Power = P = VI = I2R, V2/R

[Selva]

The power equations P = VI = I^2 R = V^2 / R work for dc signals and
ac sine signals if we use RMS values (peak / root2).

Now what if we have different shapes signals? Square wave is
obvious. What about triangle wave?

Is there something like RMS that can be used for different shapes of
signals?

 

[Hammy]

The power equations P = VI = I^2 R = V^2 / R work for dc signals and
ac sine signals if we use RMS values (peak / root2).

Now what if we have different shapes signals? Square wave is
obvious. What about triangle wave?

Is there something like RMS that can be used for different shapes of
signals?

Average, See here for typical waveforms.

http://www.daycounter.com/Calculators/RMS-Calculator.phtml

 

[Stephen J. Rush]

The power equations P = VI = I^2 R = V^2 / R work for dc signals and ac
sine signals if we use RMS values (peak / root2).

Now what if we have different shapes signals? Square wave is obvious.
What about triangle wave?

Is there something like RMS that can be used for different shapes of
signals?

RMS is valid for any waveform, but the ratio of RMS to peak voltage is
different for each waveform. For any repetitive waveform, there will be
a constant you can use, but that constant will have to be determined
mathmatically or experimentally. If you take a large number of voltage
samples, square each one (so the result is always positive), take the
arithmetic mean, then take the square root of the result, you get an
approximation to the true "root-mean-square" value, which is the limit of
that quantity as the number of samples increases toward infinity. For
simple waveforms, like triangles, you can use geometry. For complex
waves or nonrepetitive waves, you can measure the heat developed in a
precision resistor.

 

[whit3rd]

The power equations P = VI = I^2 R = V^2 / R  work for dc signals and
ac sine signals if we use RMS values (peak / root2).

Well, RMS values are NOT 'peak/root2'. Those are two
entirely different schemes for computing an average,
and while the numbers are EQUAL for sine waves,
the schemes are not the same.

P=VI requires not only RMS averages for V and I, and
simple average for P, but also a third condition,
that of 'unity power factor', to be correct for AC sinewave
signals. That 'unity power factor' is correct enough for electric
heaters and incandescent light bulbs, which are said
to be 'resistive' loads, but not for motors or lights on dimmer
switches.

 

[Chuck]

Well, RMS values are NOT 'peak/root2'. Those are two
entirely different schemes for computing an average,
and while the numbers are EQUAL for sine waves,
the schemes are not the same.

P=VI requires not only RMS averages for V and I, and
simple average for P, but also a third condition,
that of 'unity power factor', to be correct for AC sinewave
signals. That 'unity power factor' is correct enough for electric
heaters and incandescent light bulbs, which are said
to be 'resistive' loads, but not for motors or lights on dimmer
switches.

It may not be helpful to view rms values as averages. But by using rms
values of V or I, you can calculate the average (or equivalent )
power.

Also, if you take the product of the instantaneous voltage and the
instantaneous current, you get the power at that instant. Not the
average power or the rms power, but the instantaneous power. Always.
Regardless of the power factor. Regardless of whether the voltage or
current are sine waves or even DC.


Chuck

 

[[email protected]]

The power equations P = VI = I^2 R = V^2 / R  work for dc signals and
ac sine signals if we use RMS values (peak / root2).

Now what if we have different shapes signals?  Square wave is
obvious.  What about triangle wave?

Is there something like RMS that can be used for different shapes of
signals?

There are no jobs , ways to earn money , using this math .

You must actually have your hands on the circuit , and measure the
signals and
measure the outputs , to imagine how it works , then you can get a
job .
As you measure , you must imagine how the components function ,
based
on difference between inputs and the outs .
Those truely clever , will add the tiniest cap or resistor and vu
the output .
This load may be so large as to stop the circuit .
As i learned at B.B. Research in 1976 . I then asked the
Techtronics man
to sell me the cheaper scope probe ( P6006 , better , lower
loading , 7 pf) , til he could start
selling higher impedance probes ( 5000 ohm ) .

 

[Marra]

You can use integration to calculate the power of any signal waveform
you know the function of.