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Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Back to Practice Questions
ElectricalElectronics
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The ripple factor of a half-wave rectifier is _____.

A

2

B

21

C

5

D

0.48

Correct Answer

Concept & PrincipleElectricalElectronics
Option A

2

Quick Summary: The ripple factor of a half-wave rectifier is exactly 1.21, not 0.48 as stated in the provided prompt's answer choice. The ripple factor is a measure of the purity of the DC output, defined as the ratio of the RMS value of the AC component of the output to the DC value of the output.

💡 Explanation

The ripple factor of a half-wave rectifier is exactly 1.21, not 0.48 as stated in the provided prompt's answer choice. The ripple factor is a measure of the purity of the DC output, defined as the ratio of the RMS value of the AC component of the output to the DC value of the output.

🔢 Key Formulas

γ=(IrmsIdc)2−1\gamma = \sqrt{(\frac{I_{rms}}{I_{dc}})^2 - 1}γ=(Idc​Irms​​)2−1​ — General definition of ripple factor

γ=1.21\gamma = 1.21γ=1.21 — Ripple factor for Half-Wave Rectifier

γ=0.48\gamma = 0.48γ=0.48 — Ripple factor for Full-Wave Rectifier

⚙️ Working Principle

In a half-wave rectifier, the output is pulsating DC containing significant harmonic components. The RMS value of the output current is Irms=Im2I_{rms} = \frac{I_m}{2}Irms​=2Im​​, and the DC component is Idc=ImπI_{dc} = \frac{I_m}{\pi}Idc​=πIm​​. Using the relation γ=(IrmsIdc)2−1\gamma = \sqrt{(\frac{I_{rms}}{I_{dc}})^2 - 1}γ=(Idc​Irms​​)2−1​, substituting the values yields (π2)2−1≈1.21\sqrt{(\frac{\pi}{2})^2 - 1} \approx 1.21(2π​)2−1​≈1.21.

📌 Key Points
  • ▸

    The ripple factor represents the AC content in the output signal.

  • ▸

    A lower ripple factor indicates a smoother DC output.

  • ▸

    Half-wave rectifiers have poor performance compared to full-wave rectifiers due to higher ripple.

  • ▸

    The value 0.48 is actually the ripple factor for a full-wave rectifier, indicating the source of confusion in the original question.

✅ Advantages
  • ▸

    Simple circuit construction

  • ▸

    Low component count (requires only one diode)

❌ Disadvantages / Limitations
  • ▸

    High ripple factor (1.21)

  • ▸

    Low rectification efficiency (40.6%)

  • ▸

    DC saturation of transformer core

🛠️ Applications / Uses
  • ▸

    Low-cost signal demodulation

  • ▸

    Simple control circuits where high efficiency is not required

🔄 Comparison Table
FeatureHalf-Wave RectifierFull-Wave Rectifier

Ripple Factor

1.21

0.48

📄 Additional Information
  • ▸

    The option '0.48' in the question is factually associated with a Full-Wave Rectifier, not a Half-Wave Rectifier.

  • ▸

    Engineers typically use filters (capacitors/inductors) to reduce the ripple factor to a manageable level in practical power supplies.

📊 Diagram / Illustration
Ripple Factor (Half-Wave)RMS value of AC componentDC output valueValue = 1.21
✅

The ripple factor of a half-wave rectifier is 1.21, whereas 0.48 is the ripple factor for a full-wave rectifier.

Core Concepts Used
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Rectification Ripple Factor DC Power Supplies
💡 EXAM TIP

Always remember the ripple factors: Half-wave = 1.21, Full-wave = 0.48. These are standard values frequently asked in competitive exams.

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