Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 198 of 494
What is the average (DC) output voltage delivered to the load by an ideal half-wave rectifier supplied with a peak sinusoidal input voltage of \( V_m = 100\text{ V} \)?
A
31.8 V
B
50.0 V
C
63.7 V
D
70.7 V
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: 31.8 V
Concept:

The average (DC) output voltage of a half-wave rectifier is found by integrating the sinusoidal wave over a full \( 2\pi \) period, which yields \( V_{\text{dc}} = V_m / \pi \).

Formula:

$$V_{\text{dc}} = \frac{V_m}{\pi} \approx 0.3183 \times V_m$$

Solution:

  • Given: \( V_m = 100\text{ V} \).


  • \( V_{\text{dc}} = \frac{100\text{ V}}{\pi} = \frac{100}{3.1416} \approx 31.83\text{ V} \approx 31.8\text{ V} \).


Why other options are incorrect:

  • Option B: 50.0 V is \( V_m / 2 \), which is the RMS value of the half-wave output.
  • Option C: 63.7 V is \( 2V_m / \pi \), the DC output of a full-wave rectifier.
  • Option D: 70.7 V is \( V_m / \sqrt{2} \), the RMS value of a full-wave output.

Quality & Fidelity Assurance: Every question on BeambePrep is rigorously curated against the official PMDC syllabus with zero filler, zero out-of-syllabus content, and zero typos. When an authentic past paper originally contained a historical mistake or ambiguity from the examining board (such as UHS or NUMS), BeambePrep faithfully reflects the original paper while detailing the nuance and scientific consensus in the autopsy above.

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