The rms value of an alternating current
(A) Is equal to \[0.707\] times peak value.
(B) Is equal to \[0.636\] times peak value.
(C) Is equal to$\sqrt 2 $ times the peak value.
(D) None of the above
Answer
249k+ views
Hint: The alternating current is given by$I = {I_0}\sin (\omega t)$. Now this is a continuous function defined over the interval between\[{t_1}\] and${t_2}$. ${I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\int\limits_{{t_1}}^{{t_2}} {({I_0}\sin } (\omega t){)^2}dt} $.Use the trigonometric identity to eliminate and simplify. By substituting th upper and lower limits and evaluating we get${I_{rms}} = \dfrac{{{I_0}}}{{\sqrt 2 }}$ .
Complete step-by-step answer
Peak value is the maximum value of alternating quantity. It is also called as amplitude. It is denoted by ${i_o}$ or ${V_o}$ .
Root mean square value is defined as the root of the mean square of the quantity. The quantity is usually voltage or current in ac circuit for one complete cycle. It is denoted by${i_{rms}}$or${V_{rms}}$.
It is given by,
${i_{rms}} = \sqrt {\dfrac{{i_1^2 + i_2^2 + ...}}{n}} $
The alternating current is given by$I = {I_0}\sin (\omega t)$.
Defining the continuous function over the limits t and t, we get
${I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\int\limits_{{t_1}}^{{t_2}} {({I_0}\sin } (\omega t){)^2}dt} $
\[{I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{t}{2} - \dfrac{{\sin (2\omega t)}}{{4\omega }}} \right]_{{t_1}}^{{t_2}}} \]
\[{I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{t}{2}} \right]_{{t_1}}^{{t_2}}} \]
\[{I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{{{t_2} - {t_1}}}{2}} \right]} \]
$ \Rightarrow \dfrac{{{i_o}}}{{\sqrt 2 }} = 0.707{i_0}$
Hence, the rms value of an alternating current is equal to \[0.707\] times peak value.
The correct option is A.
Note: The rms value of AC is also called virtual or effective value. AC ammeter and voltmeter always measure the rms value. In houses ac is supplied at 220 volts, which is the rms value of voltage. Its peak value is 311 volts.
Complete step-by-step answer
Peak value is the maximum value of alternating quantity. It is also called as amplitude. It is denoted by ${i_o}$ or ${V_o}$ .
Root mean square value is defined as the root of the mean square of the quantity. The quantity is usually voltage or current in ac circuit for one complete cycle. It is denoted by${i_{rms}}$or${V_{rms}}$.
It is given by,
${i_{rms}} = \sqrt {\dfrac{{i_1^2 + i_2^2 + ...}}{n}} $
The alternating current is given by$I = {I_0}\sin (\omega t)$.
Defining the continuous function over the limits t and t, we get
${I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\int\limits_{{t_1}}^{{t_2}} {({I_0}\sin } (\omega t){)^2}dt} $
\[{I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{t}{2} - \dfrac{{\sin (2\omega t)}}{{4\omega }}} \right]_{{t_1}}^{{t_2}}} \]
\[{I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{t}{2}} \right]_{{t_1}}^{{t_2}}} \]
\[{I_{rms}} = \sqrt {\dfrac{1}{{{t_2} - {t_1}}}\left[ {\dfrac{{{t_2} - {t_1}}}{2}} \right]} \]
$ \Rightarrow \dfrac{{{i_o}}}{{\sqrt 2 }} = 0.707{i_0}$
Hence, the rms value of an alternating current is equal to \[0.707\] times peak value.
The correct option is A.
Note: The rms value of AC is also called virtual or effective value. AC ammeter and voltmeter always measure the rms value. In houses ac is supplied at 220 volts, which is the rms value of voltage. Its peak value is 311 volts.
Recently Updated Pages
JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Isoelectronic Definition in Chemistry: Meaning, Examples & Trends

Ionisation Energy and Ionisation Potential Explained

Iodoform Reactions - Important Concepts and Tips for JEE

Introduction to Dimensions: Understanding the Basics

Instantaneous Velocity Explained: Formula, Examples & Graphs

Trending doubts
JEE Main 2026: Exam Dates, Session 2 Updates, City Slip, Admit Card & Latest News

Hybridisation in Chemistry – Concept, Types & Applications

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Electric Field of a Uniformly Charged Ring

Derivation of Equation of Trajectory Explained for Students

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Other Pages
CBSE Class 12 Physics Question Paper 2026: Download SET-wise PDF with Answer Key & Analysis

JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Understanding the Angle of Deviation in a Prism

Understanding Centrifugal Force in Physics

