Q180 / 200 · MHT CET - Medical 2026 (24 April Shift 1) · Physics
The alternating voltage is given by $V=V_0\sin\left(\omega t+\dfrac{\pi}{6}\right)$. When will the voltage be maximum for the first time?
$\{\sin(90)^\circ=1\}$
$\{\sin(90)^\circ=1\}$
- A$\dfrac{T}{12}$
- Correct answerB$\dfrac{T}{6}$
- C$\dfrac{T}{3}$
- D$\dfrac{T}{2}$
Solution
The given alternating voltage is $V = V_0 \sin\left(\omega t + \dfrac{\pi}{6}\right)$.
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 $\dfrac{\pi}{2}$.
$\omega t + \dfrac{\pi}{6} = \dfrac{\pi}{2}$
$\omega t = \dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3}$
Substituting $\omega = \dfrac{2\pi}{T}$:
$\dfrac{2\pi}{T} t = \dfrac{\pi}{3}$
$t = \dfrac{T}{6}$
Answer: $\dfrac{T}{6}$
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 $\dfrac{\pi}{2}$.
$\omega t + \dfrac{\pi}{6} = \dfrac{\pi}{2}$
$\omega t = \dfrac{\pi}{2} - \dfrac{\pi}{6} = \dfrac{\pi}{3}$
Substituting $\omega = \dfrac{2\pi}{T}$:
$\dfrac{2\pi}{T} t = \dfrac{\pi}{3}$
$t = \dfrac{T}{6}$
Answer: $\dfrac{T}{6}$
Hence, the correct answer is option: 'B'.