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Average, RMS and Power Factor made easy!



Original poster: "by way of Terry Fritz <twftesla-at-uswest-dot-net>" <free0076-at-flinders.edu.au>


Many people seem to have trouble with the ideas of RMS current, RMS
voltage, average power and Power Factor. I will take some time here to
explain what it is all about, and hopefully any remaining doubts will be
dispelled forever and we can get back to coiling (one of the finer points
of life! =)


RMS stands for Root Mean Squared, and it can be
applied to anything at all. It will only be useful in electronics to talk
about RMS current and voltages however and average power, however. RMS
energy doesn't make much sense but there would be applications where you
might want the average energy per pulse, for example, and you might choose
RMS rather than average. But I very much doubt that anybody on this list
has a use for talking about any RMS measurements that aren't voltage or
current.

The method of finding RMS is as follows. Say I have five numbers, I can
find the mean quite easily. Just add them and divide by five. Now lets say
I want to find the mean squared value. I first square all 5 numbers, then
I find out what the mean of the new list is. So I have the Mean Squared
value of all 5 numbers. Now lets say that I want to square-root the mean
squared value, I end up with the Root Mean Squared value, or RMS value (of
the 5 numbers). If the mean of the values is equal to 0, ie some numbers
are positive, some negative, and they add up to 0, then the RMS value
equals the standard deviation as found in statistics.

Talking about a list of numbers is fine, but doesn't really apply to a
continuous function of time, such as voltage or current. I will get to
that as soon as I explain why we even use RMS.

Lets consider an AC waveform. Assume that it is periodic, like the mains
voltage. If I connect a heater and turn it on, a certain power is
dissipated. If I then use a DC supply and find the voltage that produces
exactly the same average heating (and at 50/60Hz you wouldn't notice that
the power is fluctuating) then I can say that the mains waveform has the
same heating effect as that voltage, and that's the basis of why we use
RMS. If we consider only the _average_ power dissipation, we can quote it
as say 2400W.

So how do we find the DC equivalent voltage? Lets say that I have a
resistance hooked up to a voltage or a current source (could be AC or DC).
Then the equation for instantaneous power is (i.e. at any instant in the
cycle):

P = i^2 x R   or   P = v^2 / R

Note that the instantaneous power depends on the current or voltage
squared. If we took an average of the power, since we want average power
of course, then we would get either the Mean Squared value of current,
multiplied by the resistance, or the Mean Squared value of voltage, then
divide by the resistance. Either way we get the average power. In a DC
system the Mean Squared value is just the DC value squared, since it never
changes. So we have a way of finding the DC equivalent. If instead of
finding the Mean Squared value of voltage or current and comparing it to
the DC value squared, we can just square root the whole thing. So we see
that the DC equivalent of either a voltage or current waveform is just the
RMS value of that waveform.

Now it should be clear why we talk about RMS voltage or current, we can
use it in Ohm's law, we can use it to find power, we can use it where we
like as if it were a DC circuit. But only if we are talking about
resistors. The danger is that many people don't know that when you
multiply RMS voltage by RMS current you _only_ get the average power _if_
the voltage and current waveforms are proportional to each other at each
instant in time. In other words, the two waveforms must be exactly in
phase, like in the case of a resistor where V = I x R shows that at every
instant in time V and I must be proportional by the constant R.

If you have inductors and capacitors the current and voltage are not
necessarily in phase, one signal lags behind the other. Then the actual
average power will be lower than the value you get by multiplying RMS
current by RMS voltage. It turns out that the formula for real power in an
AC circuit is:

P = V I cos(theta)                 <--- theta is often replaced with phi

The extra cos(theta) bit is called the power factor, or PF. It can vary
between 1 and -1, where 1 is called unity power factor and corresponds to
a purely resistive load. Why -1? If we were talking about a source of
energy like the wall socket, we might like to talk about a negative power
like -2400W in the heater example, because it is delivering power. The
cos(theta) term can handle that too, since a negative answer comes out if
you use an angle between 90 and 270 degrees. By the way, the angle
represents the phase difference. Just imagine the whole cycle being 360
degrees, then a quarter of a cycle out of phase (like a capacitor or
inductor) would give 90 degrees, or a PF of 0. Note that in that case NO
power flows (if the capacitor or inductor were perfect with no
resistance). Even though we have our regular voltage present across the
thing and a measurable current through it, NO POWER FLOWS!!

Note in the above example you might still get charged for the power you
aren't using... The metering systems that I have actually seen in my own
area won't compensate for the power factor, and I doubt that yours does
either, they just measure the current and charge you for it. So if you
draw 10A and your capacitor isn't getting warm you probably are still
paying for it.

That's what PF correction is all about. If you run your coil at a PF
that's less than unity, you draw more current than you need to. In the
case of a coil, it's usually an inductive load so the answer is to place a
capacitor in parallel with the circuit and keep changing the capacitance
until the capacitance best cancels the inductance in your coil (mainly
transformer inductance I suspect). The result is less current drawn from
the mains and maybe a little lower bill. It may be that you can actually
get more power into the coil if you were close to blowing a fuse, since
the current drawn has gone down. It's as if the capacitor supplies the
extra current needed by the inductance of the coil circuit. Remember that
the power factor of the current delivered by the capacitor to the coil is
close to zero and it doesn't actually supply any power, just the current
that the inductance was planning to drain from the wall.


I hope that many people had a good read and maybe some fun too =)


Darren Freeman


PS I'm studying Electrical/Electronic Engineering and everything I've said
I'm 100% confident of..