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

[mobby]

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.

 

[Averagesupernova]

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.

 

[Gokul43201]

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.

 

[Averagesupernova]

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".

 

[Gokul43201]

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.

 

[m.s.j]

I²rms = 1/T∫₀^T (a sin 2п/T t)² . dt = 2a²

I rms = √2 a





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