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Old 04-02-2016, 03:16 PM   #31
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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.
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Old 04-02-2016, 03:23 PM   #32
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@ rockfordroo

So which is the proper side to put the disconnect, hot side or neutral side?
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Old 04-02-2016, 03:47 PM   #33
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@ 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.
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Old 04-02-2016, 04:08 PM   #34
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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.
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Old 04-02-2016, 04:20 PM   #35
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Originally Posted by rockfordroo View Post
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.
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Old 04-02-2016, 04:54 PM   #36
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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.
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Old 04-03-2016, 06:21 PM   #37
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Had a chance to go by my TT in storage today. Battery switch is in the positive lead.
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Old 04-03-2016, 06:46 PM   #38
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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.
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Old 04-03-2016, 08:40 PM   #39
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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.
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Old 04-04-2016, 08:22 AM   #40
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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.
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