Need to disconnect battery before charging?1

The message from "Clive George" contains these words:

But that's not a simple case - in terms of a generator that's a very unusual case indeed!

For AC waveforms symmetrical about zero, RMS gives you the equivalent DC voltage for power calculations etc. However, if someone's already recitifed the signal, so you've got repeated halves of a sine wave, all on the same side of zero, there's no point in taking the RMS, 'cos it'd be the same as the average.

No - the RMS is still different to the average.

For a rectified AC (sinusoidal) waveform, peaking at 1v, RMS is 0.707v and simple average is 0.637v. So if you're trying to do power calculations, you'll still want to use the RMS.

cheers, clive

A full-wave rectified signal has exactly the same RMS as its unrectified equivalent, but neither is the same as the mathematical average.

in news: snipped-for-privacy@davenoise.co.uk, "Dave Plowman (News)" gurgled :

It has as much to do with DC as with any other waveform. Of course, the RMS factor for DC is unity, but that doesn't mean that it is invalid, any more than it is for a bipolar square wave.

IMHO, you're just being pedantic. Why call something a root mean square if the mean is unity? Words for words sake.

The thing everyone seems to have missed going back to the charger in question is that the RMS current output quoted is higher than the actual one - near double. Just how can this be justified?

BTW, I've measured the actual current output using a peak hold Fluke, and even into a good but discharged battery it didn't quite make the claimed 'DC' output.

MOre to the point , the output current from a battery charger isn't sinusoidal.

in news: snipped-for-privacy@davenoise.co.uk, "Dave Plowman (News)" slurred :

Yes, I was, but you seemed to be suggesting that there was something explicitly _wrong_ about quoting an RMS value for a DC, like they were measuring power in VA or something.

I didn't pick up on that at all from the earlier posts. Sounds like figure massaging on behalf of the manufacturers.

One possibility is that they are using HF/switched mode regulation of the output, in which case your fluke will almost certainly underread, and may suffer from subsampling/aliasing. I would imagine the output current would be quoted into a short - I assume your discharged battery was still low impedance. Did you try the meaurement with the fluke directly across the charger output in place of the battery?