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.

<h3>RMS Value of AC Voltage</h3>

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: \[ V_{\text{rms}} = V_0 \sqrt{\frac{1}{2}} = \frac{V_0}{\sqrt{2}} \]

<h3>Average Value of AC Voltage</h3>

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: \(\frac{V_0}{\sqrt{2}}, 0\).

Answer 2

Since the two blanks are a pair (rms value, average value), each option can be checked against two facts about a sine wave: its rms value is always less than its peak, and its average over a full symmetric cycle is always zero.

1. \( \dfrac{V_0}{2}, \dfrac{V_0}{\sqrt{2}} \): The second entry here, \( \dfrac{V_0}{\sqrt{2}} \approx 0.707V_0 \), is being proposed as the average, but a sine wave's positive and negative halves are mirror images and must cancel over a full cycle, so the true average has to be zero, not a large non-zero fraction of the peak. This option is ruled out.
2. \( \dfrac{V_0}{\pi}, \dfrac{V_0}{2} \): Neither of these matches a sine wave over a full cycle: \( \dfrac{V_0}{\pi} \) is actually the average of a sine wave over just a half cycle, not the rms, and again the average over the full cycle should be zero, not \( \dfrac{V_0}{2} \). This option is ruled out.
3. \( \dfrac{V_0}{\sqrt{2}}, 0 \): The rms value of a sine wave is a well-known standard result, \( \dfrac{V_0}{\sqrt{2}} \approx 0.707V_0 \), sitting sensibly between zero and the peak \( V_0 \), and the average being exactly 0 matches the expected cancellation of the positive and negative halves over a full cycle. Both entries check out.
4. \( V_0, \dfrac{V_0}{\sqrt{2}} \): An rms value equal to the full peak \( V_0 \) would only happen for a waveform that stays constantly at its maximum, like a square wave, not a sine wave, which spends most of its time below the peak. This option is ruled out.

Only the third pairing is consistent with both the known rms formula for a sine wave and the zero-average property of a full symmetric cycle.

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