Relation between rms voltage and peak voltage of an AC source is:

The correct answer is Option (3) → \( V_{\mathrm{rms}} = \displaystyle\frac{V_0}{\sqrt{2}} \)

The rms (root mean square) value of an alternating voltage is the square root of the mean of \( V^{2} \) over a cycle. For \( V = V_0\sin\omega t \), the mean of \( \sin^{2}\omega t \) over a cycle is \( 1/2 \):

    \( \displaystyle V_{\mathrm{rms}} = \sqrt{\overline{V_0^{2}\sin^{2}\omega t}} = \sqrt{\frac{V_0^{2}}{2}} = \frac{V_0}{\sqrt{2}} \approx 0.707\,V_0 \)

Why the other options are incorrect

• Option (1) \( V_{\mathrm{rms}} = \displaystyle\frac{2}{\pi}V_0 \) — Incorrect. \( (2/\pi)V_0 \approx 0.637\,V_0 \) is the average value over a half cycle, not the rms value.

• Option (2) \( V_{\mathrm{rms}} = V_0\sqrt{2} \) — Incorrect. The rms value cannot exceed the peak value; this multiplies by \( \sqrt{2} \) instead of dividing.

• Option (4) \( V_{\mathrm{rms}} = \displaystyle\frac{V_0}{2} \) — Incorrect. \( V_0/2 \) would need the mean of \( \sin^{2}\omega t \) to be \( 1/4 \); it is \( 1/2 \), so the root gives \( 1/\sqrt{2} \).

NCERT (2026-27):
AC Voltage Applied to a Resistor – “… Similarly, we define the rms voltage or effective voltage by

    \( \displaystyle V = \frac{v_m}{\sqrt{2}} = 0.707\,v_m \qquad (7.8) \)”