Examoogle
ExamsTest SeriesRank CheckPrevious Year PapersPassBook StoreMy BooksAI Tutor
🛒0
अA
Examoogle

India's most trusted platform for competitive exam PDF books. Expert-authored, watermark-protected, instant access.

Exams & Practice
All Exams & SyllabusMock Test SeriesPrevious Year PapersPractice Questions (MCQs)Recruitment Notifications
Quick Links
Examoogle AI TutorExam NewsBook StoreMy BooksLogin / Sign Up
Support
About UsRefund PolicyPrivacy PolicyTerms of UseContact Us
© 2026 Examoogle. India's #1 competitive exam AI tutor.
🔒 SSL Secured📱 UPI Accepted🧾 GST Invoice
Examoogle

Join 60,000+ competitive exam aspirants

or with email
By continuing, you agree to ourTerms of Service&Privacy Policy
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Back to Practice Questions
ElectricalBasic Electrical
PrevNext

The value of peak factor is

A

0

B

1

C

1.11

D

1.414

Correct Answer

Concept & PrincipleElectricalBasic Electrical
Option A

0

Quick Summary: The peak factor (also known as the crest factor) is defined as the ratio of the peak (maximum) value to the RMS value of a waveform. For a standard sinusoidal alternating current or voltage, this ratio is calculated as $\sqrt{2}$, which is approximately 1.414.

💡 Explanation

The peak factor (also known as the crest factor) is defined as the ratio of the peak (maximum) value to the RMS value of a waveform. For a standard sinusoidal alternating current or voltage, this ratio is calculated as 2\sqrt{2}2​, which is approximately 1.414.

🔢 Key Formulas

Peak  Factor=Peak  ValueRMS  ValuePeak\;Factor = \frac{Peak\;Value}{RMS\;Value}PeakFactor=RMSValuePeakValue​

RMS  Value=Peak  Value2RMS\;Value = \frac{Peak\;Value}{\sqrt{2}}RMSValue=2​PeakValue​ for sine waves

⚙️ Working Principle

For a sinusoidal waveform represented by v(t)=Vmsin⁡(ωt)v(t) = V_m \sin(\omega t)v(t)=Vm​sin(ωt), the maximum (peak) value is VmV_mVm​ and the Root Mean Square (RMS) value is Vm2\frac{V_m}{\sqrt{2}}2​Vm​​. The peak factor is the ratio of these two quantities: VmVm/2=2≈1.414\frac{V_m}{V_m / \sqrt{2}} = \sqrt{2} \approx 1.414Vm​/2​Vm​​=2​≈1.414. This parameter indicates how 'peaky' or impulsive a waveform is compared to its RMS value.

📌 Key Points
  • ▸

    Peak factor represents the crest intensity of a periodic waveform.

  • ▸

    For a perfect sinusoidal wave, the value is always 2≈1.414\sqrt{2} \approx 1.4142​≈1.414.

  • ▸

    It is a unitless ratio used to evaluate the distortion of signal waveforms.

✅ Advantages
  • ▸

    Helps in selecting components capable of handling peak voltage stress.

  • ▸

    Useful in insulation testing and dielectric strength analysis.

❌ Disadvantages / Limitations
  • ▸

    Does not provide information regarding the frequency or phase of the signal.

  • ▸

    Can be misleading for highly distorted non-sinusoidal waveforms.

🛠️ Applications / Uses
  • ▸

    Transformer and motor insulation design.

  • ▸

    Power quality analysis and waveform monitoring.

  • ▸

    Design of peak-detecting circuits.

📄 Additional Information
  • ▸

    The form factor is another important ratio, defined as the ratio of RMS value to the average value, which is approximately 1.11 for a pure sine wave (Option C).

  • ▸

    A peak factor of 1 implies a square wave, where the peak value equals the RMS value.

📊 Diagram / Illustration
Peak Factor Formula
Peak  ValuePeak\;ValuePeakValue
RMS  ValueRMS\;ValueRMSValue
VmVm/2=1.414\frac{V_m}{V_m / \sqrt{2}} = 1.414Vm​/2​Vm​​=1.414
✅

D is correct — The peak factor of a sinusoidal wave is defined as the ratio of the maximum value to the RMS value, resulting in 2≈1.414\sqrt{2} \approx 1.4142​≈1.414.

Core Concepts Used
Click any tag to open in AI Tutor
AC Fundamentals Root Mean Square (RMS) Value Waveform Analysis
💡 EXAM TIP

Always remember: Peak Factor = 1.414 and Form Factor = 1.11 for sine waves; confusing these is a common error in competitive exams.

Related Questions

ElectricalBasic Electrical
Batteries are charged by
ElectricalBasic Electrical
48 ampere-hour capacity would deliver a current of
ElectricalBasic Electrical
The lead-acid cell should never be discharged beyond
ElectricalBasic Electrical
In a lead-acid cell, lead is called as
ElectricalBasic Electrical
Undercharging of chemical batteries

Discussion (0)

Loading discussion...
PrevNext