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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

Option: 1

\mathrm{\frac{\mathrm{T}}{2} \mathrm{sec}}


Option: 2

\mathrm{\frac{\mathrm{T}}{4} \mathrm{sec}}


Option: 3

\mathrm{\frac{\mathrm{T}}{8} \mathrm{sec}}


Option: 4

\mathrm{\frac{\mathrm{T}}{12} \mathrm{sec}}


Answers (1)

best_answer

\mathrm{\text { By using } V=V_0 \sin \omega t}

\mathrm{\begin{aligned} & \Rightarrow \frac{V_0}{\sqrt{2}}=V_0 \sin \frac{2 \pi t}{T} \Rightarrow \frac{1}{\sqrt{2}}=\sin \left(\frac{2 \pi}{T}\right) t \\ & \Rightarrow \sin \frac{\pi}{4}=\sin \left(\frac{2 \pi}{T}\right) t \\ & \Rightarrow \frac{\pi}{4}=\frac{2 \pi}{T} t \Rightarrow t=\frac{T}{8} \text { sec. } \end{aligned}}

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Devendra Khairwa

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