Sean Breheny wrote: > On Thu, May 17, 2012 at 2:21 PM, Spehro Pefhany wrot= e: > > The current taken by the charging battery isn't important, because > > the net flow is OUT of the 'dead' battery when the engine is being > > cranked. Unless the 'dead' battery is internally knackered (shorted > > cell etc), it will have an OPEN CIRCUIT voltage that is more than 10V. > > It's more the internal resistance that goes up as the battery discharge= s. >=20 > I don't think this is always the case. You are assuming that the > cranking current will be much greater than the battery charge current. No, you've missed Spehro's point: The drop in the cables due to the crankin= g current will cause the voltage at the dead battery to drop well below its open-circuit voltage, so it will not be charging at all. In other words, the combination of the rescue vehicle plus the jumper cables is anything BUT "a stiff 13V voltage source", as you put it. -- Dave Tweed --=20 http://www.piclist.com PIC/SX FAQ & list archive View/change your membership options at http://mailman.mit.edu/mailman/listinfo/piclist .