
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
134.4k+ 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 Main 2025 Session 2 Form Correction (Closed) – What Can Be Edited

What are examples of Chemical Properties class 10 chemistry JEE_Main

JEE Main 2025 Session 2 Schedule Released – Check Important Details Here!

JEE Main 2025 Session 2 Admit Card – Release Date & Direct Download Link

JEE Main 2025 Session 2 Registration (Closed) - Link, Last Date & Fees

JEE Mains Result 2025 NTA NIC – Check Your Score Now!

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Wheatstone Bridge for JEE Main Physics 2025

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

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

Diffraction of Light - Young’s Single Slit Experiment

Dual Nature of Radiation and Matter Class 12 Notes: CBSE Physics Chapter 11

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Elastic Collisions in One Dimension - JEE Important Topic

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main
