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 ? ($\text{T} = \text{Time Period, } \sin 30^\circ = \frac{1}{2}$)}

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

Hint:

$t = \frac{\theta}{360^\circ} T$. For $30^\circ$, $t = \frac{30}{360}T = \frac{T}{12}$.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
An alternating e.m.f. varies sinusoidally with time.
We need to find the specific instant $t$ when the instantaneous value of the e.m.f. ($e$) reaches half of its peak maximum value ($e_0$).
Step 2: Key Formula or Approach:
The equation for the alternating e.m.f. is: \[ e = e_0 \sin(\omega t) \] The angular frequency $\omega$ is related to the time period $T$ by: \[ \omega = \frac{2\pi}{T} \] We are given the condition $e = \frac{e_0}{2}$.
Step 3: Detailed Explanation:
Substitute the given condition into the e.m.f. equation: \[ \frac{e_0}{2} = e_0 \sin(\omega t) \] Divide both sides by the peak value $e_0$: \[ \frac{1}{2} = \sin(\omega t) \] We know from the problem statement (and standard trigonometry) that $\sin(30^\circ) = \frac{1}{2}$.
In radians, $30^\circ$ is equal to $\frac{\pi}{6}$ radians.
Therefore, we can set the argument of the sine function equal to this angle: \[ \omega t = \frac{\pi}{6} \] Now, substitute the expression for angular frequency $\omega = \frac{2\pi}{T}$: \[ \left(\frac{2\pi}{T}\right) t = \frac{\pi}{6} \] We can cancel $\pi$ from both sides: \[ \frac{2}{T} t = \frac{1}{6} \] Solve for $t$: \[ t = \frac{1}{6} \times \frac{T}{2} \] \[ t = \frac{T}{12} \] Thus, it takes $T/12$ seconds for the e.m.f. to reach half of its maximum value starting from zero.
Step 4: Final Answer:
The time taken is $\frac{T}{12}$.