Your V is the Voltage drop over the superconductor, but for the complete picture consider a voltage source that adds its inherent resistance (output impedance)

Your V is the Voltage drop over the superconductor, but for the complete picture consider a voltage source that adds its inherent resistance (output impedance) to the loop and limits the overall current. A superconductor, to my limited understanding is equivalent to the idealized resistance free wires in schematics. Voltage isn't on the line but it's the difference of two electric potentials. Without Voltage drop there is no energy loss for conducting any current, as P=IV and E(t)=Pt. Edit: OTOH you might argue that then there is no Energy gradient for the current to follow, but that's only the macroscopic picture where the Voltage drop is on average zero. Another comment pointed at quantum mechanics, but I couldn't explain lossy resistors on that level either.