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))`

Explanation

To find the root mean square (RMS) value of the alternating voltage given by the equation:

\[ e = e_1 \sin(\omega t) + e_2 \cos(\omega t) \]

we can follow these steps:

### Step 1: Identify the Form of the Voltage Equation
The given voltage can be expressed in a form that resembles the general equation of a sinusoidal function. The equation can be rewritten as:

\[ e = e_1 \sin(\omega t) + e_2 \cos(\omega t) \]

### Step 2: Use the Trigonometric Identity
We can convert the expression into a single sinusoidal function using the trigonometric identity:

\[ R = \sqrt{a^2 + b^2} \]

where \( a = e_1 \) and \( b = e_2 \). Thus, we can express the voltage as:

\[ e = R \sin(\omega t + \phi) \]

where \( R = \sqrt{e_1^2 + e_2^2} \) and \( \phi = \tan^{-1}\left(\frac{e_2}{e_1}\right) \).

### Step 3: Determine the Peak Value
From the expression \( e = R \sin(\omega t + \phi) \), we identify that the peak value \( e_0 \) of the voltage is:

\[ e_0 = R = \sqrt{e_1^2 + e_2^2} \]

### Step 4: Calculate the RMS Value
The RMS value of a sinusoidal voltage is given by the formula:

\[ e_{\text{rms}} = \frac{e_0}{\sqrt{2}} \]

Substituting the peak value we found:

\[ e_{\text{rms}} = \frac{\sqrt{e_1^2 + e_2^2}}{\sqrt{2}} \]

### Final Expression
Thus, the root mean square value of the voltage is:

\[ e_{\text{rms}} = \frac{1}{\sqrt{2}} \sqrt{e_1^2 + e_2^2} \]

### Summary
The final answer to the problem is:

\[ e_{\text{rms}} = \frac{\sqrt{e_1^2 + e_2^2}}{\sqrt{2}} \]

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