The voltage of an $AC$ supply varies with time (t) as $V = 60\sin 50\pi t.\cos 50\pi t$ . The maximum voltage and frequency are respectively (where $V$ is in volt and $t$ is in second)
(A) $30\,Volt,\,100\,Hz$
(B) $60\,Volt,\,50\,Hz$
(C) $60\,Volt,\,100\,Hz$
(D) $30\,Volt,\,50\,Hz$
Answer
292.5k+ views
Hint Compare the given voltage of the alternating current with the trigonometric formula, the change in the voltage provides the answer for the maximum voltage of the alternating current. Use the formula of the angular velocity of the wave to find the maximum frequency of the alternating current.
Useful formula:
(1) The trigonometric formula is given by
$\sin 2\theta = 2\sin \theta \cos \theta $
(2) The formula of the angular velocity is given by
$\omega = 2\pi f$
Where $\omega $ is the angular velocity and $f$ is the frequency of the alternating current.
Complete step by step answer
It is given that the
Voltage of the alternating current supply, $V = 60\sin 50\pi t.\cos 50\pi t$
Let us apply the formula of the $\sin 2\theta $ in the above voltage,
$\sin 2\left( {50\pi t} \right) = 2\sin 50\pi t.\cos 50\pi t$ ---------(1)
But the given voltage is $V = 60\sin 50\pi t.\cos 50\pi t$ ------(2)
Comparing (1) and (2), all are the same except the constant before the trigonometric parameters. Hence the maximum voltage is obtained by dividing them as $\dfrac{{60}}{2} = 30\,V$ .
Hence the maximum voltage of the given voltage of the alternating current is obtained as $30\,V$ .
Let us use the formula of the angular velocity,
$\omega = 2\pi f$
Rearranging the above formula in order to find the frequency.
$f = \dfrac{\omega }{{2\pi }}$
Substituting the $\omega = 100\pi $ in the above formula, we get
$f = \dfrac{{100\pi }}{{2\pi }}$
By simplifying the above step, we get
$f = 50\pi $
Hence the maximum frequency of the alternating current of the given voltage is $50\pi $ .
Thus the option (D) is correct.
Note: The frequency will be maximum only when the angular velocity of the given wave will be maximum. The angular velocity will be maximum at twice the theta of the voltage. Hence the angular velocity is obtained by $2 \times 50\pi $ , which is equal to $100\pi $ .
Useful formula:
(1) The trigonometric formula is given by
$\sin 2\theta = 2\sin \theta \cos \theta $
(2) The formula of the angular velocity is given by
$\omega = 2\pi f$
Where $\omega $ is the angular velocity and $f$ is the frequency of the alternating current.
Complete step by step answer
It is given that the
Voltage of the alternating current supply, $V = 60\sin 50\pi t.\cos 50\pi t$
Let us apply the formula of the $\sin 2\theta $ in the above voltage,
$\sin 2\left( {50\pi t} \right) = 2\sin 50\pi t.\cos 50\pi t$ ---------(1)
But the given voltage is $V = 60\sin 50\pi t.\cos 50\pi t$ ------(2)
Comparing (1) and (2), all are the same except the constant before the trigonometric parameters. Hence the maximum voltage is obtained by dividing them as $\dfrac{{60}}{2} = 30\,V$ .
Hence the maximum voltage of the given voltage of the alternating current is obtained as $30\,V$ .
Let us use the formula of the angular velocity,
$\omega = 2\pi f$
Rearranging the above formula in order to find the frequency.
$f = \dfrac{\omega }{{2\pi }}$
Substituting the $\omega = 100\pi $ in the above formula, we get
$f = \dfrac{{100\pi }}{{2\pi }}$
By simplifying the above step, we get
$f = 50\pi $
Hence the maximum frequency of the alternating current of the given voltage is $50\pi $ .
Thus the option (D) is correct.
Note: The frequency will be maximum only when the angular velocity of the given wave will be maximum. The angular velocity will be maximum at twice the theta of the voltage. Hence the angular velocity is obtained by $2 \times 50\pi $ , which is equal to $100\pi $ .
Recently Updated Pages
JEE Main 2025-26 Electrostatics Mock Test: Free Practice Online

Electrostatics Mock Test 2025: Practice Questions & Answers

JEE Main 2026-27 Kinematics Mock Test – Free Practice Online

JEE Main Mock Test 2025-26: Chapter-Wise Practice Papers

Wheatstone Bridge – Principle, Formula, Diagram & Applications

Kinematics Mock Test 2025: Free Practice Questions & Answers

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

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

Understanding Uniform Acceleration in Physics

What Are Current and Potential Difference in Electricity?

