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?

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

Hint:

Set sin(wt) = 1/2 and use w = 2 pi / T.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The emf \(e = e_0\sin\omega t\) starts at zero. We need the first time \(e = \dfrac{e_0}{2}\).

Step 2: Solve:
\[ \sin\omega t = \frac12 \Rightarrow \omega t = \frac\pi6 \]
With \(\omega = \dfrac{2\pi}{T}\):
\[ \frac{2\pi}{T}t = \frac\pi6 \Rightarrow t = \frac T{12} \]

Step 3: Check:
Option (C). At \(T/4\) the emf is at its peak, and at \(T/8\) it is \(e_0\sin45^\circ = 0.707e_0\), not half.

Final Answer:
sin(wt) = 1/2 gives wt = pi/6, so t = T/12. \[ \boxed{\text{(C) }\dfrac{T}{12}} \]