Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 136 of 494
A half-wave rectifier is supplied by a sinusoidal AC voltage \( v(t) = 50 \sin(100\pi t)\text{ V} \). If the diode is ideal and the load resistance is \( 1.0\text{ k}\Omega \), what is the peak output current through the load?
A
50 mA
B
25 mA
C
100 mA
D
35.3 mA
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option A: 50 mA
Concept:

Peak current \( I_m \) through the load is determined by the peak secondary voltage \( V_m \) and the total loop resistance during forward conduction.

Formula:

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

Solution:

  • Given: \( V_m = 50\text{ V} \) and \( R_L = 1.0\text{ k}\Omega = 1000\ \Omega \).


  • \( I_m = \frac{50\text{ V}}{1000\ \Omega} = 0.050\text{ A} = 50\text{ mA} \).


Why other options are incorrect:

  • Option B: 25 mA is \( I_m / 2 \), which represents the average current in specific half-cycle averages.
  • Option C: 100 mA assumes a \( 500\ \Omega \) resistance.
  • Option D: 35.3 mA is the RMS current (\( 50 / \sqrt{2} \approx 35.3\text{ mA} \)), not the peak current.

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