Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 396 of 494
What is the RMS voltage (\( V_{\text{rms}} \)) of the output of an ideal half-wave rectifier supplied with a peak input voltage \( V_m \)?
A
\( V_{\text{rms}} = \frac{V_m}{\sqrt{2}} \)
B
\( V_{\text{rms}} = \frac{V_m}{\pi} \)
C
\( V_{\text{rms}} = \frac{2 V_m}{\pi} \)
D
\( V_{\text{rms}} = \frac{V_m}{2} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \( V_{\text{rms}} = \frac{V_m}{2} \)
Concept:

The Root Mean Square (RMS) value measures the effective heating value of a periodic waveform over a complete cycle.

Formula:

$$V_{\text{rms}} = \sqrt{\frac{1}{2\pi} \int_0^{2\pi} v^2(\theta) d\theta} = \frac{V_m}{2}$$

Solution:

  • In a half-wave rectifier, voltage is non-zero only from \( 0 \) to \( \pi \):


  • $$V_{\text{rms}}^2 = \frac{1}{2\pi} \int_0^\pi V_m^2 \sin^2(\theta) d\theta = \frac{V_m^2}{2\pi} \left[ \frac{\theta}{2} - \frac{\sin(2\theta)}{4} \right]_0^\pi = \frac{V_m^2}{2\pi} \left(\frac{\pi}{2}\right) = \frac{V_m^2}{4}$$


  • Taking the square root gives:


  • $$V_{\text{rms}} = \sqrt{\frac{V_m^2}{4}} = \frac{V_m}{2} = 0.5 V_m$$


Why other options are incorrect:

  • Option A: \( \frac{V_m}{\sqrt{2}} \) is the RMS value of a full sinusoidal wave or a full-wave rectified wave.
  • Option B: \( \frac{V_m}{\pi} \) is the DC (average) value of a half-wave rectified waveform.
  • Option C: \( \frac{2 V_m}{\pi} \) is the DC (average) value of a full-wave rectified waveform.

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