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 want to find the time $t$ when $e = \frac{e_0}{2}$.Substituting this into

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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 want to find the time $t$ when $e = \frac{e_0}{2}$.Substituting this into the equation: $\frac{e_0}{2} = e_0 \sin \omega t$.$\sin \omega t = \frac{1}{2}$.Since $\sin 30^{\circ} = 0.5$, we have $\omega t = 30^{\circ} = \frac{\pi}{6}$ radians.We know that $\omega = \frac{2\pi}{T}$, where $T$ is the time period.Substituting $\omega$: $(\frac{2\pi}{T}) \cdot t = \frac{\pi}{6}$.Solving for $t$: $t = \frac{\pi}{6} \cdot \frac{T}{2\pi} = \frac{T}{12}$.

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? (Given: $\sin 30^{\circ} = 0.5$)

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