>>It is correct that "dead" batteries show almost full voltage. The reason
is
>>that usually they are measured when they don't have to supply any power.
>My reply is: How come I've tested hundreds (if not thousands) of "dead"
>batteries, ALL of which registered 0V (or thereabouts, I'm smart enough to
>realize that a 3V battery measuring about .5V is as close to dead as makes
>no nevermind). I've never had a "dead" battery measure any significant
>amount of voltage.
>Maybe I'm ignorant of something, but it seems to me that "dead" batteries
>will show as "dead" to a voltmeter when measured out of circuit.
Let's put the definition of when a battery is really dead aside for a
second...
If my memory hasn't left me completely, the interior schematics of a
battery look somewhat like that:
R
+ (internal) ------\/\/\/\/-------- + (outside)
- (internal) ---------------------- - (outside)
R is the interior resistance of the battery. In a new battery, the value
of R is very small, so not much voltage is "eaten up" inside the battery
due to R if you draw current. How much voltage is lost across R can be
calculated by good ol' Ohm's law: Voltage = resistance * current.
Due to chemical processes R increases while you discharge the battery. So
if you discharge with a constant current, more and more voltage is
lost across R and less and less voltage "arrives" at the batt's terminals.
If you do *not* draw current, though, no voltage is lost
across R and the internal voltage equals the external voltage.
As common DVM's have an input resistance of 10 MegOhms or more (which
won't draw any current worth mentioning from the battery), you'll measure
nearly full voltage from batteries that are more or less used up.
The above is, by the way, the reason why batteries that won't be able to
maintain sufficient voltage in high current devices like photo flashes
anymore will still maintain this voltage in low current devices like
remote controls or clocks.
>You can measure water pressure without
>running the tap in your sink, similarly you can measure voltage without a
>load.
Right. However, that doesn't make much sense, because if I hadn't a load,
I wouldn't need any voltage ;-))
This is why battery testers, as opposed to DVM's, have a built-in load
to simulate real-life conditions.
>as the battery is used more and more (the water level in the tower
>gets lower and lower), the voltage (pressure) drops in direct proportion.
>Eventually, there is not enough voltage (pressure) to power your device
>(like a water wheel).
As described, this is an unfit comparison I think. If you measure a battery
as dead, this is, at least from a physical point of view, not due to lacking
voltage inside, but due to above described R being infinitely large.
In that case the very little load of the DVM's interior resistance is still
large enough to have nearly all of the batt's voltage disappear across R.
>Please, illuminate me if I am misinformed.
As I'm talking a bit off the cuff, if some of you out there have more
in-depth knowledge please iluminate both of us if I'm talking nonsense.
Frank
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