An alternating voltage v(t) = 220 sin 100 pt volt is applied to a purely 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. 3.3 ms
B. 7.2 ms
C. 2.2 ms
D. 5 ms

Explanation

To solve the problem, we need to find the time taken for the current to rise from half of its peak value to its peak value when an alternating voltage is applied to a purely resistive load.

### 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. **Calculate the peak current:**
- According to Ohm's law, the current \( i(t) \) can be calculated using the formula:
\[ i(t) = \frac{v(t)}{R} \]
- Substituting the voltage function:
\[ i(t) = \frac{220 \sin(100 \pi t)}{50} \]
- Simplifying this gives:
\[ i(t) = 4.4 \sin(100 \pi t) \, \text{A} \]
- Thus, the peak current \( I_0 = 4.4 \, \text{A} \).

3. **Determine half of the peak current:**
- Half of the peak current is:
\[ \frac{I_0}{2} = \frac{4.4}{2} = 2.2 \, \text{A} \]

4. **Set up the equation for the current at half the peak value:**
- We need to find the time \( t_1 \) when the current is \( 2.2 \, \text{A} \):
\[ 2.2 = 4.4 \sin(100 \pi t_1) \]
- Dividing both sides by 4.4:
\[ \frac{1}{2} = \sin(100 \pi t_1) \]

5. **Solve for \( t_1 \):**
- The sine of \( \frac{1}{2} \) corresponds to \( \frac{\pi}{6} \):
\[ 100 \pi t_1 = \frac{\pi}{6} \]
- Dividing by \( 100 \pi \):
\[ t_1 = \frac{1}{600} \, \text{s} \]

6. **Set up the equation for the peak current:**
- Now we find the time \( t_2 \) when the current is at its peak value \( 4.4 \, \text{A} \):
\[ 4.4 = 4.4 \sin(100 \pi t_2) \]
- Dividing both sides by 4.4 gives:
\[ 1 = \sin(100 \pi t_2) \]

7. **Solve for \( t_2 \):**
- The sine of \( 1 \) corresponds to \( \frac{\pi}{2} \):
\[ 100 \pi t_2 = \frac{\pi}{2} \]
- Dividing by \( 100 \pi \):
\[ t_2 = \frac{1}{200} \, \text{s} \]

8. **Calculate the time taken to rise from half to peak:**
- The time taken \( \Delta t \) to rise from \( 2.2 \, \text{A} \) to \( 4.4 \, \text{A} \) is:
\[ \Delta t = t_2 - t_1 = \frac{1}{200} - \frac{1}{600} \]
- Finding a common denominator (600):
\[ \Delta t = \frac{3}{600} - \frac{1}{600} = \frac{2}{600} = \frac{1}{300} \, \text{s} \]

9. **Convert to milliseconds:**
- Converting seconds to milliseconds:
\[ \Delta t = \frac{1}{300} \, \text{s} = 3.33 \, \text{ms} \]

### Final Answer:
The time taken for the current to rise from half of the peak value to the peak value is approximately \( 3.33 \, \text{ms} \).

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