I dunno - size counts, doesn't it?? From a previous post elsewhere:
Charging the batteries:
We have learnt that current is taken, not given. We have also learnt that current is a factor of Voltage and Resistance - the higher the voltage and/or the lower the resistance, the higher the current.
With a battery charging circuit, we have a power source (the alternator) and a resistor (the battery). These devices are, in effect, in series.
The alternator provides a constant voltage output which, for now, we will assume to be 13.6v. The battery also has voltage - let's say the battery is at 12v. The effective voltage, or Potential Difference, is 1.6v (13.6-12).
The battery's internal resistance changes with charge state, but let's assume it is around 0.1?
Charge current is therefore 16A (1.6v/0.1?) (I=V/R)
What does this mean and how does it help you. Well, if your 105A/h deep cycle battery is sitting at 12v, it has been discharged to 25% of it's capacity (any more discharge and you risk damaging it). 25% charge means you have taken out 75A of current over time. If you put it back at 16A, it will take over 4½ hours to recharge.
Not so simple; As the battery charges, the potential difference drops, the internal resistance drops slightly and, therefore, the current drops. To bring a battery up to 80 or 90% of charge state is quick (a couple of hours). The remaining 10-20% can take up to 48 hours. That's a lot of petrol in a Cruiser.
So what can you do about this? Well, you can't change the batteries internal resistance and you can't push current down the batteries throat. That leaves one option - increase the potential difference. I.e. increase the alternator voltage.
But be warned, do not exceed a charge current of 30% of the batteries A/h rating - 10% is recommended. So, for a 105A/h battery do not go over 30A charge current. If you do, the battery will boil, overheat and die a horrible death.
Also, remember that any voltage drop over the wire between the alternator and the battery will result in a lower potential difference at the battery and a subsequent lower charge current.
Time to design a dual battery system. For this exercise, we are not going to add our gadgets (fridge, etc) just yet.
Basic rules:
KISS - although some prefer bells and whistles. If it floats your boat and it works, do it.
Keep the cables as short as possible/practical
Build in a margin of "over-engineering".
But the primary rule is safety/protection/prevention.
On with the design:
1) Cable thickness
A fully revved up alternator can charge a very flat battery at 25-30A at 14.1v. We also know that a deep cycle battery needs to be charged at 14.1v or higher. So, our cable needs to a) be able to handle 30A and, b) have minimal volt drop over it.
2) Cable length
Obviously we want to keep the cable as short as possible (volt drop, cost, etc). On most of our Cruisers, we have the two batteries on either side of the radiator. The shortest route is to feed the cable between the grille and the radiator. I personally prefer to route the cable via the firewall for reasons which will become apparent later.
3) Isolator
If you are a subscriber to the KISS principle, use a solenoid. There are many other devices on the market; some work well, some don’t. If you take the solenoid option, it should be triggered only once the engine is running and not from an ignition source. More on this later
4) Safety
Fuses MUST be fitted to the +ive cable on both batteries as close to the battery as possible. The fuse rating should be higher than the maximum current expected, but well within the cable’s smoke limit.
Cable should be covered with split sleeving, securely tied and routed away from sources of abrasion. If it is routed through a hole of any material, use a grommet or suitable rubber U-channel.
5) Connections
All connections should be made with suitable crimps and the correct crimping tool. A bad connection/crimp will result in resistance which creates heat, which creates a bad connection, which creates more resistance..............And the wire corrodes.
and see here
http://www.the12volt.com/ohm/page2.asp#12
and
Cable does have resistance. The thicker it is, the lower the resistance. Length also plays a role: 1m of 1mm2 wire has the same resistance as 100m of 100mm2
and
Power is the product of Voltage and Current (P = V x I) and is measured in Watts. Power is not usually measured, it is calculated.
It is important to know the power of a device as we need to know how much current it will consume. Not many devices are labelled with the current consumption, but most do list the power rating.
As an example, SWAMBO wants to take her hairdryer in the back of your Cruiser. The hairdryer is rated at 1500W @220v.
From the formula, P=VxI; I=P/V
1500W/220v = 6.8A
OK, so the camp site you're at doesn't have 220v. You knew this before you left which is why you installed a 1500W inverter (we're living in a perfect world for the moment). How much current will flow on the primary, 12v, side of the inverter?
I=P/V
I=1500W/12v
I=125A
and
Time for a practical - the much talked about and dreaded voltage drop.
In this simple circuit, we have a battery, in the front of the vehicle, a cable running to the back and a tyre pump. The tyre pump is a Finni, of course, which is rated at 60A current draw (full load).
Our cable length is 12m (6m +ive & 6m -ive).
For now, let's assume the cable is 2mm2. Resistance per metre is 0.009?, so total cable resistance 0.108?
Using Ohm's law, we can calculate the voltage drop over the cable:
V=IxR
V=60A x 0.108
V=6.48v.
We can see that the pump is actually only going to get 5.52v (12v battery minus 6.48v lost on the cable). That pump ain't going to pump.
This excersise is actually a silly demonstration, because the cable would have melted long ago. :ncool:
As a rule of thumb, cable thickness should be specified at a minimum of 1mm2 per 10A of current required.
If we use this rule, then we should have specified 6mm2 cable. Let's reculculate voltage drop. 6mm2 Cable resistance is 0.0034? therefore total cable resistance is 0.0408?
V=IxR
V=60 x 0.0408
V=2.45v
So our pump is going to get 9.5v instead of the full 12v. It will pomp, but not at full speed.
OK, let's get silly; go big or go home: 25mm2 cable. Resistance per metre is 0.0008?. Total R is 0.0096?
V=IxR
V=60 x 0.0096
V=0.57v
Now the pomp will pomp :cheers:
Pop quiz:
Replace the pump above with a fridge. The fridge draws 3.5A when running. Calculate the voltage drop for 2mm2, 6mm2 and 25mm2 cable.
with many thanks to Jos at LCCSA ... Some folk take exception to this, and to the over-engineering / safety aspect. I don't want my fridge to not fridge, or to have a fire thanks. I go with the suggestions for 35mm and 40mm cable (BTW - I used 25mm for mine and it works fine, with a big 200amp relay and big mega-fuses either end of the +ve cable ...)
EDIT: Oh yes - I ran a cable for both +ve and -ve sides - no earthing to chassis ....