What does it mean to apply a voltage bias on a superconductor when $V = IR$ and $R = 0$ for a superconductor? I have been reading papers about superconducting nanowire single photon detectors, and in them the voltage bias the superconductor to close to $T_c$ so that a photon has enough energy to break the superconducting state and generate a voltage transient. But, how is the superconductor voltage biased if it has no resistance? Does Ohm's law apply to superconductors?
This is an arXiv paper for example that documents DC biasing the superconductor: https://arxiv.org/abs/1711.01305
My interest is in single photon detection, the paper above also covers kinetic inductance detectors, which is not my thing.
 A: In transition-edge sensors (TES), the superconducting detector is held very close to its critical temperature, so that it's resistance is a very steep function of the temperature.
The figure below, from a review paper on TES, demonstrates the resistance vs. temperature in the right panel. In operation, the sensor temperature will be kept somewhere on the steep part of the curve, here between 119 and 120 mK. When radiation is absorbed in the detector, the superconductor film temperature increases locally which can be then measured as a change in the film resistance.
Therefore, to answer the question, the superconductor in this case has a finite resistance (because it is not fully superconducting at the operating temperature), so there would be no problem with applying DC voltage to it.
I do not know which papers are referred to in the question, so it is also possible that the applied voltage is AC (which is fine for superconductors since they behave as inductors below $T_c$), or that the voltage is applied through a bias resistor.

