# current splitting in the presence of superconductors

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?

• Duplicate: physics.stackexchange.com/q/136296/57075 – Gaurav Jul 5 '15 at 6:16
• @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. – Floris Jul 6 '15 at 2:19

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.
• 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? – renormalization group Jul 7 '15 at 5:52