[Bob Johnson]

Please note that when the circuit breaker trips it does not see the full 6V or 12V. Most of the voltage drop is across the rest of the circuit. In the example below the breaker trips when when it sees a voltage of 0.6 volts regardless of the system voltage (6V or 12V). The resistance of the circuit breaker has to be quite low because you do not want to have a voltage drop across it. During normal operation you do not want to have a 1V drop across the breaker leaving only 5V to power the car. In the example below I use a resistance for the breaker of 0.02 ohms. If you are not using the headlights you will have under 5 amps which would have a voltage drop of only 0.1V. Below I state that it the number of watts that triggers the breaker watts = volts * amps. Dan said it a different way Amps * Amps * ohms. (note volts = Amps * ohms).




Here is some information:
If you calculate the resistance needed for 30 amps for a 6 and a 12 volt system you get 0.4 ohms for a 12V and 0.2 for 6V. So a dead short across a circuit breaker would require it have those resistance. But that is not how it works. The circuit breaker has a much lower resistance, typicality 0.02 ohms. Thus it will "see" a voltage of 30 X 0.02 = 0.6 volts when it triggers regardless of the overall voltage of the system. The number of watts (power) is what triggers the breaker. In this case it is 18 watts (30A X 0.6V).



Note that when the breaker triggers there is a difference in the "short" resistance for 6V vs 12V. In a 6V system it take a resistance of (6V - 0.6V) / 30 A = 0.18 Ohms for 6V and (12V - 0.6V) / 30 A = 0.38 Ohms for 12V. That would result from a 180 watt load for a 6V system and a 360 watt load for a 12V system. That is where the difference between a 6V and 12V system is seen.




Bob