On 7 May 2007 09:54:11 -0700, Frank wrote:

On May 7, 12:30 pm, " wrote:
On May 7, 12:29?pm, "Robert11" wrote:



Hello:

Have been buying a bunch of those standard 9V batteries (a few different
brands) over the past few months, and have noticed that instead of 9V they
have been measuring out, with a VOM, at around 10.0 to 10.25. ?This is under
no load.

Anyone else notice this ?

Was wondering if this might be the cause of all the problems I've been
having with my fire alarms
giving 3 to 4 chirps, around once a day ?

It's probably the same type of chirp signal that the unit gives when the
batteries are low.

Thanks,
Bob

I doubt thats it, under load the voltage drops fast,

Have you blown the fire alarms out with compressed air....

They might be overheating


Don't believe voltage changes with load. Amperage is number to
deteriorate.

That would be an IDEAL voltage source (and, of course, ideal things
aren't).

A REAL voltage source behaves as an ideal voltage source in series
with a resistor. This resistance (the cell resistance, not a separate
component) is in series with the load. In this way, voltage at the
battery terminals will vary with load. As the battery discharges, its
internal resistance increases, causing a greater voltage drop when a
load is applied.

Electromotive potential determines voltage.but

Not "determines", "is". "Voltage" is a name for electromotive
potential. This explains the use of the symbol 'E' for voltage.

If you open up a standard
9 volt battery you will find 6 1.5 volt batteries in series.
Battery's voltage will depend on materials it is made from. Alkaline
batteries will vary from the standard 1.5 volts of old carbon/zinc
batteries.

True of fully charged cells with no load, although the numbers ARE
approximate.
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