An alternating current is given by the equation `i=i_1sinomegat+i_2cosomegat`. The rms current will be

Options

A. `1/sqrt2(i_1^2+i_2^2)^(1/2)`
B. `1/sqrt2(i_1+i_2)^2`
C. `1/2(i_1^2+i_2^2)^(1/2)`
D. `1/sqrt2(i_1+i_2)`

Answer

To find the RMS current given by the equation \( i = i_1 \sin(\omega t) + i_2 \cos(\omega t) \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding RMS Current**:
The RMS (Root Mean Square) current is defined as the square root of the average of the squares of the instantaneous values of current over one complete cycle.

2. **Square the Current Expression**:
Start by squaring the given current expression:
\[ i^2 = (i_1 \sin(\omega t) + i_2 \cos(\omega t))^2 \]
Expanding this gives:
\[ i^2 = i_1^2 \sin^2(\omega t) + i_2^2 \cos^2(\omega t) + 2 i_1 i_2 \sin(\omega t) \cos(\omega t) \]

3. **Taking the Average**:
To find the RMS, we need to take the average of \( i^2 \) over one complete cycle. This is represented as:
\[ \langle i^2 \rangle = \frac{1}{T} \int_0^T i^2 \, dt \]
where \( T \) is the period of the wave.

4. **Calculate the Averages**:
- The average of \( \sin^2(\omega t) \) over one cycle is \( \frac{1}{2} \).
- The average of \( \cos^2(\omega t) \) over one cycle is also \( \frac{1}{2} \).
- The average of \( \sin(\omega t) \cos(\omega t) \) over one cycle is \( 0 \).

Therefore, we have:
\[ \langle i^2 \rangle = i_1^2 \cdot \frac{1}{2} + i_2^2 \cdot \frac{1}{2} + 2 i_1 i_2 \cdot 0 \]
Simplifying this gives:
\[ \langle i^2 \rangle = \frac{i_1^2}{2} + \frac{i_2^2}{2} \]

5. **Final Calculation of RMS Current**:
Now, we take the square root of the average:
\[ I_{\text{rms}} = \sqrt{\langle i^2 \rangle} = \sqrt{\frac{i_1^2 + i_2^2}{2}} \]

6. **Final Result**:
Thus, the RMS current can be expressed as:
\[ I_{\text{rms}} = \frac{1}{\sqrt{2}} \sqrt{i_1^2 + i_2^2} \]

### Conclusion:
The final expression for the RMS current is:
\[ I_{\text{rms}} = \frac{1}{\sqrt{2}} \sqrt{i_1^2 + i_2^2} \]