If an alternating signal is given by the equation v = 311 sin 440 t, volts, where t is in second, which of the following is correct.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → The rms value of the signal is ~220 V.
Compare the signal with the standard form \( v = v_0 \sin\omega t \): the peak value is \( v_0 = 311 \) V and the angular frequency is \( \omega = 440 \) rad s−1.
\( \displaystyle v_{\mathrm{rms}} = \frac{v_0}{\sqrt{2}} = \frac{311}{1.414} \approx 220\,\mathrm{V} \)
\( \displaystyle f = \frac{\omega}{2\pi} = \frac{440}{2\pi} \approx 70\,\mathrm{Hz} \)
Option (1) The peak value of the signal is 440 V. — Incorrect. The peak value is 311 V; 440 is the angular frequency in rad s−1.
Option (2) The average value of the signal is 311 V. — Incorrect. 311 V is the peak value; the average over a complete cycle is zero.
Option (3) The rms value of the signal is ~220 V. — Correct. As worked out above, the rms value is about 220 V.
Option (4) The frequency of source is 440 Hz. — Incorrect. 440 is \( \omega \), not f; the frequency is about 70 Hz.
NCERT (2026-27):
Ac Voltage Applied to a Resistor – “… It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of
\( \displaystyle v_m = \sqrt{2}\,V = (1.414)(220\ \mathrm{V}) = 311\ \mathrm{V} \)”