If the rms current in a 50 Hz ac circuit is 5 A, the value of the current `1//300` second after its value becomes zero is

Options

A. `5sqrt(2)A`
B. `5sqrt((3)/(2))A`
C. `(5)/(6)A`
D. `(5)/(sqrt(2))A`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the given values
We are given:
- RMS current \( I_{rms} = 5 \, A \)
- Frequency \( f = 50 \, Hz \)

### Step 2: Calculate the peak current \( I_0 \)
The relationship between the RMS current and the peak current is given by:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
From this, we can express \( I_0 \) as:
\[ I_0 = I_{rms} \times \sqrt{2} = 5 \times \sqrt{2} \, A \]

### Step 3: Calculate the angular frequency \( \omega \)
The angular frequency \( \omega \) is calculated using the formula:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 50 = 100\pi \, rad/s \]

### Step 4: Calculate the time \( t \)
We need to find the current at \( t = \frac{1}{300} \, s \).

### Step 5: Substitute values into the current equation
The instantaneous current \( I(t) \) in an AC circuit is given by:
\[ I(t) = I_0 \sin(\omega t) \]
Substituting the values we found:
\[ I\left(\frac{1}{300}\right) = 5\sqrt{2} \sin\left(100\pi \times \frac{1}{300}\right) \]

### Step 6: Simplify the argument of the sine function
Calculating the argument:
\[ 100\pi \times \frac{1}{300} = \frac{100\pi}{300} = \frac{\pi}{3} \]
Thus, we have:
\[ I\left(\frac{1}{300}\right) = 5\sqrt{2} \sin\left(\frac{\pi}{3}\right) \]

### Step 7: Calculate \( \sin\left(\frac{\pi}{3}\right) \)
The value of \( \sin\left(\frac{\pi}{3}\right) \) is:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

### Step 8: Substitute back to find \( I(t) \)
Now substituting this value back:
\[ I\left(\frac{1}{300}\right) = 5\sqrt{2} \times \frac{\sqrt{3}}{2} \]
This simplifies to:
\[ I\left(\frac{1}{300}\right) = \frac{5\sqrt{6}}{2} \, A \]

### Step 9: Final answer
Thus, the value of the current \( 1/300 \) second after it becomes zero is:
\[ I\left(\frac{1}{300}\right) \approx 6.12 \, A \]