When you have two resistors in parallel, the current splits up based on the resistances. What will happen if we have two superconductors in place of the resistors? What will happen to the current?

  • $\begingroup$ Duplicate: physics.stackexchange.com/q/136296/57075 $\endgroup$ – Gaurav Jul 5 '15 at 6:16
  • $\begingroup$ @Gaurav - what happens to current in a single superconductor does not predict what happens when you have two superconductors in parallel. I think this question is sufficiently different to remain open. $\endgroup$ – Floris Jul 6 '15 at 2:19

Current going through a superconductor (or otherwise) will form a magnetic field. The potential energy of the magnetic field depends on its size, and the permeability of its surrounding environment.

The current will be divided between the two superconductors such that the total magnetic field energy is minimal.

Actually, this effect is observable using inductors instead of superconductors. Perhaps the better answer is simply that the superconductors will behave like ideal inductors.

If you connect a big inductor in parallel with a small one, the small one will tend to short-circuit the big one. The total inductance is determined by $1/(1/L_1+1/L_2)$, similar to the parallel resistance formula. Likewise, the current through each inductor is inversely proportional to its inductance. If you leave the parallel inductors connected to a DC source, then the current will gradually be redistributed according to their resistances. This effect is usually neglected because inductors are typically used in an AC frequency domain where $IR << LdI/dt$. But it's always there.

  • $\begingroup$ … and, likewise, ideal resistors always have some inductance which will tend to dominate at sufficiently high frequencies. $\endgroup$ – Blackbody Blacklight Jul 5 '15 at 7:34
  • $\begingroup$ Thank you!So the current here is divided based on the inductances as $I_{L1}=I\dfrac{L_{2}}{L_{1}+L_{2}}$ and $I_{L2}=I\dfrac{L_{1}}{L_{1}+L_{2}}$. And does the current remain in that loop formed by the superconductors? $\endgroup$ – renormalization group Jul 7 '15 at 5:52
  • $\begingroup$ @ShriyaPai Yup. If you're satisfied with this answer, please click upvote and click the green checkmark. $\endgroup$ – Blackbody Blacklight Jul 7 '15 at 10:37
  • $\begingroup$ My upvotes don't get displayed-I'm new here. But the green checkmark is done :) $\endgroup$ – renormalization group Jul 7 '15 at 16:11

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