A voltmeter reads a sinusoidal AC signal as 500 mV (assume the meter reports RMS). What is the average value of the corresponding full-wave–rectified sine wave?

Introduction:
Different “measures” of a sine wave—RMS, peak, and average (of the rectified waveform)—serve different purposes. This problem asks for the mean of the absolute value (full-wave–rectified average) given an RMS reading.

Given Data / Assumptions:

• Measured Vrms = 500 mV for a pure sinusoid.
• Relationships: Vpeak = Vrms * sqrt(2); Vavg(rectified) = (2 / π) * Vpeak ≈ 0.637 * Vpeak.
• We are not asking the algebraic average over one full cycle (which is zero).

Concept / Approach:
Convert the RMS value to peak, then apply the rectified-average factor. This is standard in AC measurement theory and is foundational for interpreting “average responding” meters calibrated to RMS for sinusoidal inputs.

Step-by-Step Solution:
1) Compute Vpeak = 0.5 V * 1.414 ≈ 0.7071 V.2) Compute Vavg(rectified) = 0.637 * Vpeak ≈ 0.637 * 0.7071 ≈ 0.4509 V.3) Convert to millivolts: ≈ 451 mV.4) Choose the closest option: 451.0 mV.

Verification / Alternative check:
Using exact factors: (2/π) * sqrt(2) ≈ 0.9003; 0.9003 * 500 mV ≈ 450.2 mV, matching the rounded result.

Why Other Options Are Wrong:
159.0 mV and 318.5 mV: mismatch common factors; do not follow the correct sequence.

353.5 mV: equals Vrms * 0.707; not the rectified-average relation.

500.0 mV: equals Vrms; not the rectified average.

Common Pitfalls:
Confusing average of the actual sinusoid (zero) with average of its rectified magnitude, or skipping the conversion from RMS to peak before applying the 0.637 factor.

Final Answer:
451.0 mV