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

Options

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

Answer

To solve the problem, we need to find the time taken for the current in a purely resistive load to rise from half of its peak value to its peak value when an alternating voltage is applied. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Identify the given values**:
- The alternating voltage is given by \( v(t) = 220 \sin(100 \pi t) \) volts.
- The resistance \( R = 50 \, \Omega \).

2. **Determine the peak voltage**:
- The peak voltage \( V_0 = 220 \) volts.

3. **Calculate the peak current**:
- Using Ohm's law, the peak current \( I_0 \) can be calculated as:
\[ I_0 = \frac{V_0}{R} = \frac{220 \, \text{V}}{50 \, \Omega} = 4.4 \, \text{A} \]

4. **Write the current as a function of time**:
- The current \( i(t) \) can be expressed as:
\[ i(t) = I_0 \sin(100 \pi t) = 4.4 \sin(100 \pi t) \]

5. **Find the time when the current is half of the peak value**:
- Half of the peak current is:
\[ \frac{I_0}{2} = \frac{4.4}{2} = 2.2 \, \text{A} \]
- Set the equation:
\[ 2.2 = 4.4 \sin(100 \pi t_1) \]
- Simplifying gives:
\[ \sin(100 \pi t_1) = \frac{1}{2} \]
- The angle whose sine is \( \frac{1}{2} \) is \( \frac{\pi}{6} \) (30 degrees). Therefore:
\[ 100 \pi t_1 = \frac{\pi}{6} \]
- Solving for \( t_1 \):
\[ t_1 = \frac{1}{600} \, \text{s} \]

6. **Find the time when the current is at its peak value**:
- The peak current occurs when:
\[ I = I_0 \sin(100 \pi t_2) = 4.4 \, \text{A} \]
- This implies:
\[ \sin(100 \pi t_2) = 1 \]
- The angle whose sine is 1 is \( \frac{\pi}{2} \) (90 degrees). Therefore:
\[ 100 \pi t_2 = \frac{\pi}{2} \]
- Solving for \( t_2 \):
\[ t_2 = \frac{1}{200} \, \text{s} \]

7. **Calculate the time taken to rise from half to peak**:
- The time taken \( \Delta t \) is:
\[ \Delta t = t_2 - t_1 = \frac{1}{200} - \frac{1}{600} \]
- Finding a common denominator (600):
\[ \Delta t = \frac{3}{600} - \frac{1}{600} = \frac{2}{600} = \frac{1}{300} \, \text{s} \]
- Converting to milliseconds:
\[ \Delta t = \frac{1}{300} \times 1000 = 3.33 \, \text{ms} \]

### Final Answer:
The time taken for the current to rise from half of the peak value to the peak value is approximately **3.33 ms**.