Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 39 of 494
A sinusoidal AC voltage with an RMS value of \(V_{\text{rms}} = 10\text{ V}\) is connected across an ideal half-wave rectifier. What is the peak output voltage (\(V_m\)) across the load?
A
\(10.0\text{ V}\)
B
\(7.07\text{ V}\)
C
\(14.14\text{ V}\)
D
\(20.0\text{ V}\)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \(14.14\text{ V}\)
Concept:

For a sinusoidal alternating waveform, the peak voltage \(V_m\) is related to the root-mean-square voltage \(V_{\text{rms}}\) by a factor of \(\sqrt{2}\). For an ideal diode, the forward drop is zero, so the peak output voltage equals the peak input voltage.

Formula:

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

Solution:

  • Given: \(V_{\text{rms}} = 10\text{ V}\).


  • $$V_m = 10 \times \sqrt{2} = 10 \times 1.4142 = 14.14\text{ V}$$


Why other options are incorrect:

  • Option A: \(10.0\text{ V}\) is the RMS value, not the peak value.
  • Option B: \(7.07\text{ V}\) is obtained by dividing by \(\sqrt{2}\) instead of multiplying.
  • Option D: \(20.0\text{ V}\) represents the peak-to-peak amplitude (\(2 V_{\text{rms}}\)), not \(V_m\).

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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