An alternating e.m.f. is e = e_0 sin omega t. | vedclass

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(A) Given the alternating e.m.f. equation: $e = e_0 \sin \omega t$.We want to find the time $t$ when $e = \frac{e_0}{2}$.Substituting the value into t

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$\frac{T}{12}$

Vedclass · Answer

(A) Given the alternating e.m.f. equation: $e = e_0 \sin \omega t$.We want to find the time $t$ when $e = \frac{e_0}{2}$.Substituting the value into the equation: $\frac{e_0}{2} = e_0 \sin \omega t$.This simplifies to: $\sin \omega t = \frac{1}{2}$.Since $\sin 30^{\circ} = \frac{1}{2}$,we have $\omega t = 30^{\circ} = \frac{\pi}{6} 	ext{ radians}$.Substituting $\omega = \frac{2\pi}{T}$: $\left( \frac{2\pi}{T} ight) t = \frac{\pi}{6}$.Solving for $t$: $t = \frac{T}{12}$.

An alternating e.m.f. is $e = e_0 \sin \omega t$. 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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