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

Answer

To find the root mean square (r.m.s.) value of the current \( i = 2\sqrt{t} \) between \( t = 2 \) seconds and \( t = 4 \) seconds, we can follow these steps:

### Step 1: Understand the formula for r.m.s. value
The r.m.s. value of a current over a time interval can be calculated using the formula:
\[ I_{rms} = \sqrt{\frac{1}{T} \int_{t_1}^{t_2} i^2 \, dt} \]
where \( T = t_2 - t_1 \).

### Step 2: Set up the integral for \( i^2 \)
Given \( i = 2\sqrt{t} \), we first need to calculate \( i^2 \):
\[ i^2 = (2\sqrt{t})^2 = 4t \]

### Step 3: Set the limits of integration
We are given the limits \( t_1 = 2 \) seconds and \( t_2 = 4 \) seconds.

### Step 4: Calculate the integral of \( i^2 \)
Now we need to evaluate the integral:
\[ \int_{2}^{4} 4t \, dt \]
This can be simplified as:
\[ = 4 \int_{2}^{4} t \, dt \]

### Step 5: Evaluate the integral
The integral \( \int t \, dt \) is:
\[ \int t \, dt = \frac{t^2}{2} \]
Thus, we have:
\[ 4 \left[ \frac{t^2}{2} \right]_{2}^{4} = 4 \left( \frac{4^2}{2} - \frac{2^2}{2} \right) \]
Calculating this gives:
\[ = 4 \left( \frac{16}{2} - \frac{4}{2} \right) = 4 \left( 8 - 2 \right) = 4 \times 6 = 24 \]

### Step 6: Calculate the total time interval \( T \)
The total time interval \( T \) is:
\[ T = t_2 - t_1 = 4 - 2 = 2 \]

### Step 7: Calculate the r.m.s. value
Now we can calculate the r.m.s. value:
\[ I_{rms} = \sqrt{\frac{1}{T} \int_{t_1}^{t_2} i^2 \, dt} = \sqrt{\frac{1}{2} \cdot 24} = \sqrt{12} = 2\sqrt{3} \]

### Final Answer
Thus, the r.m.s. value of the current between \( t = 2 \) seconds and \( t = 4 \) seconds is:
\[ I_{rms} = 2\sqrt{3} \text{ A} \]