Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 74 of 494
If an alternating sinusoidal voltage with peak value \( V_m = 100\text{ V} \) is rectified by an ideal full-wave rectifier, the RMS value of the output voltage is approximately:
A
\( 70.7\text{ V} \)
B
\( 50.0\text{ V} \)
C
\( 63.7\text{ V} \)
D
\( 100.0\text{ V} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: \( 70.7\text{ V} \)
Concept:

Squaring the negative half-cycles in a full-wave rectifier produces the same instantaneous power profile as a full continuous AC sine wave.

Formula:

$$V_{\text{rms, full-wave}} = \frac{V_m}{\sqrt{2}}$$

Solution:

  • For \( V_m = 100\text{ V} \):


  • $$V_{\text{rms}} = \frac{100\text{ V}}{\sqrt{2}} \approx 70.71\text{ V}$$


Why other options are incorrect:

  • Option B: \( 50.0\text{ V} \) (\( V_m/2 \)) is the RMS value of a half-wave rectifier.


  • Option C: \( 63.7\text{ V} \) (\( 2V_m/\pi \)) is the average DC output voltage.


  • Option D: \( 100.0\text{ V} \) is the peak amplitude \( V_m \).

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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