Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 120 of 494
If the peak alternating voltage across the secondary winding of a full-wave rectifier transformer is \( 20.0\text{ V} \), what is the RMS value of the full-wave rectified output voltage across the load (assuming ideal diodes)?
A
10.0 V
B
14.14 V
C
28.28 V
D
7.07 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: 14.14 V
Concept:

For a full-wave rectified sine wave, the Root Mean Square (RMS) value is identical to that of the original AC sine wave because squaring the waveform eliminates the sign inversion.

Formula:

$$V_{\text{rms, full-wave}} = \frac{V_m}{\sqrt{2}} \approx 0.7071 \times V_m$$

Solution:

  • Given: Peak voltage \( V_m = 20.0\text{ V} \).


  • RMS value: \( V_{\text{rms}} = \frac{20.0}{\sqrt{2}} = \frac{20.0}{1.414} \approx 14.14\text{ V} \).


Why other options are incorrect:

  • Option A: 10.0 V is \( V_m / 2 \), which is the RMS value for a half-wave rectifier.
  • Option C: 28.28 V incorrectly multiplies the peak voltage by \( \sqrt{2} \).
  • Option D: 7.07 V is \( 10 / \sqrt{2} \), which uses half the required peak voltage.

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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