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

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

Let us analyze the given specifications of the alternating current (AC) source:
1. The rated voltage is $V_{\text{rms}}=220\text{V}$. In AC systems, the rated value refers to the root-mean-square (RMS) voltage.
2. The frequency of the AC source is $f=50\text{Hz}$.

The time period $T$ of one complete cycle of the AC source is the reciprocal of the frequency:
$T=\frac{1}{f}=\frac{1}{50}\text{s}$

An AC voltage is represented as a sinusoidal function of time:
$V\left(t\right)=V_{0}sin\left(\omega t\right)$
where $V_0$ is the peak voltage and $\omega =2\pi f$ is the angular frequency.

A complete cycle of a sine wave consists of a symmetric positive half-cycle and a symmetric negative half-cycle.
The average value of any periodic function $V\left(t\right)$ over one full time period $T$ is given by the integral:
$V_{\text{avg}}=\frac{1}{T}{\int }_0^{T}V\left(t\right)dt$

Substituting $V\left(t\right)=V_{0}sin\left(\frac{2\pi }{T}t\right)$ into the formula:
$V_{\text{avg}}=\frac{1}{T}{\int }_0^T{V}_{0}sin\left(\frac{2\pi }{T},t\right)dt$
$V_{\text{avg}}=\frac{V_0}{T}{\left[-\frac{T}{2\pi }cos\left(\frac{2\pi }{T},t\right)\right]}_0^T$
$V_{\text{avg}}=-\frac{V_0}{2\pi }\left[cos\left(2\pi \right)-cos\left(0\right)\right]$
Since $cos\left(2\pi \right)=1$ and $cos\left(0\right)=1$, we have:
$V_{\text{avg}}=-\frac{V_0}{2\pi }\left[1-1\right]=0\text{V}$

Therefore, the average value of the alternating voltage over a complete time period of $T=\frac{1}{50}\text{s}$ is exactly $0\text{V}$.