An alternating e.m.f. is given by e = e_0 sin | vedclass

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(C) The given equation for alternating e.m.f. is $e = e_0 \sin \omega t$.We are looking for the time $t$ when the e.m.f. $e$ is half of its maximum va

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$\frac{T}{12}$

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(C) The given equation for alternating e.m.f. is $e = e_0 \sin \omega t$.We are looking for the time $t$ when the e.m.f. $e$ is half of its maximum value $e_0$, i.e.,$e = \frac{e_0}{2}$.Substituting this into the equation: $\frac{e_0}{2} = e_0 \sin \omega t$.This simplifies to $\sin \omega t = \frac{1}{2}$.We know that $\sin 	heta = \frac{1}{2}$ when $	heta = \frac{\pi}{6}$.Therefore, $\omega t = \frac{\pi}{6}$.Since $\omega = \frac{2\pi}{T}$, we substitute this into the equation: $(\frac{2\pi}{T}) t = \frac{\pi}{6}$.Solving for $t$: $t = \frac{\pi}{6} 	imes \frac{T}{2\pi} = \frac{T}{12}$.

An alternating e.m.f. is given by $e = e_0 \sin \omega t$. In what time will the e.m.f. reach half its maximum value, if $e$ starts from zero?

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