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 step by step, we will follow the outlined approach based on the information provided in the video transcript.

### Step 1: Understand the relationship between RMS current and peak current
The RMS (Root Mean Square) current \( I_{rms} \) is related to the peak current \( I_0 \) by the formula:
\[ I_{rms} = \frac{I_0}{\sqrt{2}} \]
From this, we can express the peak current as:
\[ I_0 = I_{rms} \times \sqrt{2} \]

### Step 2: Substitute the given RMS current
Given that \( I_{rms} = 5 \, \text{A} \), we can calculate \( I_0 \):
\[ I_0 = 5 \times \sqrt{2} \]

### Step 3: Write the equation for instantaneous current
The instantaneous current \( I(t) \) in an AC circuit can be expressed as:
\[ I(t) = I_0 \sin(\omega t) \]
where \( \omega = 2\pi f \) and \( f \) is the frequency.

### Step 4: Calculate the angular frequency
Given that the frequency \( f = 50 \, \text{Hz} \), we can calculate \( \omega \):
\[ \omega = 2\pi \times 50 = 100\pi \, \text{rad/s} \]

### Step 5: Calculate the time \( t \)
We need to find the current \( 1/300 \) seconds after it becomes zero. The time \( t \) is:
\[ t = \frac{1}{300} \, \text{s} \]

### Step 6: Substitute values into the instantaneous current equation
Now we substitute \( I_0 \) and \( \omega t \) into the equation for \( I(t) \):
\[ I\left(\frac{1}{300}\right) = 5\sqrt{2} \sin\left(100\pi \times \frac{1}{300}\right) \]

### Step 7: Simplify the argument of the sine function
Calculating the argument:
\[ 100\pi \times \frac{1}{300} = \frac{100\pi}{300} = \frac{\pi}{3} \]
Thus, the equation becomes:
\[ I\left(\frac{1}{300}\right) = 5\sqrt{2} \sin\left(\frac{\pi}{3}\right) \]

### Step 8: Calculate \( \sin\left(\frac{\pi}{3}\right) \)
We know that:
\[ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \]
Substituting this back into the equation gives:
\[ I\left(\frac{1}{300}\right) = 5\sqrt{2} \times \frac{\sqrt{3}}{2} \]

### Step 9: Simplify the expression
This simplifies to:
\[ I\left(\frac{1}{300}\right) = \frac{5\sqrt{6}}{2} \, \text{A} \]

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