Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 223 of 494
An AC transformer secondary coil supplies \( 12.0\text{ V} \) RMS to a bridge rectifier. What is the peak voltage (\( V_m \)) delivered to the rectifier diodes (neglecting diode forward drops)?
A
12.0 V
B
8.48 V
C
16.97 V
D
24.0 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: 16.97 V
Concept:

The peak voltage \( V_m \) of a sinusoidal AC waveform is related to its RMS voltage by \( V_m = \sqrt{2} \times V_{\text{rms}} \).

Formula:

$$V_m = \sqrt{2} \times V_{\text{rms}} \approx 1.4142 \times V_{\text{rms}}$$

Solution:

  • Given: \( V_{\text{rms}} = 12.0\text{ V} \).


  • Peak voltage: \( V_m = 12.0 \times 1.4142 \approx 16.97\text{ V} \).


Why other options are incorrect:

  • Option A: 12.0 V is the RMS voltage, not the peak voltage.
  • Option B: 8.48 V is \( 12 / \sqrt{2} \), which divides by \( \sqrt{2} \) instead of multiplying.
  • Option D: 24.0 V is the peak-to-peak half-wave estimate (\( 2 \times 12 \)).

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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