Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 303 of 494
A full-wave center-tapped rectifier supplies a \( 900\ \Omega \) load resistor using two diodes, each having a forward dynamic resistance of \( r_f = 100\ \Omega \). What is the conversion efficiency \( \eta \) of this circuit?
A
73.1%
B
81.2%
C
40.6%
D
50.0%
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: 73.1%
Concept:

Rectification efficiency accounting for diode forward resistance \( r_f \) is given by \( \eta = \frac{8}{\pi^2} \left(\frac{R_L}{r_f + R_L}\right) \).

Formula:

$$\eta = \frac{0.812 \times R_L}{r_f + R_L} \times 100\%$$

Solution:

  • Given: \( R_L = 900\ \Omega \) and \( r_f = 100\ \Omega \).


  • Resistance factor: \( \frac{R_L}{r_f + R_L} = \frac{900}{100 + 900} = \frac{900}{1000} = 0.90 \).


  • Efficiency: \( \eta = 81.2\% \times 0.90 = 73.08\% \approx 73.1\% \).


Why other options are incorrect:

  • Option B: 81.2% is the ideal efficiency assuming zero diode resistance (\( r_f = 0 \)).
  • Option C: 40.6% is the maximum efficiency of a half-wave rectifier.
  • Option D: 50.0% is an arbitrary linear estimate.

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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