Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 259 of 494
A full-wave bridge rectifier is fed by a transformer secondary with a peak voltage of \( V_m = 50.0\text{ V} \). If each silicon diode has a forward voltage drop of \( V_D = 0.70\text{ V} \), what is the peak voltage across the load resistor?
A
50.0 V
B
49.3 V
C
51.4 V
D
48.6 V
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: 48.6 V
Concept:

In a bridge rectifier, two diodes conduct in series with the load during each half-cycle, introducing two forward voltage drops (\( 2 V_D \)).

Formula:

$$V_{\text{load,peak}} = V_m - 2 V_D$$

Solution:

  • Given: \( V_m = 50.0\text{ V} \) and \( V_D = 0.70\text{ V} \).


  • Total diode voltage drop: \( 2 \times 0.70\text{ V} = 1.40\text{ V} \).


  • Peak load voltage: \( V_{\text{load,peak}} = 50.0\text{ V} - 1.40\text{ V} = 48.60\text{ V} = 48.6\text{ V} \).


Why other options are incorrect:

  • Option A: 50.0 V assumes ideal diodes with zero forward voltage drop.
  • Option B: 49.3 V subtracts only one diode drop (\( 50.0 - 0.70 \)), as in a center-tapped rectifier.
  • Option C: 51.4 V incorrectly adds the diode drops to the supply 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.

Previous Question All Electronics MCQs Next Question

Want to solve full-length papers under timed exam conditions?

Practice with zero-scroll lockdown sprints, dynamic latency zone timers, live peer selection telemetry, and the automated Amber mistake recovery loop.

Enter Arena Sprint iOS App Android App