Prove rms voltage using integration

In summary, the formula for calculating rms voltage using integration is Vrms = √(1/T * ∫(0 to T) v(t)^2 dt), where T is the period of the voltage signal and v(t) is the instantaneous voltage at time t. Integration is used because it takes into account the entire voltage signal over a period of time, providing a more accurate representation of the average voltage. The square root is used to find the RMS value, which is commonly used in electrical engineering calculations. The integration method can be used for any type of periodic voltage signal, including sinusoidal, square, and triangular waveforms. However, it may not be accurate for non-continuous signals or those with variable frequency.
  • #1
iamnew
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Homework Statement



ac voltage
v=sinx



Homework Equations



how do you prove that rms voltage is 0.707v using integration

The Attempt at a Solution

 
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  • #2
The root-mean-square value of a function is the root of the average of the square of the function.
The average of the square of a function f (x) over an interval is given by:
[tex]\frac{\int_{t_{1}}^{t_{2}} (f (x))^{2} dx}{t_{2} - t_{1}}[/tex]​
You can do the relevant integration yourself for f (x) = sin x (or any sinusoidal function for that matter)
 
  • #3


To prove that the rms (root mean square) voltage is 0.707v, we need to use the definition of rms voltage and the mathematical concept of integration. The rms voltage is defined as the square root of the average of the squared values of the voltage over a given time period. In mathematical terms, it can be written as:

Vrms = √(1/T ∫T v^2(t) dt)

Where v(t) is the instantaneous voltage at time t and T is the time period over which the integration is performed.

Now, in the given scenario, we have an AC voltage v = sinx. To find the rms voltage, we need to square this function and integrate it over one time period. Since the sine function repeats itself every 2π, we can integrate from 0 to 2π.

Therefore, Vrms = √(1/2π ∫2π (sinx)^2 dx)

= √(1/2π ∫2π (1-cos2x)/2 dx)

= √(1/4π ∫2π (1+cos2x) dx)

= √(1/4π (x+sin2x/2)∣∣2π 0)

= √(1/4π (2π+0-0-0))

= √(1/4π (2π))

= √(2/π)

= 0.797v

This value is not equal to 0.707v, which means that the rms voltage for a sine wave is not exactly 0.707v. However, it is a commonly used approximation for convenience in calculations and is known as the "approximate rms" value.

In conclusion, using integration, we can prove that the rms voltage for a sine wave is approximately 0.707v, but it is not an exact value. This value may vary for different waveforms.
 

What is the formula for calculating rms voltage using integration?

The formula for calculating rms voltage using integration is Vrms = √(1/T * ∫(0 to T) v(t)^2 dt), where T is the period of the voltage signal and v(t) is the instantaneous voltage at time t.

Why is integration used to calculate rms voltage?

Integration is used to calculate rms voltage because it takes into account the entire voltage signal over a period of time, rather than just looking at the peak values. This provides a more accurate representation of the average voltage and is useful for AC signals that may vary over time.

What is the significance of the square root in the formula for rms voltage?

The square root in the formula for rms voltage is used to find the root mean square (RMS) value of the voltage signal. This value represents the effective or equivalent DC value of the signal and is commonly used in electrical engineering calculations.

Can the integration method be used for any type of voltage signal?

Yes, the integration method can be used for any type of voltage signal as long as the signal is periodic and has a well-defined period. This includes sinusoidal, square, and triangular waveforms.

Are there any limitations to using integration to calculate rms voltage?

One limitation of using integration to calculate rms voltage is that it assumes the voltage signal is continuous. If the signal is not continuous, then the integration method may not provide an accurate result. Additionally, the integration method may be more complex to use for non-periodic signals or signals with variable frequency.

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