Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 302 of 494
An AC transformer supplies \( 24.0\text{ V} \) RMS to an ideal half-wave rectifier. What is the average (DC) voltage delivered to the load resistor?
A
24.0 V
B
33.9 V
C
10.8 V
D
21.6 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: 10.8 V
Concept:

Calculate the peak secondary voltage \( V_m = \sqrt{2} V_{\text{rms}} \), then apply the half-wave DC formula \( V_{\text{dc}} = V_m / \pi \).

Formula:

$$V_m = \sqrt{2} \times 24.0\text{ V} \approx 33.94\text{ V}$$

$$V_{\text{dc}} = \frac{V_m}{\pi} = \frac{33.94\text{ V}}{3.1416} \approx 10.80\text{ V} \approx 10.8\text{ V}$$

Solution:

  • Peak voltage: \( V_m = 24.0 \times 1.4142 = 33.94\text{ V} \).


  • Average DC voltage: \( V_{\text{dc}} = \frac{33.94}{\pi} \approx 10.8\text{ V} \).


Why other options are incorrect:

  • Option A: 24.0 V is the input RMS voltage.
  • Option B: 33.9 V is the peak AC voltage \( V_m \).
  • Option D: 21.6 V is the DC output of a full-wave rectifier (\( 2V_m / \pi \)).

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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