Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 86 of 494
An ideal half-wave rectifier is connected across an AC source \( v(t) = 100\sin(100\pi t)\text{ V} \) with a load resistance of \( R_L = 100\,\Omega \). The peak instantaneous load current \( I_m \) is:
A
\( 0.50\text{ A} \)
B
\( 0.707\text{ A} \)
C
\( 0.318\text{ A} \)
D
\( 1.00\text{ A} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \( 1.00\text{ A} \)
Concept:

Peak current \( I_m \) is determined by dividing the peak instantaneous input voltage \( V_m \) by the total circuit resistance during the forward conduction interval.

Formula:

$$I_m = \frac{V_m}{R_L}$$

Solution:

  • From the input equation \( v(t) = 100\sin(100\pi t)\text{ V} \), the peak voltage is \( V_m = 100\text{ V} \).


  • $$I_m = \frac{100\text{ V}}{100\,\Omega} = 1.00\text{ A}$$


Why other options are incorrect:

  • Option A: \( 0.50\text{ A} \) is the average DC current (\( I_m/2 \)) calculated incorrectly without \( \pi \).


  • Option B: \( 0.707\text{ A} \) (\( I_m/\sqrt{2} \)) is the RMS current of a full wave.


  • Option C: \( 0.318\text{ A} \) (\( I_m/\pi \)) is the average DC output current \( I_{\text{dc}} \).

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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