Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 41 of 494
In an unfiltered full-wave rectifier supplied with a peak sinusoidal secondary voltage of \(V_m\), the average DC output voltage (\(V_{\text{dc}}\)) is given by:
A
\(\frac{V_m}{\pi}\)
B
\(\frac{2 V_m}{\pi}\)
C
\(\frac{V_m}{2\pi}\)
D
\(\pi V_m\)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \(\frac{2 V_m}{\pi}\)
Concept:

Because a full-wave rectifier inverts the negative half-cycles so that both halves contribute equally to the output, its average (DC) voltage is twice that of a half-wave rectifier.

Formula:

$$V_{\text{dc}} = \frac{1}{\pi} \int_0^\pi V_m \sin(\theta) d\theta = \frac{2V_m}{\pi}$$

Solution:

  • $$V_{\text{dc}} = \frac{2}{\pi} V_m \approx 0.636 V_m$$


  • This provides double the DC voltage level of an equivalent half-wave rectifier (\(V_m / \pi\)).


Why other options are incorrect:

  • Option A: \(V_m / \pi\) is the average DC value for a half-wave rectifier.
  • Option C: \(V_m / 2\pi\) is mathematically incorrect by a factor of 4.
  • Option D: Multiplying by \(\pi\) yields an impossible voltage higher than peak supply.

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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