What is the ratio of peak-to-effective voltage values of a sine wave?

What is the ratio of peak-to-effective voltage values of a sine wave?

• A. 1.414 to 1
• B. 1 to 0.707
• C. Both A and B
• D. Neither A nor B

Answer: (C)

The ratio of peak-to-effective voltage values of a sine wave is indeed commonly expressed as both 1.414 to 1 (which is the square root of 2) and 1 to 0.707. For a sine wave, the peak voltage (also known as the amplitude) is the maximum voltage level it reaches, while the effective voltage (often referred to as the root mean square or RMS voltage) represents the value of direct current (DC) that would produce the same amount of heat in a resistive load as the AC voltage would. The mathematical relationship between these two values is derived from the properties of sinusoidal waves. When you calculate the effective voltage from the peak voltage, the formula is: \[ V_{effective} = \frac{V_{peak}}{\sqrt{2}} \] This leads to the peak-to-effective voltage ratio being: \[ \frac{V_{peak}}{V_{effective}} = \frac{V_{peak}}{V_{peak}/\sqrt{2}} = \sqrt{2} \] This results in the ratio being approximately 1.414 to 1. On the other hand, the effective voltage can also be expressed in terms of its

The ratio of peak-to-effective voltage values of a sine wave is indeed commonly expressed as both 1.414 to 1 (which is the square root of 2) and 1 to 0.707.

For a sine wave, the peak voltage (also known as the amplitude) is the maximum voltage level it reaches, while the effective voltage (often referred to as the root mean square or RMS voltage) represents the value of direct current (DC) that would produce the same amount of heat in a resistive load as the AC voltage would. The mathematical relationship between these two values is derived from the properties of sinusoidal waves.

When you calculate the effective voltage from the peak voltage, the formula is:

[

V_{effective} = \frac{V_{peak}}{\sqrt{2}}

]

This leads to the peak-to-effective voltage ratio being:

[

\frac{V_{peak}}{V_{effective}} = \frac{V_{peak}}{V_{peak}/\sqrt{2}} = \sqrt{2}

]

This results in the ratio being approximately 1.414 to 1.

On the other hand, the effective voltage can also be expressed in terms of its