A thermocouple-type AC voltmeter indicates which electrical quantity of an applied waveform when properly calibrated?

Difficulty: Easy

Correct Answer: rms value

Explanation:


Introduction / Context:
Thermocouple voltmeters convert electrical energy into heat and then into a temperature-dependent electrical signal. Because heating depends on the mean of the square of the current (or voltage across a fixed resistance), these instruments naturally respond to the root-mean-square (rms) quantity, making them valuable for non-sinusoidal waveforms as true-RMS sensors.


Given Data / Assumptions:

  • Thermocouple sensor measures temperature rise due to Joule heating in a heater element.
  • Heating power P ∝ I^2R or V^2/R averaged over time.
  • Instrument is calibrated in rms volts for a sinusoidal reference and reads true-RMS for arbitrary waveforms within bandwidth.


Concept / Approach:
Since the sensor integrates power over time, its output is proportional to the average of v^2(t). The voltage equivalent is V_rms = sqrt( average{ v^2(t) } ). Therefore, the proper reading is rms value. Peak, average (rectified), or peak-to-peak indications correspond to other measurement principles (diode rectifier, crest detector, or oscilloscope, respectively).


Step-by-Step Solution:

Heating element: P(t) = v^2(t)/R; the thermocouple responds to ⟨P⟩.Thus meter scale is proportional to sqrt(⟨v^2(t)⟩) = V_rms.


Verification / Alternative check:

Check on a sine: V_rms = V_max/√2; instrument agrees with standard definitions.


Why Other Options Are Wrong:

Peak or peak-to-peak require different detection; average rectified is used in simple diode meters, not thermocouple.


Common Pitfalls:

Assuming ”average” equals ”rms” for all waveforms; true only for specific shapes with scale factors.


Final Answer:

rms value

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