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An alternating e.m.f. is given as $e=e_0 \sin \omega t$. In what time, the e.m.f. will have half its maximum value if $e$ starts from zero?
$\left(T=\right.$ time period, $\left.\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)$
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$\left(T=\right.$ time period, $\left.\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)$
Solution:
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Verified Answer
The correct answer is:
$\frac{T}{12}$
Let $t^{\prime}$ be the time when emf is half the maximum value
$\begin{aligned} & \therefore \frac{e_0}{2}=e_0 \sin \left(\omega t^{\prime}\right) \\ & \Rightarrow \frac{1}{2}=\sin \left(\omega t^{\prime}\right) \\ & \Rightarrow \omega t^{\prime}=\left(\frac{\pi}{6}\right) \\ & \Rightarrow\left(\frac{2 \pi}{T}\right) t^{\prime}=\frac{\pi}{6} \\ & \Rightarrow t^{\prime}=\frac{T}{12}\end{aligned}$
$\begin{aligned} & \therefore \frac{e_0}{2}=e_0 \sin \left(\omega t^{\prime}\right) \\ & \Rightarrow \frac{1}{2}=\sin \left(\omega t^{\prime}\right) \\ & \Rightarrow \omega t^{\prime}=\left(\frac{\pi}{6}\right) \\ & \Rightarrow\left(\frac{2 \pi}{T}\right) t^{\prime}=\frac{\pi}{6} \\ & \Rightarrow t^{\prime}=\frac{T}{12}\end{aligned}$
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