Suppose in my above diagram I simplified so heavily that I abstracted an entire circuit as a power source. Then the internal resistance of that "power source" would be the overall resistance of the circuit, correct?
In a short-circuit, yes! But you should avoid short-circuits
They are mainly only good for calculus.
The electrical circuit
(...mind the word circuit! ...from circle, 'cause it's a loop) then consists of the "idealized"
source (witch sources 10 V e.g.), the
internal resistance of the power source and a short-circuiting
wire (that is assumed to have no resistance in all analytic views).
If you have however a power-supply that has over-current-protection, it is designed to limit the current it will deliver in a save way. Most modern power-supplies will have this "feature".
"Constant-Current" Mode and "Constant Voltage" Mode are key features of good (decent) power supplies.
With that
(Constant-Current) feature the power-supply will (kind of) change it's internal resistance, so that only the set amount of current (say 1 A) will be sourced. Less current (when the load is less - it has a higher resistance than zero Ohms) is always working, though.
If for example you have a power-supply of 10 V that can deliver
up to 1 A, this current will basically only flow if you short-circuit it.
Driving a load with -say 100 Ohms- will result in a current of 100 mA (I=U/R) flowing through the simple circuit.
Ohms law is hard to get around in the real world
The main thing about your power-bus with several contributors (sources) is that they have to be at the same potential (voltage)! That's why photovoltaic modules have to have a device, that boosts the voltage up to a level, that fits the net voltage (230 V AC in your case[1]).
Any lower voltage makes that module essentially a Load, that will get hot and eventually die :facepalm:
By the way,
that is a really good and easy to play with online tool! :thumbup:
Here's a "circuit", you can play with
http://tinyurl.com/jjca63z
[1] AC is a little bit more into electrical engineering than this thread can handle I believe...your diagram in question clearly marked the supplies with a plus and a minus :thumbup: