In a simple circuit consists of a source and resistor, according to Ohm's law, if 10V is applied to a 10 ohm resistor, the current flow through the circuit will be 1A.
But what if the source in this circuit has a max output of 0.5A? What happens in that case?
Then you won't get 10V across the resistor, you would get 0.5 * 10 = 5V across this resistor, by V=IR. The resistance is fixed but you have varied the current so the across voltage changes. The other 5V will be across the current limiting element.
If the source acts like a bench power supply and actively limits the current to 0.5A you will have 5V across the resistor.
If the source acts like an ideal voltage source with a 20\$\Omega\$ resistor in series (10V open circuit, 0.5A into a short), the voltage across the external 10 ohm resistor will be less and thus the current will be (1/3) less than the above case.
There are other possibilities- batteries would behave somewhat differently.
If the 0.5A limit is because of a fuse the current will be zero.
If the limit is because it's an auto retry switching supply with current limiting the current will be short pulses and probably average fairly low.
A current limit for a voltage source implies that the source is not ideal and have an output resistance. Then you need to add this output resistance model to your circuit.
0.5A current limit indicates that when the terminals of the voltage source is shorted, the current through output resistance of the voltage source is 0.5A. This shows us that the value of output resistance of 10V source is 20Ω.
So, you should modify your circuit model including voltage source output resistance and analyze it as usual in case any current limit is of interest.
The answer to your question is that the current through 10Ω resistor is 0.33A.
simulate this circuit – Schematic created using CircuitLab
