A resistance of 20 ohm is connected to a source of an alternating potential V = 200 `cos (100 pi t)`. The time taken by the current to change from its peak value to rms value `alpha` is

Options

A. `2.5 xx 10^(-3) s`
B. `25 xx 10^(-3)s`
C. `0.25 s`
D. `0.20 s`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the given values
We have the following values given in the problem:
- Resistance (R) = 20 ohms
- Voltage (V) = 200 cos(100πt)

### Step 2: Calculate the peak current (I₀)
Using Ohm's Law, the current (I) can be calculated from the voltage (V) and resistance (R):
\[ I = \frac{V}{R} \]
Substituting the given values:
\[ I = \frac{200 \cos(100\pi t)}{20} \]
\[ I = 10 \cos(100\pi t) \]
Thus, the peak current (I₀) is:
\[ I₀ = 10 \, \text{A} \]

### Step 3: Calculate the RMS current (I_rms)
The RMS (Root Mean Square) value of the current is given by:
\[ I_{\text{rms}} = \frac{I₀}{\sqrt{2}} \]
Substituting the peak current:
\[ I_{\text{rms}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A} \]

### Step 4: Set up the equation to find the time (t)
We need to find the time taken for the current to change from its peak value to its RMS value. We set the RMS current equal to the instantaneous current:
\[ I_{\text{rms}} = 10 \cos(100\pi t) \]
Substituting the value of I_rms:
\[ 5\sqrt{2} = 10 \cos(100\pi t) \]
Dividing both sides by 10:
\[ \frac{\sqrt{2}}{2} = \cos(100\pi t) \]

### Step 5: Solve for t
The cosine of an angle equals \(\frac{\sqrt{2}}{2}\) at angles of \(\frac{\pi}{4}\) and \(\frac{7\pi}{4}\) (in radians). Therefore, we can write:
\[ 100\pi t = \frac{\pi}{4} \]
Solving for t:
\[ t = \frac{\frac{\pi}{4}}{100\pi} = \frac{1}{400} \, \text{s} \]

### Step 6: Convert to a more readable form
This can be simplified to:
\[ t = 2.5 \times 10^{-3} \, \text{s} \]

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

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