Understanding Sine Wave Voltages: RMS vs. Peak-to-Peak

The question asks for the peak-to-peak voltage ($V_{p-p}$) of a sine wave, given its Root Mean Square (RMS) voltage ($V_{RMS}$) is 2 VRMS. We need to understand the relationship between these voltage measurements.

Key Voltage Concepts for Sine Waves

• RMS Voltage ($V_{RMS}$): This represents the effective value of an alternating current (AC) voltage. For a sine wave, it's related to the peak voltage by the formula $V_{RMS} = \frac{V_{peak}}{\sqrt{2}}$.
• Peak Voltage ($V_{peak}$): This is the maximum voltage amplitude the sine wave reaches from its zero-crossing point.
• Peak-to-Peak Voltage ($V_{p-p}$): This is the total voltage swing between the positive peak and the negative peak of the waveform. For a standard sine wave, $V_{p-p} = 2 \times V_{peak}$.

Calculation Steps

We are provided with the RMS voltage:

$V_{RMS} = 2$ V

Step 1: Calculate the Peak Voltage ($V_{peak}$)

We use the formula relating RMS voltage to peak voltage for a sine wave:

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

Substitute the given $V_{RMS}$ value:

$V_{peak} = 2 \text{ V} \times \sqrt{2}$

To get a numerical value, we use $\sqrt{2} \approx 1.414$:

$V_{peak} \approx 2 \text{ V} \times 1.414 = 2.828$ V

Step 2: Calculate the Peak-to-Peak Voltage ($V_{p-p}$)

Now, we use the relationship between peak voltage and peak-to-peak voltage:

$V_{p-p} = 2 \times V_{peak}$

Substitute the calculated peak voltage:

$V_{p-p} = 2 \times (2 \times \sqrt{2})$ V

$V_{p-p} = 4 \times \sqrt{2}$ V

Using the numerical approximation for $V_{peak}$:

$V_{p-p} \approx 2 \times 2.828 \text{ V} = 5.656$ V

Final Answer Summary

The calculation shows that the peak-to-peak voltage ($V_{p-p}$) for a 2 VRMS sine wave is approximately 5.656 V.