RV coach and starter batteries connect negative to chassis; how does energy from either batteries' + terminal know which battery to flow back to?

Question

Consider a van or RV that has two 12V DC batteries. The starter battery is used for normal automotive stuff, most notably turning over the starter to start the engine, as well as powering loads such as the radio, headlights, and AC when the engine is off.

The coach battery is used to power loads such as a water pump, the cabin fridge, cabin lights, etc.

Both the starter and coach batteries have their negative terminals connected to the chassis. The loads powered by these two batteries have have their positive wire connected to the positive terminal of their respective battery as you would expect: The starter, radio, headlights, and AC have their positive wire connected to the starter battery; the water pump, cabin fridge, and cabin lights have their postive wire connected to the coach battery; and, all loads have their negative terminal connected to a nearby chassis bolt (which may or may not be shared with another load's negative terminal connection).

Consider the engine is off, and the radio is turned on, and the fridge is running. The radio's positive wire is connected to the starter battery +, and the fridge's positive wire is connected to the coach battery +. No bus bars or fuse boxes, for simplicity's sake.

Both the radio and the fridge have their negative wires attached to the chassis; perhaps very close to eachother, or perhaps even on the same bolt!

Now, I've always understood electrical charge as a flow from one point to another: How do the electrons flowing to power the radio know to return to the starter battery, and the electrons powering the fridge know to return to the coach battery? How don't they get mixed up?

I am asking this in the context of trying to understand how a smart shunt, connected as the first and only item to the negative terminal of the coach battery, and then to a chassis ground, is able to compute the energy usage of only those items connected to the + terminal of the coach battery and not of those items connected to the + terminal of the starter battery, since they share the same ground.

Answer 1

The answer to your question is primarily governed by what we call Kirchhoff's Circuit Laws

These laws are divided into two, but they complement each other:

1. Kirchhoff's Current Law (KCL) - the sum of all currents connected at one point (or node) is zero.

2. Kirchhoff's Voltage Law (KVL) - the sum of all voltages connected in one loop is zero.

Your question can be explained by any of these two, but to make an example, let's try to apply the Kirchhoff's Current Law since your concern is about the current distribution thru the chassis node (or ground):

In the photo, the chassis (encircled in red) is considered as the ground node (0 V), where all connections from the negative terminal of any battery are connected. By applying the Ohm's Law to the Load1 and Load2, we can calculate the corresponding currents for each of them:

L1_Current = 12 V / 12 Ohms = 1.0 A
L2_Current = 12 V / 24 Ohms = 0.5 A
Chassis Total Input Current = 1.5 A

So based on KCL, the same amount of current should go out of the chassis node, so there must be -1.5 A going out to make a total of 0 A.

But how is it divided into the two outer wires? This is where the KVL law applies.

The KVL dictates that any single circuit loop should have a total voltage of 0, and because of this it also implies that the current flowing inside that loop should never change.

So, if we know that L1_Current = 1A, then it should follow that B1_Current should be 1A as well since they are in a single loop, but we apply the negative sign to imply the opposite direction (in reference to the chassis).

Likewise, if L2_Current = 0.5 A, then B2_Current should be equal to -0.5 A.

Answer 2

If charge leaves a battery via its positive terminal (using +ve charge convention here, don't worry about electrons, the sign only confuses things), then an equal amount of charge has to enter that same battery on its negative terminal.

What do we mean 'has to'?

Let's say the battery is not connected at a negative terminal. Now somehow allow 1 coulomb of charge to flow out of the positive terminal, that's 1 amp for 1 second.

The battery will have a capacitance to ground¹ of somewhere in the 10 pF to 100 pF ballpark, let's say 30 pF.

On losing 1 C of charge, the battery will now fall to a voltage of 1C/30pF = -30 GV (-30,000,000,000 volts) with respect to ground.

That's a lot of 'encouragement' for some charge to flow in at that particular negative terminal. So much encouragement that we just say 'has to'.

If you have two batteries, both connected to ground, and one is supplying current, then that same current will be entering it by its negative lead.

Another way to look at these numbers is to say that maybe the negative lead to the battery has a 1 mΩ resistance. How long must one amp flow out of the positive terminal before the battery potential with respect to ground has dropped low enough to cause 1 amp to flow in along the negative lead? We need 1 mV drop in the battery for 1 A to flow along the lead, so with 30 pF to ground, that's a charge of 1 mV x 30 pF = 30 fC, or 30 fs with 1 amp flowing. That's pretty fast, let's just say it happens at once. The fact that it's calculated to take orders of magnitude less time than it takes light to cross the distance between the two battery terminals means the model is too simple. It does mean that battery capacitance is not the limiting factor for how fast the current in the ground lead makes up for the current exiting the supply lead.

¹ If it's installed in a vehicle, then the chassis of the vehicle. If it's sitting on your bench, then the ground wires in your garage wiring and the mildly conductive ground you're standing on. Basically the nearest conductors surrounding but not connected to the battery.

Answer 3

If you have any electrical device with two terminals, such as a lamp, it is clear that whatever current goes into the lamp on positive suppply wire must result into equal current coming out of the lamp negative supply wire.

Now, the battery is also a two terminal electrical device, so same rule applies.

So any load taking current from the secondary battery will result into same current returning to the same battery.

Answer 4

Here's the schematic.

And the equivalent circuit.

Consider the following:

1. In a closed circuit the current originating from a source must return to it.

   This condition is satisfied in both the closed circuits.

2. Kirchhoff's Current Law states that the sum of currents entering a node must be equal to the sum of those exiting it.

   This is also satisfied at both the nodes.

Thus the current through the shunt is limited to that drawn from the coach battery alone.

Answer 5

As electrons have a negative charge, they flow from the negative terminal to the positive terminal. But this is generally irrelevant to your question.

It is stated in many comments and answers here that the electrons flow in a loop (that's what a circuit is). Any particular electron flow is only involved in the circuit in which it is flowing.

It may help to think of flow as constant. There is no opportunity for an electron to change its mind, go back, and visit the other battery. Aside from the fact that they don't behave intelligently, the electrons are constantly flowing, always in one direction. When the circuit changes (such as with a shunt, although probably not the sort you're looking at), the entire flow changes.

Under some circumstances, having the two negative terminals of two batteries connected together could cause problems. None of these are likely to occur in any automotive circuit, which is why your RV is designed so.

To attach your circuit, you may connect all of the grounds together and to all of the negative terminals. Since you have 2 batteries and your device seems to be a current monitoring shunt, you will need one for each battery you wish to monitor, unless the device is designed specifically to monitor two batteries.