View Full Version : rms voltage

ELECTRONERD

04-02-2009, 07:03 PM

hey everybody!

i'm studyin' for my license and came across rms voltage stuff. after researching different references, i found that most of the newer books didn't explain it very well...not for a beginner at least! so, i came across this old ameco manual and they explained it pretty darn well!

so this is for the other kids or for anyone who needs to understand it:

imagine having a lamp. if we have 100V DC and 100V AC, will the DC or AC light up brighter? since AC is commonly associated with a sine wave (it would be best to look and see what one looks like!), there are points where there is no voltage. however, DC remains a constant voltage source. therefore, DC would light the lamp up more brilliantly. If we multiply 0.707 X 100V DC, we get 70.7V AC. So the lamp for 100V AC is actually illuminating relative to 70.7V DC. How much voltage is needed for them to have the same luminosity? To find that, we divide 100V DC by .707 and get about 141V AC. So, to make the lamps have the same brightness, we could have 100V DC or 141V AC.

Hope that clears things up!

G3TXQ

04-02-2009, 07:27 PM

Well done!

I would add just one thing to your explanation: when specifying AC voltages you need to be very clear about what you mean. For example, you talked about "100V AC" without it being clear whether you mean Peak, Peak-to-Peak, or RMS.

So I can't answer your question: "If we have 100V DC and 100V AC, will the DC or AC light up brighter?" until you tell me what you mean by 100V AC.

100Vpeak, 100Vpeak-to-peak, and 100Vrms are all very different things.

Steve

AK7EE

04-02-2009, 08:07 PM

hey everybody!

i'm studyin' for my license and came across rms voltage stuff. after researching different references, i found that most of the newer books didn't explain it very well...not for a beginner at least! so, i came across this old ameco manual and they explained it pretty darn well!

so this is for the other kids or for anyone who needs to understand it:

imagine having a lamp. if we have 100V DC and 100V AC, will the DC or AC light up brighter? since AC is commonly associated with a sine wave (it would be best to look and see what one looks like!), there are points where there is no voltage. however, DC remains a constant voltage source. therefore, DC would light the lamp up more brilliantly. If we multiply 0.707 X 100V DC, we get 70.7V AC. So the lamp for 100V AC is actually illuminating relative to 70.7V DC. How much voltage is needed for them to have the same luminosity? To find that, we divide 100V DC by .707 and get about 141V AC. So, to make the lamps have the same brightness, we could have 100V DC or 141V AC.

Hope that clears things up!

The general rule is that if don't specify the measurement, peak, peak to peak, or average, rms is assumed. Now we'll assume that your lamp will work equally well on DC and AC, 100Volts DC or 100 Volts AC (RMS assumed) will provide the current, therefore the same amount of light.

1.414 times 100 Volts AC will give you the Peak voltage 141.4 Volts. and a Peak to Peak of 2 times 141.4 or 282.8 Volts. Average is generally considered Average of the peak voltage in sin wave, therefore the average voltage in our example would 70.7 volts.

ELECTRONERD

04-03-2009, 04:14 PM

thanks guys! those are good tips to remember! Its too bad that modern references don’t explain things as good as older ones. what I don’t like is that most writers try to show off their knowledge rather than explain it so others can understand.

Simplicity is the key folks!

VE3LDJ

04-03-2009, 04:47 PM

rms of ac voltage is 0

when rms of an ac voltage is given, it is assumed to be rectified.

VA2GK

04-03-2009, 04:53 PM

rms of ac voltage is 0

when rms of an ac voltage is given, it is assumed to be rectified.

¿HUH? :confused:

VE3LDJ

04-03-2009, 05:04 PM

The RMS of an AC sine wave is equal to the RMS of the Positive voltage, MINUS the RMS of the Negative voltage. That is 0.

When the Hacks talk about AC rms voltage, they are really talking about an The rms of a fully rectified AC voltage.

G3TXQ

04-03-2009, 05:15 PM

The RMS of an AC sine wave is equal to the RMS of the Positive voltage, MINUS the RMS of the Negative voltage. That is 0.

When the Hacks talk about AC rms voltage, they are really talking about an The rms of a fully rectified AC voltage.

I just had to check the date of the posting! Never, in 45 years of engineering have I heard that definition of RMS.

Because of the "Square" part of "Root Mean Square", the positive half cycle and negative half cycle of a sine wave both contribute identical positive amounts.

You were out by 2 days - admit it!

Steve

VA2GK

04-03-2009, 05:31 PM

The RMS of an AC sine wave is equal to the RMS of the Positive voltage, MINUS the RMS of the Negative voltage. That is 0.

When the Hacks talk about AC rms voltage, they are really talking about an The rms of a fully rectified AC voltage.

I'm starting the popcorn machine.

VE3LDJ

04-03-2009, 05:44 PM

Ok, Just to see if I was wrong, (I've been told it would happen some day), I read up on it. It appears to be a terminology problem. This from Wiki (Open source and not subject to copyright).

Physical scientists often use the term "root mean square" as a synonym for standard deviation when referring to the square root of the mean squared deviation of a signal from a given baseline or fit. This is useful for electrical engineers in calculating the "AC only" RMS of a signal. Standard deviation being the root mean square of a signal's variation about the mean, rather than about 0, the DC component is removed (i.e. RMS(signal) = Stdev(signal) if the mean signal is 0)

Can't we both be right?

-Luke

K5UOS

04-03-2009, 05:46 PM

This is how I learned it.

Assume a sine wave is an infinite set of points (instantaneous values). Square these instantaneous values and take the average of all the squared values. This resultant value will be .707 of the peak value.

Simply, RMS (root - mean - squared) equals .707 peak as others have said. It has really nothing directly to do with DC.

If you add all the the instantaeous and average them the result will be .636 of the peak value. I think there is a more precise rule that says to use 360 degrees, one degree apart as your points.

I really only refer to this rule when building a DC power supply. It does not directly relate to DC but is a mathmatical property of a sine wave not necessarily limited to discussions of voltage.

If the RMS is zero then theoretically no work would be ever be done.

I remember when I was learning the relationship between dbm and milliwatts from my dad. My dad wanted to smack me! That was only one difficult concept, too!

Don K5UOS

G3TXQ

04-03-2009, 06:47 PM

Can't we both be right?

Nope.

That definition is OK but you misapplied it. The RMS value of a sinusoid is 0.707 times its peak value.

Check it out:

http://en.wikipedia.org/wiki/Root_mean_square

Steve

VE3LDJ

04-03-2009, 07:23 PM

OK.

I've always thought of rms voltage being the same as the average.

I'm wrong.

Thanks for straightening me out :o

-Luke

AC0OB

04-03-2009, 09:13 PM

Another approach.

In a purely resistive (non-reactive) circuit, and since a light bulb filament is a pure resistance, the average power of an sinewave (AC) dissipated in a resistor is Pa = 1/2.I^2.r + 1/2.I^2.r.coswt. Now since the 1/2.I^2.r.coswt term is equal to zero for one cycle, the average power loss in a resistor is 1/2.I^2.r. (The ^2 means squared where the I is multiplied by itself). The period stands for "multiply."

Now, by algebra, Pa = 1/2.I^2.r is also equal to (I/sqrt(2))^2.r is also equal to Irms^2.r. So Irms = I/sqrt(2).

Now the AC rms (sinewave) current "through" a resistor is (Irms). So the average AC power dissipated in a resistor is Pa = (Irms)^2.r.

So, for AC "Average" power calculations, the current I through the resistor is the rms value.

Irms is equal to I/sqrt(2) or this is equal to I.(1/sqrt(2)). And since

1/sqrt(2) = 0.707, then the rms current through a resistor is 0.707.I.

So the average power dissipated in a resistor is Pa = (0.707.I)^2.r and the average power delivered is equal to Pa = 0.707I.0.707V. Recalling that 0.707.I is the rms current value, and the 0.707.V is the rms voltage value.

Also, Pa = Vrms^2/r. (an AC version of Ohms Law).

Take an example:

A 100W (average power Pa) incandescant light bulb is in a 120 V 60 Hz AC circuit. What is the resistance of the bulb filament and the peak current "through" the bulb?

Model the filament as a resistor. The 120V is assumed to represent an rms voltage value.

Rearranging Pa = Vrms^2/r

r = Vrms^2/Pa = (14400 Vrms^2)/100W = 144 ohms.

Now we calculate the rms current:

Irms = Pa/Vrms = 100W/120 Vrms = 0.833 Amps rms.

The PEAK current through the resistive filament is Ipeak = sqrt(2).Irms =

1.414.Irms = 1.414.0.833 = 1.18 amps.

Likewise, the PEAK voltage across the filament is Vpeak = sqrt(2).Vrms = 1.414.Vrms = 1.414.120Vrms = 169.68 Volts.

So why can we use the same Ohms Law equations for AC (with proper modifications), because the average AC power would be the equivalent DC heating power dissipated in a resistor.

The peak currents and voltages is what you would see on an oscilliscope for a half-cycle.

KB5PN

04-04-2009, 12:53 AM

I'm starting the popcorn machine.

I'll throw my bag in the microwave as soon as I get thru scratching my head!!:D

VK6ZGO

04-05-2009, 04:07 PM

Bravo! AC0OB,a masterful explanation.

Many people fail to see the derivation of RMS voltage & current from the power produced & end up misled.

The original source of this relationship was experimental.

The early pioneers of electrical theory knew how much power was required to do a particular amount of work,such as boiling a jug of water in a given time using DC,so they applied AC at varying levels until they obtained the same result. So if 100v dc boiled the water in the same time as x volts ac,the effective ac voltage was 100volts.

They then derived the relationship between ac peak voltage & effective or RMS voltage.

73 VK6ZGO