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: \[ (V_{\text{rms}}, V_{\text{avg}}) = \left(\frac{V_0}{\sqrt{2}}, 0\right) \]