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

• mobby
In summary, the conversation discusses how the peak voltage or current is divided by square root 2 to get the root mean square value. The speaker requests a simple derivation to understand this concept better and suggests graphing a pure AC sine wave and calculating the area above zero inside the line in the graph. They also mention comparing this to a square wave and the importance of knowing the definition of RMS. The conversation ends with a suggestion to think creatively about surrounding things and other thought products.
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.

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

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Creative thinking is enjoyable, Then think about your surrounding things and other thought products. http://electrical-riddles.com

## 1. Why is it necessary to divide the peak value by the square root of 2 to get RMS?

The square root of 2 (or approximately 1.414) is derived from the relationship between the peak value and the RMS value of a sine wave. When an alternating current or voltage is represented as a sine wave, the peak value refers to the maximum value of the wave, while the RMS value refers to the effective or average value of the wave. Since the RMS value is the most important representation for the overall magnitude of the wave, it is necessary to convert the peak value to RMS by dividing it by the square root of 2.

## 2. How does dividing by the square root of 2 affect the RMS value?

Dividing the peak value by the square root of 2 essentially reduces the magnitude of the wave by a factor of 1.414. This conversion ensures that the RMS value accurately reflects the overall magnitude of the wave, taking into account both the amplitude (peak value) and the duration of the wave.

## 3. Can the square root of 2 be replaced with another value when calculating RMS?

No, the square root of 2 is a fundamental constant in the calculation of RMS for sine waves. This value is derived from the relationship between the peak value and RMS value of a sine wave and cannot be replaced with another value.

## 4. Is dividing by the square root of 2 only applicable to sine waves?

While the relationship between peak value and RMS value is specific to sine waves, the concept of dividing by the square root of 2 to convert peak value to RMS can be applied to other types of waves as well. However, the exact value may differ depending on the waveform in question.

## 5. What is the significance of the RMS value in electrical engineering?

The RMS value of an alternating current or voltage is an important measure in electrical engineering as it represents the equivalent DC (direct current) value that would produce the same amount of heat or power dissipation in a circuit. This makes it a more accurate representation of the overall magnitude of the wave and allows for easier comparison between different waveforms.

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