A resistance of `20 Omega` is connected to a source of an alternating potential `V=220 sin (100 pi t)`. The time taken by the current to change from the peak value to rms value is

Options

A. `0.2 s`
B. `0.25s`
C. `25 s`
D. `2.5 xx 10^(-3)s`

Explanation

To solve the problem, we need to determine the time taken by the current to change from its peak value to its root mean square (RMS) value. Here are the steps to find the solution:

### Step 1: Identify the peak current
The given voltage is \( V = 220 \sin(100 \pi t) \). The peak voltage \( V_0 \) is 220 V. Using Ohm's law, the peak current \( I_0 \) can be calculated as:
\[ I_0 = \frac{V_0}{R} = \frac{220}{20} = 11 \, \text{A} \]

### Step 2: Determine the RMS current
The RMS current \( I_{rms} \) is given by:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.78 \, \text{A} \]

### Step 3: Write the equation for current
The current \( I(t) \) can be expressed as:
\[ I(t) = I_0 \sin(100 \pi t) = 11 \sin(100 \pi t) \]

### Step 4: Find the time to reach peak current
To find the time \( t_1 \) when the current reaches its peak value \( I_0 \):
\[ I_0 = 11 \sin(100 \pi t_1) \]
Setting \( I(t_1) = 11 \):
\[ 11 = 11 \sin(100 \pi t_1) \implies \sin(100 \pi t_1) = 1 \]
This occurs when:
\[ 100 \pi t_1 = \frac{\pi}{2} \implies t_1 = \frac{1}{200} \, \text{s} \]

### Step 5: Find the time to reach RMS current
Next, we find the time \( t_2 \) when the current reaches its RMS value:
\[ I_{rms} = 11 \sin(100 \pi t_2) \]
Setting \( I(t_2) = \frac{11}{\sqrt{2}} \):
\[ \frac{11}{\sqrt{2}} = 11 \sin(100 \pi t_2) \implies \sin(100 \pi t_2) = \frac{1}{\sqrt{2}} \]
This occurs when:
\[ 100 \pi t_2 = \frac{\pi}{4} \implies t_2 = \frac{1}{400} \, \text{s} \]

### Step 6: Calculate the time difference
The time taken to change from peak value to RMS value is given by:
\[ \Delta t = t_2 - t_1 = \frac{1}{400} - \frac{1}{200} \]
Finding a common denominator:
\[ \Delta t = \frac{1}{400} - \frac{2}{400} = -\frac{1}{400} = -0.0025 \, \text{s} \]
Taking the absolute value:
\[ \Delta t = 0.0025 \, \text{s} = 2.5 \times 10^{-3} \, \text{s} \]

### Final Answer
The time taken by the current to change from the peak value to the RMS value is:
\[ \Delta t = 2.5 \times 10^{-3} \, \text{s} \]