The root mean square value of the alternating current is equal to

Options

A. Twice the peak value
B. Half the peak value
C. `1/sqrt(2)` times the peak value
D. Equal to the peak value

Answer

To find the root mean square (RMS) value of an alternating current (AC), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the AC Current Equation**:
The alternating current can be expressed as:
\[ I(t) = I_0 \cos(\omega t + \phi) \]
where \(I_0\) is the peak current, \(\omega\) is the angular frequency, and \(\phi\) is the phase difference.

2. **Define RMS Current**:
The root mean square (RMS) value of the current is defined as:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T I^2(t) \, dt} \]
where \(T\) is the time period of the AC signal.

3. **Substitute the Current Equation**:
Substitute \(I(t)\) into the RMS formula:
\[ I_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T (I_0 \cos(\omega t + \phi))^2 \, dt} \]

4. **Simplify the Integral**:
This simplifies to:
\[ I_{\text{rms}} = \sqrt{\frac{I_0^2}{T} \int_0^T \cos^2(\omega t + \phi) \, dt} \]

5. **Evaluate the Integral**:
The integral of \(\cos^2\) over one complete cycle (time period \(T\)) is:
\[ \int_0^T \cos^2(\omega t + \phi) \, dt = \frac{T}{2} \]
Therefore, substituting this back into the equation gives:
\[ I_{\text{rms}} = \sqrt{\frac{I_0^2}{T} \cdot \frac{T}{2}} = \sqrt{\frac{I_0^2}{2}} = \frac{I_0}{\sqrt{2}} \]

6. **Final Result**:
Thus, the root mean square value of the alternating current is:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \]