MHT CET · Physics · Alternating Current
An alternating e.m.f. is given as \(e=e_0 \sin \omega t\). In what time, the e.m.f. will have half its maximum value if \(e\) starts from zero?
\(\left(T=\right.\) time period, \(\left.\sin 30^{\circ}=\cos 60^{\circ}=0.5\right)\)
- A \(\frac{T}{12}\)
- B \(\frac{T}{8}\)
- C \(\frac{T}{16}\)
- D \(\frac{T}{4}\)
Answer & Solution
Correct Answer
(A) \(\frac{T}{12}\)
Step-by-step Solution
Detailed explanation
Let \(t^{\prime}\) be the time when emf is half the maximum value
\(\begin{aligned} & \therefore \frac{e_0}{2}=e_0 \sin \left(\omega t^{\prime}\right) \\ & \Rightarrow \frac{1}{2}=\sin \left(\omega t^{\prime}\right) \\ & \Rightarrow \omega t^{\prime}=\left(\frac{\pi}{6}\right) \\ & \Rightarrow\left(\frac{2 \pi}{T}\right) t^{\prime}=\frac{\pi}{6} \\ & \Rightarrow t^{\prime}=\frac{T}{12}\end{aligned}\)
\(\begin{aligned} & \therefore \frac{e_0}{2}=e_0 \sin \left(\omega t^{\prime}\right) \\ & \Rightarrow \frac{1}{2}=\sin \left(\omega t^{\prime}\right) \\ & \Rightarrow \omega t^{\prime}=\left(\frac{\pi}{6}\right) \\ & \Rightarrow\left(\frac{2 \pi}{T}\right) t^{\prime}=\frac{\pi}{6} \\ & \Rightarrow t^{\prime}=\frac{T}{12}\end{aligned}\)
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