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2nd battery in parallel

Brett said:
It will give you double the available current
Click to expand...

Physics and Ohm’s Law says unequivocally you are absolutely wrong.

I just showed this above.

It will not give you double the available current.

The amperage will remain the same as it was with just 1 battery, regardless if you string 100 batteries together in parallel.

Only the Amp-hr will be additive.
 
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Another issue with two lead acid batteries in parallel is that failure of a cell in one battery will cause the other battery to discharge into the failed battery. Depending upon the capacity and state of charge of the two batteries when the failure occurs, the failed cell may catastrophically fail with acid and debris making a mess of the bike (or the battery enclosure in the case of a UPS/emergency lighting system failure that I helped repair). The only safe way to connect batteries in parallel is to have overcurrent protection (fusing) between the batteries.

I have used lithium jumper packs to successfully start V8 SUV's, motorcycles, other vehicles. Some brands are good, others have insufficient capacity. I carry one if I have any concerns about the condition of the battery in the vehicle I am driving/riding.
 
scottmastrocinque said:
Physics and Ohm’s Law says unequivocally you are absolutely wrong.

I just showed this above.

It will not give you double the available current.

The amperage will remain the same as it was with just 1 battery, regardless if you string 100 batteries together in parallel.

Only the Amp-hr will be additive.
Click to expand...
It's much more subtle than this.

The current will be determined by the load and the internal resistance of the battery. It's this internal resistance that leads to the battery voltage dropping under load and is what you are testing when you put a battery on a load tester.
Adding a second (identical) battery effectivly halves the internal resistance the load sees. It won't double the current but will increase both the current and the voltage seen in operation. not easy to measure accurately without some careful work but pretty easy to simulate/calculate.
Now if you do this for a headlight bulb (50W or approx 4A) the current increase will be minimal, but if you do it for a starter motor (initial current draw maybe 100A+ before it starts to turn) then, like reducing any resistance in the circuit or ground connections, there will be a possibly significant increase in current (and hence force) which MIGHT (but only might) help with starting, it might also help stop the voltage dropping so far that the ECU on a modern vehicle will not shut down.
Physically bigger batteries, or two 6V in series, have more plate area and hence lower internal resistance leading to higher CCA ratings.

I suspect that all of the battery effects would be swamped by increasing wire sizes, minimising wire lengths and making sure all connections are as low a resistance as possible.
 
scottmastrocinque said:
No. You are completely incorrect.

Putting two voltage sources in parallel does not increase amperage in the circuit.

Ohm's law ( V=IR ), tells us the only way to increase current is to increase voltage, or decrease resistance. [Reaaranged for amperage (current) it becomes I=V/R. More current is only available by increasing V or reducing R. - Period.]

Putting two voltage sources in parallel doesn't do either of those things.

The current (amps) will remain exactly the same. What changes is the capacity (amp hr).

The voltage will also remain 12V.
Click to expand...
Cheers for this. Should have known this ...
 
The natural flow of electrons within the circuit will draw power from the good battery into the flat battery.

If you left them joined, they would eventually end up with even charge.

What you might be thinking is that if the flat battery is completely between the good battery and the circuit, that the dead battery will suck the charge in, but won't do it if it is off to one side.

Electricity doesn't work that way, it doesn't follow the path of least resistance and not down the other path, it spreads across the entire circuit.
 
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