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tech / sci.electronics.repair / Re: Can somone explain WHY positive first when jumping a car battery?

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o Can somone explain WHY positive first when jumping a car battery?Charles Lucas

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Re: Can somone explain WHY positive first when jumping a car battery?

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Subject: Re: Can somone explain WHY positive first when jumping a car battery?
From: clsnowy...@gmail.com (Charles Lucas)
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 by: Charles Lucas - Sat, 4 Feb 2023 20:06 UTC

On Friday, February 3, 2023 at 4:49:25 PM UTC-6, Michael Trew wrote:
> On 1/30/2023 16:42, Carlos E.R. wrote:
> > On 2023-01-19 02:01, Rod Speed wrote:
> >> On Thu, 19 Jan 2023 04:54:10 +1100, John Robertson <j...@flippers.com>
> >> wrote:
> >>>
> >>> After the engine starts, then disconnect the negative lead that is
> >>> AWAY FROM THE BATTERY so any spark created at the disconnect point is
> >>> unlikely to cause the battery to explode if hydrogen gas was created.
> >>
> >> No, because the battery doesn't gas unless it is being charged when
> >> fully charged already and that won't be happening with either battery.
> >
> > The "bad" battery will start charging when connected, and there will be
> > no current limiting.
> The last time I had a really flat battery, and jumping it didn't work
> (perhaps the cables weren't heavy enough gauge), I started it with
> another battery installed (cables semi-tight pushed-on)... then with the
> alternator keeping it going, I disconnected that battery, and hooked up
> the flat battery. You could instantly hear the engine bog down from the
> alternator kicking into high-gear. This was a 1960's car, BTW, with a
> modern 12v battery. Perhaps this isn't the hottest idea with a new car
> using expensive electronics, but it works fine on older rigs with a good
> alternator.

Remember, this can be done only if what this man said is true and/or the
voltage regulation circuitry can handle it. I would try using a load resistor
Run to ground at the base of the circuit to handle the load and to prevent
overload. The resistor should be of adequate size (1/4 watt at so many
mega ohms) to absorb the energy.

Good Luck,

Charles Lucas

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