Measurement and Instrumentation: practice problems and exam questions

Measuring physical quantities with electrical circuits: Wheatstone bridges, strain gauges and load cells, amplifiers that condition small sensor signals, and the loading of a measurement by the instrument.

Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Wheatstone bridge & strain gaugesbalance and node voltages · unbalanced bridges · strain gauges and load cellscalculations
  2. Data acquisition & samplingNyquist, LSB and data rates · aliasing, bits and UART frames · harmonics, SQNR and system designcalculations
  3. Op-amp circuitsinverting amplifier · non-inverting amplifier · summing amplifiercalculations
  4. Series & parallel resistors3 resistors · 4 resistors · ladder of 6 resistorscalculations
  5. Thévenin & Nortondivider network · with series resistor and load · with current source and loadcalculations

Exam tips for Measurement and Instrumentation

  • A bridge is two voltage dividers: work out each node voltage and subtract.
  • Strain is usually given in microstrain: 500 µε is 0.0005.
  • Count the active gauges: a quarter, half or full bridge changes the output by a factor of 4.

Worked examples

Wheatstone bridge & strain gaugesEasy
+−6 VR1​ = 250 ΩR2​ = 300 ΩR3​ = ?R4​ = 150 ΩabG0

The Wheatstone bridge in the diagram is balanced by adjusting R₃. R₁ = 250 Ω, R₂ = 300 Ω and R₄ = 150 Ω. Find the setting R3.

R1 = 250 ΩR2 = 300 ΩR4 = 150 Ω

Find R3 (Ω)

Show hints
  1. At balance both dividers give the same voltage: Va = Vb.
  2. R₁/R₂ = R₃/R₄.
  3. R₃ = R₁·R₄/R₂.
Show the answer and the step-by-step solution

Wheatstone bridge: two voltage dividers share one supply Vs. The left divider R1–R2 gives Va = Vs·R2/(R1 + R2) and the right divider R3–R4 gives Vb = Vs·R4/(R3 + R4); the output is Vab = Va − Vb. The bridge is balanced (Vab = 0, no current in the meter) when R1/R2 = R3/R4, that is R1·R4 = R2·R3. Seen from the output, the bridge is a Thévenin source with Rth = R1∥R2 + R3∥R4.

Balance condition

R₁·R₄ = R₂·R₃

Unknown resistor

R₃ = 250·150/300 = 125 Ω

Answer

R3 = 125 Ω

The supply voltage does not matter at balance.

Data acquisition & samplingMedium
Vin​N = 85 V2ᴺ − 101 LSB2.125 V?

An 8-bit ADC with an input range of 0 to 5 V truncates: its output code is ⌊Vin/LSB⌋ with 1 LSB = VFSR/2ᴺ. Find the decimal output code D for an input of 2.125 V.

N = 8VFSR = 5 VVin = 2.125 V

Find D

Show hints
  1. First find the size of one LSB.
  2. 1 LSB = 5/256 = 19.53 mV.
  3. Divide the input by the LSB and keep the whole part.
Show the answer and the step-by-step solution

ADC resolution: an N-bit converter divides its full-scale range VFSR into 2ᴺ codes, from 0 to 2ᴺ − 1. One step is 1 LSB = VFSR/2ᴺ: the smallest change the converter can show. A truncating ADC gives the code ⌊Vin/LSB⌋; a rounding ADC is never more than ½ LSB away from the true value, its quantisation error. Every extra bit halves the LSB and adds about 6 dB to the ideal signal-to-quantisation-noise ratio SQNR = 6.02·N + 1.76 dB for a full-scale sine. To reach a resolution r over a span, choose the smallest N with 2ᴺ ≥ span/r.

Step size

1 LSB = 5/28 = 19.53 mV

Code

Vin/LSB = 2125/19.53 = 108.8
code = ⌊108.8⌋ = 108

Answer

D = 108

In binary that is 01101100.

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

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