Question

An alternating e.m.f. is given by \(e = e_0sinωt\). In what time the e.m.f. will have half its maximum value, if 'e' starts from zero? (\(sin30^{\circ} = 0.5\))

• A. \(\frac{T}{4}\)
• B. \(\frac{T}{8}\)
• C. \(\frac{T}{12}\)
• D. \(\frac{T}{16}\)

Hint:

Set e0 sin(wt) equal to e0 divided by 2 and use w = 2 pi / T.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The emf is \(e = e_0\sin\omega t\), and \(\omega = \dfrac{2\pi}{T}\).

Step 2: Set up.
\[ \frac{e_0}{2} = e_0\sin\omega t \Rightarrow \sin\omega t = \frac{1}{2} \Rightarrow \omega t = 30^\circ = \frac{\pi}{6} \]

Step 3: Solve for t.
\[ t = \frac{\pi}{6}\times\frac{T}{2\pi} = \frac{T}{12} \]

Step 4: Check the options.
\(T/4\) is when the emf reaches the maximum value. \(T/8\) and \(T/16\) correspond to angles of \(45^\circ\) and \(22.5^\circ\).

Final Answer:
The emf is half its maximum after \(\dfrac{T}{12}\), option (C). \[ \boxed{\frac{T}{12}} \]