jumping battery to discharged battery, amp drain?

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Batteries are 12vdc standard car batteries
Lets say I have a discharged battery bank of two batteries down low enough they wont crank the engine.
I then connect a third fully charged battery to those 2, and the engine starts.
How much current is flowing from the fully charged battery to those two discharged batteries?
Is there too much internal resistance in those two discharged batteries that the current flow is very small?
 
A clamp on amp meter will tell you. No need to open the circuit. It depends on the state of the discharged batteries. I've seen car batteries when discharged present such a low resistance that an intelligent charger trips off when connected, and others which have such a high resistance they won't accept a charge. Chargers with short circuit protection can be fooled into refusing to charge a battery with low internal resistance. The charger reacts as though the clamps are shorted together and shuts down. A high resistance does not present a lot of danger, a low resistance and high current flow can cause a discharged battery to erupt. Better to use a trickle charger to bring the batteries back than to insert a fully charged battery into the circuit.
 
Lots of current. Not sure I'd ever want to bring up a fully dead battery this way, but I've jumped a few at 10-11V (?) (won't crank, lights real dim), and after maybe 5min of charging the dead vehicle will start. Gotta be 50A plus, maybe upwards of 100A (whatever full tilt on the alternator is, or close to it). Always makes me wonder about standing too close.
 
Ignore any chargers or alternators in this thread, just battery to battery only.

A charger though has a much higher output voltage so will be pumping in the amps.
A fully charged battery to a discharged battery, the voltage delta will be a lot smaller, so I think a lot less current flow.

I do have a clamp on DC amp meter I can try someday.
 
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There isn't enough voltage from a charged battery to effect much charging of a dead one. Since the battery in the car remains dead, when starting the full current of the starter has to come from the live battery at the other end of the cables.
 
Assuming that all the batteries are in good condition (not sulphated, no shorted cells, etc.), then the answer is "less than you might think."

If you connect a fully charged "12 V" battery (which is usually around 13-13.5 volts) to a discharged "12 V" battery (lets' say its pretty solidly dead at 9 V), there's only a 4-4.5 volt difference between the two. Now the "dead" battery will have a pretty low internal resistance, so there will be a surge of high current (10s of amps, limited by the resistance of the cables and connections mainly) but it will quickly fall off. That's because the "dead" battery's terminal voltage will come up to near 12 pretty much as soon as it begins accepting current (even though its still very depleted of energy), and the charged battery's terminal voltage will fall to around 12.5V pretty much as soon as you put a load on it. You can never really effectively charge one battery from another identical battery because of this- you need a charger that can push the dead battery's terminal voltage well above 13 volts in order to effectively push current into it and truly charge it. The ability of a dead battery to come up to a near-normal voltage and yet still be deeply discharged is a reflection of the lead-acid "charging threshhold" voltage, which you have to be above to effectively charge the battery. Its normally considered to be about 13.8 volts. You cannot do that with another battery, since even if its just recently charged, its open-circuit voltage will be below 13.8.
 
There's a famous segment on a Click and Clack program in which a woman calls to ask if her neighbor from eastern Europe is wise to be jumping a dead car without jumper cables by placing a charged battery right on top of the dead one, terminal to terminal. They suggested it was tempting Murphy a little too much.
 
Both batteries will strive to be at equilibrium with each other. That means that current will flow at roughly the value of:

DeltaV/impedance
 
I tested this once with a healthy fully charged 12.8v battery connected to a relatively healthy depleted 12.0v battery, and ~15 amps quickly tapered to 1.3amps or so.

Don't remember exact results but the amp flow was much less than I had expected.
 
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