A resistance of `20ohms` is connected to a source of an alternating potential `V=220sin(100pit)`. The time taken by the current to change from its peak value to r.m.s. value is

Options

A. 0.2sec
B. 0.25sec
C. `25xx10^(-3)sec`
D. `2.5xx10^(-3)sec`

Answer

To solve the problem step by step, we will break down the process of finding the time taken by the current to change from its peak value to its root mean square (r.m.s.) value.

### Step-by-Step Solution:

1. **Identify the Given Parameters:**
- Resistance, \( R = 20 \, \Omega \)
- Voltage, \( V(t) = 220 \sin(100 \pi t) \)

2. **Calculate the Peak Current:**
- The peak voltage \( V_0 = 220 \, V \).
- The peak current \( I_0 \) can be calculated using Ohm's law:
\[ I_0 = \frac{V_0}{R} = \frac{220}{20} = 11 \, A \]

3. **Write the Current Equation:**
- The current \( I(t) \) can be expressed as:
\[ I(t) = I_0 \sin(100 \pi t) = 11 \sin(100 \pi t) \]

4. **Determine the Time Period \( T \):**
- The angular frequency \( \omega = 100 \pi \).
- The time period \( T \) is given by:
\[ T = \frac{2\pi}{\omega} = \frac{2\pi}{100 \pi} = \frac{1}{50} \, s \]

5. **Identify the Time to Reach Peak and RMS Values:**
- The time to reach the peak value (from 0 to peak) is \( \frac{T}{4} \) because the sine function reaches its maximum at \( \frac{\pi}{2} \) (which corresponds to \( \frac{T}{4} \)):
\[ t_{\text{peak}} = \frac{T}{4} = \frac{1/50}{4} = \frac{1}{200} \, s \]

6. **Calculate the RMS Current:**
- The r.m.s. current \( I_{\text{rms}} \) is given by:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.78 \, A \]

7. **Determine the Time to Reach RMS Value:**
- To find the time when the current reaches the r.m.s. value, set:
\[ I(t) = I_{\text{rms}} \Rightarrow 11 \sin(100 \pi t) = \frac{11}{\sqrt{2}} \]
- Dividing both sides by 11:
\[ \sin(100 \pi t) = \frac{1}{\sqrt{2}} \]
- The sine function equals \( \frac{1}{\sqrt{2}} \) at \( 45^\circ \) or \( \frac{\pi}{4} \):
\[ 100 \pi t = \frac{\pi}{4} \Rightarrow t = \frac{1}{400} \, s \]

8. **Calculate the Time Difference:**
- The time taken for the current to change from peak to r.m.s. is:
\[ \Delta t = t_{\text{rms}} - t_{\text{peak}} = \frac{1}{400} - \frac{1}{200} \]
- Finding a common denominator:
\[ \Delta t = \frac{1}{400} - \frac{2}{400} = -\frac{1}{400} \, s \]
- Since we are interested in the absolute time taken, we take the positive value:
\[ \Delta t = \frac{1}{400} \, s = 0.0025 \, s = 2.5 \times 10^{-3} \, s \]

### Final Answer:
The time taken by the current to change from its peak value to its r.m.s. value is \( 2.5 \times 10^{-3} \, s \).