Question 306: AC fundamentals

Question

The alternating voltage is given by $\mathrm{v}=\mathrm{v}_0 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right)$ when will be the voltage maximum for first time?

Accepted Answer

$$\begin{aligned}
& \mathrm{V}=\mathrm{V}_0 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}ight) \
& \sin \left(\omega \mathrm{t}+\frac{\pi}{3}ight)=1 \quad 	ext { [For voltage to be maximum] } \
	herefore \quad & \sin \left(\omega \mathrm{t}+\frac{\pi}{3}ight)=\sin \left(\frac{\pi}{2}ight) \
& \omega \mathrm{t}+\frac{\pi}{3}=\frac{\pi}{2} \
	herefore \quad & \omega t=\frac{\pi}{6} \
& t=\frac{\pi}{6 \omega}
\end{aligned}$$
$$\begin{aligned}
	herefore \quad & =\frac{\pi 	imes T}{6 	imes 2 \pi} \quad \ldots\left(\because \omega=\frac{2 \pi}{T}ight) \
t & =\frac{T}{12}
\end{aligned}$$

Answer

$\frac{\mathrm{T}}{6}$

Answer

$\frac{\mathrm{T}}{3}$

Answer

$\frac{T}{2}$

Answer

$\frac{\mathrm{T}}{12}$

AC fundamentals

The alternating voltage is given by \mathrm{v}=\mathrm{v}_0 \sin \left(\omega \mathrm{t}+\frac{\pi}{3}\right) when will be the voltage maximum for first time?

Choose the correct answer: