It's not as drained as you think. Here's the problem: if you have a 12v battery and 0.015 Ω resistance in the cable and connectors you will get 9v to the starter if there is 200A draw at the starter. That's ok because the starter would be designed to operate at 9V (they still call it 12v) I can assure you zero percent chance the starter ever gets 12v while cranking.

Now: change nothing but the starter draw and use 400A. 0.015 Ω x 400A is now 6V loss on the wire and you will get only 6V to the starter when cranking.

I think the normal draw is 200-250A for E53 depending on number of cylinders and if it's diesel.

I've been seeing a lot of posts where people replace battery and alternator when it's really the starter that has failed.

To put things into perspective and put an exclamation point on keeping contacts clean: a high quality switch contact or plug-in pin socket is typically between 0.005 and 0.010 Ω. Basically this is why a loose battery terminal is a disabling event. It could be 0.1 Ω which means at 200A you would lose 20V (of of 12) and the high resistance contact if the battery could push the full current would be 20V · 200A = 4000W which would quickly destroy the contact.


Sent from my iPhone using Tapatalk Pro

__________________
2011 E70 • N55 (me)
2012 E70 • N63 (wife)