Notes
Thermoelectric Effect — 20 MCQs
20 MCQs covering Seebeck effect, thermocouple construction & working, Peltier effect and thermopile devices. Options are shuffled each load.
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1. Seebeck effect describes:
easyExplanation: Seebeck effect: a temperature difference between two junctions of dissimilar metals produces a thermoelectric emf. -
2. Seebeck coefficient has units of:
easyExplanation: Seebeck coefficient (thermopower) is emf per unit temperature difference, units volts per kelvin. -
3. A thermocouple is formed by:
easyExplanation: Thermocouple: two different conductors joined at two junctions; temperature difference produces emf. -
4. In a thermocouple measurement the measured emf depends on:
mediumExplanation: Thermocouple emf is integral of difference of Seebeck coefficients over temperatures; depends on ΔT and material properties. -
5. To measure temperature using a thermocouple one typically needs a reference (cold) junction because:
easyExplanation: Thermocouples produce emf proportional to temperature difference between measuring junction and reference junction; reference temperature must be known. -
6. Cold-junction (reference) compensation is required because:
mediumExplanation: If reference is not at 0°C, compensation (electronic or using an ice bath) is needed to get correct absolute temperature reading. -
7. Which of the following metals/pairs is a common thermocouple type (e.g., type K)?
mediumExplanation: Type K thermocouples use chromel (Ni–Cr) and alumel (Ni–Al) and are widely used for general-purpose measurements. -
8. Seebeck coefficient can be positive or negative depending on:
mediumExplanation: If metal A has greater thermopower than B the measured emf sign depends on which is positive; coefficients can be positive or negative. -
9. Thermopile is:
easyExplanation: Thermopile stacks thermocouples in series (often with alternate hot/cold junctions) to amplify voltage for sensing. -
10. Application of thermopile includes:
mediumExplanation: Thermopiles detect radiant energy (IR) by absorbing radiation to heat junctions and produce measurable voltage. -
11. Peltier effect is:
easyExplanation: Peltier effect: when current passes through junction of dissimilar conductors heat is absorbed or released at junction depending on current direction. -
12. Peltier coefficient has units of:
hardExplanation: Peltier coefficient Π has units of energy per charge (J/C) and is often expressed as volts. -
13. A Peltier cooler (thermoelectric cooler) works on which effect?
easyExplanation: Thermoelectric coolers use Peltier effect: DC current creates heat pumping from one side to the other producing a temperature difference. -
14. If two thermocouple materials have Seebeck coefficients S_A(T) and S_B(T), the thermocouple emf between temperatures T1 and T2 is:
hardExplanation: Thermocouple emf is integral of difference of Seebeck coefficients across temperature range; if coefficients are constant it reduces to (S_A − S_B)ΔT. -
15. If Seebeck coefficients are approximately constant over small ΔT, thermocouple emf E ≈
mediumExplanation: For small temperature span with nearly constant coefficients, emf ≈ difference in coefficients times temperature difference. -
16. Advantage of thermocouple over resistance thermometer (RTD) is:
mediumExplanation: Thermocouples cover wide ranges, respond quickly and are rugged; RTDs can be more accurate in limited ranges. -
17. Limitation of thermocouple measurement includes:
mediumExplanation: Thermocouple emf is not perfectly linear; tables or polynomial fits and reference compensation are required. -
18. Thermocouple polarity is determined by:
hardExplanation: Polarity indicates which junction is at higher potential dependent on Seebeck coefficients; instrumentation must respect polarity for correct sign. -
19. A thermopile used as a radiation detector typically connects thermocouples such that:
mediumExplanation: Thermopile arranges hot junctions at absorber (heated by radiation) and cold junctions at reference temperature to generate cumulative voltage. -
20. Example numerical: Two metals A and B have constant Seebeck coefficients S_A = 12 μV/K and S_B = 3 μV/K. If hot junction is at 150°C and reference at 50°C, approximate thermocouple emf is:
hardExplanation: ΔT = 100 K; emf ≈ (S_A − S_B)ΔT = (12−3) μV/K × 100 K = 900 μV.