For a zero-mean wave, if the RMS value is 100 V, what is the peak-to-peak amplitude?( Note: take the wave as sine or cosine)
280 V
Understanding the properties of different types of waves is fundamental in electrical engineering and physics. For a sinusoidal wave, such as a sine or cosine wave, several key amplitude values are used to describe its magnitude. These include the peak value, RMS (Root Mean Square) value, and peak-to-peak amplitude. This solution focuses on how to determine the peak-to-peak amplitude when the RMS value of a zero-mean wave is known.
A zero-mean wave is a waveform whose average value over one complete cycle is zero. Sinusoidal waves (like sine or cosine waves) that oscillate symmetrically around the horizontal axis (zero voltage line) are classic examples of zero-mean waves. This means that the positive area above the axis perfectly balances the negative area below it over a full cycle.
The RMS value (Root Mean Square value) of an alternating current (AC) or voltage is a way to express its effective value. It is particularly useful because it represents the DC voltage or current that would produce the same amount of heat in a resistive load. For a pure sinusoidal waveform, there's a specific relationship between its RMS value and its peak value.
The peak-to-peak amplitude ($V_{\text{p-p}}$) is the total voltage difference between the positive peak (maximum positive value) and the negative peak (maximum negative value) of a waveform. For a zero-mean sinusoidal wave, the negative peak has the same magnitude as the positive peak but opposite sign. Therefore, the peak-to-peak amplitude is simply twice the peak value.
Given the RMS value of a zero-mean wave as 100 V, we can calculate its peak-to-peak amplitude step-by-step:
Using the relationship between RMS and peak voltage for a sinusoidal wave:
$$V_{\text{peak}} = V_{\text{rms}} \times \sqrt{2}$$Substitute the given RMS value:
$$V_{\text{peak}} = 100 \, \text{V} \times 1.414$$ $$V_{\text{peak}} \approx 141.4 \, \text{V}$$Using the relationship between peak voltage and peak-to-peak amplitude:
$$V_{\text{p-p}} = 2 \times V_{\text{peak}}$$Substitute the calculated peak voltage:
$$V_{\text{p-p}} = 2 \times 141.4 \, \text{V}$$ $$V_{\text{p-p}} \approx 282.8 \, \text{V}$$Comparing the calculated value to the given options, 282.8 V is closest to 280 V.
| Parameter | Formula for Sinusoidal Wave | Calculated Value (for $V_{\text{rms}} = 100 \, \text{V}$) |
|---|---|---|
| RMS Value ($V_{\text{rms}}$) | Given | 100 V |
| Peak Value ($V_{\text{peak}}$) | $V_{\text{rms}} \times \sqrt{2}$ | $100 \, \text{V} \times 1.414 = 141.4 \, \text{V}$ |
| Peak-to-Peak Amplitude ($V_{\text{p-p}}$) | $2 \times V_{\text{peak}}$ | $2 \times 141.4 \, \text{V} = 282.8 \, \text{V}$ |
Therefore, the peak-to-peak amplitude of the zero-mean sinusoidal wave is approximately 280 V.
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