The time required for a 50Hz alternating current to increase from zero to `70.7%` of its peak value is-

Options

A. 2.5 ms
B. 10 ms
C. 20 ms
D. 14.14 ms

Explanation

To find the time required for a 50 Hz alternating current to increase from zero to 70.7% of its peak value, we can follow these steps:

### Step 1: Understand the relationship between peak value and the sine function
The current \( i(t) \) in an alternating current circuit can be expressed as:
\[ i(t) = I_0 \sin(\omega t) \]
where \( I_0 \) is the peak value of the current and \( \omega \) is the angular frequency.

### Step 2: Determine the value of \( 70.7\% \) of the peak value
We need to find the time when the current reaches \( 70.7\% \) of its peak value:
\[ i(t) = 0.707 I_0 \]

### Step 3: Set up the equation
Substituting into the equation:
\[ 0.707 I_0 = I_0 \sin(\omega t) \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ 0.707 = \sin(\omega t) \]

### Step 4: Find the angle corresponding to \( 0.707 \)
From trigonometric values, we know:
\[ \sin(\frac{\pi}{4}) = 0.707 \]
Thus, we can equate:
\[ \omega t = \frac{\pi}{4} \]

### Step 5: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is related to the frequency \( f \) by the formula:
\[ \omega = 2\pi f \]
Given that the frequency \( f = 50 \) Hz, we can calculate \( \omega \):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 6: Substitute \( \omega \) back into the equation
Now we substitute \( \omega \) back into our equation:
\[ 100\pi t = \frac{\pi}{4} \]

### Step 7: Solve for \( t \)
Dividing both sides by \( 100\pi \):
\[ t = \frac{\frac{\pi}{4}}{100\pi} = \frac{1}{400} \, \text{s} \]

### Step 8: Convert to milliseconds
To convert seconds to milliseconds:
\[ t = \frac{1}{400} \, \text{s} = 0.0025 \, \text{s} = 2.5 \, \text{ms} \]

### Final Answer
The time required for the alternating current to increase from zero to \( 70.7\% \) of its peak value is:
\[ \boxed{2.5 \, \text{ms}} \]

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