For an ac source rated at 220 V, 50 Hz, which of the following statements is correct?

The average value over a period of (1/50) s is 0 V.

The correct answer is Option (3) → The average value over a period of (1/50) s is 0 V.

For an AC source, the voltage is given by $V = V_0 \sin(\omega t)$ where $\omega = 2\pi f$.

Given: frequency $f = 50\,Hz$.

The time period of one complete cycle is

$T = \frac{1}{f} = \frac{1}{50}\,s$.

This means the AC voltage completes one full oscillation — from 0 → +$V_0$ → 0 → −$V_0$ → 0 — in $\frac{1}{50}$ seconds.

Over this complete cycle, the positive and negative halves are equal and opposite, so the average value of $\sin(\omega t)$ over $0$ to $T$ is zero.

∴ The average value of the AC voltage over a period $\frac{1}{50}\,s$ is $0\,V$.

Correct option: The average value over a period of (1/50) s is 0 V.