Comments on: How does jump starting a car work if current only flows when there’s a difference in voltage? Fri, 31 Mar 2017 07:13:36 +0000 hourly 1 By: chilljeet Fri, 22 May 2015 06:50:04 +0000 @Andreas, that was spot on.

By: Andreas Mon, 15 Oct 2012 16:32:54 +0000 Two minor points: The minor voltage difference is between a full and an empty car battery is not enough to charge the empty one in reasonable amount of time. Using your example and assuming an internal resistance of 25mOhm, you would have a charging current of ~7A (at that only while the voltage difference was that high), which would be a pretty meager charge current for a car battery which typically has 40Ah-50Ah.

Also, if you waited long enough then you would end up with two batteries that are half full (since the net capacity has to stay the same). The reason it works lies in the behaviour of the battery under load.

The voltage of an empty battery will break down if you try to extract a high current, partly because of the higher internal resistance, chemistry etc. Which is the reason why you might be able to still turn on your car (lights, etc.) even when you can't start it. The needed current for those small loads are small enough that the battery voltage stays high enough to still work.

When you now connect a second battery in parallel you can start the car with the empty battery it will use the energy provided by the full battery to start, running the motor and therefore the alternator will then load the empty battery.

As a sidenote: You don't want to start a car with a car that is not running, since you don't want to discharge the battery if the car that is working.