Tom=E1s =D3 h=C9ilidhe wrote:
> Olin Lathrop wrote:
>   =

>>> Think about it.  If a perfect battery with 0 internal resistance was ra=
ted
>>> for 1AH and you put 10ohms in series with it, how many AH can the batte=
ry
>>> and resistor deliver to the load?
>>>   =

>>>       =

> Depends on the resistance/impedance of the load. But if I have my device =

> and I know it has an "average resistance" of 100 ohms, then I can do the =

> maths to work it out.

OK so let's say our battery is as follows:
    Voltage =3D 3 V
    Capacity =3D 1 Ahr =3D 3 Whr
    Internal resistance =3D none (it's the perfect battery)

And our circuit is as follows:
    Resistor in series =3D 10 ohms
    Resistance of load =3D 100 ohms

The total energy in the battery is 3 Whr   or   10.8 kJ.

Looking at the ratio of load resistance to "series resistance", we can =

see that ten elevenths of the battery's energy will be expended in the =

load, while one eleventh will be expended in the "series resistor".

Therefore, we've got a total of (10.8 / 11 * 10) for the actual load, =

which is 9.818 kJ (or  2.727 Whr).

The voltage across the load will be (3 / 11 * 10), which is equal to =

2.727 V.

Therefore the Ahr for the battery, (i.e the amount of coulombs it has =

available), is 1 Ahr (but this is figure is misleading if you don't know =

what voltage you have across the load).

So scrap my original idea of Ahr and internal resistance. I'd prefer Whr =

and internal resistance. But anyway this is irrelevant now because I've =

been told that a battery's internal resistance rises as more current is =

drawn from it.



-- =

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