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I recently bought a van that has a starting and an auxillary battery, who ever installed it didn't isolate it from the starting battery. I'm planning on installing a simple switch to isolate it, but I need to know how many amps it needs to be rated for.

How can I find this out? I have a multi- meter, but worry that the amperage may be high enough to blow it even unfused (it says 10A max) if I connect it in series between the aux battery and the incoming + lead.

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  • \$\begingroup\$ Use a Guest battery switch. \$\endgroup\$ – mkeith Aug 6 '17 at 21:48
  • \$\begingroup\$ You may want to consider a relay which connects the aux battery to the car's system for charging when the engine is running and disconnects when the engine stops to save the starter battery from discharging. \$\endgroup\$ – JimmyB Aug 7 '17 at 11:13
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You are right to be cautious. The big lead-acid batteries can give out hundreds of amps.

enter image description here

Figure 1. Battery isolator or master switch.

A battery isolator is made for this task. The battery negative lead is disconnected from the battery, the end with the bolt slid onto the post and the original lead-clamp connected to the other end. You can just see in the photo the insulating layer peeping out at the rear terminal. Screwing down the green knob connects both halves together with a large contact area to prevent heating.

Your local auto parts supplier will have them.

Back to the question:

I have a multi-meter, but worry that the amperage may be high enough to blow it even unfused (it says 10A max) if I connect it in series between the aux battery and the incoming + lead.

You can't with the equipment you've got. With the isolator suggested you won't have to.

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  • \$\begingroup\$ Doesn't answer the question but solves the problem... \$\endgroup\$ – Passerby Aug 6 '17 at 22:00
  • \$\begingroup\$ Thanks. I often cut and past the whole original post into my answer and break it into quotes and answers so I don't forget what I'm doing. Fixed, I think. \$\endgroup\$ – Transistor Aug 6 '17 at 22:14
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The weaker (AUX?) of the two batteries determines the ESR ( in micro/milliohms) and using the CCA rating is a rough measure of ESR rated @7.5V from full charge @12.5V or a 5V drop.

e.g. a 10 mohm 12V battery with 12.5V no load and 11.5V with a load is expected to draw 1V/0.01Ohm= 100 Amp. THe same is true of a battery switch with a 1 V drop and the total ESR of the 2 batteries is 10mA. It is expected to switch 100A until the charges equalize.

A Battery is expected to be in the range of 10k Farads or more with low ESR, but there is also a secondary charge layer (C2*ESR2) which gives rise to the memory effects with a certain time constant T2 in normal operating range. It gets complicated, but this is your main storage with a long time constant and the low ESR is short term storage. THis is why if you fail to start a car and the voltge drops down momentarily, it resumes to the previously voltage due to the bulk larger C which has larger ESR. THis larger C2*ESR2 affects your Ah capacity but not your CCA rating although they are somewhat related.

But if you keep the batteries in their normal range of 11.5 to 12.5 when not being charged, the max Current surge should be determined by the weaker of the two batteries for CCA rating. THis may be your AUX battery.

Thus Isw/1V(drop) =CCA/5V(drop) or Isw= CCA/5V

So if CCA is 500A , the switch ought to be rated for 100A surge. You can make it cheap Ammeters rated for this which use heavy copper shunts and sensitive 50kOhm/V galvometers but often sold for 50~100$

NEVER Use a 10A DMM unless you current limit with a tungsten headlamp bulb which acts a a constant current PTC.

If anything use a 55W Headlamp in series to limit the charge current so you can use a smaller A Rated switch or 25A Relay.

schematic

simulate this circuit – Schematic created using CircuitLab

Here you can use a 5A switch

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    \$\begingroup\$ Given the technical level of the question I think you might have lost the OP at 'ESR' on the first line. \$\endgroup\$ – Transistor Aug 6 '17 at 22:44
  • \$\begingroup\$ given he is a member of the physics group, I expect more from his learning skills. \$\endgroup\$ – Sunnyskyguy EE75 Aug 6 '17 at 23:19
  • \$\begingroup\$ at least I EXPLAINED THE THEORY AND didn't turn it into a shopping answer \$\endgroup\$ – Sunnyskyguy EE75 Aug 6 '17 at 23:33
  • \$\begingroup\$ I like my old headlamp bulb solution.... \$\endgroup\$ – Sunnyskyguy EE75 Aug 6 '17 at 23:41
  • \$\begingroup\$ Thanks, Tony. I try not to give shopping answers other than to give the correct search term when the OP doesn't know. I've used the light bulb trick for current limiting on suspect mains equipment but hadn't considered leaving it in-circuit like this. That might do the job. \$\endgroup\$ – Transistor Aug 7 '17 at 8:12

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