Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 435 of 494
In an ideal half-wave rectifier delivering peak current \( I_m \) to a load resistor \( R_L \), what is the DC power (\( P_{\text{dc}} \)) dissipated in the load?
A
\( P_{\text{dc}} = \left(\frac{I_m}{2}\right)^2 R_L \)
B
\( P_{\text{dc}} = \left(\frac{2I_m}{\pi}\right)^2 R_L \)
C
\( P_{\text{dc}} = \left(\frac{I_m}{\pi}\right)^2 R_L \)
D
\( P_{\text{dc}} = \left(\frac{I_m}{\sqrt{2}}\right)^2 R_L \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: \( P_{\text{dc}} = \left(\frac{I_m}{\pi}\right)^2 R_L \)
Concept:

DC power delivered to a resistive load is calculated from the square of the average (DC) load current.

Formula:

$$P_{\text{dc}} = I_{\text{dc}}^2 R_L$$

Solution:

  • For a half-wave rectifier, the average DC current is \( I_{\text{dc}} = \frac{I_m}{\pi} \).


  • Calculating DC power:


  • $$P_{\text{dc}} = I_{\text{dc}}^2 R_L = \left(\frac{I_m}{\pi}\right)^2 R_L = \frac{I_m^2}{\pi^2} R_L$$


Why other options are incorrect:

  • Option A: \( \left(\frac{I_m}{2}\right)^2 R_L \) is the total RMS AC power dissipated in the load.
  • Option B: \( \left(\frac{2I_m}{\pi}\right)^2 R_L \) is the DC power of a full-wave rectifier.
  • Option D: \( \left(\frac{I_m}{\sqrt{2}}\right)^2 R_L \) is the total power for a full sinusoidal AC waveform.

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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