Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 121 of 494
For a half-wave rectifier fed by an AC sinusoidal voltage with peak value \( V_m = 100\text{ V} \), what is the RMS value of the output voltage across the load?
A
50.0 V
B
70.7 V
C
31.8 V
D
63.7 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: 50.0 V
Concept:

Because a half-wave rectifier outputs voltage during only half of the total period, the RMS integration yields half the peak voltage.

Formula:

$$V_{\text{rms, half-wave}} = \sqrt{\frac{1}{2\pi}\int_0^\pi (V_m \sin\theta)^2 d\theta} = \frac{V_m}{2}$$

Solution:

  • Given: \( V_m = 100\text{ V} \).


  • \( V_{\text{rms}} = \frac{100\text{ V}}{2} = 50.0\text{ V} \).


Why other options are incorrect:

  • Option B: 70.7 V is \( V_m / \sqrt{2} \), which is the RMS value of a full-wave rectifier.
  • Option C: 31.8 V is \( V_m / \pi \), which is the DC (average) value of a half-wave rectifier.
  • Option D: 63.7 V is \( 2V_m / \pi \), which is the DC (average) value of a full-wave rectifier.

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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