The question asks us to identify the correct statement about an AC source rated at 220 V and 50 Hz.

AC Source Voltage Fundamentals

An AC (Alternating Current) source provides voltage that varies sinusoidally over time. The rating of an AC source, such as 220 V, typically refers to its RMS (Root Mean Square) value. We need to understand the relationships between RMS voltage, peak voltage, and average voltage.

Understanding Voltage Definitions

• RMS Voltage ($V_{RMS}$): This is the effective voltage of an AC source. It's the equivalent DC voltage that would produce the same amount of power dissipation in a resistive load. For the given AC source, $V_{RMS} = 220$ V.
• Peak Voltage ($V_{peak}$): This is the maximum instantaneous voltage reached by the AC waveform during a cycle. For a sinusoidal waveform, the relationship between peak voltage and RMS voltage is given by $V_{peak} = V_{RMS} \times \sqrt{2}$.
• Average Voltage ($V_{avg}$): This is the average of the instantaneous voltage values over a specific time interval. For a sinusoidal waveform over a complete cycle, the average voltage is zero because the positive half-cycle cancels out the negative half-cycle.

Calculating the Period of the AC Source

The frequency ($f$) of the AC source is given as 50 Hz. The period ($T$) is the time taken for one complete cycle of the waveform and is the reciprocal of the frequency:

$ T = \frac{1}{f} $

Substituting the given frequency:

$ T = \frac{1}{50} \, \text{s} $

This period of $\frac{1}{50}$ s corresponds to one full cycle of the AC waveform.

Evaluating Peak Voltage

Let's calculate the peak voltage ($V_{peak}$) for the 220 V RMS source:

$ V_{peak} = V_{RMS} \times \sqrt{2} $

$ V_{peak} = 220 \, \text{V} \times \sqrt{2} \approx 311 \, \text{V} $

Therefore, the peak value is approximately 311 V, not 220 V.

Evaluating Average Voltage

The average value of a sinusoidal AC voltage waveform over a complete cycle (which is $\frac{1}{50}$ s in this case) is always zero. This is because the waveform is symmetrical about the time axis, with the positive area balancing the negative area.

Mathematically, for a voltage waveform $v(t) = V_{peak} \sin(\omega t)$, where $\omega = 2\pi f$, the average voltage over one period $T = \frac{2\pi}{\omega}$ is:

$ V_{avg} = \frac{1}{T} \int_0^T V_{peak} \sin(\omega t) \, dt $

The integral of $\sin(\omega t)$ over a full period is zero, hence:

$ V_{avg} = 0 \, \text{V} $

Analyzing the Options

Based on our calculations and understanding:

• Option 1: The peak value over a period of (1/50) s is 220 V. Incorrect. The peak value is $220\sqrt{2}$ V.
• Option 2: The average value over a period of (1/50) s is 220 V. Incorrect. The average value over a full cycle is 0 V.
• Option 3: The average value over a period of (1/50) s is 0 V. Correct. This aligns with the properties of sinusoidal AC waveforms over a full cycle.
• Option 4: The average value over a period of (1/50) s is $220\sqrt{2}$ V. Incorrect. $220\sqrt{2}$ V is the peak value.