Carultch
Senior Member
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- Massachusetts
I don't follow where you got that idea from.
The power waveform in your example still is a sine wave, just with an offest, phase difference, and twice the frequency as the Volts & Amps.
Defining Average Power
- Thread starter dnem
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I don't follow where you got that idea from.
The power waveform in your example still is a sine wave, just with an offest, phase difference, and twice the frequency as the Volts & Amps.
gar
Senior Member
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161115-0849 EST
dnem:
1 V * 1 A = 1 W independent of the type of load.
If you have a constant amplitude sine wave voltage source with a peak voltage of V, a resistive load of R ohm, and over an integral number of half cycles, then you can use the equation Vrms * Irms to obtain the average power over said number of half cycles. You can not just plug numbers into equations and necessarily expect to get a useful answer. You have to understand the definitions and restrictions on the components of the equation.
The instantaneous power for the same resistive load is V ( sin wt ) * (V/R) ( sin wt ) = V^2/R ( 1 - cos 2 wt ) / 2 for any instant of time. Where w is in radians and = 2*Pi*f with f in Hz and t in seconds. 2*Pi radians = 360 degrees. See https://en.wikibooks.org/wiki/Trigonometry/Graph_of_Sine_Squared . Note that the instantaneous power is a double frequency cosine wave with an average value over an integral number of cycles of 1/2 its peak value.
Thus, average power over N half cycles for a resistive load is V*I/2 where V and I are the peak values of in phase sine waves.
The interesting aspect of an RMS measurement is that for a resistive load you can use the same equation in an AC circuit as you would in a DC circuit to calculate power dissipation in the load, that is, heating effect. I can take an RMS voltage measurement, where the RMS averaging is over an adequate time, of a gaussian noise source and determine the power dissipation in a resistor. This is possible even though a gaussian signal has no periodic waveform.
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dnem:
For instantaneous values
1 V * 1 A = 1 W independent of the type of load.
If you have a constant amplitude sine wave voltage source with a peak voltage of V, a resistive load of R ohm, and over an integral number of half cycles, then you can use the equation Vrms * Irms to obtain the average power over said number of half cycles. You can not just plug numbers into equations and necessarily expect to get a useful answer. You have to understand the definitions and restrictions on the components of the equation.
The instantaneous power for the same resistive load is V ( sin wt ) * (V/R) ( sin wt ) = V^2/R ( 1 - cos 2 wt ) / 2 for any instant of time. Where w is in radians and = 2*Pi*f with f in Hz and t in seconds. 2*Pi radians = 360 degrees. See https://en.wikibooks.org/wiki/Trigonometry/Graph_of_Sine_Squared . Note that the instantaneous power is a double frequency cosine wave with an average value over an integral number of cycles of 1/2 its peak value.
Thus, average power over N half cycles for a resistive load is V*I/2 where V and I are the peak values of in phase sine waves.
The interesting aspect of an RMS measurement is that for a resistive load you can use the same equation in an AC circuit as you would in a DC circuit to calculate power dissipation in the load, that is, heating effect. I can take an RMS voltage measurement, where the RMS averaging is over an adequate time, of a gaussian noise source and determine the power dissipation in a resistor. This is possible even though a gaussian signal has no periodic waveform.
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winnie
Senior Member
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- Springfield, MA, USA
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- Electric motor research
I think that the first thing to remember is that the whole concept of using an 'average' is that you are throwing data away in order to better see the forest for the trees.
To describe an AC waveform perfectly, you would need an infinitely long list of numbers, describing the voltage at every instant in time. Such a perfect description would well be meaningless if all you wanted to know was how well a lamp would light up.
So we come up with a set of calculations which reduce this huge pile of data to a few numbers that better describe the whole picture...but to get to this we need to throw away the supposedly irrelevant details.
There happen to be different 'measures of central tendency' which we calculate using different equations...and each has its place where it makes sense.
For example, the common 'average' where you sum up your set of values and divide by the number of values. You can extend this concept to continuous functions, and you will find that the _average_ value of a perfect sine wave, taken over a full cycle, is _zero_. Not particularly useful if you want to know how well the bulb will light, but actually a very useful measurement if you are concerned about transformer saturation.
When you apply a voltage to a resistive load, the power delivered to the load is proportional to the square of the voltage. Because of this, the 'average' most used for voltage measurements is the 'RMS' average. You start with your voltage values, square them, take the common average of all these squared values, and then take the square root of this common average.
When power is delivered to a load is not continuous, generally the common average is used to describe the power delivered over time.
So we measure RMS current and RMS voltage, but mean (common average) power. Assuming a resistive load, RMS current * RMS voltage gives average power.
If the load is not resistive, or the waveforms are not perfect sinusoids, then the information thrown away to generate the various averages might actually be relevant, and you have a choice; either you walk away from the averages or you have to add additional terms to describe the missing information.
For example, you can use RMS voltage and RMS current plus 'power factor' to deal with reactive loads. Or you can use RMS voltage plus 'crest factor' to deal with non-sinusoidal waveforms.
-Jon
To describe an AC waveform perfectly, you would need an infinitely long list of numbers, describing the voltage at every instant in time. Such a perfect description would well be meaningless if all you wanted to know was how well a lamp would light up.
So we come up with a set of calculations which reduce this huge pile of data to a few numbers that better describe the whole picture...but to get to this we need to throw away the supposedly irrelevant details.
There happen to be different 'measures of central tendency' which we calculate using different equations...and each has its place where it makes sense.
For example, the common 'average' where you sum up your set of values and divide by the number of values. You can extend this concept to continuous functions, and you will find that the _average_ value of a perfect sine wave, taken over a full cycle, is _zero_. Not particularly useful if you want to know how well the bulb will light, but actually a very useful measurement if you are concerned about transformer saturation.
When you apply a voltage to a resistive load, the power delivered to the load is proportional to the square of the voltage. Because of this, the 'average' most used for voltage measurements is the 'RMS' average. You start with your voltage values, square them, take the common average of all these squared values, and then take the square root of this common average.
When power is delivered to a load is not continuous, generally the common average is used to describe the power delivered over time.
So we measure RMS current and RMS voltage, but mean (common average) power. Assuming a resistive load, RMS current * RMS voltage gives average power.
If the load is not resistive, or the waveforms are not perfect sinusoids, then the information thrown away to generate the various averages might actually be relevant, and you have a choice; either you walk away from the averages or you have to add additional terms to describe the missing information.
For example, you can use RMS voltage and RMS current plus 'power factor' to deal with reactive loads. Or you can use RMS voltage plus 'crest factor' to deal with non-sinusoidal waveforms.
-Jon
That's why I posted that its been a while, and please correct me if I wrong.
The power waveform is actually the square of a sine wave, since its a voltage X current. (Or you could look it as I squared times R, or even V squared/ R). All give you the same thing- a sine wave that is squared.
I was thinking the square of a sine wave would have a slightly different than a pure sine wave. But after your post, and thinking about it some more, I found sin2 x = (1-cos(2x)/2), so sure enough, the square of a sine wave is still a sine wave.
And there is your double frequency and offset right there in the identity. cos 2x is the double freq., and 1-cos is the offset.
Not true
if you use only the real components of V and I you do not get accurate P
the complex portion of the product of j gives you a real number, ie, power
as was his assertion
if you understood my previous post you would not have made your incorrect statement
Yes true.
If you understood........nah, I won't go there.
If you let:
V = v1 + v2j
I = i1 + i2j, I*= i1 - i2j
P does not equal v1 x i1
S=v1xi1 - (v1xi2)j + (v2xi1)j + (v2xi2)
therefore
P=real components v1xi1 + reactive components v2xi2 (j x j = -1)
Q = [(v2xi1) - (v1xv2)]j comprised of both real and reactive components
so v1 x i1 (real v and i components alone) do not describe Power
Show us
mathematically
not with a silly 'power is power' nugget lol
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Reactive components do not consume or dissipate power so they are our of the equation.
I thought you might have known that.
really? I just proved otherwise
P = v x i cos ang
ang requires the real AND reactive components to determine
ang = arctan reactive/real
can't determine P without knowing the reactive AND real components
absurd lol
The product of real i and real v not equal P
mivey
Senior Member
I think the problem is that in post #6 you called them real and reactive components instead of real and imaginary parts. They are not equivalent terms and that is where you went sideways.it is not I who lack understanding
or the self awareness to KNOW I don't understand
The real and reactive V&I are both sinusoidal and thus both have real and imaginary parts.
Hope that helps.
for electrical power they are equivilent
the 'imaginary' component in the product of v x i is the reactive power
you can't describe P with only the real components of v and i
the real and imaginary of v or i determine phase ang
the difference the power factor
Yes, you can and should.
mivey
Senior Member
I agree.
I just don't agree that the imaginary portion of V or I is called the reactive portion. It is confusing the issue and Besoeker is thus correct on a technicality but he would agree of course that your previous proof in post #27 is mathematically sound (if we ignore the wrong terminology).
wrong
assume
v = 1 + xj
i = 2 + yj
x NE Y NE 0
what is the P?
should be easy you know real v and i
if
v = 1 + 1j
z = 2 + 4j
i = ?
does the i have a reactive (not imaginary component
v = 1 + 0j
z = 2 + 0j
i = ?
any reactive component?
v = 1 + 3 j
z = 2 + 4j
i = ?
does i have a reactive component?
acting on X = 4j ohm
and real or active on R = 2 ohm???
if the load z is 0 + 10j and v = 100 + 0j
is the current reactive or imaginary? Not 'real' or active for certain
will an amp clamp 'see' it?
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- Placerville, CA, USA
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- Retired PV System Designer
I interpret the bold text as trying to say that the complex portions will contribute to the real part of the complex product. The complex product is another way of getting the same number as the dot product of the two vectors involved.
If you base your zero phase point on the incoming voltage waveform, then the voltage waveform will have only a real component.
Once you have established that, then the current will be at some phase angle relative to the voltage, and only the real part of the complex number that represents the current will contribute to the power.
If you set your zero phase reference point so that both voltage and current are imaginary, then indeed the imaginary part will have to be taken into consideration when calculating real power.
mivey
Senior Member
Are you having fun yanking his chain or
what? Slow news day?
V = 1 + j2
I = 7 + j9
S = P + Q = V x I* = (1 + j2) x (7 - j9) = 25 + j5
P = 25 W
Q = 5 var
except for your understanding of it
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