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. `5 sqrt2 A`
B. `5 sqrt(3//2 ) A`
C. `5//6 A`
D. `5 // sqrt2 A`

Explanation

To find the value of the current in an AC circuit at a specific time after it becomes zero, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- RMS Current (I_RMS) = 5 A
- Frequency (f) = 50 Hz
- Time (T) = 1/300 seconds

2. **Calculate the Angular Frequency (ω)**:
\[ \omega = 2\pi f \]
Substituting the given frequency:
\[ \omega = 2\pi \times 50 = 100\pi \text{ radians/second} \]

3. **Calculate the Peak Current (I₀)**:
The peak current (I₀) can be calculated using the relationship between RMS current and peak current:
\[ I₀ = \sqrt{2} \times I_{RMS} \]
Substituting the given RMS current:
\[ I₀ = \sqrt{2} \times 5 = 5\sqrt{2} \text{ A} \]

4. **Write the Current Equation**:
The instantaneous current (I) in an AC circuit can be expressed as:
\[ I(t) = I₀ \sin(\omega t) \]
Substituting the values of I₀ and ω:
\[ I(t) = 5\sqrt{2} \sin(100\pi t) \]

5. **Substitute the Time (T = 1/300 seconds)**:
Now, we substitute T into the equation:
\[ I\left(\frac{1}{300}\right) = 5\sqrt{2} \sin\left(100\pi \times \frac{1}{300}\right) \]
Simplifying the argument of the sine function:
\[ 100\pi \times \frac{1}{300} = \frac{100\pi}{300} = \frac{\pi}{3} \]

6. **Calculate the Sine Value**:
Now, we find the sine of \(\frac{\pi}{3}\):
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]

7. **Substitute Back to Find I**:
Substitute the sine value back into the current equation:
\[ I\left(\frac{1}{300}\right) = 5\sqrt{2} \times \frac{\sqrt{3}}{2} \]
Simplifying this gives:
\[ I\left(\frac{1}{300}\right) = \frac{5\sqrt{6}}{2} \text{ A} \]

### Final Answer:
The value of the current \(1/300\) seconds after it becomes zero is:
\[ I\left(\frac{1}{300}\right) = \frac{5\sqrt{6}}{2} \text{ A} \]