Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 72 of 494
An alternating sinusoidal voltage with a peak value of \( V_m = 100\text{ V} \) is applied to a full-wave rectifier. The average direct voltage (\( V_{\text{dc}} \)) obtained across the output load is:
A
\( \dfrac{100}{\pi}\text{ V} \)
B
\( \dfrac{200}{\pi}\text{ V} \)
C
\( \dfrac{50}{\pi}\text{ V} \)
D
\( \dfrac{100}{\sqrt{2}}\text{ V} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option B: \( \dfrac{200}{\pi}\text{ V} \)
Concept:

Because a full-wave rectifier conducts during both half-cycles, its average DC output voltage is exactly double that of a half-wave rectifier.

Formula:

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

Solution:

  • Substituting \( V_m = 100\text{ V} \):


  • $$V_{\text{dc}} = \frac{2(100)}{\pi} = \frac{200}{\pi}\text{ V} \approx 63.66\text{ V}$$


Why other options are incorrect:

  • Option A: \( 100/\pi\text{ V} \) is the average output for a half-wave rectifier.


  • Option C: \( 50/\pi\text{ V} \) is half of the half-wave output.


  • Option D: \( 100/\sqrt{2}\text{ V} \approx 70.7\text{ V} \) is the RMS value of the output waveform.

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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