Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 123 of 494
For a full-wave rectifier with a peak output voltage \( V_m = 100\text{ V} \), the average (DC) voltage delivered to the load resistor is approximately:
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 C: 63.7 V
Concept:

A full-wave rectifier provides two symmetric conduction pulses per period, doubling the DC average value compared to a half-wave rectifier.

Formula:

$$V_{\text{dc}} = \frac{2 V_m}{\pi} \approx 0.6366 \times V_m$$

Solution:

  • Given: \( V_m = 100\text{ V} \).


  • \( V_{\text{dc}} = \frac{2 \times 100}{\pi} = \frac{200}{3.1416} \approx 63.7\text{ V} \).


Why other options are incorrect:

  • Option A: 31.8 V is \( V_m / \pi \), the average DC value of a half-wave rectifier.
  • Option B: 50.0 V is the RMS value of a half-wave rectifier.
  • Option D: 70.7 V is the RMS 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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