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RTD and Thermocouple Quiz

Test yourself with 20 multiple choice questions. Check your answers to get a score and an explanation for every question.

20 questionsInstant scoreExplanationsInterview practice

RTD and Thermocouple Quiz: 20 Questions

Choose one answer for each question, then press Check answers. Explanations appear after checking.

1. Resistance of a Pt100 at 0 °C is:

Answer: 100 Ω. Pt100 means 100 Ω at 0 °C. Learn more: Pt100

2. Resistance of a Pt100 at 100 °C is about:

Answer: 138.5 Ω. R = 100 (1 + 0.39083 − 0.00578) = 138.51 Ω. Learn more: Pt100

3. Standard alpha coefficient of IEC 60751 platinum is:

Answer: 0.00385. α = 0.00385 /°C. Learn more: Alpha Coefficient

4. Which RTD connection fully removes lead resistance error?

Answer: 4 wire. Voltage is sensed on leads carrying almost no current. Learn more: RTD Wiring (2, 3 and 4 Wire)

5. IEC 60751 class A tolerance at 100 °C is:

Answer: ± 0.35 °C. ± (0.15 + 0.002 × 100) = ± 0.35 °C. Learn more: RTD Tolerance Class
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6. Thermocouple voltage depends on:

Answer: Temperature difference between hot and cold junctions. Seebeck voltage is generated along the gradient. Learn more: Thermocouple

7. Cold junction compensation adds:

Answer: The table voltage at terminal temperature. Add voltages, not temperatures. Learn more: Cold Junction Compensation

8. Type K thermocouple output near room temperature is about:

Answer: 41 µV/°C. Seebeck coefficient about 41 µV/°C. Learn more: Seebeck Effect

9. Which thermocouple type is best below 0 °C?

Answer: Type T. Type T copper constantan suits low temperatures. Learn more: Thermocouple Types

10. Type K class 1 tolerance at 800 °C is:

Answer: ± 3.2 °C. 0.004 × 800 = 3.2 °C, larger than 1.5 °C. Learn more: Thermocouple Tolerance Class
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11. Using copper cable from a thermocouple head to the DCS causes:

Answer: Error, since the cold junction moves to the unmeasured head. Extension or compensating cable must continue to the CJC point. Learn more: Thermocouple Extension Cable

12. T90 of a first order sensor is about:

Answer: 2.3 time constants. ln 10 ≈ 2.3. Learn more: Response Time

13. Thermowell failures are mostly caused by:

Answer: Vortex induced vibration. Shedding near the natural frequency causes fatigue. Learn more: Wake Frequency

14. Thermowell design is checked with:

Answer: ASME PTC 19.3 TW. It covers frequency limits and stresses. Learn more: Thermowell

15. An NTC thermistor resistance:

Answer: Falls as temperature rises. Negative temperature coefficient. Learn more: NTC Thermistor
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16. A pyrometer reading a polished metal surface reads low because of:

Answer: Low emissivity. Low emitting surfaces radiate less and reflect surroundings. Learn more: Emissivity

17. Temperature class T4 limits surface temperature to:

Answer: 135 °C. T1 450, T2 300, T3 200, T4 135, T5 100, T6 85 °C. Learn more: Temperature Class

18. Self heating error of an RTD increases with:

Answer: Measuring current. Power = I² R. Learn more: RTD Self Heating

19. A temperature transmitter ranged 0 to 200 °C outputs at 75 °C:

Answer: 10 mA. 4 + 16 × 0.375 = 10 mA. Learn more: Temperature Transmitter

20. 0 °C in kelvin is:

Answer: 273.15 K. T(K) = t(°C) + 273.15. Learn more: Temperature Scales
0 of 20 answered
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About this Quiz

20 multiple choice questions on rtd and thermocouple, the kind asked in instrumentation interviews, technician exams and plant training. Each answer comes with a short explanation, so the quiz works as a revision sheet too.

Show answer key
  1. 100 Ω: Pt100 means 100 Ω at 0 °C. Learn more: Pt100
  2. 138.5 Ω: R = 100 (1 + 0.39083 − 0.00578) = 138.51 Ω. Learn more: Pt100
  3. 0.00385: α = 0.00385 /°C. Learn more: Alpha Coefficient
  4. 4 wire: Voltage is sensed on leads carrying almost no current. Learn more: RTD Wiring (2, 3 and 4 Wire)
  5. ± 0.35 °C: ± (0.15 + 0.002 × 100) = ± 0.35 °C. Learn more: RTD Tolerance Class
  6. Temperature difference between hot and cold junctions: Seebeck voltage is generated along the gradient. Learn more: Thermocouple
  7. The table voltage at terminal temperature: Add voltages, not temperatures. Learn more: Cold Junction Compensation
  8. 41 µV/°C: Seebeck coefficient about 41 µV/°C. Learn more: Seebeck Effect
  9. Type T: Type T copper constantan suits low temperatures. Learn more: Thermocouple Types
  10. ± 3.2 °C: 0.004 × 800 = 3.2 °C, larger than 1.5 °C. Learn more: Thermocouple Tolerance Class
  11. Error, since the cold junction moves to the unmeasured head: Extension or compensating cable must continue to the CJC point. Learn more: Thermocouple Extension Cable
  12. 2.3 time constants: ln 10 ≈ 2.3. Learn more: Response Time
  13. Vortex induced vibration: Shedding near the natural frequency causes fatigue. Learn more: Wake Frequency
  14. ASME PTC 19.3 TW: It covers frequency limits and stresses. Learn more: Thermowell
  15. Falls as temperature rises: Negative temperature coefficient. Learn more: NTC Thermistor
  16. Low emissivity: Low emitting surfaces radiate less and reflect surroundings. Learn more: Emissivity
  17. 135 °C: T1 450, T2 300, T3 200, T4 135, T5 100, T6 85 °C. Learn more: Temperature Class
  18. Measuring current: Power = I² R. Learn more: RTD Self Heating
  19. 10 mA: 4 + 16 × 0.375 = 10 mA. Learn more: Temperature Transmitter
  20. 273.15 K: T(K) = t(°C) + 273.15. Learn more: Temperature Scales
Tip: take the quiz before an interview, note the questions you missed and read the linked article for those topics.
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Frequently Asked Questions

Is this quiz free?

Yes. No sign up is needed and you can retake it as often as you like.

Are the answers explained?

Yes. After you check your answers, each question shows the correct answer and a short explanation.

Where can I find more questions?

See the Interview Q&A section of instrumentationblog.in and the linked article below.

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