An alternating voltage is given by: `e = e_(1) sin omega t + e_(2) cos omega t`. Then the root mean square value of voltage is given by:

Options

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

Answer

To find the root mean square (RMS) value of the given alternating voltage \( e = e_1 \sin(\omega t) + e_2 \cos(\omega t) \), we can follow these steps:

### Step 1: Rewrite the expression
We start with the expression for the alternating voltage:
\[ e = e_1 \sin(\omega t) + e_2 \cos(\omega t) \]

### Step 2: Convert cosine to sine
To combine the terms, we can express \( \cos(\omega t) \) in terms of sine:
\[ \cos(\omega t) = \sin\left(\omega t + \frac{\pi}{2}\right) \]
Thus, we can rewrite the voltage as:
\[ e = e_1 \sin(\omega t) + e_2 \sin\left(\omega t + \frac{\pi}{2}\right) \]

### Step 3: Use the formula for RMS value
The RMS value of a function \( e(t) \) over one complete cycle is given by:
\[ E_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T e^2(t) \, dt} \]
where \( T \) is the period of the function.

### Step 4: Calculate \( e^2(t) \)
Now, we calculate \( e^2(t) \):
\[ e^2(t) = \left(e_1 \sin(\omega t) + e_2 \cos(\omega t)\right)^2 \]
Expanding this:
\[ e^2(t) = e_1^2 \sin^2(\omega t) + e_2^2 \cos^2(\omega t) + 2 e_1 e_2 \sin(\omega t) \cos(\omega t) \]

### Step 5: Integrate over one period
Next, we integrate \( e^2(t) \) over one period \( T \):
\[ \int_0^T e^2(t) \, dt = \int_0^T e_1^2 \sin^2(\omega t) \, dt + \int_0^T e_2^2 \cos^2(\omega t) \, dt + 2 e_1 e_2 \int_0^T \sin(\omega t) \cos(\omega t) \, dt \]

Using the identities:
- \( \int_0^T \sin^2(\omega t) \, dt = \frac{T}{2} \)
- \( \int_0^T \cos^2(\omega t) \, dt = \frac{T}{2} \)
- \( \int_0^T \sin(\omega t) \cos(\omega t) \, dt = 0 \)

We find:
\[ \int_0^T e^2(t) \, dt = e_1^2 \frac{T}{2} + e_2^2 \frac{T}{2} + 0 = \frac{T}{2} (e_1^2 + e_2^2) \]

### Step 6: Divide by the period and take the square root
Now, substituting back into the RMS formula:
\[ E_{\text{rms}} = \sqrt{\frac{1}{T} \cdot \frac{T}{2} (e_1^2 + e_2^2)} = \sqrt{\frac{1}{2} (e_1^2 + e_2^2)} \]

### Final Result
Thus, the root mean square value of the voltage is:
\[ E_{\text{rms}} = \sqrt{\frac{e_1^2 + e_2^2}{2}} \]