Clarification on RMS and frequency

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

I've read lots about RMS and "almost" understood. I think your description was the best I've read. Thanks.

only

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

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

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

Graham

frequency

and

take

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

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

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

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.

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 what about over frequency?

cheers, RR

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

Graham

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.

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)

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).

supply

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.

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.

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.