Question

At what time (from zero) will the alternating voltage become $\frac{1}{\sqrt{2}}$ times of its peak value where $T$ is the periodic time?

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

Answer 1

The correct option is C $\frac{T}{8}s$

It is given that $V=\frac{1}{\sqrt{2}}{V}_0$ ,where V is instantaneous voltage and V₀ is peak value of voltage.

$V_0\sin \omega t=\frac{1}{\sqrt{2}}{V}_0$

$\sin \frac{2\pi }{T}t=\frac{1}{\sqrt{2}}$

$\frac{2\pi }{T}t=\frac{\pi }{4}$

$t=\frac{T}{8}s$