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Draining the Battery

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Old 11-23-2002
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Draining the Battery

i just got to thinking about how much of a drain all the aftermarket parts have on the batt.. probably more than the alternator can supply

i mean, with:
???W OEM HU
300W RMS 1000W peak sub amp
100W x 2 headlights
100W x 2 foglamps
???W x 4 neon undercar lights
???W x 2 OEM parking lights
+ all the other OEM power draws

am i going to be able to start my car the next morning??? lol

i dont actually have all of this stuff right now, and short of buying it all and hooking up an ammeter, how can i tell if the alternator will be providing enough power?


also.. whats the advantage of an extra battery? where would i put it? and would it be hooked in series of parallel with the main batt?
Old 11-24-2002
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Another battery would just mean you have more current to draw from. You should be ok with that setup. Just don't leave all that stuff on for hours on end with the car off. I think there is only one way you can connect 2 batteries so I don't think you can do parallel. Technically batteries do have internal resistance in them but until the power of the battery becomes low the resistance can be negligable. If there is no resistance it wouldn't matter how you connected your batteries as long as they completed a circuit. If you ever getting dimming in your lights that's usually a good indication that you are drawing too much current from you battery. In that case just turn down your stereo or turn off some lights.
Old 11-24-2002
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[hr] think there is only one way you can connect 2 batteries so I don't think you can do parallel.[hr]
you are half right... there is only one way to connect. thing is, that way IS in parallel, unless you wanna run your car at 24V [IMG]i/expressions/face-icon-small-wink.gif[/IMG]

parallel=same voltage
series=same current

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[hr]i dont actually have all of this stuff right now, and short of buying it all and hooking up an ammeter, how can i tell if the alternator will be providing enough power?[hr]
do the math. Voltage=current * resistence. Power=Voltage * current

if i have a 12W light, running at 12 volts, it takes one amp.

**note:this rule does not apply directly to rms values

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[hr]Technically batteries do have internal resistance in them but until the power of the battery becomes low the resistance can be negligable[hr]
resistence stays the same all the time, it's the current that changes (see above equations). when the battery starts to drain
Old 11-25-2002
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[hr]resistence stays the same all the time, it's the current that changes (see above equations). when the battery starts to drain [hr]
Ah, but resistance does not stay the same the whole time. Here's an experiement to test this: Say you have 2 fully charged AA 1.5V batteries (3V total) and 2 almost drained 1.5V batteries... Make 2 seperate circuits connecting both pairs of batteries to a light bulb. If you have a voltmeter attached to both circuits, you will see the voltage decrease ever so slightly for the fully charged batteries but for the almost drained batteries the voltage will have a huge drop. Thus, resistance does not stay the same inside the battery all the time, but the current does change but that is due to in the increase in resistance.

I doubt anyone will actually experiment with this for proof (sadly to say I have already). Oh well... [IMG]i/expressions/face-icon-small-smile.gif[/IMG]
Old 11-25-2002
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2 batteries would also be too much draw for one stock tiny alternator.... you lose some power when you have the 2, youd need an isolator and all that crap, and it might actually cost you as much as a high output alternator
but for what im thinking, youll be fine with that setup
Old 11-25-2002
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you need to figure out what the amperage draw is. That sounds like you're not anywhere close to needing major electrical system upgrades.. whiterabbit is running 1860W RMS from stock electricals and having no problems.
Old 11-25-2002
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thanks for all the replys everyone...

i guess it looks like ill be ok without any type of upgrades
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