On Feb 27, 10:42 am, "Mariachi" wrote:
So you would actually have less voltage drop over the interal resistance of both
parallel batteries, therefore leaving more voltage left for whatever
series loads there are.
But this is only because your current demand on each battery is but a
fraction of what you would otherwise demand of a single battery. For
example, let's assume an "internal resistance" (in quotes because it's
not really "resistance" per se--it is simply modeled that way to aid
in understanding) of 0.5 Ohms per battery. Given a battery whose
resting voltage is 12.0 and a load drawing 2A, you would get an
internal voltage drop of:
Vi = I * R
= 2 * 0.5
= 1 Volt
So, drawing 2 Amperes from a 12 Volt battery would yield 11 Volts at
the battery terminals (12 V - 1 V).
Now put two identical batteries in parallel with the same load. The
current draw from each battery would be halved, so the voltage drop
across each battery would read 11.5 Volts (12 V - 0.5 V):
Vi = 2/2 * 0.5
= 1 * 0.5
= 0.5
See also:
http://en.wikipedia.org/wiki/Interna...ance#Batteries
As a side note, this internal resistance is the reason why a "dead
battery" can measure a full 12.8 Volts--a voltmeter presents an
"infinite" load on the battery and therefore draws no current.
However, when you put a load on the battery, the internal resitance
gobbles-up all the power and leaves you with next-to-nothing to power
your electrical devices.
Therefore you would actually measure a minute
increase in voltage across the two parallel batteries because the
total internal resistance of both batteries in parallel would be less
than one internal resistance of one battery.
Correct.
And I'm not arguing that the alternator (within its limit) supplies
all the current when the car is turned on, but why is that so? Just
curious...
I believe the answer there is simply "laziness." An alternator
delivers current much easier than a battery.
-dan