The alternating voltage is given by V=V_0 (ω t+ (π)/(6)). When will the voltage be maximum for the first time? \ (90)^ ^ =1\

Correct answer: B. (T)/(6) The given alternating voltage is V = V_0 (ω t + (π)/(6)). For the voltage to be maximum, the value of the sine function must be 1 . (ω t + (π)/(6)) =1 The first maximum occurs when the phase angle is (π)/(2) . ω t + (π)/(6) = (π)/(2) ω t = (π)/(2) - (π)/(6) = (π)/(3) Substituting ω = (2π)/(T): (2π)/(T) t = (π)/(3) t = (T)/(6) Answer: (T)/(6)

MHT CET 202624 AprilMorning Shift

Exam: MHT CET
Year: 2026
Date: 24 April
Shift: Morning Shift
Subject: Physics
Chapter: Alternating Current

MHT CET 2026 · 24 April · Morning Shift · Physics · Alternating Current

The alternating voltage is given by V=V_0 (ω t+ (π)/(6)). When will...

The alternating voltage is given by

V=V_0\sin \left(\omega

\dfrac{\pi}{6}\right). When will the voltage be maximum for the first time?<br>

\{\sin(90)^{}^{\circ}=1\}

Options

A $\dfrac{T}{12}$
B $\dfrac{T}{6}$
C $\dfrac{T}{3}$
D $\dfrac{T}{2}$

Correct answer

B. $\dfrac{T}{6}$

Step-by-step solution

The given alternating voltage is

V = V_0 \sin \left(\omega

t +

\dfrac{\pi}{6}\right).<br><br>For the voltage to be maximum, the value of the sine function must be 1

\sin\left(\omega

t +

\dfrac{\pi}{6}\right) =1

The first maximum occurs when the phase angle is

\dfrac{\pi}{2}

.

\omega

t + \dfrac{\pi}{6} = \dfrac{\pi}{2}<br><br>\omega

t =

\dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3}<br><br>Substituting \omega = \dfrac{2\pi}{T}:<br><br>\dfrac{2\pi}{T}

t =

\dfrac{\pi}{3}<br><br>

t =

\dfrac{T}{6}<br><br>Answer: \dfrac{T}{6}

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