Battery Disconnect Switch

[270S]

The reason the battery negative cable is disconnected on a car first is because if you "accidently" touch tools to ground when un-attaching the negative cable nothing happens (unless you touch the positive post). If you disconnect the positive first and touch ground (via battery supports metal case, wrench, ....) you will have a short and kaboom. If you disconnect the negative first and the during positive disconnect (typically battery replacement) you touch ground with the wrench, nothing should happen - you will just make the frame, ... a positive. If you have a short somewhere, then there will be a problem.


Regardless of either positive or negative cable removal first, care must be exercised to not have the wrench come in contact with both battery terminals by accident.


Do not confuse cable disconnect or attachment order procedure with the installation location of a disconnect switch. They are two different frankly unrelated scenarios.

[rockfordroo]

Originally Posted by Asrabbit

After watching the videos it looks like the negative side is the way to go. I do wonder however if the frame is used for a ground anywhere? I guess, there really would still not be a path back to the battery. A question I really thought would be easy, just does not seem all that easy. I guess I can go one way and always change it later on if I need to.

I know my brakes are grounded through the frame. Not sure about anything else, though.

Come to think of it, I'm not sure how the battery negative gets to the frame, but it must.

[MillerTime]

Originally Posted by 25FKS

The reason the battery negative cable is disconnected on a car first is because if you "accidently" touch tools to ground when un-attaching the negative cable nothing happens (unless you touch the positive post). If you disconnect the positive first and touch ground (via battery supports metal case, wrench, ....) you will have a short and kaboom.


Regardless of either positive or negative cable removal first, care must be exercised to not have the wrench come in contact with both battery terminals by accident.


Do not confuse cable disconnect or attachment order procedure with the installation location of a disconnect switch. They are two different frankly unrelated scenarios.

I understand, we are on the same side. I'm in the auto collision repair industry. But 99.99 % of vehicles dont have disconnect switches. Thats why i stated what i have.

I've also seen whole ground side wiring harnesses fried from not being able to handle the amperage. Most ground wires can only handle the REMAINING amp load AFTER a load draw. This is why when jumping a vehicle you need to connect the ground to the frame.
Also all the fuses are on the positive side on 12 volt systems. Polarity makes a big difference.

I know alot that last paragraph doesn't apply much to the topic at hand.....

[rockfordroo]

Originally Posted by MillerTime

Most ground wires can only handle the REMAINING amp load AFTER a load draw.

Don't understand this statement. You will get a VOLTAGE drop through a load or string of loads, but CURRENT flow (amps) is constant through any load or string of loads, so I don't understand what you mean by "remaining."

[wahoonc]

Ifn I recall my edumucation in DC correctly the electrons actually flow from the negative side TO the positive side. You can disconnect from either side and it will stop the flow of them pesky electrons. I used to own British cars with positive ground electrical systems. Seldom had a corrosion problem on them.

I have seen many an automotive electrical system (as well as the whole vehicle) destroyed because of lack of a proper ground bonding strap between various components. In most vehicles the positive cables are the larger ones because of the starter draw, which may be why the put the cutoff on the positive side. I suspect the switch is on the positive side because many times the switch is also used to switch between multiple batteries.

Aaron

[MillerTime]

Originally Posted by rockfordroo

Don't understand this statement. You will get a VOLTAGE drop through a load or string of loads, but CURRENT flow (amps) is constant through any load or string of loads, so I don't understand what you mean by "remaining."

Ok i may have mis-worded some of that. But.....
You don't have voltage drop though, or shouldn't, but amps will be used. My house voltage will stay 110-120v regardless of me using 10 amps or 400 amps. (Yes i have a 400 amp panel).
Your statement of voltage drop is not true or you could never have 6 120v lights on the same circuit, and if any led lights are used you really cant have very much voltage change or they wont work.
You can loose some voltage over a long distance of under guaged wire with the same amp load as is designed for a shorter distance or lesser load.

Back to the point. I meant that the positive side is designed to carry all the load or is protected with fuses. The negative side usually has many smaller wires going to a heavy frame, engine block or something else. Its not designed to handle amperage loads like the positive is.

[WolfWhistle]

Originally Posted by Asrabbit

After watching the videos it looks like the negative side is the way to go. I do wonder however if the frame is used for a ground anywhere? I guess, there really would still not be a path back to the battery. A question I really thought would be easy, just does not seem all that easy. I guess I can go one way and always change it later on if I need to.

A, lots of good comments and maybe some slight confusion. A few points because I find this an interesting subject and hope to help.
1- you cannot determine if a battery is 100% full by its voltage - only a hydrometer can determine this. Flybob's chart indicates what a typical specific gravity should be. You can get a cheap hydrometer to know for sure.
2- a 12 volt battery will only hold 13.1v or so (after a float charge has been applied) for seconds. You must let it rest before taking a reading. If it has just been charged it will be artificially high; but of a load just removed it will be artificially low. Your 12.6 v indicates you were nearly full. Two of my 6 v batteries in series is also about 12.6 (to .7) volts; resting full.
3- when removing cables from battery (or batteries) remove the negative first for reason given by 25FKS. I actually have two banks with a selector / disconnect and I chose to use the negative side, because of a chance of cables being grounded where I mounted it. I can select 1, 2, both or none. My negative and positive battery cables are all the same wire size and as short as possible (#2/0).
4- All 12volt device negative or return path is via wiring that attaches to the frame at various points. Whether right at the device or a a point of distribution. But, you only want one cable from the battery to the frame ground.
5- if your charging source cannot produce 14.6 v and stay there long enough you will never get a fully charged battery. It could take days rather than hours to charge with a poor charger. It isn't all about the amps.
6- the last 10% of the charging process takes HOURS. Thats why solar is so great for topping them off.
7- A good charger needs the ability to equalize.
8- there is always voltage drop across any load and through any length of cable. Or, you have a super conductor.
WW

[ScottBrownstein]

Originally Posted by MillerTime

...The negative side usually has many smaller wires going to a heavy frame, engine block or something else. Its not designed to handle amperage loads like the positive is.

That might be true for all of the individual negative cables that go from each device to the chassis..They are sized to match the current of each specific device. HOWEVER, the battery (or batteries) have a single main negative cable that is as big as the main positive (or larger) and goes to a single point on the chassis. It is designed to handle all of the loads put together, as is the main positive lead. Both are usually #4 or #6. The negative lead, however, is usually only a foot or two in length before it attaches to the frame. The frame is equivalent to a really, really big wire!!!

[MillerTime]

Originally Posted by ScottBrownstein

That might be true for all of the individual negative cables that go from each device to the chassis..They are sized to match the current of each specific device. HOWEVER, the battery (or batteries) have a single main negative cable that is as big as the main positive (or larger) and goes to a single point on the chassis. It is designed to handle all of the loads put together, as is the main positive lead. Both are usually #4 or #6. The negative lead, however, is usually only a foot or two in length before it attaches to the frame. The frame is equivalent to a really, really big wire!!!

True. But i have seen a vehicle wiring harnesse cooked on the ground only side from being jump started from a battery post instead of grounding to the frame or engine.
It was an expensive mistake and also fried the computer. I think the harness was $800+ and a computer was $400-500. Dont know what made all the stars align just wrong, but they did!

[rockfordroo]

Originally Posted by MillerTime

Ok i may have mis-worded some of that. But.....
You don't have voltage drop though, or shouldn't, but amps will be used. My house voltage will stay 110-120v regardless of me using 10 amps or 400 amps. (Yes i have a 400 amp panel).
Your statement of voltage drop is not true or you could never have 6 120v lights on the same circuit, and if any led lights are used you really cant have very much voltage change or they wont work.
You can loose some voltage over a long distance of under guaged wire with the same amp load as is designed for a shorter distance or lesser load.

Back to the point. I meant that the positive side is designed to carry all the load or is protected with fuses. The negative side usually has many smaller wires going to a heavy frame, engine block or something else. Its not designed to handle amperage loads like the positive is.

I believe you are confusing loads in parallel with loads in series. I would assume your "6 120v lights" are in PARALLEL. So each light drops 120volts, and each passes some equal amount of current (lets say 2 amps for sake of argument) for a total of 12 amps. So each light circuit draws 2 amps, but the current through the breaker is 6 x 2 = 12 amps. Since V = I x R (Ohms Law) OR R = V/I, the resistance of the lights is R = V/I = 120/2 = 60 ohms.

If you put the two lamps in SERIES, the total resistance is now 60 + 60 = 120 ohms. Therefore the total current through the lights is I = V/R = 120/120 = 1 amp. BUT, the voltage across just one of the lights is V=I x R = 1 x 60 = 60 volts (remember that R for one light is 60 ohms).

So in series, they each drop 1/2 of the voltage (120/2 = 60v), but the current through both of them would be the same, 1 amp. Note the current is also lower because we've doubled the original resistance across the 120 volt source.

Of course, the wire back to the battery ultimately carries the current of ALL loads.

[M&K Hoffman]

The only thing I will add to the conservation is that my Windjammer has a factory installed battery disconnect and it breaks the positive cable with the break-away switch being wired in ahead of the disconnect.

[boondocking]

@ rockfordroo

So which is the proper side to put the disconnect, hot side or neutral side?

[rockfordroo]

Originally Posted by boondocking

@ rockfordroo

So which is the proper side to put the disconnect, hot side or neutral side?

I don't really have a strong opinion. If you're working in an engine bay, the negative side argument seems to make sense. But in an RV situation, where you're not really working around the battery, either one seems equally good (or bad, as the case may be).

I don't know which one the factory used on my Mini Lite, but someone on another similar post just stated his factory-installed switch is on the positive side.

[WolfWhistle]

Factory will wire or attempt to wire a BCO to the positive +B side. I have also seen a factory BCO that did nothing, due to a mistake. It is common to wire breakaway on the battery side of the BCO for safety reasons. But a BCO IMO should cut off every thing else.
WW

[MillerTime]

Originally Posted by rockfordroo

I believe you are confusing loads in parallel with loads in series. I would assume your "6 120v lights" are in PARALLEL. So each light drops 120volts, and each passes some equal amount of current (lets say 2 amps for sake of argument) for a total of 12 amps. So each light circuit draws 2 amps, but the current through the breaker is 6 x 2 = 12 amps. Since V = I x R (Ohms Law) OR R = V/I, the resistance of the lights is R = V/I = 120/2 = 60 ohms.

If you put the two lamps in SERIES, the total resistance is now 60 + 60 = 120 ohms. Therefore the total current through the lights is I = V/R = 120/120 = 1 amp. BUT, the voltage across just one of the lights is V=I x R = 1 x 60 = 60 volts (remember that R for one light is 60 ohms).

So in series, they each drop 1/2 of the voltage (120/2 = 60v), but the current through both of them would be the same, 1 amp. Note the current is also lower because we've doubled the original resistance across the 120 volt source.

Of course, the wire back to the battery ultimately carries the current of ALL loads.

I am explaining this wrong i guess. What i mean i guess is this .... . ie...120v, 15amp circuit, if each load draws 3 amps, 120v you can only use 5- 3 amp loads without over loading the circuit.

Im confusing my self now!

Op......the factory installs battery disconnects on the positive side.
-all of my disconnects are on the positive side of a battery.
- i take off my battery terminals, negative first

You can install it anyway you want!

Positive side is protected against overload by fuses or breakers, negative is not.
That's basically my point....i think.

[rockfordroo]

Originally Posted by MillerTime

I am explaining this wrong i guess. What i mean i guess is this .... . ie...120v, 15amp circuit, if each load draws 3 amps, 120v you can only use 5- 3 amp loads without over loading the circuit.

Im confusing my self now!

Op......the factory installs battery disconnects on the positive side.
-all of my disconnects are on the positive side of a battery.
- i take off my battery terminals, negative first

You can install it anyway you want!

Positive side is protected against overload by fuses or breakers, negative is not.
That's basically my point....i think.

OK - got it. You are speaking of the loads you may be plugging into your 120V outlets. These will all be in parallel, therefore the voltage drop across each one is 120 volts and the currents of each one will be additive at the common line, i.e., AT THE BREAKER. So if you had 3 different loads each pulling, say, 4 amps, plugged into your 120V outlets, there would be 12 amps through the breaker.

Same is true of your batteries negative and positive cables. Even though several of your DC circuits may be pulling only a few amps each, those amps ADD UP and all of them go through the positive and negative cables to the battery.

[rockfordroo]

Had a chance to go by my TT in storage today. Battery switch is in the positive lead.

[bubbles]

Originally Posted by rockfordroo

Had a chance to go by my TT in storage today. Battery switch is in the positive lead.

In my motorhome it is the same way and that is why there are parasitic loads that will drain the battery even though it has been disconnected by a disconnect switch on the positive side. Of course this allows certain systems to still function through switches because they are fused directly to the battery. Not saying it shouldn't be that way just stating the way it typically is.

[rockfordroo]

Originally Posted by bubbles

In my motorhome it is the same way and that is why there are parasitic loads that will drain the battery even though it has been disconnected by a disconnect switch on the positive side. Of course this allows certain systems to still function through switches because they are fused directly to the battery. Not saying it shouldn't be that way just stating the way it typically is.

Positive/negative shouldn't make a difference with respect to parasitic loads if it's in the direct line to the battery. Any parasitic loads still need both a positive and a negative feed to the battery. Cut either one with a battery cutoff switch and it's no longer a parasitic load.

[bubbles]

Originally Posted by rockfordroo

Positive/negative shouldn't make a difference with respect to parasitic loads if it's in the direct line to the battery. Any parasitic loads still need both a positive and a negative feed to the battery. Cut either one with a battery cutoff switch and it's no longer a parasitic load.

And that's why I said those parasitic loads are fused directly off the battery (LP detector is an example of one without an on/off switch). The awning for example would not be a parasitic load even though it is fused directly off the battery because it uses a switch that is normally open. Slides are another example with the same configuration. The disconnect removes 12 volt power to those systems that would not be required should a disconnect fail in the disconnected condition. Disconnecting the negative side would render all 12 volt items inoperative even though they may be fused directly off the battery. That's why the positive side is opened via the disconnect switch. That's all I was trying to point out to the input I responded to. To me, if someone needed a complete disconnect I would disconnect the negative side.