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let's say i have an open circuit with a 5V voltage source, with the positive terminal connected to node A and the negative terminal connected to node B (which are not connected to anything, hence an open circuit).

If take B as a reference point, A should have a potential of 5V.

let's say I have a 2 ohm resistor connected between the positive terminal and A.

The circuit is open, there is no voltage drop across this resistor, hence A still has a potential of 5V.

if I connect a current source across that resistor, say 5A, then ohm's law tells us that the resistor has a voltage drop of 10V.

What i want to know is that does A now have a potential of 5V-10V = -5V? does the voltage drop across the resistor caused by the current cause the potential of the battery's positive terminal to drop across the resistor?

I chose an open circuit as an example, to make the calculation easy, but if the circuit were closed, would this voltage drop in resistor caused by the current source ADD to the voltage drop in the resistor caused by the current flowing in the main, voltage source loop?