Question

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

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

Hint:

AC voltmeters and ammeters always measure and display the \textbf{RMS} value unless otherwise specified.

Answer 1

The Correct Option is C

The problem requires finding the root mean square (RMS) and average value of an AC voltage described by \(V = V_0 \sin \omega t\) over a full cycle.

Let's calculate step by step:

1. RMS Value:

The RMS value for a sinusoidal voltage is calculated using the formula:

\(V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T (V_0 \sin \omega t)^2 \, dt}\)

Where \(T\) is the period of the sine function.

For one complete cycle, integrate from 0 to \(T\):

\(V_{\text{rms}} = \sqrt{\frac{1}{T} \int_0^T V_0^2 \sin^2 \omega t \, dt}\)

The formula \(\int_0^T \sin^2 \omega t \, dt = \frac{T}{2}\) can be used here.

Thus, substituting \(\frac{T}{2}\) gives us:

\(V_{\text{rms}} = \sqrt{\frac{V_0^2}{T} \cdot \frac{T}{2}} = \frac{V_0}{\sqrt{2}}\)

2. Average Value:

The average value of a full cycle of a sinusoidal waveform:

\(V_{\text{avg}} = \frac{1}{T} \int_0^T V_0 \sin \omega t \, dt\)

This integral over one complete cycle of a sine function is zero:

\(V_{\text{avg}} = 0\)

Conclusion:

From the computed values, the RMS and average values of the AC voltage \(V = V_0 \sin \omega t\) over a complete cycle are:

RMS Value: \(\frac{V_0}{\sqrt{2}}\)

Average Value: \(0\)

The correct answer is: \(\left( \frac{V_0}{\sqrt{2}}, 0 \right)\)