Relation between r.m.s. voltage and instantaneous voltage of an AC

Options

A. `V_(0) = V_("RMS")//sqrt""2`
B. `V_("RMS") = V_(0)//sqrt""2`
C. `V_("RMS") = 0.70V_(0)`
D. Both (C) and (D)

Explanation

To find the relation between the root mean square (RMS) voltage and the instantaneous voltage of an alternating current (AC), we can follow these steps:

### Step 1: Understand the Definitions
- **Instantaneous Voltage (V)**: This is the voltage at any given moment in time during the AC cycle. It can be represented as:
\[ V(t) = V_0 \sin(\omega t + \phi) \]
where \( V_0 \) is the peak (maximum) voltage, \( \omega \) is the angular frequency, \( t \) is time, and \( \phi \) is the phase angle.

- **RMS Voltage (V_rms)**: This is a measure of the effective value of the AC voltage. It is defined as the square root of the average of the squares of the instantaneous voltages over one complete cycle.

### Step 2: Calculate RMS Voltage
The RMS voltage can be calculated using the formula:
\[ V_{rms} = \sqrt{\frac{1}{T} \int_0^T [V(t)]^2 dt} \]
where \( T \) is the time period of one complete cycle.

### Step 3: Substitute the Instantaneous Voltage
Substituting the expression for \( V(t) \):
\[ V_{rms} = \sqrt{\frac{1}{T} \int_0^T [V_0 \sin(\omega t + \phi)]^2 dt} \]

### Step 4: Simplify the Integral
Using the identity \( \sin^2(x) = \frac{1 - \cos(2x)}{2} \):
\[ V_{rms} = \sqrt{\frac{V_0^2}{T} \int_0^T \frac{1 - \cos(2(\omega t + \phi))}{2} dt} \]

### Step 5: Evaluate the Integral
The integral of \( \cos(2(\omega t + \phi)) \) over one complete cycle is zero, so:
\[ V_{rms} = \sqrt{\frac{V_0^2}{T} \cdot \frac{T}{2}} = \sqrt{\frac{V_0^2}{2}} = \frac{V_0}{\sqrt{2}} \]

### Step 6: Final Relation
Thus, we arrive at the final relation:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} \approx 0.707 V_0 \]

### Conclusion
The relation between RMS voltage and peak voltage is:
- \( V_{rms} = \frac{V_0}{\sqrt{2}} \) or approximately \( V_{rms} \approx 0.707 V_0 \).