Question

An alternating voltage is given by $E=100sin(\omega +\frac{\pi }{6})V$ The voltage will be maximum for the first time when is [$T$ = periodic time)

• A. $\frac{T}{12}$
• B. $\frac{T}{2}$
• C. $\frac{T}{6}$
• D. $\frac{T}{3}$

Answer 1

Answer: (C)

Given, alternating voltage is
$E=100\sin \left(\omega t+\frac{\pi }{6}\right)V$
The voltage will be maximum, when
$\sin \left(\omega t+\frac{\pi }{6}\right)=1$
$\Rightarrow \sin \left(\omega t+\frac{\pi }{6}\right)=\sin \frac{\pi }{2}$
$\Rightarrow \omega t+\frac{\pi }{6}=\frac{\pi }{2}$
$\Rightarrow \omega t=\frac{\pi }{3}$
$\Rightarrow t=\frac{\pi }{3\omega }=\frac{\pi \times T}{3\times 2\pi }\left[\because \omega =\frac{2\pi }{T}\right]$
$=\frac{T}{6}$
Thus, the voltage will be maximum for the first time when $t=\frac{T}{6}$.