Since the resistance of a superconducting circuit is exactly 0, it would try to draw infinite current from a power source, since P=I^2*R. For a finite power source, I-> infinity as R->0. Correct? If that is true, then how is current delivered to the circuit in a controlled way that is safe for the power source? I'm guessing the circuit is brought to near critical temperature, then current is introduced, then it is shorted, and then it is quickly brought below the critical temperature before too much current has dissipated.
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Not a full answer, but I'd like to bring the following relevant point to your attention. A superconducting magnet is very different from a dead short. Consider an MRI machine in a hospital: it produces a magnetic induction of about 1.5 teslas over a human body-sized volume - say $0.1{\rm m^3}$. Using the formula $U = \frac{1}{2\,\mu_0}\,B^2$ for the energy density, I calculate that the energy needed to set that field up is about $90{\rm kJ}$. So it's going to take even a hefty $20{\rm kW}$ power supply several seconds to set that up. As an electrical circuit element, the system is going to look like a pretty hefty inductor. Naturally, the power supply will need a control system to back off any applied voltage when the coil reaches its operating current.
answered Oct 17, 2015 at 12:23