Question

An alternating e.m.f. is given by \( e = e_0 \sin \omega t \). In what time the e.m.f. will have half its maximum value, if \( e_0 \) starts from zero? (T = Time period)

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

Hint:

To find when the e.m.f. is half its maximum value in an alternating current, use the equation \( \sin \omega t = \frac{1}{2} \).

Answer 1

The Correct Option is A

Step 1: Finding the maximum value.
The maximum value of \( e \) is \( e_0 \). The e.m.f. is \( e_0 \sin \omega t \), and we need to find the time when it is half of its maximum value, i.e., \( \frac{e_0}{2} \).
Step 2: Solving for time.
We need to solve for \( t \) when \( e = \frac{e_0}{2} \): \[ \frac{e_0}{2} = e_0 \sin \omega t \Rightarrow \sin \omega t = \frac{1}{2} \] The solution to this equation is \( \omega t = \frac{\pi}{6} \), so the time is \[ t = \frac{T}{12} \] Step 3: Conclusion.
Thus, the time when the e.m.f. is half its maximum value is \( \frac{T}{12} \), which corresponds to option (A).