Question

The rms and the average value of an AC voltage \( V = V_0 \sin \omega t \) over a cycle respectively will be:

• A. \( \dfrac{V_0}{2}, \dfrac{V_0}{\sqrt{2}} \)
• B. \( \dfrac{V_0}{\pi}, \dfrac{V_0}{2} \)
• C. \( \dfrac{V_0}{\sqrt{2}}, 0 \)
• D. \( V_0, \dfrac{V_0}{\sqrt{2}} \)

Hint:

For sine wave AC:

RMS = \( V_0/\sqrt{2} \)
Average over full cycle = 0
Average over half cycle = \( 2V_0/\pi \)

Answer 1

The Correct Option is C

To determine the root mean square (RMS) and average value of the AC voltage \( V = V_0 \sin \omega t \) over a cycle, we need to understand the mathematical definitions and calculations for these values.

RMS Value of AC Voltage

1. The RMS value of an AC waveform is defined as the square root of the mean of the squares of all instantaneous values during one complete cycle.
2. For a sinusoidal waveform \( V = V_0 \sin \omega t \), the RMS value is given by the formula: \[ V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T (V_0 \sin \omega t)^2 \, dt} \]
3. Solving the integral, \[ V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T V_0^2 \sin^2 \omega t \, dt} = V_0 \sqrt{\frac{1}{T} \int_0^T \frac{1 - \cos 2\omega t}{2} \, dt} \]
4. This evaluates to: \[ V_{\text{rms}} = V_0 \sqrt{\frac{1}{2}} = \frac{V_0}{\sqrt{2}} \]

Average Value of AC Voltage

1. The average value of a full cycle of a symmetrical AC signal like a sine wave over one complete cycle is zero because the positive and negative halves of the sine wave cancel each other out.
2. If we calculate, we find: \[ V_{\text{avg}} = \frac{1}{T} \int_0^T V_0 \sin \omega t \, dt = 0 \]

Thus, the RMS and average values of the AC voltage, respectively, are:

• \(\frac{V_0}{\sqrt{2}}\) for RMS value.
• \(0\) for average value.

Therefore, the correct option is: \(\frac{V_0}{\sqrt{2}}, 0\).