An altemating current is given by `1 = 1_(1) cos omega t + 1_(2) sin omega t.` The RMS value of current is given by

Options

A. `(l_(1)+l_(2))/(sqrt(2))`
B. `((l_(1) + l_(2))^(2))/(2)`
C. `sqrt((l_(1)^(2)+l_(2)^(2))/(2)`
D. `sqrt(l_(1)^(2)+l_(2)^(2))/(2)`

Answer

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

### Step-by-Step Solution:

1. **Identify the given current equation**:
The current is given as:
\[ i = i_1 \cos(\omega t) + i_2 \sin(\omega t) \]

2. **Convert to a standard form**:
We can express the current in a form that allows us to identify the peak current. We can use the trigonometric identity:
\[ A \cos(\omega t) + B \sin(\omega t) = R \sin(\omega t + \phi) \]
where \( R = \sqrt{A^2 + B^2} \) and \( \tan(\phi) = \frac{B}{A} \).

Here, \( A = i_1 \) and \( B = i_2 \). Thus, we have:
\[ R = \sqrt{i_1^2 + i_2^2} \]

3. **Determine the peak current**:
From the above conversion, we can express the current as:
\[ i = R \sin(\omega t + \phi) \]
where \( R = \sqrt{i_1^2 + i_2^2} \).

4. **Calculate the RMS value**:
The RMS value of a sinusoidal current is given by:
\[ I_{\text{RMS}} = \frac{I_0}{\sqrt{2}} \]
where \( I_0 \) is the peak current. Therefore, substituting \( I_0 = R \):
\[ I_{\text{RMS}} = \frac{R}{\sqrt{2}} = \frac{\sqrt{i_1^2 + i_2^2}}{\sqrt{2}} \]

5. **Final expression for RMS current**:
Thus, the RMS value of the current can be expressed as:
\[ I_{\text{RMS}} = \frac{\sqrt{i_1^2 + i_2^2}}{\sqrt{2}} \]

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