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 provided voltage function is:

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

The angular frequency \( \omega \) is derived from the sine function's argument:

\[ \omega = 100\pi. \]

The period \( T \) of this sinusoidal function is calculated as:

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

To determine the duration for the voltage (and consequently, the current, assuming a purely resistive load) to increase from half its peak value to its maximum value, we examine the time required for a sine wave to transition from \( \frac{1}{2} \) of its maximum to its maximum.

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

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 time to milliseconds:

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