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

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Solve this multiple-choice question and choose the correct option.

Accepted Answer

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

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

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