Whoa, a bit heavy for a Friday night. Sorry I don't have my regs book at home any more as I now work in the office but I'll have a go.
Fuses blow simply by overheating and take time. The larger the current the less time they need. Close protection (a cartridge
fuse) should blow within 4 hours at 1.5 times the rated current. Breakers typically have a thermal part for overload and a magnetic part for short circuit.
This gives a typical characteristic for fuses of various ratings.
link
Time in seconds is the vertical axis and current on the horizontal. It shows a 60A
fuse will take about 1200A to blow in 0.1 seconds and about 6 seconds at 100A.
We protect cables by coordinating the protection to what the cable will withstand. A PVC cable will normally run at 70deg max conductor temperature at full load although I usually choose them not to get any where near that otherwise people start getting worried. The modern XLPE cables run at 90deg.
When a fault occurs, the current will increase and cause the cable to start heating up. PVC cable can rise to a conductor temperature of 145deg before being damaged. So we want the
fuse to blow before the cable reaches 145 deg.
The impedance of the circuit (posh word for resistance) limits the fault current. We want enough fault current to blow the
fuse within 0.4 seconds for sockets and 5 seconds for lights etc., (to protect people), but we want the fault current limited to below the amount the cables and switchgear and relays, contactors etc will withstand for the duration of the fault (to protect the equipment).
Circuit breakers are good for overloads (a small amount of excess current) but fuses are better for short circuits (a large amount of excess current). We can use a
fuse to back up a circuit breaker so the breaker trips on overload but the
fuse trips on short circuit. That way the fault withstand current of the breaker may be less than the prospective (max possible) fault current in the circuit.
This coordination is just a small part of what I have to do when designing the electrics for large buildings.
The peak current with the 2 batteries and alernator etc. depends how it is wired. Basically you will have a network with 3 sources of potential(voltage), a load and various resistances. Everthing will have some resistance, some in the alternator, some in each cable, some in the load, some in each
battery etc. Ohms law can be used to determine how much will flew in each part of the circuit. Amps = volts divided by resistance(ohms)
The relay will pass charging current from the alternator to the leisure
battery, and the invertor will be connected to the leisure
battery. If a heavy load such as an invertor is connected to the leisure
battery it will draw current from each of the alternator, vehicle
battery and leisure
battery.
The current from the alternator and vehicle
battery will pass through the relay, the current from the leisure
battery will not. If the leisure
battery is flat, it will contribute very little current to the invertor, if any, and will most likely draw charging current from the alternator and vehicle
battery.
So the relay will pass the current flowing to the invertor and the charging current to the leisure
battery.
The more current that is drawn, the more the voltage will drop in the vehicle
battery and alternator and cables (due to resistance) and this will lower the current drawn by the invertor and leisure
battery charging.
If there is a fault, the current will be drawn from the alternator and both batteries. The prospective fault current of a
battery wil be the voltage divded by the sum of the resistance of the cables, fault and internal resistance of the
battery. The alternator will also have a fault current. *As long as each
battery and the alternator are individually fused with fuses with a high rupturing current (HRC) such as cartridge fuses, they should blow before cables or equipment become damaged as long as they are all correctly coordinated by the system designer.