An alternating voltage `V(t)=220sin100 pi t` 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. 2.2 ms
B. 7.2 ms
C. 5 ms
D. 3.3 ms

Answer

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 parameters:**
- The alternating voltage is given by \( V(t) = 220 \sin(100 \pi t) \) volts.
- The resistance \( R = 50 \, \Omega \).

2. **Calculate the peak voltage:**
- The peak voltage \( V_0 = 220 \, \text{V} \).

3. **Calculate the peak current using Ohm's Law:**
- The peak current \( I_0 \) can be calculated using the formula:
\[ I_0 = \frac{V_0}{R} = \frac{220}{50} = 4.4 \, \text{A} \]

4. **Determine half of the peak current:**
- Half of the peak current \( I_{half} = \frac{I_0}{2} = \frac{4.4}{2} = 2.2 \, \text{A} \).

5. **Set up the equation for current:**
- The current \( I(t) \) in the circuit is given by:
\[ I(t) = \frac{V(t)}{R} = \frac{220 \sin(100 \pi t)}{50} = 4.4 \sin(100 \pi t) \]

6. **Find the time \( t_1 \) when the current is half of the peak value:**
- Set \( I(t_1) = 2.2 \):
\[ 2.2 = 4.4 \sin(100 \pi t_1) \]
- Simplifying gives:
\[ \frac{1}{2} = \sin(100 \pi t_1) \]
- Taking the inverse sine:
\[ 100 \pi t_1 = \frac{\pi}{6} \quad \Rightarrow \quad t_1 = \frac{1}{600} \, \text{s} \]

7. **Find the time \( t_2 \) when the current is at its peak value:**
- Set \( I(t_2) = 4.4 \):
\[ 4.4 = 4.4 \sin(100 \pi t_2) \]
- This simplifies to:
\[ 1 = \sin(100 \pi t_2) \]
- Taking the inverse sine:
\[ 100 \pi t_2 = \frac{\pi}{2} \quad \Rightarrow \quad t_2 = \frac{1}{200} \, \text{s} \]

8. **Calculate the time difference \( \Delta t \):**
- The time taken for the current to rise from half of the peak value to the peak value is:
\[ \Delta t = t_2 - t_1 = \frac{1}{200} - \frac{1}{600} \]
- 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:
\[ \Delta t = \frac{1}{300} \times 1000 = 3.33 \, \text{ms} \approx 3.3 \, \text{ms} \]

### Final Answer:
The time taken for the current to rise from half of the peak value to the peak value is approximately **3.3 ms**.