An alternating e.m.f. is given by e = e_{0} s | vedclass

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(A) The given equation for the alternating e.m.f. is $e = e_{0} \sin(\omega t)$.We know that $\omega = \frac{2\pi}{T}$,so the equation becomes $e = e_

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$T/12$

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(A) The given equation for the alternating e.m.f. is $e = e_{0} \sin(\omega t)$.We know that $\omega = \frac{2\pi}{T}$,so the equation becomes $e = e_{0} \sin\left(\frac{2\pi t}{T}ight)$.We want to find the time $t$ when $e = \frac{e_{0}}{2}$.Substituting this into the equation: $\frac{e_{0}}{2} = e_{0} \sin\left(\frac{2\pi t}{T}ight)$.$\frac{1}{2} = \sin\left(\frac{2\pi t}{T}ight)$.Since $\sin(30^{\circ}) = \sin\left(\frac{\pi}{6}ight) = \frac{1}{2}$,we have $\frac{2\pi t}{T} = \frac{\pi}{6}$.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? ($T$ = Time period)

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