Reason for dividing peak value by sq. root 2 to get RMS

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Discussion Overview

The discussion revolves around the mathematical derivation of why the peak voltage or current is divided by the square root of 2 to obtain the root mean square (RMS) value. Participants explore various approaches to understanding this concept, including comparisons with square waves and definitions of RMS.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant requests a simple derivation to understand the relationship between peak value and RMS value, indicating a desire for clarity on the concept.
  • Another participant suggests graphing a pure AC sine wave and comparing it to a square wave to explore the area under the curves, hinting at a connection to RMS calculations.
  • A different participant argues that comparing with a square wave is unnecessary, stating that the RMS amplitude can be derived directly from the peak amplitude through the area under the curve.
  • Some participants emphasize the importance of the square wave in demonstrating that RMS voltage equates to the voltage that would produce the same heating effect in a resistor as a DC voltage, suggesting a conceptual basis for the comparison.
  • One participant mentions that sampling a sine wave, squaring the samples, averaging them, and taking the square root will yield the RMS value, reiterating the definition of RMS.
  • A mathematical expression for RMS is presented, showing the relationship between the RMS current and the peak amplitude, although the context of this derivation is not fully explored.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of comparing the sine wave to a square wave, with some arguing it is essential for understanding while others consider it redundant. The discussion remains unresolved regarding the best approach to derive and understand the RMS value.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the definitions of RMS and the mathematical steps involved in deriving the relationship between peak and RMS values. Some participants may overlook the broader implications of power dissipation in their explanations.

mobby
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I NEED TO GET A DERIVATION SO THAT I CAN MATHEMATICALLY UNDERSTAND WHY:
The peak voltage or current is divided by square root 2 to get the root mean square value.please i need a simple derivation just to understand,i will be very greatful for ur consideration.thankyou.
 
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Graph a pure AC sine wave and calculate the area above zero inside the line in your graph. Now do this to a square wave and compare the two. What do you think you will find? I'd be more specific but this sounds like homework. In fact I've probably already told you too much.
 
No need to actually compare with a square wave, is there? All one needs to do is find the RMS amplitude relative to the peak amplitude (which will naturally appear in the expression for the area under the curve).

TO mobby: Start with the definition of the RMS value, and show us what you find.
 
Yes there is a reason we have to compare it to a square wave. Since RMS voltage is the voltage it takes to heat a resistor by the same amount of DC voltage, we have to use a square wave. Since a square wave is always at peak it is the same as DC. That was my point. Maybe I'm not reading the OP question correctly. The conclusion will naturally be the same. It sounds to me like the OP wants to know why and is not simply satisfied with the sqrt(2) formula.
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You can also sample the sine wave (or any waveform) at a number of points, square each sample, average them, and then take the square root of the average and you will find the same thing. I would hope so because that is the definition of RMS which many people seem to forget and simply refer to it as "RMS".
 
Averagesupernova said:
Yes there is a reason we have to compare it to a square wave. Since RMS voltage is the voltage it takes to heat a resistor by the same amount of DC voltage, we have to use a square wave. Since a square wave is always at peak it is the same as DC. That was my point.
You could compare to a square wave, but that is completely redundant. You will be setting the RMS value of the square wave (also its own peak value) to the peak value of the sine wave, but this peak value will naturally arise in the calculation of the area under the curve for the sine wave (so the square wave bit is unnecessary).

That the RMS voltage is the one that dissipates the same power as what comes from an equal dc signal is a (somewhat trivially) derived result. The definition of an RMS value does not require you to even know what power is. It certainly is important to know, but not required for the calculation.
 
Last edited:
I2rms = 1/T∫0T (a sin 2п/T t)2 . dt = 2a2

I rms = √2 a

:wink:



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