Does a current source override the v=ir rule?
You can think of a current source as a smart voltage source. It creates a voltage across itself so that the desired current is reached in the attached circuit.

So, looking at your circuit...







simulate this circuit &ndash; Schematic created using CircuitLab

Ask yourself, what voltage would need to be at point B in order to have 5A passing thorough R1

Well, from OHM's law, the voltage drop across R1 must be 5,000V, so point B must be at 10V - 5,000  = -4,990V.

So the current source is generating a large negative voltage at its top.

Another way you can look at it is, for 5A to flow through this circuit, the circuit equivalent resistance must be \$2\Omega\$. The current source then must have an effective resistance of \$-998\Omega\$. Yes that is correct, the current source has a negative resistance in this instance.