Relating half-wave average to rms: If the half-cycle average voltage of a sine wave is 12 V (i.e., average of a half-wave-rectified sine over one full period), what is the corresponding rms voltage of the half-wave-rectified waveform?

Introduction:Different “averages” and “effective” (rms) values are used for sine-derived waveforms. For rectified signals, the average (mean) and rms are not the same. Converting between them requires remembering the constants that relate peak, average, and rms for half-wave rectified sinusoids.

Given Data / Assumptions:

• Half-wave-rectified sine wave.
• Half-cycle average (over a full period) is 12 V.
• Ideal rectification and sinusoid.

Concept / Approach:

For a half-wave-rectified sine: average value Vavg_half = Vp / π. Its rms value is Vrms_half = Vp / 2. Therefore, if we first find Vp from the average, we can compute the rms directly.

Step-by-Step Solution:

Verification / Alternative check:

Cross-relations: For full-wave rectified, Vrms_full = Vp / √2 and Vavg_full = 2 * Vp / π; those do not apply here. The half-wave constants used are standard identities and produce a self-consistent result.

Why Other Options Are Wrong:

• 13.33 V / 7.64 V / 8.48 V: Do not follow the Vp / 2 formula once Vp is inferred from the given average.
• 24.00 V: Would imply Vp = 48 V and average = 15.28 V, inconsistent with the 12 V average.

Common Pitfalls:

• Using Vrms = Vp / √2 (valid for unrectified sine or full-wave, not half-wave rms).
• Confusing average of rectified with average of the original sinusoid (which is zero over a full period).

Final Answer:

18.84 V