Lasse wrote:

Hi all! I've been told that the lead battery is a nonlinear
source. That must mean that it doesn't obet to
Ohms law. Correct ?

Technically, nothing obeys ohm's law perfectly. A lead acid battery
is pretty ohmic over some range of output current for brief periods of
time.

Example for the above question: I have a 12V battery, and
I have a resistor of 1000 ohms connected to it. I should
expect that the current in my circuit becomes:
12/1000=0.012 A. Does it ?

For a while, yes (to a good approximation). Eventually, the voltage
will droop and the current will be reduced in proportion ot the
voltage. If the resistor was a tenth of an ohm (and the current was
about 100 amps) the voltage would droop immediately, because the
internal resistance of the battery is effectively in series with the
load resistor, and the 12 volts is divided up between these two
resistances.

If not, what will the voltage over the resistor(an ohmic device) be?
Say that the lead battery injects 1Amp in the resistor.

The battery does not inject a given level of current, it applies a
voltage across the resistor and the resistor allows a certain current
to pass through.

According to Ohms law, the voltage over the resistor becomes:
U=R*I=1000*1=1000V. But it is connected to a 12V source!
Who wins(which voltage will be measured) ?
And why ?
Kindest regards,
Lasse

P.S. Could you look into a thread at electromagnetics that I've started.
There I ask for hint(or book) for a microscopic explanation to why
magnetic poles attract or repell. What is maxwells explanation by the way ?


--
John Popelish