Question 277: AC fundamentals

Question

An alternating e.m.f. is given by $e=e_0 \sin \omega t$. In how much time the e.m.f. will have half its maximum value, if e starts from zero ?
$$ \left(\mathrm{T}=	ext { Time Period, } \sin 30^{\circ}=\frac{1}{2}ight) $$

Accepted Answer

We are asked:
An alternating emf is given by
$ e = e_0 \sin (\omega t). $
We want the time when emf becomes
half its maximum value
for the
first time starting from zero
.
Step 1: Condition for half maximum.
Maximum value = $e_0$.
Half of that = $	frac{e_0}{2}$.
So:
$ e_0 \sin (\omega t) = 	frac{e_0}{2} \quad \Rightarrow \quad \sin (\omega t) = 	frac{1}{2}. $
Step 2: Solve for $\omega t$.
$ \sin (\omega t) = 	frac{1}{2} \quad \Rightarrow \quad \omega t = \frac{\pi}{6}, \, \frac{5\pi}{6}, \dots $
The
first time after starting from 0
is:
$ \omega t = \frac{\pi}{6}. $
Step 3: Express in terms of time period $T$.
$ T = \frac{2\pi}{\omega} \quad \Rightarrow \quad \omega = \frac{2\pi}{T}. $
So
$ t = \frac{\pi/6}{\omega} = \frac{\pi/6}{2\pi/T} = \frac{T}{12}. $
Ã¢ÂÂ Final Answer:
The emf reaches half its maximum value in
$ \boxed{	frac{T}{12}}. $
Correct Option: C.

Answer

$\frac{\mathrm{T}}{8}$

Answer

$\frac{\mathrm{T}}{4}$

Answer

$\frac{T}{12}$

Answer

$\frac{T}{16}$

AC fundamentals

An alternating e.m.f. is given by e=e_0 \sin \omega t. In how much time the e.m.f. will have half its maximum value, if e starts from zero ? $ \left(\mathrm{T}=\text { Time Period, } \sin 30^{\circ}=\frac{1}{2}\right)

Choose the correct answer: