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MHT CET 2021 (24 Sep, Shift 2) · AC Basics: Peak, RMS & Average Values

AC Basics: Peak, RMS & Average Values question from MHT CET 2021 (24 Sep, Shift 2)

Q32MHT CET 2021 · 24 SEP · SHIFT 2Medium

AC Basics: Peak, RMS & Average Values

An alternating e.m.f. is e=e0sin⁡ωte = e_0 \sin \omega t. In what time the e.m.f. will have half its maximum value, if 'ee' starts from zero? (T=T = time period, sin⁡30∘=0.5\sin 30^\circ = 0.5)

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Answer: (a)

  1. e=e02⇒sin⁡ωt=12⇒ωt=π6e = \frac{e_0}{2} \Rightarrow \sin\omega t = \frac{1}{2} \Rightarrow \omega t = \frac{\pi}{6}
  2. 2πTt=π6⇒t=T12\frac{2\pi}{T}t = \frac{\pi}{6} \Rightarrow t = \frac{T}{12}
  3. So the answer is (a) T12\frac{T}{12}.
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