I'm currently taking a class called AC Principles and have a few questions. Our instructor has been teaching us about sign waves and I'm having a hard time understanding peak voltage.

How do we know that the peak voltage of a 120VAC sign wave comes to 170 volts?

And how does the figure ".707" relate to the 170 volts?

We'll be getting into 3-phase in the next class and I don't think I have grasped this concept yet.

Peak Voltage: Peak voltage tell you how far the voltage swings, either positive or negative, from the point of reference. Peak voltage is only a moderately useful way of measuring voltage when trying to express the amount of work that will be done when driving a specified load.

RMS Voltage: RMS voltage is absolutely the most common way to measure/quantify AC voltage. It is also the most useful. Because AC voltage is constantly changing and is at or near the highest and lowest points in the cycle for only a tiny fraction of the cycle, the peak voltage is not a good way to determine how much work can be done by an AC power source.

The RMS voltage of a pure‡ sine wave is approximately .707*peak voltage. If you read voltage with a voltmeter you are generally given the RMS voltage of the wave form.

So for your example the sine wave goes from 0 to 170V in the positive and then goes down to 170V in the negative. To find RMS (peak Voltage) x (0.707) this will give you 120V.

If you hooked up an osiliscope to a 120volt receptacle you would see the sinewave and would notice that the highest voltage that is recorded is 170volts. This is Peak Voltage. Its only there for a really, really short time. So use the RMS (RMS is .707 of Peak Voltage)

[This message has been edited by RobbieD (edited 01-29-2006).]

Re: Peak Voltage #61614 01/29/0607:46 PM01/29/0607:46 PM

Thank you for that explanation. I really do appreciate the help. So if I know 170 volts is the peak voltage level for a 120 volt sine wave, would I also use the .707 figure to find the peak level of a 24OV sine wave? How 'bout 480?

Re: Peak Voltage #61615 01/29/0607:52 PM01/29/0607:52 PM

This is true for a sine wave, not necessarily for other wave shapes. It has to do with the square root of 2, a bit of trig trickery (not unlike smoke & mirrors).

Hook up a scope to an AC circuit and see the peak values are + & - 170V. Then divide 170 by the square root of 2 (1.4142) and you get 120V (approximately) RMS value. RMS is the effective, or DC equilivant value, work wise.

Radar

There are 10 types of people. Those who know binary, and those who don't.

Re: Peak Voltage #61617 01/30/0604:08 AM01/30/0604:08 AM

The RMS value of a sine wave voltage generates the same amount heat for a resistor at the same DC value.

e.g. 120 VoltsRMS AC generates the same amount of heat as 120 Vdc, regardles of the higher peak value. ( and lower values near the zero crossing )

Robbie gives you a very good explanation for it anyway.

In case of 3Ø power the value V3 or root 3 (1.73) becomes more important as you will learn at tech. later on. Phase voltage between phase and neutral. is 120 volts in your case. Line voltage is 120*1.73=208 volts, between 2 phases.

Hope it helps a little regards Ray.

The product of rotation, excitation and flux produces electricty.

Re: Peak Voltage #61618 01/30/0611:49 AM01/30/0611:49 AM

The .707 is the square root of the integral from 0 to 2π of sin(θ)^2 dθ, divided by 2π. Which is, as RobbieD said, the RMS value of the waveform. That integral comes out to 1/sqrt(2), which is, rounded to three digits, 0.707.

(Edited to turn off smilies, which were showing up in the middle of the math.)

[This message has been edited by SolarPowered (edited 01-30-2006).]

Re: Peak Voltage #61619 01/30/0607:59 PM01/30/0607:59 PM