Question

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

• A. The peak value over a period of (1/50) s is 220 V.
• B. The average value over a period of (1/50) s is 220 V.
• C. The average value over a period of (1/50) s is 0 V.
• D. The average value over a period of (1/50) s is 220\(\sqrt{2}\) V.

Hint:

Always remember that the standard voltage rating for AC circuits is the RMS value. For a sine wave, the average value over a full cycle is 0, and the average value over a half cycle is \(\frac{2V_0}{\pi}\). Don't confuse RMS, peak, and average values.

Answer 1

The Correct Option is C

Step 1: Conceptual Basis: The specified voltage for an AC source, such as household electricity, represents its Root Mean Square (RMS) value. Analysis of this sinusoidal AC voltage necessitates understanding its peak and average values across a complete cycle.

Step 2: Governing Principles: For a sinusoidal AC voltage: - The RMS voltage relates to the peak voltage (\(V_0\)) via the formula \(V_{rms} = \frac{V_0}{\sqrt{2}}\). - Frequency (\(f\)) is the inverse of the time period (\(T\)), expressed as \(T = 1/f\). - The average value of a sinusoidal function over one full cycle is zero.

Step 3: Detailed Analysis: Provided Information: RMS voltage, \(V_{rms} = 220 \, \text{V}\). Frequency, \(f = 50 \, \text{Hz}\). The time period for this AC source is calculated as \(T = 1/f = 1/50 \, \text{s}\). The properties under examination are within this duration, which constitutes a single full cycle. Evaluation of Statements: 1. Peak value over (1/50) s period is 220 V. The actual peak value is derived as \(V_0 = V_{rms} \times \sqrt{2} = 220\sqrt{2}\) V. This statement is erroneous. 2. Average value over (1/50) s period is 220 V. A duration of (1/50) s represents one complete cycle. The average of a sine wave over a full cycle is inherently zero. The value 220 V is the RMS value, not the average. This statement is incorrect. 3. Average value over (1/50) s period is 0 V. A sinusoidal AC waveform (for voltage or current) exhibits symmetry around the time axis. The positive half-cycle's area is numerically equal to the negative half-cycle's area. Consequently, the average value across one full cycle is consistently zero. This statement is accurate. 4. Average value over (1/50) s period is 220\(\sqrt{2}\) V. This value represents the peak voltage, not the average value. This statement is incorrect.

Step 4: Conclusion: The singular correct assertion is that the average value over a full period amounts to 0 V.