Hi thanks for your help. I also received the following info from yahoo answers, and have simulated the circuit on workbench, see attached and any comments welcome : -
There is some ambiquity in this question as to what is meant by output voltage. It could refer to the open circuit (unloaded) output voltage or to the output voltage under load. These differ by the voltage dropped across the 1 ohm referred to as the supply system's internal resistance. Since the voltage under load clearly has to be 110V based on the both the lighting and battery voltage requirements, I will presume we are talking about the open circuit voltage. Note that I am also presuming that the lighting and battery circuits are in parallel with each other and that the battery is already charged to an open circuit voltage of 110V.
In the absence of other details about the battery, we can only presume that 110V across the battery is the point where the battery just begins to charge. Stated another way, there is no current flow into the battery with 110V across it, but above 110V, charge current will begin to flow. The battery will supply current only if the load voltage drops below 110V. When running, the alternator sees that this doesn't happen. At exactly 110V, it is like the battery isn't even there.
Note also that since the lighting has 110V across it at this point, there is no voltage drop across the 2 Ohm internal battery resistance. Thus, there is no current being supplied by the battery either. All of the current from the lighting is coming from the alternator.
Since the alternator is providing all of the current (10A to the lighting and 0A to the battery), there must be 10A going thru the 1 Ohm supply system internal resistance. Using Ohm's Law V = I*R, we get V= (10A)*(1 Ohm) = 10V across the supply system internal resistance. In order for the alternator to supply 110V to the lighting and the battery, it must supply 110V + 10V = 120V open circuit voltage (unloaded).
