Question

An alternating voltage is given by \( e = 8 \sin(628.4 t) \). 
Find:
(i) Peak value of e.m.f. 
(ii) Frequency of e.m.f. 
(iii) Instantaneous value of e.m.f. at time \( t = 10 \, {ms} \) 

Hint:

-The peak value of an alternating voltage is the highest voltage it reaches, which is the same as the amplitude of the sine wave.

-The frequency of an alternating voltage is related to the angular frequency by the formula:

\[ f = \frac{\omega}{2\pi} \]

The angular frequency \( \omega \) is measured in radians per second.

- The instantaneous value of an alternating voltage can be calculated by substituting the time into the voltage equation. When the argument of the sine function is a multiple of \( 2\pi \), the instantaneous value becomes zero.

Answer 1

The given equation for the alternating voltage is: \[ e = 8 \sin(628.4 t) \] where \( 8 \) represents the amplitude of the voltage, and \( 628.4 \) is the angular frequency (\( \omega \)).
(i) Peak value of e.m.f.
The peak value of an alternating voltage is the maximum voltage that the wave attains. By inspecting the equation, it is clear that the peak value of the e.m.f. corresponds to the amplitude of the sine function. 
Therefore: \[ {Peak value of e.m.f.} = 8 \, {V}. \]

(ii) Frequency of e.m.f.

The angular frequency \( \omega \) is related to the frequency \( f \) by the equation:

\[ \omega = 2 \pi f \]

Given that \( \omega = 628.4 \), we can solve for \( f \):

\[ f = \frac{\omega}{2\pi} = \frac{628.4}{2\pi} = 100 \, \text{Hz}. \]

Thus, the frequency of the e.m.f. is \( 100 \, \text{Hz} \).

(iii) Instantaneous value of e.m.f. at \( t = 10 \, \text{ms} \)

To find the instantaneous value of the e.m.f. at time \( t = 10 \, \text{ms} \), we substitute \( t = 10 \times 10^{-3} \, \text{s} \) into the equation:

\[ e = 8 \sin(628.4 \times 10 \times 10^{-3}) = 8 \sin(6.284) \]

Now, \( 6.284 \) is very close to \( 2\pi \), and we know that:

\[ \sin(2\pi) = 0 \]

Thus:

\[ \sin(6.284) \approx 0 \]

So the instantaneous value of the e.m.f. at \( t = 10 \, \text{ms} \) is:

\[ e = 8 \times 0 = 0 \, \text{V}. \]

Therefore, the instantaneous value of the e.m.f. at \( t = 10 \, \text{ms} \) is \( 0 \, \text{V} \).