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. `2.5 xx10^(-3)`sec

Answer

To solve the problem step by step, we need to find the time taken by the current to change from its peak value to its RMS value.

### Step 1: Identify the given values
- Resistance, \( R = 20 \, \Omega \)
- Voltage, \( V(t) = 220 \sin(100 \pi t) \)

### Step 2: Calculate the peak current
The peak current \( I_0 \) can be calculated using Ohm's law:
\[ I_0 = \frac{V_0}{R} \]
Where \( V_0 \) is the peak voltage. Here, \( V_0 = 220 \, V \).
\[ I_0 = \frac{220}{20} = 11 \, A \]

### Step 3: Find the RMS current
The RMS (Root Mean Square) value of the current \( I_{rms} \) is given by:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{11}{\sqrt{2}} \approx 7.78 \, A \]

### Step 4: Set up the equations for the current
The current as a function of time is given by:
\[ I(t) = 11 \sin(100 \pi t) \]

### Step 5: Find the time \( t_1 \) when the current reaches its peak value
The peak value occurs when:
\[ I(t_1) = I_0 = 11 \]
Thus,
\[ 11 = 11 \sin(100 \pi t_1 \implies \sin(100 \pi t_1) = 1 \]
The sine function equals 1 at:
\[ 100 \pi t_1 = \frac{\pi}{2} \implies t_1 = \frac{1}{200} \, s \]

### Step 6: Find the time \( t_2 \) when the current reaches its RMS value
Set the current equal to the RMS value:
\[ I(t_2) = I_{rms} = \frac{11}{\sqrt{2}} \]
Thus,
\[ \frac{11}{\sqrt{2}} = 11 \sin(100 \pi t_2 \implies \sin(100 \pi t_2) = \frac{1}{\sqrt{2}} \]
The sine function equals \( \frac{1}{\sqrt{2}} \) at:
\[ 100 \pi t_2 = \frac{\pi}{4} \implies t_2 = \frac{1}{400} \, s \]

### Step 7: Calculate the time difference \( \Delta t \)
The time taken to change from peak value to RMS value is:
\[ \Delta t = t_1 - t_2 = \frac{1}{200} - \frac{1}{400} \]
Finding a common denominator:
\[ \Delta t = \frac{2}{400} - \frac{1}{400} = \frac{1}{400} \, s \]

### Step 8: Convert to standard form
\[ \Delta t = 0.0025 \, s = 2.5 \times 10^{-3} \, s \]

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

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