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# Determine the RMS amplitude of this sinusoidal waveform

1. ## Determine the RMS amplitude of this sinusoidal waveform

I know this is a simple question but can someone help explain this?

Determine the RMS amplitude of this sinusoidal waveform, as displayed by an oscilloscope with a vertical sensitivity of 0.2 volts per division:   Reply With Quote

2. Originally Posted by Ronwilson1801 I know this is a simple question but can someone help explain this?

Determine the RMS amplitude of this sinusoidal waveform, as displayed by an oscilloscope with a vertical sensitivity of 0.2 volts per division: The thicker lines in the middle that split the graph into quadrants are the 0 lines, representing voltage (horizontal) and time (vertical). The wave peaks just above the second horizontal line up from the middle, if the lines are 0.2 volts each, then the peak is a little over 0.4 volts. If you look very closely there are 5 diving lines per division (0.2 / 5 = 0.04), so closer to 0.44 volts peak.

To find RMS, take the peak value and multiply by 0.707. In this case 0.44 x 0.707 = 0.31108.  Reply With Quote

3. Originally Posted by SecondGen The thicker lines in the middle that split the graph into quadrants are the 0 lines, representing voltage (horizontal) and time (vertical). The wave peaks just above the second horizontal line up from the middle, if the lines are 0.2 volts each, then the peak is a little over 0.4 volts. If you look very closely there are 5 diving lines per division (0.2 / 5 = 0.04), so closer to 0.44 volts peak.

To find RMS, take the peak value and multiply by 0.707. In this case 0.44 x 0.707 = 0.31108.
very helpfull thanks for the info  Reply With Quote

4. Originally Posted by SecondGen The thicker lines in the middle that split the graph into quadrants are the 0 lines, representing voltage (horizontal) and time (vertical). The wave peaks just above the second horizontal line up from the middle, if the lines are 0.2 volts each, then the peak is a little over 0.4 volts. If you look very closely there are 5 diving lines per division (0.2 / 5 = 0.04), so closer to 0.44 volts peak.

To find RMS, take the peak value and multiply by 0.707. In this case 0.44 x 0.707 = 0.31108.
RMS = Peak Voltage / sqrt(2)

1/sqrt(2) = 0.707

Just so you know why you are multiplying by 0.707. Unfortunately we can't put actual equations here to show an actual square root symbol.  Reply With Quote

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Reputation ## Understanding why we want RMS helps understand the diference between peak and RMS val

One of the reasons we developed RMS values for voltage and current is we wanted a value for alternating current that equals the effect of direct current. One idea was to simply take the average or mean value of the sine wave of alternating current but how does one find the area under a curve; the answer was to represent the curve by a series of rectangles whose height touches the sine wave; the area of a rectangle is easy to calculate. As these rectangles get thinner (and hence more numerous), they more accurately represent the curve of a sine wave so all that is left to do is use extremely thin and very numerous rectangles and then find their average. Unfortunately, for a sine wave, these average to zero as half are positive and half are negative. The next iteration was to square each rectangle so all the negative rectangles become positive (and all the positive rectangles stay positive). Now we can add up all the squared rectangles and then take the average. All that is left to do is to undue the squaring by taking the square root. That is how we go to taking the square Root of the Mean of the Squares or RMS. For a sine wave, this becomes taking the Peak and dividing by the square root of 2 which is appr. 1.414 or you can multiply the peak by the inverse of 1.414 which is 0.707. There are many good videos of the actual math involved which involves taking an integral and they show you how the result becomes the peak divided by the square root of 2.  Reply With Quote

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