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

Explanation

To solve the problem of finding the time taken by the current to change from its peak value to its root mean square (r.m.s.) value, we can follow these steps:

### Step 1: Identify the given values
- Resistance (R) = 20 ohms
- Voltage (V) = 220 sin(100πt)

### Step 2: Calculate the peak voltage (V_peak)
The peak voltage (V_peak) is the coefficient of the sine function in the voltage equation:
\[ V_{\text{peak}} = 220 \, \text{V} \]

### Step 3: Calculate the peak current (I_peak)
Using Ohm's law, the peak current can be calculated as:
\[ I_{\text{peak}} = \frac{V_{\text{peak}}}{R} = \frac{220}{20} = 11 \, \text{A} \]

### Step 4: Calculate the r.m.s. current (I_rms)
The r.m.s. current can be calculated using the relationship:
\[ I_{\text{rms}} = \frac{I_{\text{peak}}}{\sqrt{2}} \]
Substituting the peak current:
\[ I_{\text{rms}} = \frac{11}{\sqrt{2}} \approx 7.78 \, \text{A} \]

### Step 5: Determine the time taken to go from peak to r.m.s. current
The current in an AC circuit can be expressed as:
\[ I(t) = I_{\text{peak}} \sin(100\pi t) \]
To find the time when the current reaches the r.m.s. value, we set:
\[ I(t) = I_{\text{rms}} \]
Substituting the values:
\[ 11 \sin(100\pi t) = \frac{11}{\sqrt{2}} \]

### Step 6: Simplify the equation
Dividing both sides by 11:
\[ \sin(100\pi t) = \frac{1}{\sqrt{2}} \]

### Step 7: Solve for time (t)
The sine function equals \( \frac{1}{\sqrt{2}} \) at:
\[ 100\pi t = \frac{\pi}{4} \quad \text{and} \quad 100\pi t = \frac{3\pi}{4} \]
From the first equation:
\[ t = \frac{\pi/4}{100\pi} = \frac{1}{400} \, \text{s} \]

### Step 8: Find the time taken to go from peak to r.m.s.
The time taken to go from peak to r.m.s. is the difference between the two times:
- Time for \( \frac{\pi}{4} \): \( t_1 = \frac{1}{400} \, \text{s} \)
- Time for \( \frac{3\pi}{4} \): \( t_2 = \frac{3}{400} \, \text{s} \)

Thus, the time taken to go from peak to r.m.s. is:
\[ \Delta t = t_2 - t_1 = \frac{3}{400} - \frac{1}{400} = \frac{2}{400} = \frac{1}{200} \, \text{s} \]

### Final Answer
The time taken by the current to change from its peak value to its r.m.s. value is \( \frac{1}{200} \, \text{s} \).

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