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

Concept: For a sinusoidal alternating voltage: \[ V = V_0 \sin \omega t \] Key results:

RMS value: \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]
Average value over a complete cycle: \[ V_{\text{avg}} = 0 \]

Step 1: RMS value. The RMS value is defined as: \[ V_{\text{rms}} = \sqrt{\frac{1}{T}\int_0^T V^2 dt} \] For sine wave: \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \]
Step 2: Average over a full cycle. Over one full cycle:

Positive half cancels negative half
Net average becomes zero
\[ V_{\text{avg}} = 0 \]
Step 3: Conclusion. Thus: \[ (V_{\text{rms}}, V_{\text{avg}}) = \left(\frac{V_0}{\sqrt{2}}, 0\right) \]

Answer 2

Both the rms value and the average value of \( V = V_0\sin\omega t \) can actually be reasoned out from the shape of the sine curve itself, without carrying out the integrals explicitly.

Average value, using symmetry: Over one full cycle, the sine wave spends exactly half its time above the axis (positive) and half below (negative), and the positive half is a perfect mirror image of the negative half, just flipped in sign. Every positive contribution to the average is exactly cancelled by an equal, opposite negative contribution somewhere else in the cycle. So the average over a full cycle works out to exactly zero.

RMS value, using the identity for sin-squared: Squaring the voltage removes the sign, so \( V^2 = V_0^2\sin^2\omega t \) is always positive and no longer averages to zero. Using the standard identity \( \sin^2\theta = \dfrac{1-\cos 2\theta}{2} \), the average of \( \sin^2\omega t \) over a full cycle is exactly \( \dfrac{1}{2} \), since the \( \cos 2\omega t \) part is itself a symmetric wave that averages to zero, leaving just the constant \( \dfrac{1}{2} \).

So the mean of \( V^2 \) over a cycle is \( V_0^2 \times \dfrac{1}{2} \), and taking the square root to undo the earlier squaring gives:

\[ V_{\text{rms}} = \sqrt{\frac{V_0^2}{2}} = \frac{V_0}{\sqrt{2}} \]

So, without solving a single integral directly, symmetry gives the average as zero, and the standard sin-squared identity gives the rms value as \( \dfrac{V_0}{\sqrt{2}} \).

Therefore, the correct answer is \( \dfrac{V_0}{\sqrt{2}}, 0 \).