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:

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

Answer 2

The problem provides an alternating voltage applied to a purely resistive load and asks for the time it takes for the current to rise from half of its peak value to the peak value.

Concept Used:

For a purely resistive AC circuit, the current \( I(t) \) is in phase with the voltage \( V(t) \). The relationship is given by Ohm's law:

\[ I(t) = \frac{V(t)}{R} \]

Given a sinusoidal voltage \( V(t) = V_{peak} \sin(\omega t) \), the current will be \( I(t) = \frac{V_{peak}}{R} \sin(\omega t) = I_{peak} \sin(\omega t) \), where \( I_{peak} \) is the peak or maximum current.

To find the time interval, we need to solve the trigonometric equation for the current at two different instants and find the difference between them.

Step-by-Step Solution:

Step 1: Write down the given equations and parameters.

The applied voltage is given by:

\[ V(t) = 220 \sin(100 \pi t) \, \text{V} \]

By comparing this with the standard form \( V(t) = V_{peak} \sin(\omega t) \), we can identify:

• Peak voltage, \( V_{peak} = 220 \) V
• Angular frequency, \( \omega = 100 \pi \) rad/s

The resistance of the load is \( R = 50 \, \Omega \).

Step 2: Determine the equation for the current.

Using Ohm's law for the resistive circuit:

\[ I(t) = \frac{V(t)}{R} = \frac{220 \sin(100 \pi t)}{50} \] \[ I(t) = 4.4 \sin(100 \pi t) \, \text{A} \]

From this equation, the peak current is \( I_{peak} = 4.4 \) A.

Step 3: Find the time \( t_1 \) when the current is half of its peak value.

We need to solve for \( t_1 \) when \( I(t_1) = \frac{I_{peak}}{2} \).

\[ 4.4 \sin(100 \pi t_1) = \frac{4.4}{2} \] \[ \sin(100 \pi t_1) = \frac{1}{2} \]

For the current to be rising, we consider the first instance in the cycle where this occurs. This corresponds to the phase angle being in the first quadrant.

\[ 100 \pi t_1 = \frac{\pi}{6} \] \[ t_1 = \frac{\pi}{6 \times 100 \pi} = \frac{1}{600} \, \text{s} \]

Step 4: Find the time \( t_2 \) when the current reaches its peak value.

We need to solve for \( t_2 \) when \( I(t_2) = I_{peak} \).

\[ 4.4 \sin(100 \pi t_2) = 4.4 \] \[ \sin(100 \pi t_2) = 1 \]

The first time the current reaches its positive peak corresponds to the phase angle being \( \frac{\pi}{2} \).

\[ 100 \pi t_2 = \frac{\pi}{2} \] \[ t_2 = \frac{\pi}{2 \times 100 \pi} = \frac{1}{200} \, \text{s} \]

Step 5: Calculate the time difference.

The time taken for the current to rise from half of the peak value to the peak value is \( \Delta t = t_2 - t_1 \).

\[ \Delta t = \frac{1}{200} \, \text{s} - \frac{1}{600} \, \text{s} \]

Final Computation & Result:

To subtract the fractions, we find a common denominator, which is 600.

\[ \Delta t = \frac{3}{600} - \frac{1}{600} = \frac{2}{600} \, \text{s} \]

Simplifying the fraction:

\[ \Delta t = \frac{1}{300} \, \text{s} \]

This is approximately 3.33 ms.

The time taken for the current to rise from half of the peak value to the peak value is \( \frac{1}{300} \) s or approximately 3.33 ms.