Physics Electronics PMDC Conceptual Practice
PMDC Verified Question 206 of 494
If an AC sinusoidal voltage with peak value \( V_m = 100\text{ V} \) is applied separately to an ideal half-wave rectifier and an ideal full-wave rectifier, what are the RMS output voltages across their load resistors?
A
35.3 V and 70.7 V
B
70.7 V and 100 V
C
50.0 V and 70.7 V
D
31.8 V and 63.7 V
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option C: 50.0 V and 70.7 V
Concept:

The RMS value for a half-wave rectified waveform is \( V_{\text{rms}} = V_m / 2 \), and for a full-wave rectified waveform it is \( V_{\text{rms}} = V_m / \sqrt{2} \).

Formula:

$$V_{\text{rms, half-wave}} = \frac{V_m}{2} = \frac{100\text{ V}}{2} = 50.0\text{ V}$$

$$V_{\text{rms, full-wave}} = \frac{V_m}{\sqrt{2}} = \frac{100\text{ V}}{1.414} \approx 70.7\text{ V}$$

Solution:

  • Half-wave RMS: \( V_{\text{rms}} = \frac{100}{2} = 50.0\text{ V} \).


  • Full-wave RMS: \( V_{\text{rms}} = \frac{100}{\sqrt{2}} \approx 70.7\text{ V} \).


Why other options are incorrect:

  • Option A: 35.3 V is \( 50 / \sqrt{2} \).
  • Option B: 100 V is the peak voltage, not the RMS value.
  • Option D: 31.8 V and 63.7 V are the DC (average) output values, not the RMS values.

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