An AC current is given by `I=I_(0)+I_(1)` sin wt then its rms value will be

Options

A. `sqrt(I_(0)^(2)+0.5I_(1)^(2))`
B. `sqrt(I_(0)^(2)+0.5I_(0)^(2))`
C. `0`
D. `I_(0)//sqrt(2)`

Explanation

To find the RMS (Root Mean Square) value of the given AC current \( I = I_0 + I_1 \sin(\omega t) \), we will follow these steps:

### Step 1: Understand the RMS Value Formula
The RMS value of a function over one complete cycle is given by the formula:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T I^2 \, dt} \]
where \( T \) is the period of the function.

### Step 2: Substitute the Current Expression
Substituting the expression for current into the formula, we have:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T (I_0 + I_1 \sin(\omega t))^2 \, dt} \]

### Step 3: Expand the Square
Expanding the square gives:
\[ (I_0 + I_1 \sin(\omega t))^2 = I_0^2 + 2I_0 I_1 \sin(\omega t) + I_1^2 \sin^2(\omega t) \]

### Step 4: Integrate Each Term
Now we will integrate each term separately over the interval from \( 0 \) to \( T \):
1. The integral of \( I_0^2 \) over one period \( T \):
\[ \int_0^T I_0^2 \, dt = I_0^2 T \]
2. The integral of \( 2I_0 I_1 \sin(\omega t) \) over one period \( T \):
\[ \int_0^T 2I_0 I_1 \sin(\omega t) \, dt = 0 \quad \text{(since the integral of sine over one complete cycle is zero)} \]
3. The integral of \( I_1^2 \sin^2(\omega t) \) over one period \( T \):
\[ \int_0^T I_1^2 \sin^2(\omega t) \, dt = I_1^2 \cdot \frac{T}{2} \quad \text{(the average value of } \sin^2 \text{ over one cycle is } \frac{1}{2}) \]

### Step 5: Combine the Results
Putting it all together, we have:
\[ \int_0^T (I_0 + I_1 \sin(\omega t))^2 \, dt = I_0^2 T + 0 + \frac{I_1^2 T}{2} \]
Thus,
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \left( I_0^2 T + \frac{I_1^2 T}{2} \right)} = \sqrt{I_0^2 + \frac{I_1^2}{2}} \]

### Step 6: Final Expression
Therefore, the RMS value of the given AC current is:
\[ I_{\text{rms}} = \sqrt{I_0^2 + \frac{I_1^2}{2}} \]