Question

An alternating voltage \( V(t) = 220 \sin 100 \pi t \) volt is applied to a purely resistive load of 50 \( \Omega \). The time taken for the current to rise from half of the peak value to the peak value is:

• A. 5 ms
• B. 3.3 ms
• C. 7.2 ms
• D. 2.2 ms

Answer 1

The Correct Option is B

The given voltage function is:

\[ V(t) = 220 \sin(100\pi t). \]

The angular frequency \( \omega \) can be identified from the argument of the sine function:

\[ \omega = 100\pi. \]

The period \( T \) of the sinusoidal function is given by:

\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{100\pi} = \frac{1}{50} \text{ seconds} = 20 \, \text{ms}. \]

To find the time taken for the voltage (and hence the current, since the load is purely resistive) to rise from half of the peak value to the peak value, we consider the time interval needed for a sine wave to go from \( \frac{1}{2} \) of its maximum to its maximum value.

For a sine function, this interval corresponds to a phase change of \( \frac{\pi}{6} \) radians (from \( \sin(\theta) = \frac{1}{2} \) to \( \sin(\theta) = 1 \)).

Thus, the time \( t \) for this phase change is:

\[ t = \frac{\pi/6}{\omega} = \frac{\pi/6}{100\pi} = \frac{1}{600} \text{ seconds}. \]

Converting this to milliseconds:

\[ t = \frac{1}{600} \times 1000 = 3.33 \, \text{ms}. \]