NAC battery question

Thu Dec 14 02:51:45 EST 2006

```At 01:52 AM 12/14/2006, you wrote:
> > Let's say 1W average output and 50% efficiency.  1W@ 8 ohms is about
> > 0.33A (i^2r=P).
> > So 0.33A for a 700A-H battery sounds like 2100 hours x 50% = 1050 hours.
>
>Make that 525 hours for stereo.
>
>Plus the source device.  And the light bulbs in both, if any.
>
>And I think the efficiency factor might be off by an order of magmitude.
>   Still good for a day or two of continuous running.
>
>A more typical 400 AH battery that is a couple of years old should still
>hold up for a third of that, say 8-16 hours...

Way off, running a stock car stereo at average volumes can whack a battery
in less than 6 hours (I've done it.)

One thing not being considered is there is double loss on the efficiency,
there would be a loss at the inverter, and a loss at the receiver.  So to
get 1 watt of output, you would need 4 watts coming into the receiver, and
more than likely this would be a stereo set-up, so there would be 8 watts
coming into the inverter to get 2 watts of output.   Plus the wattage
required to run the receiver.  I'm guessing about 40 watts for this with
the inverter needing double this to make up for losses.  So I would guess
you would be pulling about 100 watts total to run the thing at average
volumes, 100 watts at 12V would be (I=P/V) = 8.25 amps being pulled out of
the battery (you'll being pulling juice out of the battery at 12V
regardless of the speaker impedance, the receiver/amp will step up the
voltage to higher than 12V.)  Furthermore, I think you are confusing Cold
Cranking Amps (between 400-700 CCA) with Amp-Hours (more typically 40-70
A-h for an average battery.)  Now take 8.25 amps out of a 40-70 amp-hour
battery and you get a running time of 5-8.5 hours.

George Selby

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