In a certain circuit current changes with time according to `i=2sqrt(t)` RMS value of current between t=2s to t=4s will be

Options

A. 3A
B. `3sqrt(3)A`
C. `2sqrt(3)A`
D. `(2-sqrt(2))A`

Answer

To find the RMS (Root Mean Square) value of the current \( i = 2\sqrt{t} \) between \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \), we will follow these steps:

### Step 1: Understand the RMS formula
The RMS value of a current over a time interval is given by the formula:
\[ I_{RMS} = \sqrt{\frac{1}{T} \int_{t_1}^{t_2} i^2 \, dt} \]
where \( T = t_2 - t_1 \) is the duration of the time interval.

### Step 2: Define the time interval
In this case, \( t_1 = 2 \, \text{s} \) and \( t_2 = 4 \, \text{s} \). Therefore, the duration \( T \) is:
\[ T = 4 - 2 = 2 \, \text{s} \]

### Step 3: Substitute the current function
The current is given as \( i = 2\sqrt{t} \). We need to find \( i^2 \):
\[ i^2 = (2\sqrt{t})^2 = 4t \]

### Step 4: Set up the integral
Now we substitute \( i^2 \) into the RMS formula:
\[ I_{RMS} = \sqrt{\frac{1}{2} \int_{2}^{4} 4t \, dt} \]

### Step 5: Simplify the integral
Factor out the constant:
\[ I_{RMS} = \sqrt{\frac{4}{2} \int_{2}^{4} t \, dt} = \sqrt{2 \int_{2}^{4} t \, dt} \]

### Step 6: Calculate the integral
The integral \( \int t \, dt \) is:
\[ \int t \, dt = \frac{t^2}{2} \]
Now evaluate it from 2 to 4:
\[ \int_{2}^{4} t \, dt = \left[ \frac{t^2}{2} \right]_{2}^{4} = \frac{4^2}{2} - \frac{2^2}{2} = \frac{16}{2} - \frac{4}{2} = 8 - 2 = 6 \]

### Step 7: Substitute back into the RMS formula
Now substitute the value of the integral back:
\[ I_{RMS} = \sqrt{2 \cdot 6} = \sqrt{12} \]

### Step 8: Simplify the result
Finally, simplify \( \sqrt{12} \):
\[ I_{RMS} = \sqrt{4 \cdot 3} = 2\sqrt{3} \]

### Final Answer
The RMS value of the current between \( t = 2 \, \text{s} \) and \( t = 4 \, \text{s} \) is:
\[ \boxed{2\sqrt{3} \, \text{A}} \]