Average and RMS Value

well thanks for the kind and obvious advice but google isnt helping me with this specific problem..Everyone explains the math but not the difference in the physical sense.

This small paragraph from Fanco's Fundamental of EC is what is confusing me.

Consider the effect of dissipation of heat in a fixed resistance placed across the supply.

http://www.sayedsaad.com/fundmental/4_Alternating%20Current.htm

Average value - for half cycle of AC voltage .

RMS value - Equivalent DC voltage for applied AC voltage

Thanks guys but one thing is still not getting clear to me..

If i have a simple resistive circuit with 2A of sinusoidal current flowing through it. The average current for the half cycle is 0.63x2 = 1.27 A. Now according to the following lines

the charge transferred by this 2A peak current for the half cycle will be equal to that transferred by a 1.27A DC current. (please correct me if i am wrong) for the second half cycle(the negative one) we will have the same current flowing so the charge transferred again will be same as -1.27A DC current.
For a resistive load, heat will be dissipated the same amount irrespective of the current direction so shouldn't 0.636xIp be the effective DC current instead of 0.707xIp?

No, because in a fixed resistance, the heat dissipated is the square of the voltage applied divided by the resitance.

ok let me rephrase the question in a more logical way

i have the above two waveforms A and B (suppose peak voltage and current to be Vp and Ip for both cases ). I have 2 questions:

1. What is the DC equivalent current and voltage for both cases?

2. The area under the graphs is same for both cases so shouldn't the DC equivalents be the same as well?

i think answer to these questions will clear my confusion.

Well, what is the equation of the waveform, because they look like semicircles from here? Such a waveform would be rich in harmonics.

If the equation isn't known, then what does the frequency analyser have to say about the waveform?

The RMS voltage can still be determined by putting the waveform through a known, fixed resistance, and measuring the heat that comes out.

_{<mutters rude remarks about modern education systems, and unsubscribes>}

sorry i forgot to mention

for simplicity consider them as a complete sin wave and a half cycle sin wave..

Ok i think i have not been able to make myself clear. I know why we use rms for AC signals and why simply averaging them is not ok because we would get a 0.

The point i don't understand is why don't we get a relation like below for a completely pure sin wave

Area underneath half sin wave = (Area underneath full sin wave)/2

Mathematically it is clear when we integrate the sin wave for period T and T/2 we get the area as 0.707 times amplitude and 0.63 times amplitude respectively.

But I just want to understand this in an intuitive way.

Looking at your response here you actually answered your own question. Re read your question. It's all in .707 as you have stated. Stop trying to put the tooth paste back in the tube.

The heart of the problem is that power is a nonlinear function of either voltage or current. Doubling the voltage across or doubling the current through a resistive load does *not* double the power dissipation for a fixed load. This is why integrating the area underneath a voltage or current sine wave to calculate that wave's effective "average" value does *not* yield the average power dissipated in the load. Here are some practical examples to help clarify the distinction between average and RMS:

(1) Suppose you were using AC to energize an electrical heating element. It would be useful to rate the amount of AC voltage impressed across this heating element in terms of equivalent DC voltage that would dissipate the same amount of heat over time. Since we're talking about *power* here, the mathematical equivalence must be based on the square of the instantaneous voltage, hence the Root-Mean-Square calculation.

(2) Now suppose we were energizing a permanent magnet motor with rectified AC to generate torque. It would be useful to rate the amount of AC current through the armature winding of this motor in terms of equivalent DC current that would generate the same amount of torque. Since the torque is a linear function of armature current for a permanent magnet motor, the mathematical equivalence would be a much simpler calculation of area integrated under the AC current curve (i.e. the DC average value), with no squaring or square-rooting.

To summarize, when we're dealing with a phenomenon that is linearly related to voltage or current (e.g. example #2), the AC-DC equivalence is based on the mathematical average of the sine wave's instantaneous values; when we're dealing with a phenomemon related to power, which is quadratically related to voltage or current (e.g. example #1), the AC-DC eqivalence must be based on a root-mean-square calculation.

Incidentally, this is why analog meter movements don't read AC voltage or current accurately when the wave shape is non-sinusoidal. PMMC meter movements generate torque linearly with current, just like the armature of the PM motor in example #2. This means analog meter movements are naturally responsive to the *average* value of an AC wave, not its RMS value. Meter manufacturers compensate for this fact by skewing the calibration of the meter so it will read accurate RMS values when supplied with a pure sine wave. This means, though, that if you send a non-sinusoidal waveform to an analog meter, it will read in error because the ratio between RMS and average is not the same as it would be for a sine wave, and all the meter can respond to is the wave's average value.

Why not try a physical experiment. Take a well filtered DC supply and a 10 Ohm 20 W wirewound resistor, put 10 volts DC on it for 5 minutes and measure its temperature on a few spots over time. Then let it cool and take a variable AC supply and and an oscilloscope and gradually increase the voltage until the temperature reads the same as the DC case. Now use your oscilloscope and meters to view the waveform and measure the current and voltage, then do the same thing with half-wave and full-wave diodes. Also record the readings on a true RMS and peak reading meters if you have them.

It won't be exact but it should be close enough to get a feel for what is happening physically compared to the theoretical calculations. Class dismissed.

For sine wave function

Vrms = 0.70710678118654752440084436210485 * Vp

Vavr = 0.63661977236758134307553505349006 * Vp

Average value and rms value need not be same. These are two different ways of understanding a physical parameter in electrical circuit.

Power is proportional to the square of the voltage. So to get the DC voltage equivalent (the rms value) you have to average the square of the ac waveform and then take the square root.

thank you everyone for ur time

The best explanation I can find is here:
http://www.nuffieldfoundation.org/practical-physics/explaining-rms-voltage-and-current

Basically, your 0.637Vp is the time average of a half-wave rectifier voltage output, and it is indeed what will be measured by a DC voltmeter at that point. However, the current flow and power output resulting from a sinusoidal wave relate to the RMS value of 0.707Vp.

Looking at Wikipedia the second equation is the mathematical definition. If you do the mathematics for a sine wave, you end up with Vp/sqrt(2) as the RMS value.

What happens in common DMM's, the value is rectified and averaged. Easy to do with a capacitive filter and multipled by a fudge factor to yield the RMS value.

Root - means square root

Mean - means average

Squared - means ^2

Squaring a waveform and taking the squareroot tends to flip the negative stuff. The average, that you are taking is a sinewave where all of the negative excursions are now inverted. The resulting waveform all lies above the x-axis. That's what your taking the average of.

TRMS meters usually have the ability to include the DC part. e.g. TRMS AC or TRMS (AC+DC)

Technically the TRMS of 12 VDC is 12 Vrms and will be when you use the mode (DC+AC)