Question

At what time (From zero) the alternating voltage becomes $\frac{1}{\sqrt{2}}$ times of it's peak value. Where T is the periodic time

• A. $\frac{T}{2}\sec$
• B. $\frac{T}{4}\sec$
• C. $\frac{T}{8}\sec$
• D. $\frac{T}{12}\sec$

Answer 1

Answer: (C)

$\frac{T}{8}\sec$

Answer 2

By using $V = V_{0}\sin\omega t$ ⇒ $\frac{V_{0}}{\sqrt{2}} = V_{0}\sin\frac{2\pi t}{T}$⇒

$\frac{1}{\sqrt{2}} = \sin\left( \frac{2\pi}{T} \right)t$ ⇒ $\sin\frac{\pi}{4} = \sin\left( \frac{2\pi}{T} \right)t$

⇒ $\frac{\pi}{4} = \frac{2\pi}{T}t$ ⇒ $t = \frac{T}{8}sec.$