The equation of an alternating voltage is $V = 100\sqrt 2 \sin 100\pi t$ volt. The RMS value of voltage and frequency will be respectively
A) $100\,V,\,50\,Hz$
B) $50\,V,\,100\,Hz$
C) $150\,V,\,50\,Hz$
D) $200\,V,\,50\,Hz$
Answer
298.5k+ views
Hint: To solve the question, you must need to find the rms value of voltage and the value of frequency. Remember that RMS value of current and voltage is nothing but the value corresponding to the peak value of current or voltage divided by square root of two. As for the frequency, the question has already given us the value of angular frequency. There is a direct relation between the angular frequency and the frequency, you can use that to find the answer.
Complete step by step answer:
As explained in the hint section of the solution to the question, we first need to find the RMS value of voltage and then find the value of frequency using the given equation of voltage.
Let us have a look at the equation of voltage which is given to us to be:
$V = 100\sqrt 2 \sin 100\pi t$
If we compare it to the general equation of voltage which is given as:
$V = {V_o}\sin \omega t$
Where, $V$ is the voltage at time $t$
${V_o}$ is the peak voltage, or the maximum value of the voltage possible and,
$\omega $ is the angular frequency
After comparing, we can safely say that:
$
{V_o} = 100\sqrt 2 \\
\omega = 100\pi \\
$
Now, if we try to recall facts about RMS value of voltage, we can easily recall that it is nothing but the ratio of the peak value of voltage with square root of two, mathematically, it can be represented as:
${V_{rms}} = \dfrac{{{V_o}}}{{\sqrt 2 }}$
If we substitute the given value of peak value of voltage, we get:
${V_{rms}} = \dfrac{{100\sqrt 2 }}{{\sqrt 2 }}$
Upon simplifying, we get:
${V_{rms}} = 100\,V$
Now that we have found out the RMS value of voltage, we need to find the value of frequency. We already know that there is a direct relation between angular frequency $\left( \omega \right)$ and the frequency of the circuit, this relation can be given as:
$f = \dfrac{\omega }{{2\pi }}$
If we substitute the value of angular frequency as given in the question, we get:
$f = \dfrac{{100\pi }}{{2\pi }}$
After solving, we get:
$f = 50\,Hz$
Hence, we got the values of RMS value of voltage and frequency as:
$
{V_{rms}} = 100\,V \\
f = 50\,Hz \\
$
So, we can easily see that the correct option is the option (A) as the value matches what we found out.
Note: Many students get confused and try to find the RMS value of frequency since the question is worded like it. But there is nothing as the RMS value of frequency. Also, some students get confused and instead of dividing the peak value of voltage by square root of two, they multiply it and thus reach a completely wrong answer.
Complete step by step answer:
As explained in the hint section of the solution to the question, we first need to find the RMS value of voltage and then find the value of frequency using the given equation of voltage.
Let us have a look at the equation of voltage which is given to us to be:
$V = 100\sqrt 2 \sin 100\pi t$
If we compare it to the general equation of voltage which is given as:
$V = {V_o}\sin \omega t$
Where, $V$ is the voltage at time $t$
${V_o}$ is the peak voltage, or the maximum value of the voltage possible and,
$\omega $ is the angular frequency
After comparing, we can safely say that:
$
{V_o} = 100\sqrt 2 \\
\omega = 100\pi \\
$
Now, if we try to recall facts about RMS value of voltage, we can easily recall that it is nothing but the ratio of the peak value of voltage with square root of two, mathematically, it can be represented as:
${V_{rms}} = \dfrac{{{V_o}}}{{\sqrt 2 }}$
If we substitute the given value of peak value of voltage, we get:
${V_{rms}} = \dfrac{{100\sqrt 2 }}{{\sqrt 2 }}$
Upon simplifying, we get:
${V_{rms}} = 100\,V$
Now that we have found out the RMS value of voltage, we need to find the value of frequency. We already know that there is a direct relation between angular frequency $\left( \omega \right)$ and the frequency of the circuit, this relation can be given as:
$f = \dfrac{\omega }{{2\pi }}$
If we substitute the value of angular frequency as given in the question, we get:
$f = \dfrac{{100\pi }}{{2\pi }}$
After solving, we get:
$f = 50\,Hz$
Hence, we got the values of RMS value of voltage and frequency as:
$
{V_{rms}} = 100\,V \\
f = 50\,Hz \\
$
So, we can easily see that the correct option is the option (A) as the value matches what we found out.
Note: Many students get confused and try to find the RMS value of frequency since the question is worded like it. But there is nothing as the RMS value of frequency. Also, some students get confused and instead of dividing the peak value of voltage by square root of two, they multiply it and thus reach a completely wrong answer.
Recently Updated Pages
If the magnetizing field on a ferromagnetic material class 12 physics JEE_Main

A point charge is placed at the corner of a cube The class 12 physics JEE_Main

What changes occur if the monochromatic light used class 12 physics JEE_Main

The unit of specific conductance is A Ohm B Ohmmetre class 12 physics JEE_Main

Which lens is used in magnifying glass A Concave lens class 12 physics JEE_Main

Sir C V Raman won the Nobel Prize in which year A 1928 class 12 physics JEE_Main

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

Understanding the Electric Field of a Uniformly Charged Ring

Understanding Atomic Structure for Beginners

Electron Gain Enthalpy and Electron Affinity Explained

Derivation of Equation of Trajectory Explained for Students

How to Convert a Galvanometer into an Ammeter or Voltmeter

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

JEE Advanced Percentile vs Marks 2026: JEE Main Cutoff, AIR & IIT Admission Guide

JEE Advanced 2026 Notification Out with Exam Date, Registration (Extended), Syllabus and More

What Are Current and Potential Difference in Electricity?

Understanding Uniform Acceleration in Physics

JEE Advanced Weightage Chapter Wise 2026 for Physics, Chemistry, and Mathematics

