital conversion, linearization and modulation.
  3. Terminating (readout) stage: presents the result to the observer or controller. It may be an indicator, recorder, display or data logger.

Thermocouple temperature indicator

  • Sensor: the thermocouple produces a small emf (mV) proportional to the difference between hot and cold junction temperatures.
  • Signal conditioning: a cold-junction compensation circuit adds the reference emf, an amplifier raises the mV signal, a filter removes noise, and linearization corrects the nonlinear emf-temperature relation. An ADC converts it to digital form.
  • Readout: a digital display or chart recorder shows the temperature in ∘^\circC.

Calibration against a known standard temperature links the readout to the measurand.

  • Practice · 3+2 marks

(a) The power dissipated in a resistor is calculated from P=V2/RP = V^2/R. The voltage is measured as 120 V with a limiting error of ±0.5%\pm 0.5\% and the resistance is 50 Ω\Omega with a guaranteed accuracy of ±1%\pm 1\%. Calculate the power and its limiting error in watt and in percent.
(b) Differentiate between systematic error and random error, giving one cause of each.

Answer

Power and its limiting error

Given: V=120V = 120 V (±0.5%\pm 0.5\%), R=50 ΩR = 50\ \Omega (±1%\pm 1\%).

P=V2R=120250=288 WP = \frac{V^2}{R} = \frac{120^2}{50} = 288\ \text{W}

For a quantity P=V2R−1P = V^2 R^{-1}, limiting (worst-case) relative errors add, each multiplied by its power:

δPP=2 δVV+δRR=2(0.5)+1=2%\frac{\delta P}{P} = 2\,\frac{\delta V}{V} + \frac{\delta R}{R} = 2(0.5) + 1 = 2\% δP=0.02×288=5.76 W\delta P = 0.02 \times 288 = 5.76\ \text{W}

Answer: P=288±5.76P = 288 \pm 5.76 W (limiting error ±2%\pm 2\%).

If the errors are independent, the probable error is (2×0.5)2+12=1.41%\sqrt{(2\times 0.5)^2 + 1^2} = 1.41\%, about ±4.1\pm 4.1 W.

Systematic and random errors

PointSystematic errorRandom error
NatureConstant or varies in a known way, same signVaries unpredictably, positive or negative
Cause exampleZero error, wrong calibration, loading effectElectrical noise, friction, vibration, observer variation
EffectShifts all readings (affects accuracy)Scatters readings (affects precision)
ReductionCalibration, correction, better methodAveraging many readings, statistics

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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