Question 37

The Correct Answer is D **Explanation for Option D (root-mean-square voltage):** Most AC voltmeters, whether analog (using a D'Arsonval movement with an internal rectifier circuit) or digital (Digital Multimeters or DMMs), are designed and calibrated to display the **root-mean-square (RMS) voltage**. The RMS value of an AC waveform is the *effective* voltage—it represents the DC voltage level that would produce the same amount of heat (power dissipation) in a resistive load as the measured AC voltage. This standardization is crucial because power calculations rely directly on the RMS voltage ($\text{Power} = V_{RMS}^2 / R$). While many standard meters assume the waveform is a perfect sinusoid and calculate the RMS value based on a relationship with the average rectified voltage, the displayed result is always intended to represent the RMS value. High-end meters specifically advertise the ability to measure "True RMS," meaning they accurately calculate the RMS value regardless of the waveform shape (not just for sine waves). **Explanation of Incorrect Options:** * **A) peak-to-peak voltage:** While the peak-to-peak value ($V_{p-p}$) is important for understanding signal amplitude and determining saturation limits, it is generally double the peak voltage. Standard voltmeters do not display this value directly as the primary measurement because it doesn't directly relate to the power delivered. * **B) average voltage:** For a purely symmetrical AC waveform (like a sine wave, square wave, or triangle wave) the true average voltage over one complete cycle is zero. Meters generally measure the *average rectified value* (the average of the absolute value of the voltage), but they then multiply this value by a specific "form factor" (1.11 for a sine wave) so that the final displayed number corresponds to the RMS value, not the average rectified voltage itself. * **C) peak voltage only:** The peak voltage ($V_{peak}$) is the maximum instantaneous voltage reached during a cycle. While related to RMS voltage by $V_{RMS} = V_{peak} / \sqrt{2}$ (for sine waves), voltmeters are calibrated to show the RMS value because it is the standard measure of effective power delivery, not just the momentary maximum amplitude.