# Clarification on RMS and frequency

Discussion in 'Electronic Basics' started by RR, Dec 4, 2005.

1. ### RRGuest

Hi,

As I understand it, RMS is calculated on the area under the sine or cosine
wave for the AC supply and the peak voltage.

It happens to work out at 0.707 of peak, as I understand, for mains supply.

The frequency (50Hz or 60Hz, 100Hz, or 1Hz) makes no difference to this if
the peak is unchanging. Is that correct?

I've attempted to follow the calculations here:
http://www.alpharubicon.com/altenergy/understandingAC.htm
and the frequency seems to always reduce to a factor of 0.5.

So, why does an appliance designed for 240volts 50Hz care whether you supply
240volts 60Hz or 240volts 100Hz, for that matter?

Or does merely depend on the type of appliance? I can see how a motor that
relies on the frequency would care, but some appliances should not care
less.

What have I missed?

tia,
RR

2. ### BobGuest

RMS is easy to understand once you've learned what it's supposed to
represent.

The purpose of the RMS value is to assign a single value to a time-varying
waveform, and that value is the value that would produce the same power,
into resistor.

For example, if you apply a steady 2 volts to a 5 ohm resistor then that
resistor will dissipate 4/5 watts, because P=V^2/R. Square the voltage,
divide by the resistance, and you have the power.

What if you have a time-varying voltage into that 5 ohm resistor and you
wanted to find the "effective" voltage (i.e., a single value for the voltage
that would result if it were a steady voltage being applied)? You could find
the effective voltage two ways:

FIrst, apply the varying voltage to the resistor, and then:

Measure the power being dissipated (maybe using a calorimeter), then
rearrange P=V^2/R and solve for V.

or

By knowing what the varying waveform was (mathematically or graphically),
you first square the waveform (each point on the voltage vs. time curve),
then find the 'mean' of the squared waveform, and finally take the square
root of that mean. The reason you need to square the voltage vs. time
waveform, again, is because the power is proportional to the square of the
voltage. This 'root of the mean of the square' technique will give you the
effective voltage -- aka, the RMS voltage (root, mean, square).

As for your second question, the frequency matters, to some appliances, only
if they have some frequency-sensitive components:

motors
transformers
capacitors
inductors
and so on.

Bob

3. ### John GGuest

An appliance with NO inductance or capacitance will not care. i.e. a
simple heater or a simple lamp.

But most other appliances are more complicated and then raising or in
fact lowering the frequency will have undesirable effects.

4. ### RRGuest

was the best I've read. Thanks.
So, that covers just about everything except an electric bar heater (incl
kettles, ovens, etc.), as most appliances would have one or more of those
items.

Switch mode power supplies (power packs) have inductors and capacitor and
transformers, but they are typically "smart" enough to deal with a range of
inputs. Correct?

cheers,
RR

5. ### Pooh BearGuest

With regard to frequency, if the appliance has a conventional line frequency
power transformer, the magnetising current is frequency related. If the
transformer in a 60Hz piece of equipment wasn't also designed to operate
correctly at 50Hz too it may have problems ( the core may indeed saturate and
the transformer burn out - this tends to be a marginal issue and may even take
hours to happen ).

Graham

6. ### Pooh BearGuest

Yes. Switch mode supplies don't care about the line frequency.

Graham

7. ### RRGuest

Thanks for that.

I'm in the opposite situation. My appliances are designed for 50Hz
(Australia) and my backup generator is outputting 51.9 HZ.

I need to increase it's voltage output (it's only giving me 215V instead of
230-240V) and that will increase the frequency to about 54-55Hz. (It doesn't
have an AVR).

Do power transformers and motors get damaged if they have a frequency fed in
that's a little too high?

I was told that low frequency is a problem but slightly higher frequency is
OK.

tia,
RR

8. ### John GGuest

I think my signature line may be pertinent.

A few cycles here or there will not generally matter except if there is
a speed dependant motor eg. a synchronous clock ( not common to-day)

9. ### Jasen BettsGuest

the clock runs fast as do any other synchronous motors etc...

some devices aren't effected. (lightbulbs, radiant heaters, etc...)
transformers designed for 60Hz run hot at 50Hz, going the other way is less
troublesome.

Bye.
Jasen

10. ### Andrew HolmeGuest

Some equipment is designed to operate on anything between 50 and 60Hz, to
suit both the US and European standards. For example, I have a wireless set
in front of me with a plate screwed to the front which says: "48 - 62 ~" In
the US, this would run off 110V at 60Hz, but I'm feeding it 110V at 50Hz
here in the UK, and it works fine - although the transfomer does hum a bit.

As a general rule, if you're not sure about the specification of an
appliance, I think higher frequency would be safer because it produces a
lower magnetising current in the transformer.

Unless the equipment uses the mains frequency for speed control or timing, I
can't see how a few cycles here or there would make a lot of difference.

11. ### RRGuest

Lots of different problems, but all related to similar goals.

We're moving to a large block of land, and thinking of possibly not
connecting to the grid.

So, solar and wind power into batteries then back out through inverter(s).
Hence my question titled "Switch between power sources".

But, you can't run washing machines from batteries (at least, it doesn't
seem very practical).

So, that's where the generator comes in....

Of course, a few cycles ain't going to matter, but we might be talking 10%
above the appliance's rating (55Hz instead of 50Hz). Hence, my concern and
question.

Pooh Bear said that being under-frequency will tend to burn out things. But

cheers,
RR

12. ### Pooh BearGuest

Yup, basically that's right. You should have no trouble.

Graham

13. ### Bob MyersGuest

Sort of, but you'll get a better idea of what it's really all about simply
by going through what "RMS" stands for in the first place.

The problem is one of determining the "effective" voltage, current, or
whatever of an alternating source; in other words - and to use the most
popular example - if I pass an AC current through a resistor, how much
power is dissipated in that resistor? How can I compare AC to DC in
this sense?

You clearly can't use the peak voltage or current - the waveform isn't
at the peak but for an instant, so obviously doing a power or some
other such calculation based on that value would be wrong. The next
idea would probably be to try to find the "average" value of the
waveform, but that winds up even worse - if you average any "pure"
AC (meaning that it is symmetrical, regardless of the form of the wave,
and spends as much time above zero as below), you get a result of
exactly zero. That's obviously not right, either, since the resistor DOES
heat up.

So instead, we start by squaring the function that describes the AC
waveform; if you square such a wave, everything winds up above
zero, right? Then find the average, or mean, of the squared waveform
(so now you have a constant value), and to correct for the squaring
operation you did in the first place, find the square root of that average
value.

In short, you are finding the (square) Root of the Mean of the Square.
"RMS," see?

It happens to work out to 0.707 peak for a pure sinusoid; other
waveforms wind up with different fractions of the peak value, and
this is really only a shortcut which can be applied to those examples
where one of these "regular" waveforms is in question. But the
above process - square the wave, average it, and take the square
root of the result - applies to all.

Obviously, frequency does not enter into this at all - RMS is 0.707
of peak for ANY sinusoid, regardless of frequency.
The final result is just a numeric value - it has no frequency (that
is the result of averaging the squared waveform - an average has
no frequency, since it's a constant value, right?). You seem to get
into some unexpected frequencies during all of this because the
squared waveform is itself a periodic wave with a frequency 2X
the original (at least for anything but a 50% duty-cycle square
wave).

When an appliance "cares" about the line frequency, one of two
things are generally at issue:

1. The appliance relies on the frequency of the line in order to
run at the proper speed - e.g., the motor in an electric clock.

2. The appliance contains components (typically, transformers or
similar magnetics) which are designed to operate at one frequency,
and will be less efficient at others.

Bob M.

14. ### John GGuest

Bob,
Where you a trained teacher?
Thats one of the best replies I have seen here to a very basic question.

Sometimes the posters get carried away with their own imagined expertise
and launch off into long explanations that are technically way over the
head of the original poster and sometimes not even accurate.

15. ### Bob MyersGuest

First of all, thanks for the nice comment - but no, I was never exactly
trained
as a teacher, other than on-the-job experience. I spent several years as a
part-time electronics instructor at a local community college, and my job
today often involves tutorial sorts of presentations. It's the simplest
stuff
that can often be the most challenging to try to get across, though, that's
for
sure.
Unfortunately, I'd have to agree with you there - too often, we DO seem
to forget that this is sci.electronics.BASICS, not
sci.prove-your-understanding
-of-electronics-to-the-Nth-decimal-place. That sort of thing is often
counter-productive, anyway - far better to get an almost-intuitive,
"back-of-the-envelope" grasp on what's going on than to shoot for the
most precise, state-of-the-art answer possible first time out.

Bob M.